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2006-I-6
| 2,006
| 6
|
Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$
| 360
|
I
|
[
"Numbers of the form $0.\\overline{abc}$ can be written as $\\frac{abc}{999}$. There are $10\\times9\\times8=720$ such numbers. Each digit will appear in each place value $\\frac{720}{10}=72$ times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is $\\frac{45\\times72\\times111}{999}= 360.",
"Alternatively, for every number, $0.\\overline{abc}$, there will be exactly one other number, such that when they are added together, the sum is $0.\\overline{999}$, or, more precisely, 1. As an example, $.\\overline{123}+.\\overline{876}=.\\overline{999} \\Longrightarrow 1$. Thus, the solution can be determined by dividing the total number of permutations by 2. The answer is $\\frac{10 \\cdot 9 \\cdot 8}{2} = \\frac{720}{2}= 360"
] |
2006-I-7
| 2,006
| 7
|
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is $\frac{11}{5}$ . Find the ratio of shaded region $D$ to the area of shaded region $A$ . [asy] size(6cm); defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S); [/asy]
| 408
|
I
|
[
"Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof. Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at $0, 1, 2 \\ldots$. The base of region $\\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\\mathcal{D}$ is on the line $x = 7$. Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem. Since the area of the triangle is equal to $\\frac{1}{2}bh$, \\[\\frac{\\textrm{Region\\ }\\mathcal{C}}{\\textrm{Region\\ }\\mathcal{B}} = \\frac{11}{5} = \\frac{\\frac 12(5-s)^2 - \\frac 12(4-s)^2}{\\frac 12(3-s)^2 - \\frac12(2-s)^2}\\] Solve this to find that $s = \\frac{5}{6}$. Using the same reasoning as above, we get $\\frac{\\textrm{Region\\ }\\mathcal{D}}{\\textrm{Region\\ }\\mathcal{A}} = \\frac{\\frac 12(7-s)^2 - \\frac 12(6-s)^2}{\\frac 12(1-s)^2}$, which is $408.",
"Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$. Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$. Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\\left(\\frac{x+1}{x}\\right)}^2$. Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$. Repeating this process, we get that the area of B is $2x+3$, the area of C is $2x+7$, and the area of D is $2x+11$. We can now use the given condition that the ratio of C and B is $\\frac{11}{5}$. $\\frac{11}{5} = \\frac{2x+7}{2x+3}$ gives us $x = \\frac{1}{6}$ So now we compute the ratio of D and A, which is $\\frac{(2)(\\frac{1}{6}) + 11}{(\\frac{1}{6})^2} = 408. Edit: fixed misplaced brackets",
"Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\\frac{x}{x+a} =\\frac{y}{y+b} \\implies bx = ya.$ Therefore the ratio we seek is $\\frac{66(ay)}{11xy} =\\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \\implies ab = 6bx \\implies a = 6x$ so the final ratio is $6 \\cdot 68 = 408",
"Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\\frac{x}{x+a} =\\frac{y}{y+b} \\implies bx = ya.$ Therefore the ratio we seek is $\\frac{66(ay)}{11xy} =\\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \\implies ab = 6bx \\implies a = 6x$ so the final ratio is $6 \\cdot 68 = 408"
] |
2006-I-8
| 2,006
| 8
|
Hexagon $ABCDEF$ is divided into five rhombuses, $P, Q, R, S,$ and $T$ , as shown. Rhombuses $P, Q, R,$ and $S$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $T$ . Given that $K$ is a positive integer, find the number of possible values for $K.$ [asy] // TheMathGuyd size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy]
| 89
|
I
|
[
"Let $x$ denote the common side length of the rhombi, and let $y$ denote one of the smaller interior angles of rhombus $\\mathcal{P}$. Then $x^2\\sin(y)=\\sqrt{2006}$. We also see that $K=x^2\\sin(2y) \\implies K=2x^2\\sin y \\cdot \\cos y \\implies K = 2\\sqrt{2006}\\cdot \\cos y$. Thus $K$ can be any positive integer in the interval $(0, 2\\sqrt{2006})$. Since $2\\sqrt{2006} = \\sqrt{8024}$ and $89^2 = 7921 < 8024 < 8100 = 90^2$, $K$ can be any integer between $1$ and $89$, inclusive, so the number of positive values for $K$ is $089.",
"Call the side length of each rhombus $w$. $w$ is the width of the rhombus. Call the height h, where $wh=\\sqrt{2006}$. The height of rhombus T would be 2h, and the width would be $\\sqrt{w^2-h^2}$. Substitute the first equation to get $\\sqrt{\\frac{2006}{h^2}-h^2}$. Then the area of the rhombus would be $2h * \\sqrt{\\frac{2006}{h^2}-h^2}$. Combine like terms to get $2 * \\sqrt{2006-h^4}$. This expression equals an integer K. $2006-h^4$ specifically must be in the form $n^2/4$. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of $n^2$ for $2006-h^4$. Now, quick testing shows that $44^2 < 2006$ and $45^2>2006$, but we must also test $44.5^2$, because the product of two will make it an integer. $44.5^2$ is also less than $2006$, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us $44*2+1=$ $089 -jackshi2006",
"[asy] size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); label(\"$A$\",A,2*N); label(\"$B$\",B,2*N); label(\"$C$\",C,2*E); label(\"$D$\",D,2*S); label(\"$E$\",EE,2*S); label(\"$F$\",F,2*W); label(\"$G$\",(0.47,-1.55),NW); label(\"$H$\",(3.73,-1.55),NE); label(\"$I$\",I,2*N); label(\"$J$\",J,2*S); label(\"$K$\",K,2*SW); draw(F--C); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); draw((2.1,0)--(2.1,-3.2)); [/asy] To determine the possible values of $[GIHJ],$ we must determine the maximum and minimum possible areas. In the case where the $4$ rhombi are squares, we have $[GIHJ]=0,$ implying the minimum possible positive-integer-valued area is $1.$ Denote the length $HC=a$ and $KH=b.$ We have \\[KI=\\sqrt{a^2-b^2}\\] by the Pythagorean Theorem, which implies \\[[IBCH]=a\\sqrt{a^2-b^2}=\\sqrt{2006}\\] and \\[[GIHJ]=2b\\sqrt{a^2-b^2}.\\] The first equation yields \\[\\sqrt{a^2-b^2}=\\frac{\\sqrt{2006}}{a}.\\] Plugging into the second, we have \\[[GIHJ]=2\\sqrt{2006}\\frac{b}{a}.\\] The maximal value of $\\frac{b}{a}$ occurs when the height of $ABCDEF$ is minimized, which means \\[\\frac{b}{a}\\leq 1.\\] Plugging back up, we have \\[[GIHJ]\\leq 2\\sqrt{2006}=\\sqrt{8024}.\\] We have \\[\\lfloor \\sqrt{8024} \\rfloor = 89,\\] thus our answer is \\[89-1+1=089.\\]"
] |
2006-I-9
| 2,006
| 9
|
The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$
| 46
|
I
|
[
"\\[\\log_8 a_1+\\log_8 a_2+\\ldots+\\log_8 a_{12}= \\log_8 a+\\log_8 (ar)+\\ldots+\\log_8 (ar^{11}) \\\\ = \\log_8(a\\cdot ar\\cdot ar^2\\cdot \\cdots \\cdot ar^{11}) = \\log_8 (a^{12}r^{66})\\] So our question is equivalent to solving $\\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers. $a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so $a^{2}r^{11}=2^{1003}$. The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$: \\begin{eqnarray*}(2^x)^2\\cdot(2^y)^{11}&=&2^{1003}\\\\ 2^{2x}\\cdot 2^{11y}&=&2^{1003}\\\\ 2x+11y&=&1003\\\\ y&=&\\frac{1003-2x}{11} \\end{eqnarray*} For $y$ to be an integer, the numerator must be divisible by $11$. This occurs when $x=1$ because $1001=91*11$. Because only even integers are being subtracted from $1003$, the numerator never equals an even multiple of $11$. Therefore, the numerator takes on the value of every odd multiple of $11$ from $11$ to $1001$. Since the odd multiples are separated by a distance of $22$, the number of ordered pairs that work is $1 + \\frac{1001-11}{22}=1 + \\frac{990}{22}=46$. (We must add 1 because both endpoints are being included.) So the answer is $046.",
"Using the above method, we can derive that $a^{2}r^{11} = 2^{1003}$. Now, think about what happens when r is an even power of 2. Then $a^{2}$ must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so $2^{1}$, $2^{3}$, $2^{5}$ .... all work for r, until r hits $2^{93}$, when it gets greater than $2^{1003}$, so the greatest value for r is $2^{91}$. All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields $046.",
"Using the method from Solution 1, we get $\\log_8a^{12}r^{66}=2006 \\implies a^{12}r^{66}=8^{2006}=2^{6018}$. Since $a$ and $r$ both have to be powers of $2$, we can rewrite this as $12x+66y=6018$. $6018 \\equiv 66 \\equiv 6\\pmod{12}$. So, when we subtract $12$ from $6018$, the result is divisible by $66$. Evaluating that, we get $(1,91)$ as a valid solution. Since $66 \\cdot 2 = 12 \\cdot 11$, when we add $11$ to the value of $a$, we can subtract $2$ from the value of $r$ to keep the equation valid. Using this, we get $(1,91),(12,89),(23,87), \\cdots (541,1)$. In order to count the number of ordered pairs, we can simply count the number of $y$ values. Every odd number from $1$ to $91$ is included, so we have $046 solutions. -Phunsukh Wangdu"
] |
2006-I-10
| 2,006
| 10
|
Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ [asy] unitsize(0.50cm); draw((0,-1)--(0,6)); draw((-1,0)--(6,0)); draw(shift(1,1)*unitcircle); draw(shift(1,3)*unitcircle); draw(shift(1,5)*unitcircle); draw(shift(3,1)*unitcircle); draw(shift(3,3)*unitcircle); draw(shift(3,5)*unitcircle); draw(shift(5,1)*unitcircle); draw(shift(5,3)*unitcircle); [/asy]
| 65
|
I
|
[
"[asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label(\"x\", (7,0)); label(\"y\", (0,7)); draw((5/3,0)--(23/6,6.5),blue);[/asy] Solution 1 The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\\left(1,\\frac 12\\right)$ and $\\left(\\frac 32,2\\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = 065. This solution is an alternate explanation to solution 1. -jackshi2006",
"The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\\left(1,\\frac 12\\right)$ and $\\left(\\frac 32,2\\right)$, which can be easily solved to be $6x = 2y + 5$. Thus, $a^2 + b^2 + c^2 = 065. This solution is an alternate explanation to solution 1. -jackshi2006",
"Assume that if unit squares are drawn circumscribing the circles, then the line will divide the area of the concave hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the x-axis as $(x, 0)$. The line divides the region into 2 sections. The left piece is a trapezoid, with its area $\\frac{1}{2}((x) + (x+1))(3) = 3x + \\frac{3}{2}$. The right piece is the addition of a trapezoid and a rectangle, and the areas are $\\frac{1}{2}((1-x) + (2-x))(3)$ and $2 \\cdot 1 = 2$, totaling $\\frac{13}{2} - 3x$. Since we want the two regions to be equal, we find that $3x + \\frac 32 = \\frac {13}2 - 3x$, so $x = \\frac{5}{6}$. We have that $\\left(\\frac 56, 0\\right)$ is a point on the line of slope 3, so $y - 0 = 3\\left(x - \\frac 56\\right) \\Longrightarrow 6x = 2y + 5$. Our answer is $2^2 + 5^2 + 6^2 = 65$. We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1. Solution 3 This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: $6x = 2y + 5$. The answer is then $6^2 + 2^2 + 5^2 = 065. This solution is an alternate explanation to solution 1. -jackshi2006",
"This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: $6x = 2y + 5$. The answer is then $6^2 + 2^2 + 5^2 = 065. This solution is an alternate explanation to solution 1. -jackshi2006"
] |
2006-I-12
| 2,006
| 12
|
Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x,$ where $x$ is measured in degrees and $100< x< 200.$
| 906
|
I
|
[
"Observe that $2\\cos 4x\\cos x = \\cos 5x + \\cos 3x$ by the sum-to-product formulas. Defining $a = \\cos 3x$ and $b = \\cos 5x$, we have $a^3 + b^3 = (a+b)^3 \\rightarrow ab(a+b) = 0$. But $a+b = 2\\cos 4x\\cos x$, so we require $\\cos x = 0$, $\\cos 3x = 0$, $\\cos 4x = 0$, or $\\cos 5x = 0$. Hence we see by careful analysis of the cases that the solution set is $A = \\{150, 126, 162, 198, 112.5, 157.5\\}$ and thus $\\sum_{x \\in A} x = 906."
] |
2006-I-13
| 2,006
| 13
|
For each even positive integer $x,$ let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.
| 899
|
I
|
[
"Given $g : x \\mapsto \\max_{j : 2^j | x} 2^j$, consider $S_n = g(2) + \\cdots + g(2^n)$. Define $S = \\{2, 4, \\ldots, 2^n\\}$. There are $2^0$ elements of $S$ that are divisible by $2^n$, $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \\ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$. Thus \\begin{align*} S_n &= 2^0\\cdot2^n + 2^0\\cdot2^{n-1} + 2^1\\cdot2^{n-2} + \\cdots + 2^{n-2}\\cdot2^1\\\\ &= 2^n + (n-1)2^{n-1}\\\\ &= 2^{n-1}(n+1).\\end{align*} Let $2^k$ be the highest power of $2$ that divides $n+1$. Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$. For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$, we have five choices for $k$, namely $k=0,2,4,6,8$. If $k=0$, then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square. In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^{k}$ is an odd perfect square. If $k=2$, then $n<1000$ implies that $\\frac{n+1}{4} \\le 250$, so we have $n+1 = 4, 4 \\cdot 3^2, \\ldots, 4 \\cdot 13^2, 4\\cdot 3^2 \\cdot 5^2$. If $k=4$, then $n<1000$ implies that $\\frac{n+1}{16} \\le 62$, so $n+1 = 16, 16 \\cdot 3^2, 16 \\cdot 5^2, 16 \\cdot 7^2$. If $k=6$, then $n<1000$ implies that $\\frac{n+1}{64}\\le 15$, so $n+1=64,64\\cdot 3^2$. If $k=8$, then $n<1000$ implies that $\\frac{n+1}{256}\\le 3$, so $n+1=256$. Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\\cdot 3^2 \\cdot 5^2 - 1 = 899.",
"First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\\sum_{k=1}^{2^{n-1}}g(2k). = \\sum_{k=1}^{2^{n-1}}2g(k) = 2\\sum_{k=1}^{2^{n-1}}g(k) = 2\\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\\sum_{k=1}^{2^{n-2}}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\\cdot (n-1)+2^n\\cdot S_0=2^{n-1}\\cdot (n-1)+2^{n-1}\\cdot2=2^{n-1}\\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$. However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= 899",
"At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations. We first simplify all the messy notation in the $S_n$ term. Note that the problem asks us to find the smallest value of $n<1000$ such that there exists an integer $k$ that satisfies \\[k^2 = g(2) + g(4) + \\cdots + g(2^n)\\]. Since there is no obvious way to approach this problem, we start by experimenting with small values of $n$ to evaluate some $S_n$. We play with these values: \\[S_1 = g(2) = 2\\] \\[S_2 = g(2) + g(4) = 2+4 = 6\\] \\[S_3 = g(2) + g(4) + g(6) + g(8) = 16\\] \\[S_4 = g(2) + g(4) + g(6) + g(8) + g(10) +g(12)+g(14)+g(16) = 40\\] We are certainly not going to expand all of this out... so let's look for patterns from these $4$ values! Using a little bit of ingenuity, we note that \\[S_2 = 2+4 = S_1 + 4\\] \\[S_3 = 2+4+2+8 = 8+8 = S_2 + S_1 + 8\\] \\[S_4 = 2+4+2+8+2+4+2+16 = S_3 + S_2 + S_1 + 16\\] Aha! We see powers of two in each of our terms! Therefore, we can say that \\[S_2 = S_1 + 2^2\\] \\[S_3 = S_2+S_1 + 2^3\\] \\[S_4 = S_3 + S_2 + S_1 + 2^4\\] Hooray! We have a recursion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it. On the actual AIME, go with whatever patterns you see, because most likely those are the patterns that the test-makers want the students to see. So we may generalize a formula for $S_n$: \\[S_n = 2^n + S_{n-1} + S_{n-2} + \\cdots + S_2 + S_1\\] Uh oh... this formula is not in closed form. Looks like we'll have to use our recursion to develop one manually. We do so by using our recursion for $S_{n-1}$: \\[S_n = 2^n + (2^{n-1} + S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1) + S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1\\] \\[S_n = 2^n + 2^{n-1} + 2 (S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1\\] Let's simplify a bit further, where we use our recursion for $S_{n-2}$. \\[S_n = 2^n + 2^{n-1} +(2S_{n-2}) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] \\[S_n = 2^n + 2^{n-1} + 2(2^{n-2}) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] \\[S_n = 2^n + 2^{n-1} + 2^{n-1} + 4(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form: \\[S_n = 2^n + \\underbrace{2^{n-1}}_{n-2} + 2^{n-2}(S_1)\\] \\[S_n = 2^n + 2^{n-1}(n-2) + 2^{n-1}\\] \\[S_n = 2^n + 2^{n-1}(n-1)\\] \\[S_n = 2^{n-1}(2 + (n-1)\\] \\[S_n = 2^{n-1}(n+1)\\] We have our closed form, so now we can find the largest value of $n$ such that $S_n$ is a perfect square. In order for $S_n$ to be a perfect square, we must have $n-1$ even and $n+1$ be a perfect square. Since $n<1000$, we have $n+1 < 1001$. We first try $n+1 = 31^2 = 961$(since it is the smallest square below $1000$, which gives us $n=960$. But $n-1$ isn't even, so we discard this value. Next, we try the second smallest value, which is $n = 30^2 = 900$, which tells us that $n=899$. $n-1$ is indeed even, and $n+1$ is a perfect square, so the largest value of $n$ such that $S_n$ is a perfect square is $899$. Our answer is $899",
"At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations. We first simplify all the messy notation in the $S_n$ term. Note that the problem asks us to find the smallest value of $n<1000$ such that there exists an integer $k$ that satisfies \\[k^2 = g(2) + g(4) + \\cdots + g(2^n)\\]. Since there is no obvious way to approach this problem, we start by experimenting with small values of $n$ to evaluate some $S_n$. We play with these values: \\[S_1 = g(2) = 2\\] \\[S_2 = g(2) + g(4) = 2+4 = 6\\] \\[S_3 = g(2) + g(4) + g(6) + g(8) = 16\\] \\[S_4 = g(2) + g(4) + g(6) + g(8) + g(10) +g(12)+g(14)+g(16) = 40\\] We are certainly not going to expand all of this out... so let's look for patterns from these $4$ values! Using a little bit of ingenuity, we note that \\[S_2 = 2+4 = S_1 + 4\\] \\[S_3 = 2+4+2+8 = 8+8 = S_2 + S_1 + 8\\] \\[S_4 = 2+4+2+8+2+4+2+16 = S_3 + S_2 + S_1 + 16\\] Aha! We see powers of two in each of our terms! Therefore, we can say that \\[S_2 = S_1 + 2^2\\] \\[S_3 = S_2+S_1 + 2^3\\] \\[S_4 = S_3 + S_2 + S_1 + 2^4\\] Hooray! We have a recursion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it. On the actual AIME, go with whatever patterns you see, because most likely those are the patterns that the test-makers want the students to see. So we may generalize a formula for $S_n$: \\[S_n = 2^n + S_{n-1} + S_{n-2} + \\cdots + S_2 + S_1\\] Uh oh... this formula is not in closed form. Looks like we'll have to use our recursion to develop one manually. We do so by using our recursion for $S_{n-1}$: \\[S_n = 2^n + (2^{n-1} + S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1) + S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1\\] \\[S_n = 2^n + 2^{n-1} + 2 (S_{n-2} + S_{n-3} + \\cdots + S_2 + S_1\\] Let's simplify a bit further, where we use our recursion for $S_{n-2}$. \\[S_n = 2^n + 2^{n-1} +(2S_{n-2}) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] \\[S_n = 2^n + 2^{n-1} + 2(2^{n-2}) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1) + 2(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] \\[S_n = 2^n + 2^{n-1} + 2^{n-1} + 4(S_{n-3} + S_{n-4} + \\cdots + S_2 + S_1)\\] We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form: \\[S_n = 2^n + \\underbrace{2^{n-1}}_{n-2} + 2^{n-2}(S_1)\\] \\[S_n = 2^n + 2^{n-1}(n-2) + 2^{n-1}\\] \\[S_n = 2^n + 2^{n-1}(n-1)\\] \\[S_n = 2^{n-1}(2 + (n-1)\\] \\[S_n = 2^{n-1}(n+1)\\] We have our closed form, so now we can find the largest value of $n$ such that $S_n$ is a perfect square. In order for $S_n$ to be a perfect square, we must have $n-1$ even and $n+1$ be a perfect square. Since $n<1000$, we have $n+1 < 1001$. We first try $n+1 = 31^2 = 961$(since it is the smallest square below $1000$, which gives us $n=960$. But $n-1$ isn't even, so we discard this value. Next, we try the second smallest value, which is $n = 30^2 = 900$, which tells us that $n=899$. $n-1$ is indeed even, and $n+1$ is a perfect square, so the largest value of $n$ such that $S_n$ is a perfect square is $899$. Our answer is $899",
"First, we set intervals. Say that a number $N$ falls strictly within $[2^k, 2^{k+1}]$. $N<2^{k+1}=2^k+2^k$ We can generalize this: If a number is in the form $N=2^k+2^{R}O$ where $O$ is a positive odd number, $R<k$: $N<2^{k+1}=2^k+2^k\\Longrightarrow O<2^{k-R}$ so there are $2^{k-R-1}$ numbers that satisfy this form. For all numbers that satisfy this form, $g(N)=2^R$. The sum of $g(N)$ for all $N$ in this form and interval is $2^{k-1}$. $R$ can vary between $1$ and $k-1$, so the total sum is $\\underbrace{2^{k-1}+2^{k-1}\\cdots 2^{k-1}}_{k-1}=(k-1)2^{k-1}$. The domain of the function we are trying to find is all even integers in the interval $[2^1, 2^n]$ so there are $n-1$ values of $k$. Now we have the sum $\\sum_{k=1}^{n-1}{(k-1)2^{k-1}}=\\sum_{i=1}^{n-2}{k\\cdot2^{k}}$. However, we did not consider powers of two yet(since our interval was strictly between powers of 2), so we have to add $\\sum_{k=1}^{n}{2^k}$. Note that $\\sum_{i=1}^{n-2}{k\\cdot2^{k}}=(2^1+2^2\\cdots 2^{n-2})+(2^2+2^3\\cdots 2^{n-2})\\cdots +(2^{n-3}+2^{n-2})+2^{n-2}=2^1(2^{n-2}-1)+2^2(2^{n-3}-1)\\cdots +2^{n-3}(2^2-1)+2^{n-2}(2^1-1)=(n-2)(2^{n-1})-\\sum_{k=1}^{n-2}{2^k}=(n-2)(2^{n-1})-2(2^{n-2}-1)=(n-3)(2^{n-1})+2$. Adding $\\sum_{k=1}^{n}{2^k}=2(2^n-1)$, we get $(n+1)(2^{n-1})$. If this sum is a perfect square, $n\\not\\equiv0\\pmod2$, since that would make $2^{n-1}$ not a perfect square, and $n+1$ odd so it cannot contribute a factor of 2 to make the power of 2 a perfect square. We want the least odd number less than $1000$ such that $(n+1)$ is an even perfect square. The greatest even square less than $1000$ is $30^2=900$ so $n+1=900 \\Longrightarrow n=899"
] |
2006-I-14
| 2,006
| 14
|
A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$ )
| 183
|
I
|
[
"[asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"$T$\",T,N);label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,NE);label(\"$S$\",S,NW);label(\"$O$\",O,SW);label(\"$M$\",M,NE); label(\"$4$\",(S+T)/2,NW);label(\"$1$\",(S+A)/2,NW);label(\"$5$\",(B+T)/2,NE);label(\"$4$\",(O+T)/2,W); dot(S);dot(O); [/asy] We will use $[...]$ to denote volume (four letters), area (three letters) or length (two letters). Let $T$ be the top of the tripod, $A,B,C$ are end points of three legs. Let $S$ be the point on $TA$ such that $[TS] = 4$ and $[SA] = 1$. Let $O$ be the center of the base equilateral triangle $ABC$. Let $M$ be the midpoint of segment $BC$. Let $h$ be the distance from $T$ to the triangle $SBC$ ($h$ is what we want to find). We have the volume ratio $\\frac {[TSBC]}{[TABC]} = \\frac {[TS]}{[TA]} = \\frac {4}{5}$. So $\\frac {h\\cdot [SBC]}{[TO]\\cdot [ABC]} = \\frac {4}{5}$. We also have the area ratio $\\frac {[SBC]}{[ABC]} = \\frac {[SM]}{[AM]}$. The triangle $TOA$ is a $3-4-5$ right triangle so $[AM] = \\frac {3}{2}\\cdot[AO] = \\frac {9}{2}$ and $\\cos{\\angle{TAO}} = \\frac {3}{5}$. Applying Law of Cosines to the triangle $SAM$ with $[SA] = 1$, $[AM] = \\frac {9}{2}$ and $\\cos{\\angle{SAM}} = \\frac {3}{5}$, we find: $[SM] = \\frac {\\sqrt {5\\cdot317}}{10}.$ Putting it all together, we find $h = \\frac {144}{\\sqrt {5\\cdot317}}$. $\\lfloor 144+\\sqrt{5 \\cdot 317}\\rfloor =144+ \\lfloor \\sqrt{5 \\cdot 317}\\rfloor =144+\\lfloor \\sqrt{1585} \\rfloor =144+39=183.",
"We note that $AO=3$. From this we can derive that the side length of the equilateral is $3\\sqrt{3}$. We now use 3D coordinate geometry. \\[A = (0,0,0)\\] \\[B = (3\\sqrt{3},0,0)\\] \\[C = (\\frac{3\\sqrt{3}}{2}, \\frac{9}{2}, 0)\\] \\[T = (\\frac{3\\sqrt{3}}{2}, \\frac{3}{2}, 4)\\] \\[S= (\\frac{3\\sqrt{3}}{10}, \\frac{3}{10}, \\frac{4}{5})\\] We know three points of plane $SCB$ hence we can write out the equation for the plane. Plane $SCB$ can be expressed as \\[4\\sqrt{3}x+4y+39z-36=0.\\] Applying the distance between a point and a plane formula. \\[\\frac{ax+by+cz+d}{\\sqrt{a^{2}+b^{2}+c^{2}}} = \\frac{4\\sqrt{3} \\cdot \\frac{3\\sqrt{3}}{2} + 4\\cdot \\frac{3}{2} + 39 \\cdot 4 -36}{\\sqrt{(4\\sqrt{3})^2+4^2+39^2}} = \\frac{144}{\\sqrt{1585}}\\] \\[\\lfloor m+\\sqrt{n}\\rfloor = \\lfloor 144+\\sqrt{1585}\\rfloor = 183\\] Solution by SimonSun",
"Diagram borrowed from Solution 1 [asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"$T$\",T,N);label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,NE);label(\"$S$\",S,NW);label(\"$O$\",O,SW);label(\"$M$\",M,NE); label(\"$4$\",(S+T)/2,NW);label(\"$1$\",(S+A)/2,NW);label(\"$5$\",(B+T)/2,NE);label(\"$4$\",(O+T)/2,W); dot(S);dot(O); [/asy] Apply Pythagorean Theorem on $\\bigtriangleup TOB$ yields \\[BO=\\sqrt{TB^2-TO^2}=3\\] Since $\\bigtriangleup ABC$ is equilateral, we have $\\angle MOB=60^{\\circ}$ and \\[BC=2BM=2(OB\\sin MOB)=3\\sqrt{3}\\] Apply Pythagorean Theorem on $\\bigtriangleup TMB$ yields \\[TM=\\sqrt{TB^2-BM^2}=\\sqrt{5^2-(\\frac{3\\sqrt{3}}{2})^2}=\\frac{\\sqrt{73}}{2}\\] Apply Law of Cosines on $\\bigtriangleup TBC$ we have \\[BC^2=TB^2+TC^2-2(TB)(TC)\\cos BTC\\] \\[(3\\sqrt{3})^2=5^2+5^2-2(5)^2\\cos BTC\\] \\[\\cos BTC=\\frac{23}{50}\\] Apply Law of Cosines on $\\bigtriangleup STB$ using the fact that $\\angle STB=\\angle BTC$ we have \\[SB^2=ST^2+BT^2-2(ST)(BT)\\cos STB\\] \\[SB=\\sqrt{4^2+5^2-2(4)(5)\\cos BTC}=\\frac{\\sqrt{565}}{5}\\] Apply Pythagorean Theorem on $\\bigtriangleup BSM$ yields \\[SM=\\sqrt{SB^2-BM^2}=\\frac{\\sqrt{1585}}{10}\\] Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\\frac{\\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \\[TP^2=TS^2-SP^2=TM^2-PM^2\\] \\[4^2-x^2=(\\frac{\\sqrt{73}}{2})^2-(\\frac{\\sqrt{1585}}{10}-x)^2\\] Cancelling out the $x^2$ term and solving gets $x=\\frac{181}{2\\sqrt{1585}}$. Finally, by Pythagorean Theorem, \\[TP=\\sqrt{TS^2-SP^2}=\\frac{144}{\\sqrt{1585}}\\] so $\\lfloor m+\\sqrt{n}\\rfloor=183 ~ Nafer",
"Diagram borrowed from Solution 1 [asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);draw(T--O--M,ddash); label(\"$T$\",T,N);label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,NE);label(\"$S$\",S,NW);label(\"$O$\",O,SW);label(\"$M$\",M,NE); label(\"$4$\",(S+T)/2,NW);label(\"$1$\",(S+A)/2,NW);label(\"$5$\",(B+T)/2,NE);label(\"$4$\",(O+T)/2,W); dot(S);dot(O); [/asy] Apply Pythagorean Theorem on $\\bigtriangleup TOB$ yields \\[BO=\\sqrt{TB^2-TO^2}=3\\] Since $\\bigtriangleup ABC$ is equilateral, we have $\\angle MOB=60^{\\circ}$ and \\[BC=2BM=2(OB\\sin MOB)=3\\sqrt{3}\\] Apply Pythagorean Theorem on $\\bigtriangleup TMB$ yields \\[TM=\\sqrt{TB^2-BM^2}=\\sqrt{5^2-(\\frac{3\\sqrt{3}}{2})^2}=\\frac{\\sqrt{73}}{2}\\] Apply Law of Cosines on $\\bigtriangleup TBC$ we have \\[BC^2=TB^2+TC^2-2(TB)(TC)\\cos BTC\\] \\[(3\\sqrt{3})^2=5^2+5^2-2(5)^2\\cos BTC\\] \\[\\cos BTC=\\frac{23}{50}\\] Apply Law of Cosines on $\\bigtriangleup STB$ using the fact that $\\angle STB=\\angle BTC$ we have \\[SB^2=ST^2+BT^2-2(ST)(BT)\\cos STB\\] \\[SB=\\sqrt{4^2+5^2-2(4)(5)\\cos BTC}=\\frac{\\sqrt{565}}{5}\\] Apply Pythagorean Theorem on $\\bigtriangleup BSM$ yields \\[SM=\\sqrt{SB^2-BM^2}=\\frac{\\sqrt{1585}}{10}\\] Let the perpendicular from $T$ hits $SBC$ at $P$. Let $SP=x$ and $PM=\\frac{\\sqrt{1585}}{10}-x$. Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \\[TP^2=TS^2-SP^2=TM^2-PM^2\\] \\[4^2-x^2=(\\frac{\\sqrt{73}}{2})^2-(\\frac{\\sqrt{1585}}{10}-x)^2\\] Cancelling out the $x^2$ term and solving gets $x=\\frac{181}{2\\sqrt{1585}}$. Finally, by Pythagorean Theorem, \\[TP=\\sqrt{TS^2-SP^2}=\\frac{144}{\\sqrt{1585}}\\] so $\\lfloor m+\\sqrt{n}\\rfloor=183 ~ Nafer"
] |
2006-I-15
| 2,006
| 15
|
Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$
| 27
|
I
|
[
"Solution 1 Suppose $b_{i} = \\frac {x_{i}}3$. We have \\[\\sum_{i = 1}^{2006}b_{i}^{2} = \\sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \\sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\\] So \\[\\sum_{i = 0}^{2005}b_{i} = \\frac {b_{2006}^{2} - 2006}2\\] Now \\[\\sum_{i = 1}^{2006}b_{i} = \\frac {b_{2006}^{2} + 2b_{2006} - 2006}2\\] Therefore \\[\\left|\\sum_{i = 1}^{2006}b_{i}\\right| = \\left|\\frac {(b_{2006} + 1)^{2} - 2007}2\\right|\\geq \\left|\\frac{2025 - 2007}{2}\\right| = 9\\] This lower bound can be achieved if we use $b_1 = -1$, $b_2 = 0$, $b_3 = -1$, $b_4 = 0$, and so on until $b_{1962} = 0$, after which we let $b_k = b_{k - 1} + 1$ so that $b_{2006} = 44$. So \\[\\left|\\sum_{i = 1}^{2006}x_{i}\\right|\\geq 027. ~ Spacesam",
"Suppose $b_{i} = \\frac {x_{i}}3$. We have \\[\\sum_{i = 1}^{2006}b_{i}^{2} = \\sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \\sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\\] So \\[\\sum_{i = 0}^{2005}b_{i} = \\frac {b_{2006}^{2} - 2006}2\\] Now \\[\\sum_{i = 1}^{2006}b_{i} = \\frac {b_{2006}^{2} + 2b_{2006} - 2006}2\\] Therefore \\[\\left|\\sum_{i = 1}^{2006}b_{i}\\right| = \\left|\\frac {(b_{2006} + 1)^{2} - 2007}2\\right|\\geq \\left|\\frac{2025 - 2007}{2}\\right| = 9\\] This lower bound can be achieved if we use $b_1 = -1$, $b_2 = 0$, $b_3 = -1$, $b_4 = 0$, and so on until $b_{1962} = 0$, after which we let $b_k = b_{k - 1} + 1$ so that $b_{2006} = 44$. So \\[\\left|\\sum_{i = 1}^{2006}x_{i}\\right|\\geq 027. ~ Spacesam",
"First, we state that iff $x_{i - 1}\\ge0$, $|x_i| = |x_{i - 1}| + 3$ and iff $x_{i - 1} < 0$, $|x_i| = |x_{i - 1}| - 3$. Now suppose $x_i = x_j$ for some $0\\le i < j\\le2006$. Now, this means that $|x_i| = |x_j|$, and so the number of positive numbers in the set $\\{x_i,x_{i + 1},\\ldots,x_{j - 1}\\}$ equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -3. If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be $x_k$ and $x_{k + 1}$. Since one is positive and the other is negative, $|x_{k + 2}| = |x_{k + 1}|\\pm3 = |x_k|\\pm3\\mp3 = |x_k| = |x_{k - 1} + 3|$. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those). Now, take all of the repeating subsequences out of the original sequence. The only thing that will be left will be a sequence $\\{0,3,6,9,\\cdots,3k\\}$ for some even $k$. Since we started with 2006 terms, we removed $2006 - k$ (an even number) with an average of -3/2. Thus, the sum of both this remaining sequence and the removed stuff is $(2006 - k)( - 3/2) + \\sum_{i = 1}^k3k = (3/2)(k - 2006 + k(k + 1)) = 3/2(k^2 + 2k - 2006)$. This must be minimized, so we find the roots: $k^2 + 2k = 2006\\implies (k + 1)^2 = 2007$ and $44^2 = 1936 < 2007 < 2025 = 45^2$. Plugging in $k = 44$ yields $(3/2)(2025 - 2007) = 27$ (and $k = 42$ yields $- 237$, a worse result). Thus, $027. ~ Spacesam",
"We know $|x_k| = |x_{k - 1} + 3|$. We get rid of the absolute value by squaring both sides: ${x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9$. So we set this up: \\begin{align*} {x_1}^2 - {x_0}^2 & = 6{x_0} + 9 \\\\ {x_2}^2 - {x_1}^2 & = 6{x_1} + 9 \\\\ & \\vdots \\\\ {x_{2007}}^2 - {x_{2006}}^2 & = 6{x_{2006}} + 9 \\end{align*} There are $2007$ equations. Sum them. We get: ${x_{2007}}^2 = 6\\left(x_1 + x_2 + \\cdots + x_{2006}\\right) + 9\\cdot{2007}$ So $|x_1 + x_2 + \\cdots + x_{2006}| = \\tfrac 16 \\left|{x_{2007}}^2 - 9\\cdot{2007}\\right|$ We know $3\\ |\\ x_{2007}$ and we want to minimize $\\left|{x_{2007}}^2 - 9\\cdot{2007}\\right|$, so $x_{2007}$ must be $3\\cdot{45}$ for it to be minimal ($45^2 = 2025$ which is closest to $2007$). We can achieve this with $x_k=3k$ till $x_{45}=135$ and then alternating $x_{46}=132$, $x_{47}=135$ and so on ... Then $x_{2k}=132$ and $x_{2k+1}=135$ for all $k>22$. Since $2007$ is odd, we have $x_{2007}=135$. This means that $|x_1 + x_2 + \\cdots + x_{2006}| = \\tfrac 16 \\left|9(2025 - 2007)\\right| = 027. ~ Spacesam",
"Playing around with a couple numbers, we see that we can generate the sequence $0, 3, -6, 3, -6, \\cdots$, and we can also generate the sequence $3, 6, 9, 12, \\cdots$ after each $-6$ value. Thus, we will apply this to try and find some bounds. We can test if the first $1000$ pairs of numbers each sum up to $-3$, and the rest form an arithmetic sequence, if the first $990$ pairs sum up to $-3$, and so on. When we get to $980$, we find that $980(-3) + 3 + 6 + \\cdots + 3\\cdot 46 = 303$. If we shift the number of pairs up by $1$, we get $981(-3) + 3 + 6 + \\cdots + 3\\cdot 44 = 027. ~ Spacesam",
"We will work our way from $|x_0|$ to $|x_0+x_1|$ to $|x_0+x_1+x_2+\\dots+x_{2006}|$. Let the sum $|x_0+x_1+x_2+\\dots+x_i|=S_i$. Seeing as the value for the sum we want is always nonnegative, the best pseudo-strategy would be to stay as close to $0$ as possible as we increase $i$ to eventually get to $i=2006$. It turns out that a greedy algorithm works here. Let us start with some smaller values of $i$. Note that we can describe $x_{k+1}$ in terms of $x_k$ in the following way: take $x_k$, add $3$, and either multiply by $-1$ or not. We know that $S_0=0$. When we get to $S_1$, we add in $x_1$, which is either $3$ or $-3$. It does not make a difference which one we choose, so we can choose $3$ for convenience. Now, $S_1=3$. $x_2$ is either $\\pm6$; adding $6$ would take us farther away from $0$, but adding $-6$ is an okay move. Thus, let $x_2=-6$, so $S_2=-3$. $x_3$ is $\\pm3$. Adding $3$ is clearly the right choice here, which takes us to $0$. Thus, $x_3=3$ and $S_3=0$. Let us look at how this greedy algorithm performs in general. $\\begin{tabular}[t]{c|ccccccc} S_i&3a&-3&3a-3&-6&3a-6&-9&\\cdots\\\\ x_i&3a&-3a-3&3a&-3a-3&3a&-3a-3&\\cdots\\\\ \\end{tabular}$ (Error compiling LaTeX. Unknown error_msg) (the table doesn't work; if you desire, please to go https://artofproblemsolving.com/texer/zzyacvnp to view) The sums alternate between separate arithmetic sequences. One of them is of particular interest to us: specifically, the one that goes $3a,3a-3,3a-6,\\cdots$, because it will eventually become $0$. One can then derive using simple algebra how long it will take to reach zero, depending on $a$; it turns out that it takes $2a+1$ for each cycle to complete. We can then see that $S_{k^2-1}=0$ for positive integers $k$; thus, $S_{1935}=0$. We can then go through our algorithm, and it turns out that $S_{2006}=027. ~Technodoggo (sorry for the rushed solution!)"
] |
2006-II-1
| 2,006
| 1
|
In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$ .
| 46
|
II
|
[
"Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$. The diagonal $BF=\\sqrt{AB^2+AF^2}=\\sqrt{x^2+x^2}=x\\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\\frac{x^2}{2}\\cdot 2$, and the area of the rectangle BCEF equals $x\\cdot x\\sqrt{2}=x^2\\sqrt{2}$ Then we have to solve the equation $2116(\\sqrt{2}+1)=x^2\\sqrt{2}+x^2$. $2116(\\sqrt{2}+1)=x^2(\\sqrt{2}+1)$ $2116=x^2$ $x=46$ Therefore, $AB$ is $46. ~removal of extraneous zeros by K124659",
"Because $\\angle B$, $\\angle C$, $\\angle E$, and $\\angle F$ are congruent, the degree-measure of each of them is ${{720-2\\cdot90}\\over4}= 135$. Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$. Then $BF=x\\sqrt2$. Thus \\begin{align*} 2116(\\sqrt2+1)&=[ABCDEF]\\\\ &=2\\cdot {1\\over2}x^2+x\\cdot x\\sqrt2=x^2(1+\\sqrt2), \\end{align*}so $x^2=2116$, and $x=46}\",F,W); [/asy] ~minor asymptote edit by Yiyj1 ~removal of extraneous zeros by K124659"
] |
2006-II-2
| 2,006
| 2
|
The lengths of the sides of a triangle with positive area are $\log_{10} 12$ , $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$ .
| 893
|
II
|
[
"By the Triangle Inequality and applying the well-known logarithmic property $\\log_{c} a + \\log_{c} b = \\log_{c} ab$, we have that $\\log_{10} 12 + \\log_{10} n > \\log_{10} 75$ $\\log_{10} 12n > \\log_{10} 75$ $12n > 75$ $n > \\frac{75}{12} = \\frac{25}{4} = 6.25$ Also, $\\log_{10} 12 + \\log_{10} 75 > \\log_{10} n$ $\\log_{10} 12\\cdot75 > \\log_{10} n$ $n < 900$ Combining these two inequalities: \\[6.25 < n < 900\\] Thus $n$ is in the set $(6.25 , 900)$; the number of positive integer $n$ which satisfies this requirement is $893."
] |
2006-II-3
| 2,006
| 3
|
Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$ .
| 49
|
II
|
[
"Note that the product of the first $100$ positive odd integers can be written as $1\\cdot 3\\cdot 5\\cdot 7\\cdots 195\\cdot 197\\cdot 199=\\frac{1\\cdot 2\\cdots200}{2\\cdot4\\cdots200} = \\frac{200!}{2^{100}\\cdot 100!}$ Hence, we seek the number of threes in $200!$ decreased by the number of threes in $100!.$ There are $\\left\\lfloor \\frac{200}{3}\\right\\rfloor+\\left\\lfloor\\frac{200}{9}\\right\\rfloor+\\left\\lfloor \\frac{200}{27}\\right\\rfloor+\\left\\lfloor\\frac{200}{81}\\right\\rfloor =66+22+7+2=97$ threes in $200!$ and $\\left\\lfloor \\frac{100}{3}\\right\\rfloor+\\left\\lfloor\\frac{100}{9}\\right\\rfloor+\\left\\lfloor \\frac{100}{27}\\right\\rfloor+\\left\\lfloor\\frac{100}{81}\\right\\rfloor=33+11+3+1=48$ threes in $100!$ Therefore, we have a total of $97-48=049 threes. For more information, see also prime factorizations of a factorial.",
"We count the multiples of $3^k$ below 200 and subtract the count of multiples of $2\\cdot 3^k$: \\[\\left\\lfloor \\frac{200}{3}\\right\\rfloor - \\left\\lfloor \\frac{200}{6}\\right\\rfloor +\\left\\lfloor\\frac{200}{9}\\right\\rfloor - \\left\\lfloor \\frac{200}{18}\\right\\rfloor +\\left\\lfloor \\frac{200}{27}\\right\\rfloor - \\left\\lfloor \\frac{200}{54}\\right\\rfloor+\\left\\lfloor\\frac{200}{81}\\right\\rfloor - \\left\\lfloor \\frac{200}{162}\\right\\rfloor\\] \\[= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.\\]",
"We can use a modified version of Legendre's Formula. First, we count the number of multiples of 3 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is the same as the number of multiples of 3 in the sequence 3, 9, 15, 21, ..., 192, 195. There are clearly 33 terms in this sequence. Next, we count the number of multiples of 9 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is the same as the number of multiples of 9 in the sequence 9, 18, 27, 36, ...., 189, 198 - but there's a catch. Note that every other member of this sequence isn't odd and thus is not part of the product of the first 100 odd integers, so our new sequence is actually 9, 27, 45...189. Divide every term by 9 to get a new sequence; 1, 3, 5...21, which clearly has 11 terms. Next, we similarly count the number of multiples of 27 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is just 27, 81, 135, 189, so 4 multiples here. Finally, we count the number of multiples of 81 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. There is only one such multiple, 81. Any power of 3 above 81 doesn't fit into our sequence. Finally, we have 33+11+4+1=49. Our final answer is 49. (Note that I oversimplified this a lot, in real life we wouldn't have to list out the sequences as tediously as I did).",
"We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\\equiv 2 \\pmod{3}$ in set S will be divisible by 3. In other words, every number $\\equiv 3 \\pmod{6}$. This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. $33 + 11 + 4 + 1 = 49. -jackshi2006"
] |
2006-II-5
| 2,006
| 5
|
When rolling a certain unfair six-sided die with faces numbered $1, 2, 3, 4, 5$ , and $6$ , the probability of obtaining face $F$ is greater than $\frac{1}{6}$ , the probability of obtaining the face opposite is less than $\frac{1}{6}$ , the probability of obtaining any one of the other four faces is $\frac{1}{6}$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $\frac{47}{288}$ . Given that the probability of obtaining face $F$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$
| 29
|
II
|
[
"Without loss of generality, assume that face $F$ has a 6, so the opposite face has a 1. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}$, totaling $4 \\cdot \\frac{1}{36} = \\frac{1}{9}$. Subtracting all these probabilities from $\\frac{47}{288}$ leaves $\\frac{15}{288}=\\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$: \\[A(6)\\cdot B(1)+B(6)\\cdot A(1)=\\frac{5}{96}\\] Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so \\begin{align*}A(1)\\cdot A(6)+A(1)\\cdot A(6)&=\\frac{5}{96}\\\\ A(1)\\cdot A(6)&=\\frac{5}{192}\\end{align*} Also, we know that $A(2)=A(3)=A(4)=A(5)=\\frac{1}{6}$ and that the total probability must be $1$, so: \\[A(1)+4 \\cdot \\frac{1}{6}+A(6)=\\frac{6}{6} \\Longrightarrow A(1)+A(6)=\\frac{1}{3}\\] Combining the equations: \\begin{align*}A(6)\\left(\\frac{1}{3}-A(6)\\right)&=\\frac{5}{192}\\\\ 0 &= 192 \\left(A(6)\\right)^2 - 64 \\left(A(6)\\right) + 5\\\\ A(6)&=\\frac{64\\pm\\sqrt{64^2 - 4 \\cdot 5 \\cdot 192}}{2\\cdot192} =\\frac{5}{24}, \\frac{1}{8}\\end{align*} We know that $A(6)>\\frac{1}{6}$, so it can't be $\\frac{1}{8}$. Therefore, the probability is $\\frac{5}{24}$ and the answer is $5+24=29, for example, and replaced the others with variables too, but the notation would have been harder to follow.",
"We have that the cube probabilities to land on its faces are $\\frac{1}{6}$, $\\frac{1}{6}$, $\\frac{1}{6}$, $\\frac{1}{6}$ ,$\\frac{1}{6}+x$ ,$\\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \\[4 \\cdot \\left(\\frac{1}{6} \\right)^2+2 \\left(\\frac{1}{6}+x \\right) \\left(\\frac{1}{6}-x \\right)=\\frac{47}{288}\\] multiplying by 288 we get: \\[32+16(1-6x)(6x+1)=47 \\Longrightarrow 16(1-36x^2)=15\\] dividing by 16 and rearranging we get: \\[\\frac{1}{16}=36x^2 \\longrightarrow x=\\frac{1}{24}\\] so the probability F which is greater than $\\frac{1}{6}$ is equal $\\frac{1}{6}+\\frac{1}{24}=\\frac{5}{24}\\longrightarrow 24+5=29",
"Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is\\[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\\]Let the probability of obtaining face $F$ be $(1/6)+x$. Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$. Therefore \\begin{align*} {{47}\\over{288}}&= 4\\left({1\\over6}\\right)^2+2\\left({1\\over6}+x\\right) \\left({1\\over6}-x\\right)\\cr&= {4\\over36}+2\\left({1\\over36}-x^2\\right)\\cr&= {1\\over6}-2x^2. \\end{align*} Then $2x^2=1/288$, and so $x=1/24$. The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$, and $m+n=29.",
"Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is\\[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\\]Let the probability of obtaining face $F$ be $(1/6)+x$. Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$. Therefore \\begin{align*} {{47}\\over{288}}&= 4\\left({1\\over6}\\right)^2+2\\left({1\\over6}+x\\right) \\left({1\\over6}-x\\right)\\cr&= {4\\over36}+2\\left({1\\over36}-x^2\\right)\\cr&= {1\\over6}-2x^2. \\end{align*} Then $2x^2=1/288$, and so $x=1/24$. The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$, and $m+n=29."
] |
2006-II-6
| 2,006
| 6
|
Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$
| 12
|
II
|
[
"[asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label(\"$A$\", A, NW); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$D$\", D, NE); label(\"$E$\", Ep, SE); label(\"$F$\", F, E); label(\"$A'$\", Ap, W); label(\"$C'$\", Cp, SW); label(\"$D'$\", Dp, E); label(\"$s$\", Ap--B, W); label(\"$1$\", A--D, N); [/asy] Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\\frac{AA'}{A'D'} = \\frac{D'C'}{C'E} \\Longrightarrow \\frac{1 - s}{s} = \\frac{s}{1 - s - CE}$. Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$. $\\angle EAF$ is $60$ degrees, so $\\angle BAE = \\frac{90 - 60}{2} = 15$. Thus, $\\cos 15 = \\cos (45 - 30) = \\frac{\\sqrt{6} + \\sqrt{2}}{4} = \\frac{1}{AE}$, so $AE = \\frac{4}{\\sqrt{6} + \\sqrt{2}} \\cdot \\frac{\\sqrt{6} - \\sqrt{2}}{\\sqrt{6} - \\sqrt{2}} = \\sqrt{6} - \\sqrt{2}$. Since $\\triangle AEF$ is equilateral, $EF = AE = \\sqrt{6} - \\sqrt{2}$. $\\triangle CEF$ is a $45-45-90 \\triangle$, so $CE = \\frac{AE}{\\sqrt{2}} = \\sqrt{3} - 1$. Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \\sqrt{3} - s)$, so $(3 - \\sqrt{3})s = 2 - \\sqrt{3}$. Therefore, $s = \\frac{2 - \\sqrt{3}}{3 - \\sqrt{3}} \\cdot \\frac{3 + \\sqrt{3}}{3 + \\sqrt{3}} = \\frac{3 - \\sqrt{3}}{6}$, and $a + b + c = 3 + 3 + 6 = 12. The solution continues as above.",
"Since $\\triangle AFE$ is equilateral, $\\overline{AE} = \\overline{AF}$. It follows that $\\overline{FC} = \\overline{EC}$. Let $\\overline{FC} = x$. Then, $\\overline{EF} = x\\sqrt{2}$ and $\\overline{DF} = 1-x$. $\\overline{AF} = \\sqrt{1+(1-x)^2} = x\\sqrt{2}$. Square both sides and combine/move terms to get $x^2+2x-2 = 0$. Therefore $x = -1 + \\sqrt{3}$ and $x = -1 - \\sqrt{3}$. The second solution is obviously extraneous, so $x = -1 + \\sqrt{3}$. Now, consider the square ABCD to be on the Cartesian Coordinate Plane with $A = (0,0)$. Then, the line containing $\\overline{AF}$ has slope $\\frac{1}{2-\\sqrt{3}}$ and equation $y = \\frac{1}{2-\\sqrt{3}}x$. The distance from $\\overline{DC}$ to $\\overline{AF}$ is the distance from $y = 1$ to $y = \\frac{1}{2-\\sqrt{3}}x$. Similarly, the distance from $\\overline{AD}$ to $\\overline{AF}$ is the distance from $x = 0$ to $y = \\frac{1}{2-\\sqrt{3}}x$. For some value $x = s$, these two distances are equal. $(s-0) = (1 - (\\frac{1}{2-\\sqrt{3}})s)$ Solving for s, $s = \\frac{3 - \\sqrt{3}}{6}$, and $a + b + c = 3 + 3 + 6 = 12$.",
"Suppose $\\overline{AB} = \\overline{AD} = x.$ Note that $\\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\\angle BAE = \\angle DAF = 15.$ Note that if $\\overline{AD} = x$ and $\\angle BAE = 15$, then $\\overline{AA'}=\\frac{x}{\\tan(15)}.$ Also note that \\[AB = 1 = \\overline{AA'} + \\overline{A'B} = \\frac{x}{\\tan(15)} + x\\] Using the fact $\\tan(15) = 2-\\sqrt{3}$, this yields \\[x = \\frac{1}{3+\\sqrt{3}} = \\frac{3-\\sqrt{3}}{6} \\rightarrow 3 + 3 + 6 = 12\\]",
"Since $AEF$ is equilateral, $AE=EF$. Let $BE=x$. By the Pythagorean theorem, $1+x^2=2(1-x)^2$. Simplifying, we get $x^2-4x+1=0$. By the quadratic formula, the roots are $2 \\pm \\sqrt{3}$. Since $x<1$, we discard the root with the \"+\", giving $x=2-\\sqrt{3}$. [asy] real n; n=0.26794919243; real m; m=0.2113248654; draw((0,0)--(0,n)--(1,0)--(0,0)); draw((0,m)--(m,m)--(m,0)); label((0,0), \"$B$\",SW); label((0,n), \"$E$\",SW); label((0,m), \"$M$\",SW); label((1,0), \"$A$\",SW); label((m,0), \"$N$\",SW); label((m,m), \"$K$\",NE); [/asy] Let the side length of the square be s. Since $MEK$ is similar to $ABE$, $s=\\frac{2-\\sqrt{3}-s}{2-\\sqrt{3}}$. Solving, we get $s=\\frac{3-\\sqrt{3}}{6}$ and the final answer is $012.",
"Since $AEF$ is equilateral, $AE=EF$. Let $BE=x$. By the Pythagorean theorem, $1+x^2=2(1-x)^2$. Simplifying, we get $x^2-4x+1=0$. By the quadratic formula, the roots are $2 \\pm \\sqrt{3}$. Since $x<1$, we discard the root with the \"+\", giving $x=2-\\sqrt{3}$. [asy] real n; n=0.26794919243; real m; m=0.2113248654; draw((0,0)--(0,n)--(1,0)--(0,0)); draw((0,m)--(m,m)--(m,0)); label((0,0), \"$B$\",SW); label((0,n), \"$E$\",SW); label((0,m), \"$M$\",SW); label((1,0), \"$A$\",SW); label((m,0), \"$N$\",SW); label((m,m), \"$K$\",NE); [/asy] Let the side length of the square be s. Since $MEK$ is similar to $ABE$, $s=\\frac{2-\\sqrt{3}-s}{2-\\sqrt{3}}$. Solving, we get $s=\\frac{3-\\sqrt{3}}{6}$ and the final answer is $012."
] |
2006-II-7
| 2,006
| 7
|
Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.
| 738
|
II
|
[
"Solution 1 There are $\\left\\lfloor\\frac{999}{10}\\right\\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting). Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \\cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \\cdot 2 = 162$. Therefore, there are $999 - (99 + 162) = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"There are $\\left\\lfloor\\frac{999}{10}\\right\\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting). Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \\cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \\cdot 2 = 162$. Therefore, there are $999 - (99 + 162) = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"Let $a = \\overline{cde}$ and $b = \\overline{fgh}$ be 3 digit numbers: cde +fgh ---- 1000 $e$ and $h$ must add up to $10$, $d$ and $g$ must add up to $9$, and $c$ and $f$ must add up to $9$. Since none of the digits can be 0, there are $9 \\times 8 \\times 8=576$ possibilites if both numbers are three digits. There are two other scenarios. $a$ and $b$ can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are $9 \\times 8 \\times 2=144$ possibilities (the two accounting for whether $a$ or $b$ has three digits) and for the second case there are $9 \\times 2=18$ possibilities. Thus, thus total possibilities for $(a,b)$ is $576+144+18=738$. Solution 3 We first must notice that we can find all the possible values of $a$ between $1$ and $500$ and then double that result. When $1 < a < 100$ there are $9\\times9 = 81$ possible solution for $a$ so that neither $a$ nor $b$ has a zero in it, counting $1$ through $9$, $11$ through $19$, ..., $81$ through $89$. When $100 < a < 200$ there are $9\\times8 =72$ possible solution for a so that neither a nor b has a zero in it, counting $111$ through $119$, $121$ through $129$, ..., $181$ through $189$. This can clearly be extended to $100k < a < 100(k+1)$ where $k$ is an integer and $0 <k < 9$. Thus for $100 < a < 500$ there are $72\\times4$ = $288$ possible values of $a$. Thus when $1 < a < 500$ there are $288 + 81 =369$ possible values of $a$ and $b$. Doubling this yields $369\\times2= 738$. Solution 4 (Similar to Solution 2) We proceed by casework on the number of digits of $a.$ Case 1: Both $a$ and $b$ have three digits We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \\cdot 8 \\cdot 9 = 576$ possible values in this case. Case 2: $a$ (or $b$) has two digits If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$). There thus are $8 \\cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \\cdot 2 = 144$ possible values in this case. Case 3: $a$ (or $b$) has one digit This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \\cdot 2 = 18$ possible values for this case. Adding them all up, we get $576 + 144 + 18 = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"We first must notice that we can find all the possible values of $a$ between $1$ and $500$ and then double that result. When $1 < a < 100$ there are $9\\times9 = 81$ possible solution for $a$ so that neither $a$ nor $b$ has a zero in it, counting $1$ through $9$, $11$ through $19$, ..., $81$ through $89$. When $100 < a < 200$ there are $9\\times8 =72$ possible solution for a so that neither a nor b has a zero in it, counting $111$ through $119$, $121$ through $129$, ..., $181$ through $189$. This can clearly be extended to $100k < a < 100(k+1)$ where $k$ is an integer and $0 <k < 9$. Thus for $100 < a < 500$ there are $72\\times4$ = $288$ possible values of $a$. Thus when $1 < a < 500$ there are $288 + 81 =369$ possible values of $a$ and $b$. Doubling this yields $369\\times2= 738$. Solution 4 (Similar to Solution 2) We proceed by casework on the number of digits of $a.$ Case 1: Both $a$ and $b$ have three digits We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \\cdot 8 \\cdot 9 = 576$ possible values in this case. Case 2: $a$ (or $b$) has two digits If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$). There thus are $8 \\cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \\cdot 2 = 144$ possible values in this case. Case 3: $a$ (or $b$) has one digit This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \\cdot 2 = 18$ possible values for this case. Adding them all up, we get $576 + 144 + 18 = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"We proceed by casework on the number of digits of $a.$ Case 1: Both $a$ and $b$ have three digits We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \\cdot 8 \\cdot 9 = 576$ possible values in this case. Case 2: $a$ (or $b$) has two digits If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$). There thus are $8 \\cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \\cdot 2 = 144$ possible values in this case. Case 3: $a$ (or $b$) has one digit This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \\cdot 2 = 18$ possible values for this case. Adding them all up, we get $576 + 144 + 18 = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"We proceed by casework on the number of digits of $a.$ Case 1: Both $a$ and $b$ have three digits We now use constructive counting. For the hundreds digit of $a,$ we see that there are $8$ options - the numbers $1$ through $8.$ (If $a = 9,$ that means that $b$ will be a two digit number, and if $a = 0,$ $a$ will have two digits). Similarly, the tens digit can be $1-8$ as well because a tens digit of $0$ is obviously prohibited and a tens digit of $9$ will lead to a tens digit of $0$ in the other number. The units digit can be anything from $1-9.$ Hence, there are $8 \\cdot 8 \\cdot 9 = 576$ possible values in this case. Case 2: $a$ (or $b$) has two digits If $a$ has two digits, the only restrictions are that the units digit must not be $0$ and the tens digit must not be $9$ (because then that would lead to $b$ beginning with $90...$). There thus are $8 \\cdot 9 = 72$ possibilities for $a,$ and we have to multiply by $2$ because there are the same number of possibilities for $b.$ Thus, there are $72 \\cdot 2 = 144$ possible values in this case. Case 3: $a$ (or $b$) has one digit This is easy -- $a$ can be anything from $1$ to $9,$ for a total of $9$ possible values. We multiply this by $2$ to account for the single digit $b$ values, so we have $9 \\cdot 2 = 18$ possible values for this case. Adding them all up, we get $576 + 144 + 18 = 738 with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim",
"For every $a \\in [1, 999]$, $(a, 1000 - a)$ is a potential candidate for a solution, barring the cases where $a$ or $1000 -a$ has zero digits. First, let's consider all viable $a$ that do not have a zero digit. As there are 9 non-zero digits, we have: $9^1$ 1-digit numbers without a zero digit $9^2$ 2-digit numbers without a zero digit $9^3$ 3-digit numbers without a zero digit However, we have still overlooked the cases where $1000 - a$ contains zero digits: Case 1: If the one's digit of $1000 - a$ is zero, then $a$ will trivially end in a zero, which we've already excluded. Case 2: If the ten's digit of $1000 - a$ is zero, with digital representation $\\overline{X0Y}$, then $a$ has the digital representation $\\overline{[9-X]9[10-Y]}$. $X$ and $Y$ can each take on any value in $[1, 9]$ to produce a value in our set of potential $a$. There are thus $9^2$ cases that we overlooked, where $a$ had no zero digits, but $1000 - a$ did. Adding up the cases with $a \\in [1, 999]$ with no zero digits and removing the cases with $1000 - a$ with zero digits gives us \\[(9^1 + 9^2 + 9^3) - 9^2 = 738\\] ~SaifHakim"
] |
2006-II-8
| 2,006
| 8
|
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color. Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed? [asy] size(50); pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]
| 336
|
II
|
[
"If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles. There are 6 possible colors for the center triangle. There are ${6\\choose3} = 20$ possible choices for the three outer triangles, if all three have different colors. There are $6\\cdot 5 = 30$ (or $2 {6\\choose2}$) possible choices for the three outer triangles, if two are one color and the third is a different color. There are ${6\\choose1} = 6$ possible choices for the three outer triangles, if all three are the same color. Thus, in total we have $6\\cdot(20 + 30 + 6) = 336 total possibilities.",
"We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count. Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions. Case 2: 120 degree rotation. Note that the three \"outer\" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this. Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this. Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all. Now, we add our cases up: $1296+36+36+648=2016$. We have to divide by 6, so $2016/6=336 distinguishable ways to color the triangle.",
"We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count. Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions. Case 2: 120 degree rotation. Note that the three \"outer\" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this. Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this. Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all. Now, we add our cases up: $1296+36+36+648=2016$. We have to divide by 6, so $2016/6=336 distinguishable ways to color the triangle.",
"There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \\[6 \\binom{5 + 3}{3} = 6\\binom{8}{3} = 336.\\] -MP8148 Explanation of the bijection by WIlliamgolly: Let 1,2,3,4,5,6 be the colors, and WLOG assume that the middle triangle has a color of 6. Now, the color bijection can be formed as follows: Pick the colors to the immediate right of a divider. If there is no color to the immediate right of a divider, then that color is 6. For example, ||12345| would represent the colors 1,1,6 as the colors chosen. Note for any three colors, there is only one way to fix it on the triangle, thus forming our stars and bars bijection.",
"There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \\[6 \\binom{5 + 3}{3} = 6\\binom{8}{3} = 336.\\] -MP8148 Explanation of the bijection by WIlliamgolly: Let 1,2,3,4,5,6 be the colors, and WLOG assume that the middle triangle has a color of 6. Now, the color bijection can be formed as follows: Pick the colors to the immediate right of a divider. If there is no color to the immediate right of a divider, then that color is 6. For example, ||12345| would represent the colors 1,1,6 as the colors chosen. Note for any three colors, there is only one way to fix it on the triangle, thus forming our stars and bars bijection."
] |
2006-II-9
| 2,006
| 9
|
Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope, and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y),$ and that $x=p-q\sqrt{r},$ where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$
| 27
|
II
|
[
"Call the centers $O_1, O_2, O_3$, the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$, and $s_2$ on $\\mathcal{C}_2$), and the intersection of each common internal tangent to the X-axis $r, s$. $\\triangle O_1r_1r \\sim \\triangle O_2r_2r$ since both triangles have a right angle and have vertical angles, and the same goes for $\\triangle O_2s_2s \\sim \\triangle O_3s_1s$. By proportionality, we find that $O_1r = 4$; solving $\\triangle O_1r_1r$ by the Pythagorean theorem yields $r_1r = \\sqrt{15}$. On $\\mathcal{C}_3$, we can do the same thing to get $O_3s_1 = 4$ and $s_1s = 4\\sqrt{3}$. The vertical altitude of each of $\\triangle O_1r_1r$ and $\\triangle O_3s_1s$ can each by found by the formula $c \\cdot h = a \\cdot b$ (as both products equal twice of the area of the triangle). Thus, the respective heights are $\\frac{\\sqrt{15}}{4}$ and $2\\sqrt{3}$. The horizontal distance from each altitude to the intersection of the tangent with the x-axis can also be determined by the Pythagorean theorem: $\\sqrt{15 - \\frac{15}{16}} = \\frac{15}{4}$, and by 30-60-90: $6$. From this information, the slope of each tangent can be uncovered. The slope of $t_1 = \\frac{\\Delta y}{\\Delta x} = \\frac{\\frac{\\sqrt{15}}{4}}{\\frac{15}{4}} = \\frac{1}{\\sqrt{15}}$. The slope of $t_2 = -\\frac{2\\sqrt{3}}{6} = -\\frac{1}{\\sqrt{3}}$. The equation of $t_1$ can be found by substituting the point $r (4,0)$ into $y = \\frac{1}{\\sqrt{15}}x + b$, so $y = \\frac{1}{\\sqrt{15}}x - \\frac{4}{\\sqrt{15}}$. The equation of $t_2$, found by substituting point $s (16,0)$, is $y = \\frac{-1}{\\sqrt{3}}x + \\frac{16}{\\sqrt{3}}$. Putting these two equations together results in the desired $\\frac{1}{\\sqrt{15}}x - \\frac{4}{\\sqrt{15}} = -\\frac{1}{\\sqrt{3}}x + \\frac{16}{\\sqrt{3}}$ $\\Longrightarrow x = \\frac{16\\sqrt{5} + 4}{\\sqrt{5} + 1} \\cdot \\frac{\\sqrt{5}-1}{\\sqrt{5}-1}$ $= \\frac{76 - 12\\sqrt{5}}{4}$ $= 19 - 3\\sqrt{5}$. Thus, $p + q + r = 19 + 3 + 5 \\Longrightarrow 027."
] |
2006-II-10
| 2,006
| 10
|
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 831
|
II
|
[
"Solution 1 The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$. We let this probability be $p$; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$. Of these three cases ($|A| > |B|, |A| < |B|, |A|=|B|$), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). There are ${5\\choose k}$ ways to $A$ to have $k$ victories, and ${5\\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$, $1-2p = \\frac{1}{2^{5} \\times 2^{5}}\\left(\\sum_{k=0}^{5} {5\\choose k}^2\\right) = \\frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \\frac{126}{512}.$ Thus $p = \\frac 12 \\left(1-\\frac{126}{512}\\right) = \\frac{193}{512}$. The desired probability is the sum of the cases when $|A| \\ge |B|$, so the answer is $\\frac{126}{512} + \\frac{193}{512} = \\frac{319}{512}$, and $m+n = 831.",
"The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$. We let this probability be $p$; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$. Of these three cases ($|A| > |B|, |A| < |B|, |A|=|B|$), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases). There are ${5\\choose k}$ ways to $A$ to have $k$ victories, and ${5\\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$, $1-2p = \\frac{1}{2^{5} \\times 2^{5}}\\left(\\sum_{k=0}^{5} {5\\choose k}^2\\right) = \\frac{1^2+5^2+10^2+10^2+5^2+1^2}{1024} = \\frac{126}{512}.$ Thus $p = \\frac 12 \\left(1-\\frac{126}{512}\\right) = \\frac{193}{512}$. The desired probability is the sum of the cases when $|A| \\ge |B|$, so the answer is $\\frac{126}{512} + \\frac{193}{512} = \\frac{319}{512}$, and $m+n = 831.",
"You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games. If $A$ wins 0 games, then $B$ must win 0 games and the probability of this is $\\frac{{5 \\choose 0}}{2^5} \\frac{{5 \\choose 0}}{2^5} = \\frac{1}{1024}$. If $A$ wins 1 games, then $B$ must win 1 or less games and the probability of this is $\\frac{{5 \\choose 1}}{2^5} \\frac{{5 \\choose 0}+{5 \\choose 1}}{2^5} = \\frac{30}{1024}$. If $A$ wins 2 games, then $B$ must win 2 or less games and the probability of this is $\\frac{{5 \\choose 2}}{2^5} \\frac{{5 \\choose 0}+{5 \\choose 1}+{5 \\choose 2}}{2^5} = \\frac{160}{1024}$. If $A$ wins 3 games, then $B$ must win 3 or less games and the probability of this is $\\frac{{5 \\choose 3}}{2^5} \\frac{{5 \\choose 0}+{5 \\choose 1}+{5 \\choose 2}+{5 \\choose 3}}{2^5} = \\frac{260}{1024}$. If $A$ wins 4 games, then $B$ must win 4 or less games and the probability of this is $\\frac{{5 \\choose 4}}{2^5} \\frac{{5 \\choose 0}+{5 \\choose 1}+{5 \\choose 2}+{5 \\choose 3}+{5 \\choose 4}}{2^5} = \\frac{155}{1024}$. If $A$ wins 5 games, then $B$ must win 5 or less games and the probability of this is $\\frac{{5 \\choose 5}}{2^5} \\frac{{5 \\choose 0}+{5 \\choose 1}+{5 \\choose 2}+{5 \\choose 3}+{5 \\choose 4}+{5 \\choose 5}}{2^5} = \\frac{32}{1024}$. Summing these 6 cases, we get $\\frac{638}{1024}$, which simplifies to $\\frac{319}{512}$, so our answer is $319 + 512 = 831.",
"We can apply the concept of generating functions here. The generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$. Thus, the total generating function for number of games he wins is ${(1 + 0x)(1 + x)^{5}} = (1 + 5x^{1} + 10x^{2} + 10x^{3} + 5x^{4} + x^{5})$. The generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$. Thus, the generating function for $A$ is $1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$. The probability that $B$ wins 0 games is $\\frac{1}{32}$. Since the coefficients for all $x^{n}$ where $n \\geq 1$ sums to 32, the probability that $A$ wins more games is $\\frac{32}{32}$. Thus, the probability that $A$ has more wins than $B$ is $\\frac{1}{32} \\times \\frac{32}{32} + \\frac{5}{32} \\times \\frac{31}{32} + \\frac{10}{32} \\times \\frac{26}{32} + \\frac{10}{32} \\times \\frac{16}{32} + \\frac{5}{32} \\times \\frac{6}{32} +\\frac{1}{32} \\times \\frac{1}{32} = \\frac{638}{1024} = \\frac{319}{512}$. Thus, $319 + 512 = 831.",
"After the first game, there are $10$ games we care about-- those involving $A$ or $B$. There are $3$ cases of these $10$ games: $A$ wins more than $B$, $B$ wins more than $A$, or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\\binom{5}{0}^2+\\binom{5}{1}^2+\\binom{5}{2}^2+\\binom{5}{3}^2+\\binom{5}{4}^2+\\binom{5}{5}^2 = \\binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity. There are therefore $\\frac{1024-252}{2} = 386$ possibilities for each of the other two cases. If $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\\frac{638}{1024} = \\frac{319}{512}$, and $m+n=831."
] |
2006-II-11
| 2,006
| 11
|
A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000.
| 834
|
II
|
[
"Solution 1 Define the sum as $s$. Since $a_n\\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, the sum will be: $s = a_{28} + \\sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\\\ s = a_{28} + \\left(\\sum^{30}_{k=4} a_{k} - \\sum^{29}_{k=3} a_{k}\\right) - \\left(\\sum^{28}_{k=2} a_{k}\\right)\\\\ s = a_{28} + (a_{30} - a_{3}) - \\left(\\sum^{28}_{k=2} a_{k}\\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\\\ s = -s + a_{28} + a_{30}$ Thus $s = \\frac{a_{28} + a_{30}}{2}$, and $a_{28},\\,a_{30}$ are both given; the last four digits of their sum is $3668$, and half of that is $1834$. Therefore, the answer is $834. Solution by zeroman; clarified by srisainandan6",
"Define the sum as $s$. Since $a_n\\ = a_{n + 3} - a_{n + 2} - a_{n + 1}$, the sum will be: $s = a_{28} + \\sum^{27}_{k=1} (a_{k+3}-a_{k+2}-a_{k+1}) \\\\ s = a_{28} + \\left(\\sum^{30}_{k=4} a_{k} - \\sum^{29}_{k=3} a_{k}\\right) - \\left(\\sum^{28}_{k=2} a_{k}\\right)\\\\ s = a_{28} + (a_{30} - a_{3}) - \\left(\\sum^{28}_{k=2} a_{k}\\right) = a_{28} + a_{30} - a_{3} - (s - a_{1})\\\\ s = -s + a_{28} + a_{30}$ Thus $s = \\frac{a_{28} + a_{30}}{2}$, and $a_{28},\\,a_{30}$ are both given; the last four digits of their sum is $3668$, and half of that is $1834$. Therefore, the answer is $834. Solution by zeroman; clarified by srisainandan6",
"Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms: $a_{1}\\equiv 1 \\pmod {1000} \\\\ a_{2}\\equiv 1 \\pmod {1000} \\\\ a_{3}\\equiv 1 \\pmod {1000} \\\\ a_{4}\\equiv 3 \\pmod {1000} \\\\ a_{5}\\equiv 5 \\pmod {1000} \\\\ \\cdots \\\\ a_{25} \\equiv 793 \\pmod {1000} \\\\ a_{26} \\equiv 281 \\pmod {1000} \\\\ a_{27} \\equiv 233 \\pmod {1000} \\\\ a_{28} \\equiv 307 \\pmod {1000}$ Adding all the residues shows the sum is congruent to $834. Solution by zeroman; clarified by srisainandan6",
"Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms: $a_{1}\\equiv 1 \\pmod {1000} \\\\ a_{2}\\equiv 1 \\pmod {1000} \\\\ a_{3}\\equiv 1 \\pmod {1000} \\\\ a_{4}\\equiv 3 \\pmod {1000} \\\\ a_{5}\\equiv 5 \\pmod {1000} \\\\ \\cdots \\\\ a_{25} \\equiv 793 \\pmod {1000} \\\\ a_{26} \\equiv 281 \\pmod {1000} \\\\ a_{27} \\equiv 233 \\pmod {1000} \\\\ a_{28} \\equiv 307 \\pmod {1000}$ Adding all the residues shows the sum is congruent to $834. Solution by zeroman; clarified by srisainandan6",
"All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\\frac{1}{2}, 0, \\frac{1}{2})$, at least for the first few terms. From this, we have that $\\sum_{k=1}^{28}{a_k} = \\frac{a_{28}+a_{30}}{2} \\equiv{834. Solution by zeroman; clarified by srisainandan6",
"All terms in the sequence are sums of previous terms, so the sum of all terms up to a certain point must be some linear combination of the first three terms. Also, we are given $a_{28}, a_{29},$ and $a_{30}$, so we can guess that there is some way to use them in a formula. Namely, we guess that there exists some $p, q, r$ such that $\\sum_{k=1}^{n}{a_k} = pa_n+qa_{n+1}+ra_{n+2}$. From here, we list out the first few terms of the sequence and the cumulative sums, and with a little bit of substitution and algebra we see that $(p, q, r) = (\\frac{1}{2}, 0, \\frac{1}{2})$, at least for the first few terms. From this, we have that $\\sum_{k=1}^{28}{a_k} = \\frac{a_{28}+a_{30}}{2} \\equiv{834. Solution by zeroman; clarified by srisainandan6"
] |
2006-II-12
| 2,006
| 12
|
Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overline{AD}.$ Let $F$ be the intersection of $l_1$ and $l_2.$ Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A.$ Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r},$ where $p, q,$ and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r.$
| 865
|
II
|
[
"[asy] size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP(\"A\",A,N,s)--MP(\"B\",B,W,s)--MP(\"C\",C,E,s)--cycle);D(O); D(B--MP(\"D\",D,W,s)--MP(\"F\",F,s)--MP(\"E\",E1,E,s)--C); D(A--F);D(B--MP(\"G\",G,SW,s)--C); MP(\"11\",(A+E1)/2,NE);MP(\"13\",(A+D)/2,NW);MP(\"l_1\",(D+F)/2,SW);MP(\"l_2\",(E1+F)/2,SE); [/asy] Notice that $\\angle{E} = \\angle{BGC} = 120^\\circ$ because $\\angle{A} = 60^\\circ$. Also, $\\angle{GBC} = \\angle{GAC} = \\angle{FAE}$ because they both correspond to arc ${GC}$. So $\\Delta{GBC} \\sim \\Delta{EAF}$. \\[[EAF] = \\frac12 (AE)(EF)\\sin \\angle AEF = \\frac12\\cdot11\\cdot13\\cdot\\sin{120^\\circ} = \\frac {143\\sqrt3}4.\\] Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \\frac {BC^2}{AF^2}\\cdot[EAF] = \\frac {12}{11^2 + 13^2 - 2\\cdot11\\cdot13\\cdot\\cos120^\\circ}\\cdot\\frac {143\\sqrt3}4 = \\frac {429\\sqrt3}{433}$. Therefore, the answer is $429+433+3=865.",
"Solution by e_power_pi_times_i/edited by srisainandan6 Let the center of the circle be $O$ and the origin. Then, $A (0,2)$, $B (-\\sqrt{3}, -1)$, $C (\\sqrt{3}, -1)$. $D$ and $E$ can be calculated easily knowing $AD$ and $AE$, $D (-\\dfrac{13}{2}, \\dfrac{-13\\sqrt{3}+4}{2})$, $E (\\dfrac{11}{2}, \\dfrac{-11\\sqrt{3}+4}{2})$. As $DF$ and $EF$ are parallel to $AE$ and $AD$, $F (-1, -12\\sqrt{3}+2)$. $G$ and $A$ is the intersection between $AF$ and circle $O$. Therefore $G (-\\dfrac{48\\sqrt{3}}{433}, -\\dfrac{862}{433})$. Using the Shoelace Theorem, $[CBG] = \\dfrac{429\\sqrt{3}}{433}$, so the answer is $865.",
"Solution by e_power_pi_times_i/edited by srisainandan6 Let the center of the circle be $O$ and the origin. Then, $A (0,2)$, $B (-\\sqrt{3}, -1)$, $C (\\sqrt{3}, -1)$. $D$ and $E$ can be calculated easily knowing $AD$ and $AE$, $D (-\\dfrac{13}{2}, \\dfrac{-13\\sqrt{3}+4}{2})$, $E (\\dfrac{11}{2}, \\dfrac{-11\\sqrt{3}+4}{2})$. As $DF$ and $EF$ are parallel to $AE$ and $AD$, $F (-1, -12\\sqrt{3}+2)$. $G$ and $A$ is the intersection between $AF$ and circle $O$. Therefore $G (-\\dfrac{48\\sqrt{3}}{433}, -\\dfrac{862}{433})$. Using the Shoelace Theorem, $[CBG] = \\dfrac{429\\sqrt{3}}{433}$, so the answer is $865.",
"Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$. Since $BAC$ is equilateral with angle of $60^{\\circ}$, angle $D$ is $120^{\\circ}$. Use law of cosines to find $AF = \\sqrt{433}$. Then use law of sines to find angle $BAG$ and $GAC$. Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$. Next we use law of cosine on triangles $BAG$ and $GAC$, solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$. Next, $AG = 2\\cdot R \\cos{\\theta}$, where R is radius of circle $= 2$ and $\\theta =$ angle $BAG$. We already know sine of the angle so find cosine, hence we have found $AG$. At this point it is system of equation yielding $CG = \\frac{26\\sqrt{3}}{\\sqrt{433}}$ and $BG = \\frac{22\\sqrt{3}}{\\sqrt{433}}$. Given $[CBG] = \\frac{BC \\cdot CG \\cdot BG}{4R}$, and $BC = 2\\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \\frac{429\\sqrt{3}}{433}$, to give answer = $865.",
"Note that $AB=2\\sqrt3$, $DF=11$, and $EF=13$. If we take a homothety of the parallelogram with respect to $A$, such that $F$ maps to $G$, we see that $\\frac{[ABG]}{[ACG]}=\\frac{11}{13}$. Since $\\angle AGB=\\angle AGC=60^{\\circ}$, from the sine area formula we have $\\frac{BG}{CG}=\\frac{11}{13}$. Let $BG=11k$ and $CG=13k$. By Law of Cosines on $\\triangle BGC$, we have \\[12=k^2(11^2+11\\cdot13+13^2)=433k^2\\implies k^2=\\frac{12}{433}\\] Thus, $[CBG]=\\frac12 (11k)(13k)\\sin 120^{\\circ} = \\frac{\\sqrt3}{4}\\cdot 143\\cdot \\frac{12}{433}=\\frac{429\\sqrt3}{433}\\implies865. ~rayfish",
"First, by properties of cyclic quadrilterals, $<BGC = 180-<BAC=120$. Next, we notice quadrilateral AEFD is a parallelogram, and opposites angles are supplementary, and therefore $<ADF = 180-<BAC = 120$. Finally, by inscribed angles, $<BAG = <BCG$. Thus we can conclude $\\triangle CBG \\sim \\triangle AFD$ by AA similarity. Because $\\triangle FAE \\cong \\triangle AFD$, $\\triangle FAE \\sim \\triangle CBG$ aswell. So, we can write $\\frac{BC}{AF} = \\frac{BG}{DF} \\to \\frac{2\\sqrt{3}}{\\sqrt{433}} = \\frac{BG}{11}$ since $AF = \\sqrt{433}$ by the Law of Cosines. This gives $BG = \\frac{22\\sqrt{3}}{\\sqrt{433}}$ and doing an identical process yields $CG = \\frac{26\\sqrt{3}}{\\sqrt{433}}$. Thus, $[CBG] = \\frac{26\\sqrt{3}}{\\sqrt{433}} \\cdot \\frac{22\\sqrt{3}}{\\sqrt{433}} \\cdot \\sin 120 = \\frac{429\\sqrt{3}}{433} \\to p+q+r = 865. ~BossLu99"
] |
2006-II-13
| 2,006
| 13
|
How many integers $N$ less than 1000 can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$ ?
| 15
|
II
|
[
"Let the first odd integer be $2n+1$, $n\\geq 0$. Then the final odd integer is $2n+1 + 2(j-1) = 2(n+j) - 1$. The odd integers form an arithmetic sequence with sum $N = j\\left(\\frac{(2n+1) + (2(n+j)-1)}{2}\\right) = j(2n+j)$. Thus, $j$ is a factor of $N$. Since $n\\geq 0$, it follows that $2n+j \\geq j$ and $j\\leq \\sqrt{N}$. Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$. This means $N=p_1^2p_2^2$ or $N=p_1p_2^4$. Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$. Instead we do some casework: If $N$ is odd, then $j$ must also be odd. For every odd value of $j$, $2n+j$ is also odd, making this case valid for all odd $j$. Looking at the forms above and the bound of $1000$, $N$ must be \\[(3^2\\cdot5^2),\\ (3^2\\cdot7^2),\\ (3^4\\cdot5),\\ (3^4\\cdot7),\\ (3^4\\cdot 11)\\] Those give $5$ possibilities for odd $N$. If $N$ is even, then $j$ must also be even. Substituting $j=2k$, we get \\[N = 4k(n+k) \\Longrightarrow \\frac{N}{4} = k(n+k)\\] Now we can just look at all the prime factorizations since $(n+k)$ cover the integers for any $k$. Note that our upper bound is now $250$: \\[\\frac{N}{4} = (2^2\\cdot3^2),(2^2\\cdot5^2),(2^2\\cdot7^2), (3^2\\cdot5^2), (2^4\\cdot3), (2^4\\cdot5), (2^4\\cdot7), (2^4\\cdot11), (2^4\\cdot13), (3^4\\cdot2)\\] Those give $10$ possibilities for even $N$. The total number of integers $N$ is $5 + 10 = 15.",
"Let the largest odd number below the sequence be the $q$th positive odd number, and the largest odd number in the sequence be the $p$th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$. We then perform casework based on the parity of $p-q$. If $p-q$ is odd, then $p^2-q^2$ must always be odd. Therefore, to have 5 pairs of odd factors, we must have either $9$ (in which case the number is a perfect square) or $10$ factors. Considering the upper bound, the only way this can happen is $p_1^4\\cdot{p_2}$ or $p_1^2\\cdot{p_2^2}$. N must then be one of \\[(3^2\\cdot5^2),\\ (3^2\\cdot7^2),\\ (3^4\\cdot5),\\ (3^4\\cdot7),\\ (3^4\\cdot 11)\\] So, there are $5$ solutions when $(p+q)(p-q)$ is odd. If $p-q$ is even, then $(p+q)(p-q)$ must have at least two factors of $2$, so we can rewrite the expression as $4(k)(k-q)$ where $k=\\frac{p+q}{2}$. We can disregard the $4$ by dividing by $4$ and restricting our upper bound to $250$. Since $k$ and $q$ don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have \\[(2^2\\cdot3^2),(2^2\\cdot5^2),(2^2\\cdot7^2), (3^2\\cdot5^2), (2^4\\cdot3), (2^4\\cdot5), (2^4\\cdot7), (2^4\\cdot11), (2^4\\cdot13), (3^4\\cdot2)\\] as the possibilities. Therefore, there are $10+5=015 ~sigma"
] |
2006-II-14
| 2,006
| 14
|
Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer.
| 63
|
II
|
[
"Let $K = \\sum_{i=1}^{9}{\\frac{1}{i}}$. Examining the terms in $S_1$, we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$. Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2 = 20K + 1$. In general, we will have that $S_n = (n10^{n-1})K + 1$ because each digit will appear $10^{n - 1}$ times in each place in the numbers $1, 2, \\ldots, 10^{n} - 1$, and there are $n$ total places. The denominator of $K$ is $D = 2^3\\cdot 3^2\\cdot 5\\cdot 7$. For $S_n$ to be an integer, $n10^{n-1}$ must be divisible by $D$. Since $10^{n-1}$ only contains the factors $2$ and $5$ (but will contain enough of them when $n \\geq 3$), we must choose $n$ to be divisible by $3^2\\cdot 7$. Since we're looking for the smallest such $n$, the answer is $063."
] |
2006-II-15
| 2,006
| 15
|
Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$
| 9
|
II
|
[
"Let $\\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are in the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theorem we that \\[x = \\sqrt{y^2 - h_x^2} + \\sqrt{z^2 - h_x^2}\\] As a function of $h_x$, the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \\frac1{16}$ and so $h_x = \\frac{1}4$ and similarly $h_y = \\frac15$ and $h_z = \\frac16$. The area of the triangle must be the same no matter how we measure it; therefore $x\\cdot h_x = y\\cdot h_y = z \\cdot h_z$ gives us $\\frac x4 = \\frac y5 = \\frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$. The semiperimeter of the triangle is $s = \\frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have \\[A = \\sqrt{15A \\cdot 7A \\cdot 5A \\cdot 3A} = 15A^2\\sqrt{7}\\] Thus, $A = \\frac{1}{15\\sqrt{7}}$ and $x + y + z = 30A = \\frac2{\\sqrt{7}}$ and the answer is $2 + 7 = 009. But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.",
"Let $\\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are in the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theorem we that \\[x = \\sqrt{y^2 - h_x^2} + \\sqrt{z^2 - h_x^2}\\] As a function of $h_x$, the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \\frac1{16}$ and so $h_x = \\frac{1}4$ and similarly $h_y = \\frac15$ and $h_z = \\frac16$. The area of the triangle must be the same no matter how we measure it; therefore $x\\cdot h_x = y\\cdot h_y = z \\cdot h_z$ gives us $\\frac x4 = \\frac y5 = \\frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$. The semiperimeter of the triangle is $s = \\frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have \\[A = \\sqrt{15A \\cdot 7A \\cdot 5A \\cdot 3A} = 15A^2\\sqrt{7}\\] Thus, $A = \\frac{1}{15\\sqrt{7}}$ and $x + y + z = 30A = \\frac2{\\sqrt{7}}$ and the answer is $2 + 7 = 009. But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.",
"We can rewrite the equations as follows: \\begin{align*} \\frac{1}{4}x &= \\sqrt{\\left(y+\\frac{1}{4}\\right) \\left(y-\\frac{1}{4}\\right)\\left(\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right)}+\\sqrt{\\left(z+\\frac{1}{4}\\right) \\left(z-\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right)} \\\\ \\frac{1}{5} y &= \\sqrt{\\left(z+\\frac{1}{5}\\right) \\left(z-\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right)}+\\sqrt{\\left(x+\\frac{1}{5}\\right) \\left(x-\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right)} \\\\ \\frac{1}{6} z &= \\sqrt{\\left(x+\\frac{1}{6}\\right) \\left(x-\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right)}+\\sqrt{\\left(y+\\frac{1}{6}\\right) \\left(y-\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right)} \\end{align*} Take the first equation. The first square root is the area of a triangle with side lengths $y,y,\\frac{1}{2}$ by Heron’s Formula. Similarly, the second square root is the area of a triangle with side lengths $z,z,\\frac{1}{2}$. If we connect these two triangles together at the $\\frac{1}{2}$ side, we obtain a kite. The area of the kite is $\\frac{1}{4}x$, and since the first diagonal is $\\frac{1}{2}$, the second diagonal is $\\frac{2\\cdot\\frac{1}{4}x}{\\frac{1}{2}}=x$. If we draw this diagonal, we obtain two triangles with side lengths $x,y,z$. Let this triangle have area $A$. Then $\\frac{1}{4}x=2A$; extend this for the other two equations. We can substitute into the first equation to obtain a value for $x$, and the answer is $\\frac{2}{\\sqrt{7}}\\Rightarrow009. ~eevee9406",
"We can rewrite the equations as follows: \\begin{align*} \\frac{1}{4}x &= \\sqrt{\\left(y+\\frac{1}{4}\\right) \\left(y-\\frac{1}{4}\\right)\\left(\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right)}+\\sqrt{\\left(z+\\frac{1}{4}\\right) \\left(z-\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right) \\left(\\frac{1}{4}\\right)} \\\\ \\frac{1}{5} y &= \\sqrt{\\left(z+\\frac{1}{5}\\right) \\left(z-\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right)}+\\sqrt{\\left(x+\\frac{1}{5}\\right) \\left(x-\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right) \\left(\\frac{1}{5}\\right)} \\\\ \\frac{1}{6} z &= \\sqrt{\\left(x+\\frac{1}{6}\\right) \\left(x-\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right)}+\\sqrt{\\left(y+\\frac{1}{6}\\right) \\left(y-\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right) \\left(\\frac{1}{6}\\right)} \\end{align*} Take the first equation. The first square root is the area of a triangle with side lengths $y,y,\\frac{1}{2}$ by Heron’s Formula. Similarly, the second square root is the area of a triangle with side lengths $z,z,\\frac{1}{2}$. If we connect these two triangles together at the $\\frac{1}{2}$ side, we obtain a kite. The area of the kite is $\\frac{1}{4}x$, and since the first diagonal is $\\frac{1}{2}$, the second diagonal is $\\frac{2\\cdot\\frac{1}{4}x}{\\frac{1}{2}}=x$. If we draw this diagonal, we obtain two triangles with side lengths $x,y,z$. Let this triangle have area $A$. Then $\\frac{1}{4}x=2A$; extend this for the other two equations. We can substitute into the first equation to obtain a value for $x$, and the answer is $\\frac{2}{\\sqrt{7}}\\Rightarrow009. ~eevee9406",
"Note that none of $x,y,z$ can be zero. Each of the equations is in the form \\[a=\\sqrt{b^2-d^2}+\\sqrt{c^2-d^2}\\] Isolate a radical and square the equation to get \\[b^2-d^2=a^2-2a\\sqrt{c^2-d^2}+c^2-d^2\\] Now cancel, and again isolate the radical, and square the equation to get \\[a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2\\] Rearranging gives \\[a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2\\] Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$), which implies \\[-4x^2\\cdot\\frac1{16}=-4y^2\\cdot\\frac1{25}=-4z^2\\cdot\\frac1{36}\\] Or \\[x: y: z=4: 5: 6 \\implies x=\\frac{4y}5 \\textrm{ and } z=\\frac{6y}5\\] Plug these values into the middle equation to get \\[\\frac{256y^4+625y^4+1296y^4}{625}=\\frac{800y^4}{625}+\\frac{1800y^4}{625}+\\frac{1152y^4}{625}-\\frac{100y^2}{625}\\] Simplifying gives \\[1575y^4=100y^2 \\textrm{ but } y \\neq 0 \\implies y^2=\\frac4{63} \\textrm{ or } y=\\frac2{3\\sqrt7}\\] Substituting the value of $y$ for $x$ and $z$ gives \\[x+y+z = \\frac{4y+5y+6y}5 = 3y = 3 \\cdot \\frac{2}{3\\sqrt7} = \\frac{2}{\\sqrt7}\\] And thus the answer is $009 ~phoenixfire",
"Note that none of $x,y,z$ can be zero. Each of the equations is in the form \\[a=\\sqrt{b^2-d^2}+\\sqrt{c^2-d^2}\\] Isolate a radical and square the equation to get \\[b^2-d^2=a^2-2a\\sqrt{c^2-d^2}+c^2-d^2\\] Now cancel, and again isolate the radical, and square the equation to get \\[a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2\\] Rearranging gives \\[a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2\\] Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$), which implies \\[-4x^2\\cdot\\frac1{16}=-4y^2\\cdot\\frac1{25}=-4z^2\\cdot\\frac1{36}\\] Or \\[x: y: z=4: 5: 6 \\implies x=\\frac{4y}5 \\textrm{ and } z=\\frac{6y}5\\] Plug these values into the middle equation to get \\[\\frac{256y^4+625y^4+1296y^4}{625}=\\frac{800y^4}{625}+\\frac{1800y^4}{625}+\\frac{1152y^4}{625}-\\frac{100y^2}{625}\\] Simplifying gives \\[1575y^4=100y^2 \\textrm{ but } y \\neq 0 \\implies y^2=\\frac4{63} \\textrm{ or } y=\\frac2{3\\sqrt7}\\] Substituting the value of $y$ for $x$ and $z$ gives \\[x+y+z = \\frac{4y+5y+6y}5 = 3y = 3 \\cdot \\frac{2}{3\\sqrt7} = \\frac{2}{\\sqrt7}\\] And thus the answer is $009 ~phoenixfire"
] |
2007-I-1
| 2,007
| 1
|
How many positive perfect squares less than $10^6$ are multiples of 24?
| 83
|
I
|
[
"The prime factorization of $24$ is $2^3\\cdot3$. Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$. This means that each square is in the form $(12c)^2$, where $12 c$ is a positive integer less than $\\sqrt{10^6}$. There are $\\left\\lfloor \\frac{1000}{12}\\right\\rfloor = 083 solutions.",
"The perfect squares divisible by $24$ are all multiples of $12$: $12^2$, $24^2$, $36^2$, $48^2$, etc... Since they all have to be less than $10^6$, or $1000^2$, the closest multiple of $12$ to $1000$ is $996$ ($12*83$), so we know that this is the last term in the sequence. Therefore, we know that there are $083."
] |
2007-I-2
| 2,007
| 2
|
A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second. At a certain time, one of these three persons is exactly halfway between the other two. At that time, find the distance in feet between the start of the walkway and the middle person.
| 52
|
I
|
[
"Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is some point when Al is halfway between Cy and Bob. At this time $s$, we have that $\\frac{8(s-4)+10(s-2)}{2}=6s$ After solving, $s=\\frac{26}{3}$. At this time, Al has traveled $6\\cdot\\frac{26}{3}=52$ feet. We could easily check that Al is in the middle by trying all three possible cases. $\\frac{6s + 8(s-4)}{2} = 10(s-2)$ yields that $s = \\frac 43$, which can be disregarded since both Bob and Cy hadn't started yet. $\\frac{6s + 10(s-2)}{2} = 8(s-4)$ yields that $-10=-32$, a contradiction. Thus, the answer is $052."
] |
2007-I-3
| 2,007
| 3
|
The complex number $z$ is equal to $9+bi$ , where $b$ is a positive real number and $i^{2}=-1$ . Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?
| 15
|
I
|
[
"Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$. Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \\ldots + (9\\cdot 18 + 81)bi - b^3i$. Setting these two equal, we get that $18bi = 243bi - b^3i$, so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$. Since $b > 0$, the solution is $015."
] |
2007-I-4
| 2,007
| 4
|
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are $60$ , $84$ , and $140$ years. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again?
| 105
|
I
|
[
"Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \\frac{t \\pi}{30}$, $b(t) = \\frac{t \\pi}{42}$, $c(t) = \\frac{t \\pi}{70}$. In order for the planets and the central star to be collinear, $a(t)$, $b(t)$, and $c(t)$ must differ by a multiple of $\\pi$. Note that $a(t) - b(t) = \\frac{t \\pi}{105}$ and $b(t) - c(t) = \\frac{t \\pi}{105}$, so $a(t) - c(t) = \\frac{ 2 t \\pi}{105}$. These are simultaneously multiples of $\\pi$ exactly when $t$ is a multiple of $105$, so the planets and the star will next be collinear in $105 years."
] |
2007-I-5
| 2,007
| 5
|
The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer. For how many integer Fahrenheit temperatures between $32$ and $1000$ inclusive does the original temperature equal the final temperature?
| 539
|
I
|
[
"Solution 1 Examine $F - 32$ modulo 9. If $F - 32 \\equiv 0 \\pmod{9}$, then we can define $9x = F - 32$. This shows that $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(F-32)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x) + 32\\right] \\Longrightarrow F = 9x + 32$. This case works. If $F - 32 \\equiv 1 \\pmod{9}$, then we can define $9x + 1 = F - 32$. This shows that $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(F-32)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x + 1) + 32\\right] \\Longrightarrow$$F = \\left[9x + \\frac{9}{5}+ 32 \\right] \\Longrightarrow F = 9x + 34$. So this case doesn't work. Generalizing this, we define that $9x + k = F - 32$. Thus, $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(9x + k)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x + \\left[\\frac{5}{9}k\\right]) + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5} \\left[\\frac{5}{9}k \\right] \\right] + 9x + 32$. We need to find all values $0 \\le k \\le 8$ that $\\left[ \\frac{9}{5} \\left[ \\frac{5}{9} k \\right] \\right] = k$. Testing every value of $k$ shows that $k = 0, 2, 4, 5, 7$, so $5$ of every $9$ values of $k$ work. There are $\\lfloor \\frac{1000 - 32}{9} \\rfloor = 107$ cycles of $9$, giving $5 \\cdot 107 = 535$ numbers that work. Of the remaining $6$ numbers from $995$ onwards, $995,\\ 997,\\ 999,\\ 1000$ work, giving us $535 + 4 = 539 solutions. -alanisawesome2018",
"Examine $F - 32$ modulo 9. If $F - 32 \\equiv 0 \\pmod{9}$, then we can define $9x = F - 32$. This shows that $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(F-32)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x) + 32\\right] \\Longrightarrow F = 9x + 32$. This case works. If $F - 32 \\equiv 1 \\pmod{9}$, then we can define $9x + 1 = F - 32$. This shows that $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(F-32)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x + 1) + 32\\right] \\Longrightarrow$$F = \\left[9x + \\frac{9}{5}+ 32 \\right] \\Longrightarrow F = 9x + 34$. So this case doesn't work. Generalizing this, we define that $9x + k = F - 32$. Thus, $F = \\left[\\frac{9}{5}\\left[\\frac{5}{9}(9x + k)\\right] + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5}(5x + \\left[\\frac{5}{9}k\\right]) + 32\\right] \\Longrightarrow F = \\left[\\frac{9}{5} \\left[\\frac{5}{9}k \\right] \\right] + 9x + 32$. We need to find all values $0 \\le k \\le 8$ that $\\left[ \\frac{9}{5} \\left[ \\frac{5}{9} k \\right] \\right] = k$. Testing every value of $k$ shows that $k = 0, 2, 4, 5, 7$, so $5$ of every $9$ values of $k$ work. There are $\\lfloor \\frac{1000 - 32}{9} \\rfloor = 107$ cycles of $9$, giving $5 \\cdot 107 = 535$ numbers that work. Of the remaining $6$ numbers from $995$ onwards, $995,\\ 997,\\ 999,\\ 1000$ work, giving us $535 + 4 = 539 solutions. -alanisawesome2018",
"Notice that $\\left[ \\frac{9}{5} \\left[ \\frac{5}{9} k \\right] \\right] = k$ holds if $k=\\left[ \\frac{9}{5}x\\right]$ for some integer $x$. Thus, after translating from $F\\to F-32$ we want count how many values of $x$ there are such that $k=\\left[ \\frac{9}{5}x\\right]$ is an integer from $0$ to $968$. This value is computed as $\\left[968*\\frac{5}{9}\\right]+1 = 539 solutions. -alanisawesome2018",
"Let $c$ be a degree Celsius, and $f=\\frac 95c+32$ rounded to the nearest integer. Since $f$ was rounded to the nearest integer we have $|f-((\\frac 95)c+32)|\\leq 1/2$, which is equivalent to $|(\\frac 59)(f-32)-c|\\leq \\frac 5{18}$ if we multiply by $5/9$. Therefore, it must round to $c$ because $\\frac 5{18}<\\frac 12$ so $c$ is the closest integer. Therefore there is one solution per degree Celsius in the range from $0$ to $(\\frac 59)(1000-32) + 1=(\\frac 59)(968) + 1=538.8$, meaning there are $539$ solutions. Solution 4 Start listing out values for $F$ and their corresponding values of $C$. You will soon find that every 9 values starting from $F$ = 32, there is a pattern: $F=32$: Works $F=33$: Doesn't work $F=34$: work $F=35$: Doesn’t work $F=36$: Works $F=37$: Works $F=38$: Doesn’t work $F=39$: Works $F=40$: Doesn’t work $F=41$: Works There are $969$ numbers between $32$ and $1000$, inclusive. This is $107$ sets of $9$, plus $6$ extra numbers at the end. In each set of $9$, there are $5$ “Works,” so we have $107\\cdot5 = 535$ values of $F$ that work. Now we must add the $6$ extra numbers. The number of “Works” in the first $6$ terms of the pattern is $4$, so our final answer is $535 + 4 = 539$ solutions that work. Submitted by warriorcats Solution 5(similar to solution 3 but faster solution if you have no time) Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $539 solutions. -alanisawesome2018",
"Start listing out values for $F$ and their corresponding values of $C$. You will soon find that every 9 values starting from $F$ = 32, there is a pattern: $F=32$: Works $F=33$: Doesn't work $F=34$: work $F=35$: Doesn’t work $F=36$: Works $F=37$: Works $F=38$: Doesn’t work $F=39$: Works $F=40$: Doesn’t work $F=41$: Works There are $969$ numbers between $32$ and $1000$, inclusive. This is $107$ sets of $9$, plus $6$ extra numbers at the end. In each set of $9$, there are $5$ “Works,” so we have $107\\cdot5 = 535$ values of $F$ that work. Now we must add the $6$ extra numbers. The number of “Works” in the first $6$ terms of the pattern is $4$, so our final answer is $535 + 4 = 539$ solutions that work. Submitted by warriorcats Solution 5(similar to solution 3 but faster solution if you have no time) Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $539 solutions. -alanisawesome2018",
"Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $539 solutions. -alanisawesome2018",
"Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $539 solutions. -alanisawesome2018"
] |
2007-I-6
| 2,007
| 6
|
A frog is placed at the origin on the number line , and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of $3$ , or to the closest point with a greater integer coordinate that is a multiple of $13$ . A move sequence is a sequence of coordinates which correspond to valid moves, beginning with $0$ , and ending with $39$ . For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog?
| 169
|
I
|
[
"Solution 1 Let us keep a careful tree of the possible number of paths around every multiple of $13$. From $0 \\Rightarrow 13$, we can end at either $12$ (mult. of 3) or $13$ (mult. of 13). Only $1$ path leads to $12$ Continuing from $12$, there is $1 \\cdot 1 = 1$ way to continue to $24$ There are $1 \\cdot \\left(\\frac{24-15}{3} + 1\\right) = 4$ ways to reach $26$. There are $\\frac{12 - 0}{3} + 1 = 5$ ways to reach $13$. Continuing from $13$, there are $5 \\cdot 1 = 5$ ways to get to $24$ There are $5 \\cdot \\left(\\frac{24-15}{3} + 1 + 1\\right) = 25$ ways (the first 1 to make it inclusive, the second to also jump from $13 \\Rightarrow 26$) to get to $26$. Regrouping, work from $24 | 26\\Rightarrow 39$ There are $1 + 5 = 6$ ways to get to $24$ Continuing from $24$, there are $6 \\cdot \\left(\\frac{39 - 27}{3}\\right) = 24$ ways to continue to $39$. There are $4 + 25 = 29$ ways to reach $26$. Continuing from $26$, there are $29 \\cdot \\left(\\frac{39-27}{3} + 1\\right) = 145$ (note that the 1 is not to inclusive, but to count $26 \\Rightarrow 39$). In total, we get $145 + 24 = 169$. In summary, we can draw the following tree, where in $(x,y)$, $x$ represents the current position on the number line, and $y$ represents the number of paths to get there: $(12,1)$ $(24,1)$ $(39,4)$ $(26,4)$ $(39,20)$ $(13,5)$ $(24,5)$ $(39,20)$ $(26,25)$ $(39,125)$ Again, this totals $4 + 20 + 20 + 125 = 169$. Solution 2 We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else. For the first stage the possible paths are $(0,13)$, $(0,3,13)$, $(0,3,6,13)$, $(0,3,6,9,13)$, $(0,3,6,9,12,13)$, and $(0,3,6,9,12)$. That is a total of 6. For the second stage the possible paths are $(26)$, $(15,26)$, $(15,18,26)$, $(15,18,21,26)$, $(15,18,21,24,26)$, and $(15,18,21,24)$. That is a total of 6. For the third stage the possible paths are $(39)$, $(27,39)$, $(27,30,39)$, $(27,30,33,39)$, and $(27,30,33,36,39)$. That is a total of 5. However, we cannot jump from $12 \\Rightarrow 26$ (this eliminates 5 paths) or $24 \\Rightarrow 39$ (this eliminates 6 paths), so we must subtract $6 + 5 = 11$. The answer is $6*6*5 - 11=169$ Solution 3 Another way would be to use a table representing the number of ways to reach a certain number $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$ How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = 169.",
"Let us keep a careful tree of the possible number of paths around every multiple of $13$. From $0 \\Rightarrow 13$, we can end at either $12$ (mult. of 3) or $13$ (mult. of 13). Only $1$ path leads to $12$ Continuing from $12$, there is $1 \\cdot 1 = 1$ way to continue to $24$ There are $1 \\cdot \\left(\\frac{24-15}{3} + 1\\right) = 4$ ways to reach $26$. There are $\\frac{12 - 0}{3} + 1 = 5$ ways to reach $13$. Continuing from $13$, there are $5 \\cdot 1 = 5$ ways to get to $24$ There are $5 \\cdot \\left(\\frac{24-15}{3} + 1 + 1\\right) = 25$ ways (the first 1 to make it inclusive, the second to also jump from $13 \\Rightarrow 26$) to get to $26$. Regrouping, work from $24 | 26\\Rightarrow 39$ There are $1 + 5 = 6$ ways to get to $24$ Continuing from $24$, there are $6 \\cdot \\left(\\frac{39 - 27}{3}\\right) = 24$ ways to continue to $39$. There are $4 + 25 = 29$ ways to reach $26$. Continuing from $26$, there are $29 \\cdot \\left(\\frac{39-27}{3} + 1\\right) = 145$ (note that the 1 is not to inclusive, but to count $26 \\Rightarrow 39$). In total, we get $145 + 24 = 169$. In summary, we can draw the following tree, where in $(x,y)$, $x$ represents the current position on the number line, and $y$ represents the number of paths to get there: $(12,1)$ $(24,1)$ $(39,4)$ $(26,4)$ $(39,20)$ $(13,5)$ $(24,5)$ $(39,20)$ $(26,25)$ $(39,125)$ Again, this totals $4 + 20 + 20 + 125 = 169$. Solution 2 We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else. For the first stage the possible paths are $(0,13)$, $(0,3,13)$, $(0,3,6,13)$, $(0,3,6,9,13)$, $(0,3,6,9,12,13)$, and $(0,3,6,9,12)$. That is a total of 6. For the second stage the possible paths are $(26)$, $(15,26)$, $(15,18,26)$, $(15,18,21,26)$, $(15,18,21,24,26)$, and $(15,18,21,24)$. That is a total of 6. For the third stage the possible paths are $(39)$, $(27,39)$, $(27,30,39)$, $(27,30,33,39)$, and $(27,30,33,36,39)$. That is a total of 5. However, we cannot jump from $12 \\Rightarrow 26$ (this eliminates 5 paths) or $24 \\Rightarrow 39$ (this eliminates 6 paths), so we must subtract $6 + 5 = 11$. The answer is $6*6*5 - 11=169$ Solution 3 Another way would be to use a table representing the number of ways to reach a certain number $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$ How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = 169.",
"We divide it into 3 stages. The first occurs before the frog moves past 13. The second occurs before it moves past 26, and the last is everything else. For the first stage the possible paths are $(0,13)$, $(0,3,13)$, $(0,3,6,13)$, $(0,3,6,9,13)$, $(0,3,6,9,12,13)$, and $(0,3,6,9,12)$. That is a total of 6. For the second stage the possible paths are $(26)$, $(15,26)$, $(15,18,26)$, $(15,18,21,26)$, $(15,18,21,24,26)$, and $(15,18,21,24)$. That is a total of 6. For the third stage the possible paths are $(39)$, $(27,39)$, $(27,30,39)$, $(27,30,33,39)$, and $(27,30,33,36,39)$. That is a total of 5. However, we cannot jump from $12 \\Rightarrow 26$ (this eliminates 5 paths) or $24 \\Rightarrow 39$ (this eliminates 6 paths), so we must subtract $6 + 5 = 11$. The answer is $6*6*5 - 11=169$ Solution 3 Another way would be to use a table representing the number of ways to reach a certain number $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$ How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = 169.",
"Another way would be to use a table representing the number of ways to reach a certain number $\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$ How we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$, we can reach it from $13, 15, 18, 21, 24$, so we add all those values to get the value for $26$. For $27$, it is only reachable from $24$ or $26$, so we have $29 + 6 = 35$. The answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = 169.",
"I believe this is an easier way of organizing the solution to reduce the possibility of mistakes. This is a highly visual solution, so it's much easier to record than a tree or table. Here is my diagram to help you see what I did: https://drive.google.com/file/d/1Gk_cziYvoeg--uVTap5FGOds3SEfiDAW/view?usp=sharing 0 3 6 9 12 15 18 21 24 27 30 33 36 39 +- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+- - -+ + + + 13 26 39 (stations 1, 2, and 3, respectively) Using graph paper, draw a number line from 0-39. On one line, dot every multiple of 3. Then on a line below it, dot every multiple of 13. This way you can clearly see which goes before or after which. To make it easier to understand, I'll compare these jumps to a train system. Imagine that every multiple of 3 is a short transit, while the multiples of 13 are long transits; because of the possibility to skip a large section in one move. As we continue, picture each \"transit\" of 13 to be an option, like a switch. If you are the frog that are riding these trains, you would probably think like this: \"I could use the first long transit, skip the second, and use the third. Or, I could skip both first and second and use only the third. etc.) From now on I'll be calling the multiples of 13: 13, 26, and 39, as stations 1, 2, and 3 for clarity. Thinking like this organizes the problem efficiently into $2^3$ cases, where you could choose which \"long transits\" to ride. We will start with the harder cases and move downwards. Write out each of the 8 cases to record each one. Once we do the hardest case, which is the first one, every preceding one will be easier with the knowledge we gather. Case #1: 111 You have 5 locations to choose to jump to station 1. From the start (0), 3, 6, 9, or 12. The same goes for station #2, with 13, 15, 18, 21, or 24. However, station #3 is tricky. You can jump from 26, 27, 30, or 33, but if you jumped from 36, then that could qualify as skipping jumping station #3, because station #3 is both a multiple of 3 and a multiple of 13. For example, if you jumped from 36 as your last move, then those moves are essentially the same as case 110. So, $5*5*4=100$. Case #2: 110 Start with what you did on case #1, but when you reach station 2 continue only jumping multiples of 3 to 39. $5*5=25$ Case #3: 101 We have the 5 original choices, and when you reach station #1, move to 27 on the number line, slightly ahead of station #2. Here is another tricky part. Since we did not start on station 2 like we did in case #1, there are only 3 cases instead of 4. $5*3=15$ Case #4: 100 This one is simple. $5$ Case #5: 011 Jump multiples of 3 to get to 15, where you will have 4 locations to jump to station 2, and another 4 choices to reach station 3. $4*4=16$ Case #6: 010 This one is also simple. Once you reach 15 there are only 4 choices. $4$ Case #7: 001 It only gets simpler. You can only jump at 27, 30, or 33. $3$ Case #8: 000 Simple jumps. $1$ You have reached the end! $100+25+15+5+16+4+3+1=169. -jackshi2006",
"Let $f(n)$ be the number of ways one can get to $39$ starting at position $n.$ We wish to compute $f(0).$ Now it's just a long simplifications until you get to $f(36) = 1.$ We have \\[f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).\\] Most of these steps are valid since at any $n$ that is a multiple of $3$ we can either go to the next multiple of $3$ or we can skip to the next multiple of $13$ which is simply $13.$ From these equations we have deduced $f(0) = f(15) + 5f(12).$ Continuing we have \\[f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).\\] Finally, note that $f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5$ since at any point we can either go to the next multiple of $3$ or go to the next multiple of $13$ which happens to be $39.$ Therefore $f(26) = 5.$ Similarly we find $f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4$ so the end answer is $5 \\cdot 29 + 6 \\cdot 4 = 169 ways to jump to 39 through various journeys in hyperspace. -alanisawesome2018",
"Let $f(n)$ be the number of ways one can get to $39$ starting at position $n.$ We wish to compute $f(0).$ Now it's just a long simplifications until you get to $f(36) = 1.$ We have \\[f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).\\] Most of these steps are valid since at any $n$ that is a multiple of $3$ we can either go to the next multiple of $3$ or we can skip to the next multiple of $13$ which is simply $13.$ From these equations we have deduced $f(0) = f(15) + 5f(12).$ Continuing we have \\[f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).\\] Finally, note that $f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5$ since at any point we can either go to the next multiple of $3$ or go to the next multiple of $13$ which happens to be $39.$ Therefore $f(26) = 5.$ Similarly we find $f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4$ so the end answer is $5 \\cdot 29 + 6 \\cdot 4 = 169 ways to jump to 39 through various journeys in hyperspace. -alanisawesome2018",
"Another way you can visualize the problem is by thinking of points $13$, $26$, and $39$ as planets and all multiples of 3 as points at which your spaceship can jump to hyperspace. Given that you wish to visit planet $39$, you can choose to visit planets $13$ or $26$ along the way. Case 1: Neither There are $4$ ways to jump to $39$ from hyperspace. Case 2: Only planet $13$ There are $5$ ways to jump to planet $13$ and $4$ ways to jump to planet $39$. $20$ ways total. Case 3: Only planet $26$. There are $4$ ways to jump to planet $26$ and $5$ ways to jump to planet $39$. $20$ ways total. Case 4: Both There are $5$ ways to jump to $13$, $5$ ways to jump to $26$, and $5$ ways to jump to $39$. $125$ ways total. Therefore, there are $4+20+20+125=169 ways to jump to 39 through various journeys in hyperspace. -alanisawesome2018",
"Another way you can visualize the problem is by thinking of points $13$, $26$, and $39$ as planets and all multiples of 3 as points at which your spaceship can jump to hyperspace. Given that you wish to visit planet $39$, you can choose to visit planets $13$ or $26$ along the way. Case 1: Neither There are $4$ ways to jump to $39$ from hyperspace. Case 2: Only planet $13$ There are $5$ ways to jump to planet $13$ and $4$ ways to jump to planet $39$. $20$ ways total. Case 3: Only planet $26$. There are $4$ ways to jump to planet $26$ and $5$ ways to jump to planet $39$. $20$ ways total. Case 4: Both There are $5$ ways to jump to $13$, $5$ ways to jump to $26$, and $5$ ways to jump to $39$. $125$ ways total. Therefore, there are $4+20+20+125=169 ways to jump to 39 through various journeys in hyperspace. -alanisawesome2018"
] |
2007-I-7
| 2,007
| 7
|
Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )$ Find the remainder when $N$ is divided by 1000. ( $\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$ , and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$ .)
| 477
|
I
|
[
"The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not $\\log_{\\sqrt{2}} k$ is an integer. The change of base formula shows that $\\frac{\\log k}{\\log \\sqrt{2}} = \\frac{2 \\log k}{\\log 2}$. For the $\\log 2$ term to cancel out, $k$ is a power of $2$. Thus, $N$ is equal to the sum of all the numbers from 1 to 1000, excluding all powers of 2 from $2^0 = 1$ to $2^9 = 512$. The formula for the sum of an arithmetic sequence and the sum of a geometric sequence yields that our answer is $\\left[\\frac{(1000 + 1)(1000)}{2} - (1 + 2 + 2^2 + \\ldots + 2^9)\\right] \\mod{1000}$. Simplifying, we get $\\left[1000\\left(\\frac{1000+1}{2}\\right) -1023\\right] \\mod{1000} \\equiv [500-23] \\mod{1000} \\equiv 477 \\mod{1000}.$ The answer is $477"
] |
2007-I-8
| 2,007
| 8
|
The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$ ?
| 30
|
I
|
[
"Solution 1 We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic. Let this root be $a$. We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \\frac{-k}{5}$. We then know that $\\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\\frac{k^{2}}{25}+(k-29)\\left(\\frac{-k}{5}\\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$, so $k=30$ is the highest. We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $030 we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = 030.\\] This can easily be checked to see that it does indeed work, and we're done! ~Ilikeapos",
"We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic. Let this root be $a$. We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \\frac{-k}{5}$. We then know that $\\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\\frac{k^{2}}{25}+(k-29)\\left(\\frac{-k}{5}\\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$, so $k=30$ is the highest. We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $030 we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = 030.\\] This can easily be checked to see that it does indeed work, and we're done! ~Ilikeapos",
"Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\\left(x^2 + \\left(k - \\frac{43}{2}\\right)x + \\frac{k}{2}\\right)$. Therefore, we can write four equations (and we have four variables), $a + m = 29 - k$, $a + n = \\frac{43}{2} - k$, $am = -k$, and $an = \\frac{k}{2}$. The first two equations show that $m - n = 29 - \\frac{43}{2} = \\frac{15}{2}$. The last two equations show that $\\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = 030 we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = 030.\\] This can easily be checked to see that it does indeed work, and we're done! ~Ilikeapos",
"Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations: \\begin{align} a = 2c \\\\ b = -d \\\\ 2c(k - 29) - d = c(2k - 43) + 2d \\\\ -d(k - 29) - 2ck = d(2k - 43) + ck \\end{align} Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $030 we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = 030.\\] This can easily be checked to see that it does indeed work, and we're done! ~Ilikeapos",
"Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$ Thus, we see that we have \\[r^2 + (k - 29)r - k = 0,\\] \\[2r^2 + (2k - 43)r + k = 0.\\] Adding the two equations gives us \\[3r^2 + (3k - 72)r = 0 \\implies r = 0, 24 - k.\\] Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \\implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that \\[Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \\implies k = 030.\\] This can easily be checked to see that it does indeed work, and we're done! ~Ilikeapos"
] |
2007-I-9
| 2,007
| 9
|
In right triangle $ABC$ with right angle $C$ , $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ . Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius of either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 737
|
I
|
[
"Solution 1 Label the points as in the diagram above. If we draw $\\overline{O_1A}$ and $\\overline{O_2B}$, we form two right triangles. As $\\overline{AF}$ and $\\overline{AD}$ are both tangents to the circle, we see that $\\overline{O_1A}$ is an angle bisector. Thus, $\\triangle AFO_1 \\cong \\triangle ADO_1$. Call $x = AD = AF$ and $y = EB = BG$. We know that $x + y + 2r = 34$. If we call $\\angle CAB = \\theta$, then $\\angle DAO_1 = \\frac{180 - \\theta}{2}$. Apply the tangent half-angle formula ($\\tan \\frac{\\theta}{2} = \\sqrt{\\frac{1 - \\cos \\theta}{1 + \\cos \\theta}}$). We see that $\\frac rx = \\tan \\frac{180 - \\theta}{2} = \\sqrt{\\frac{1 - \\cos (180 - \\theta)}{1 + \\cos (180 - \\theta)}}$$= \\sqrt{\\frac{1 + \\cos \\theta}{1 - \\cos \\theta}}$. Also, $\\cos \\theta = \\frac{30}{34} = \\frac{15}{17}$. Thus, $\\frac rx = \\sqrt{\\frac{1 + \\frac{15}{17}}{1 - \\frac{15}{17}}}$, and $x = \\frac{r}{4}$. Similarly, we find that $y = r/\\sqrt{\\frac{1 + \\frac{8}{17}}{1 - \\frac{8}{17}}} = \\frac{3r}{5}$. Therefore, $x + y + 2r = \\frac{r}{4} + \\frac{3r}{5} + 2r = \\frac{57r}{20} = 34 \\Longrightarrow r = \\frac{680}{57}$, and $p + q = 737$. Solution 2 Use a similar solution to the aforementioned solution. Instead, call $\\angle CAB = 2\\theta$, and then proceed by simplifying through identities. We see that $\\frac rx = \\tan \\left(\\frac{180 - 2\\theta}{2}\\right) = \\tan (90 - \\theta)$. In terms of $r$, we find that $x = \\frac{r}{\\cot \\theta} = \\frac{r\\sin \\theta}{\\cos \\theta}$. Similarly, we find that $y = \\frac{r \\sin(45 - \\theta)}{\\cos (45 - \\theta)}$. Substituting, we find that $r\\left(\\frac{\\sin \\theta}{\\cos \\theta} + \\frac{\\sin(45 - \\theta)}{\\cos (45 - \\theta)} + 2\\right) = 34$. Under a common denominator, $r\\left(\\frac{\\sin \\theta \\cos (45 - \\theta) + \\cos \\theta \\sin (45 - \\theta)}{\\cos \\theta \\cos (45 - \\theta)} + 2\\right) = 34$. Trigonometric identities simplify this to $r\\left(\\frac{\\sin\\left((\\theta) + (45 - \\theta)\\right)}{\\frac 12 \\left(\\cos (\\theta + 45 - \\theta) + \\cos (\\theta - 45 + \\theta) \\right)} + 2\\right) = 34$. From here, it is possible to simplify: $r\\left(\\frac{2 \\sin 45}{\\cos 45 + \\cos 2\\theta \\cos 45 + \\sin 2\\theta \\sin 45} +2\\right) = 34$ $r\\left(\\frac{2}{\\frac{17}{17} + \\frac{8}{17} + \\frac{15}{17}} + 2\\right) = 34$ $r\\left(\\frac{57}{20}\\right) = 34$ Our answer is $34 \\cdot \\frac{20}{57} = \\frac{680}{57}$, and $p + q = 737$. Solution 3 Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly $EB=GB$. Let $EB=x$. Draw the two perpendicular radii to G and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\\frac{15}{17}$ the cosine of angle EBG is $-\\frac{15}{17}$. Since the measure of the angle opposite to EBG is the complement of this one, its cosine is $\\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\\frac{30x^{2}}{17}=r^{2}+r^{2}-\\frac{30r^{2}}{17}$ This tells us that $r=4x$. Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle D. Draw the radii to F and D and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $8x+2.4x+x=34$. Solving we find that $4x=\\frac{680}{57}$ so our answer is 737. Solution 4 By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \\frac{K}{s-c}= \\frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\\frac{40-r}{40}= \\frac{O_{1}O_{2}}{AB}= \\frac{2r}{34}\\quad \\Rightarrow \\quad r = \\frac{680}{57}$. Hence, the final answer is $680+57 = 737$. Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"Label the points as in the diagram above. If we draw $\\overline{O_1A}$ and $\\overline{O_2B}$, we form two right triangles. As $\\overline{AF}$ and $\\overline{AD}$ are both tangents to the circle, we see that $\\overline{O_1A}$ is an angle bisector. Thus, $\\triangle AFO_1 \\cong \\triangle ADO_1$. Call $x = AD = AF$ and $y = EB = BG$. We know that $x + y + 2r = 34$. If we call $\\angle CAB = \\theta$, then $\\angle DAO_1 = \\frac{180 - \\theta}{2}$. Apply the tangent half-angle formula ($\\tan \\frac{\\theta}{2} = \\sqrt{\\frac{1 - \\cos \\theta}{1 + \\cos \\theta}}$). We see that $\\frac rx = \\tan \\frac{180 - \\theta}{2} = \\sqrt{\\frac{1 - \\cos (180 - \\theta)}{1 + \\cos (180 - \\theta)}}$$= \\sqrt{\\frac{1 + \\cos \\theta}{1 - \\cos \\theta}}$. Also, $\\cos \\theta = \\frac{30}{34} = \\frac{15}{17}$. Thus, $\\frac rx = \\sqrt{\\frac{1 + \\frac{15}{17}}{1 - \\frac{15}{17}}}$, and $x = \\frac{r}{4}$. Similarly, we find that $y = r/\\sqrt{\\frac{1 + \\frac{8}{17}}{1 - \\frac{8}{17}}} = \\frac{3r}{5}$. Therefore, $x + y + 2r = \\frac{r}{4} + \\frac{3r}{5} + 2r = \\frac{57r}{20} = 34 \\Longrightarrow r = \\frac{680}{57}$, and $p + q = 737$. Solution 2 Use a similar solution to the aforementioned solution. Instead, call $\\angle CAB = 2\\theta$, and then proceed by simplifying through identities. We see that $\\frac rx = \\tan \\left(\\frac{180 - 2\\theta}{2}\\right) = \\tan (90 - \\theta)$. In terms of $r$, we find that $x = \\frac{r}{\\cot \\theta} = \\frac{r\\sin \\theta}{\\cos \\theta}$. Similarly, we find that $y = \\frac{r \\sin(45 - \\theta)}{\\cos (45 - \\theta)}$. Substituting, we find that $r\\left(\\frac{\\sin \\theta}{\\cos \\theta} + \\frac{\\sin(45 - \\theta)}{\\cos (45 - \\theta)} + 2\\right) = 34$. Under a common denominator, $r\\left(\\frac{\\sin \\theta \\cos (45 - \\theta) + \\cos \\theta \\sin (45 - \\theta)}{\\cos \\theta \\cos (45 - \\theta)} + 2\\right) = 34$. Trigonometric identities simplify this to $r\\left(\\frac{\\sin\\left((\\theta) + (45 - \\theta)\\right)}{\\frac 12 \\left(\\cos (\\theta + 45 - \\theta) + \\cos (\\theta - 45 + \\theta) \\right)} + 2\\right) = 34$. From here, it is possible to simplify: $r\\left(\\frac{2 \\sin 45}{\\cos 45 + \\cos 2\\theta \\cos 45 + \\sin 2\\theta \\sin 45} +2\\right) = 34$ $r\\left(\\frac{2}{\\frac{17}{17} + \\frac{8}{17} + \\frac{15}{17}} + 2\\right) = 34$ $r\\left(\\frac{57}{20}\\right) = 34$ Our answer is $34 \\cdot \\frac{20}{57} = \\frac{680}{57}$, and $p + q = 737$. Solution 3 Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly $EB=GB$. Let $EB=x$. Draw the two perpendicular radii to G and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\\frac{15}{17}$ the cosine of angle EBG is $-\\frac{15}{17}$. Since the measure of the angle opposite to EBG is the complement of this one, its cosine is $\\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\\frac{30x^{2}}{17}=r^{2}+r^{2}-\\frac{30r^{2}}{17}$ This tells us that $r=4x$. Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle D. Draw the radii to F and D and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $8x+2.4x+x=34$. Solving we find that $4x=\\frac{680}{57}$ so our answer is 737. Solution 4 By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \\frac{K}{s-c}= \\frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\\frac{40-r}{40}= \\frac{O_{1}O_{2}}{AB}= \\frac{2r}{34}\\quad \\Rightarrow \\quad r = \\frac{680}{57}$. Hence, the final answer is $680+57 = 737$. Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"Use a similar solution to the aforementioned solution. Instead, call $\\angle CAB = 2\\theta$, and then proceed by simplifying through identities. We see that $\\frac rx = \\tan \\left(\\frac{180 - 2\\theta}{2}\\right) = \\tan (90 - \\theta)$. In terms of $r$, we find that $x = \\frac{r}{\\cot \\theta} = \\frac{r\\sin \\theta}{\\cos \\theta}$. Similarly, we find that $y = \\frac{r \\sin(45 - \\theta)}{\\cos (45 - \\theta)}$. Substituting, we find that $r\\left(\\frac{\\sin \\theta}{\\cos \\theta} + \\frac{\\sin(45 - \\theta)}{\\cos (45 - \\theta)} + 2\\right) = 34$. Under a common denominator, $r\\left(\\frac{\\sin \\theta \\cos (45 - \\theta) + \\cos \\theta \\sin (45 - \\theta)}{\\cos \\theta \\cos (45 - \\theta)} + 2\\right) = 34$. Trigonometric identities simplify this to $r\\left(\\frac{\\sin\\left((\\theta) + (45 - \\theta)\\right)}{\\frac 12 \\left(\\cos (\\theta + 45 - \\theta) + \\cos (\\theta - 45 + \\theta) \\right)} + 2\\right) = 34$. From here, it is possible to simplify: $r\\left(\\frac{2 \\sin 45}{\\cos 45 + \\cos 2\\theta \\cos 45 + \\sin 2\\theta \\sin 45} +2\\right) = 34$ $r\\left(\\frac{2}{\\frac{17}{17} + \\frac{8}{17} + \\frac{15}{17}} + 2\\right) = 34$ $r\\left(\\frac{57}{20}\\right) = 34$ Our answer is $34 \\cdot \\frac{20}{57} = \\frac{680}{57}$, and $p + q = 737$. Solution 3 Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly $EB=GB$. Let $EB=x$. Draw the two perpendicular radii to G and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\\frac{15}{17}$ the cosine of angle EBG is $-\\frac{15}{17}$. Since the measure of the angle opposite to EBG is the complement of this one, its cosine is $\\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\\frac{30x^{2}}{17}=r^{2}+r^{2}-\\frac{30r^{2}}{17}$ This tells us that $r=4x$. Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle D. Draw the radii to F and D and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $8x+2.4x+x=34$. Solving we find that $4x=\\frac{680}{57}$ so our answer is 737. Solution 4 By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \\frac{K}{s-c}= \\frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\\frac{40-r}{40}= \\frac{O_{1}O_{2}}{AB}= \\frac{2r}{34}\\quad \\Rightarrow \\quad r = \\frac{680}{57}$. Hence, the final answer is $680+57 = 737$. Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"Let the point where CB's extension hits the circle be G, and the point where the hypotenuse hits that circle be E. Clearly $EB=GB$. Let $EB=x$. Draw the two perpendicular radii to G and E. Now we have a cyclic quadrilateral. Let the radius be length $r$. We see that since the cosine of angle ABC is $\\frac{15}{17}$ the cosine of angle EBG is $-\\frac{15}{17}$. Since the measure of the angle opposite to EBG is the complement of this one, its cosine is $\\frac{15}{17}$. Using the law of cosines, we see that $x^{2}+x^{2}+\\frac{30x^{2}}{17}=r^{2}+r^{2}-\\frac{30r^{2}}{17}$ This tells us that $r=4x$. Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle D. Draw the radii to F and D and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$. Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, $8x+2.4x+x=34$. Solving we find that $4x=\\frac{680}{57}$ so our answer is 737. Solution 4 By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \\frac{K}{s-c}= \\frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\\frac{40-r}{40}= \\frac{O_{1}O_{2}}{AB}= \\frac{2r}{34}\\quad \\Rightarrow \\quad r = \\frac{680}{57}$. Hence, the final answer is $680+57 = 737$. Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"By Pythagoras, $AB = 34$. Let $I_{C}$ be the $C$-excenter of triangle $ABC$. Then the $C$-exradius $r_{C}$ is given by $r_{C}= \\frac{K}{s-c}= \\frac{240}{40-34}= 40$. The circle with center $O_{1}$ is tangent to both $AB$ and $AC$, which means that $O_{1}$ lies on the external angle bisector of $\\angle BAC$. Therefore, $O_{1}$ lies on $AI_{C}$. Similarly, $O_{2}$ lies on $BI_{C}$. Let $r$ be the common radius of the circles with centers $O_{1}$ and $O_{2}$. The distances from points $O_{1}$ and $O_{2}$ to $AB$ are both $r$, so $O_{1}O_{2}$ is parallel to $AB$, which means that triangles $I_{C}AB$ and $I_{C}O_{1}O_{2}$ are similar. The distance from $I_{C}$ to $AB$ is $r_{C}= 40$, so the distance from $I_{C}$ to $O_{1}O_{2}$ is $40-r$. Therefore, $\\frac{40-r}{40}= \\frac{O_{1}O_{2}}{AB}= \\frac{2r}{34}\\quad \\Rightarrow \\quad r = \\frac{680}{57}$. Hence, the final answer is $680+57 = 737$. Solution 5 Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"Start with a scaled 16-30-34 triangle. Inscribe a circle. The height, $h,$ and radius, $r,$ are found via $A=\\frac{1}{2}\\times 16s\\times 30s=\\frac{1}{2}\\times 34s\\times h=\\frac{1}{2}\\times rp,$ where $p$ is the perimeter. Cut the figure through the circle and perpendicular to the hypotenuse. Slide the two pieces in opposite directions along the hypotenuse until they are one diameter of the circle apart. Complete the two partial circles. The linear dimensions of the new triangle are $\\frac{46s}{34s}=\\frac{23}{17}$ times the size of the original. The problem's 16-30-34 triangle sits above the circles. Equate heights and solve for $r=6s$: $\\frac{240s}{17}\\times\\frac{23}{17} = \\frac{240}{17}+12s$ $20s\\times 23 = 20\\times 17+s\\times 17\\times 17$ $s = \\frac{340}{171}$ $r = 6s = \\frac{680}{57}$ The answer is $737$. Solution 6 Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"Using homothety in the diagram above, as well as the auxiliary triangle, leads to the solution. Solution 7 A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"A different approach is to plot the triangle on the Cartesian Plane with $C$ at $(0,0)$, $A$ at $(0,30)$, and $B$ at $(16,0)$. We wish to find the coordinates of $O_1$ and $O_2$ in terms of the radius, which will be expressed as $r$ in the rest of this solution. When we know the coordinates, we will set the distance between the 2 points equal to $2r$. All points $r$ units away from $\\overline{AB}$ are on the line with slope $-\\frac{15}{8}$, and y-intercept $30+ \\frac{17}{8} r$ $O_1$ will have x-coordinate $r$ and likewise $O_2$ will have y-coordinate $r$ plugging this into the equation for the line mentioned in the sentence above gives us: $O_1 = \\left(r,\\frac14 r+30\\right)$ and $O_2 = \\left(\\frac35 r+16,r\\right)$ By the distance formula and the fact that the circles and tangent, we have: $\\left(16-\\frac25 r\\right)^2 + \\left(30-\\frac34 r\\right)^2 = (2r)^2$ which simplifies into the quadratic equation: $1311 r^2 + 23120 r - 462400 = 0$ And by the quadratic equation, the solutions are: $\\frac{-23120 \\pm 54400}{2622}$ The solution including the \"$-$\" is extraneous so we have the radius equal to $\\frac{31280}{2622}$ Which simplifies to $\\frac{680}{57}$. The sum of the numerator and the denominator is $737. ~Leonard_my_dude~",
"It is known that $O_1O_2$ is parallel to AB. Thus, extending $O_1F$ and $GO_2$ to intersect at H yields similar triangles $O_1O_2H$ and BAC, so that $O_1O_2 = 2r$, $O_1H = \\frac{16r}{17}$, and $HO_2 = \\frac{30r}{17}$. It should be noted that $O_2G = r$. Also, FHGC is a rectangle, and so AF = $\\frac{47r}{17} - 30$ and similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: \\[2r + \\frac{47r}{17} - 30 + \\frac{33r}{17} - 16 = 34\\] Thus, r = $\\frac{680}{57}$, and the answer is $737. ~Leonard_my_dude~",
"It is known that $O_1O_2$ is parallel to AB. Thus, extending $O_1F$ and $GO_2$ to intersect at H yields similar triangles $O_1O_2H$ and BAC, so that $O_1O_2 = 2r$, $O_1H = \\frac{16r}{17}$, and $HO_2 = \\frac{30r}{17}$. It should be noted that $O_2G = r$. Also, FHGC is a rectangle, and so AF = $\\frac{47r}{17} - 30$ and similarly for BG. Because tangents to a circle are equal, the hypotenuse can be expressed in terms of r: \\[2r + \\frac{47r}{17} - 30 + \\frac{33r}{17} - 16 = 34\\] Thus, r = $\\frac{680}{57}$, and the answer is $737. ~Leonard_my_dude~",
"Let the radius of the circle be $r$. It can be seen that $\\Delta FHO_{1}$ and $\\Delta O_{2}GJ$ are similar to $\\Delta ACB$, and the length of the hypotenuses are $\\frac{17}{8}r$ and $\\frac {17}{15}r$, respectively. Then, the entire length of $HJ$ is going to be $(\\frac{17}{8}+\\frac{17}{15}+2)r = \\frac{631}{120}r$. The length of the hypotenuse of $\\Delta ACB$ is 34, so the length of the height to $AB$ is $\\frac{16*30}{34} = \\frac{240}{17}$. Thus, the height to $\\Delta HCJ$ is going to be $\\frac{240}{17} + r$. $\\Delta HCJ$ is similar to $\\Delta ACB$ so we have the following: $\\frac{\\frac{631}{120}r}{34} = \\frac{\\frac{240}{17} + r}{\\frac{240}{17}}$. Cross multiplying and simplifying, we get that $r = \\frac{680}{57}$ so the answer is $737. ~Leonard_my_dude~"
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2007-I-10
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In a $6 \times 4$ grid ( $6$ rows, $4$ columns), $12$ of the $24$ squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by $1000$ . AIME I 2007-10.png
| 860
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I
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[
"Solution 1 Consider the first column. There are ${6\\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer. Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases: All three balls are in the same column. In this case, there are $3$ choices for which column that is. From here, the bottom half of the board is fixed. Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have $3 \\cdot 2 \\cdot 3 \\cdot 3 = 54$. All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There is 1 ball in each of the columns 1, 2, and 3 now, so in the 3x3 where we still have choices, each row and column has one square that is not filled in. Similar to the upper 3x3, there are 6 ways to do this. So there are $3 \\cdot 2 \\cdot 6 = 36$ ways. Therefor there are $20(3+54+36) = 1860$ different shadings, and the solution is $860.",
"Consider the first column. There are ${6\\choose3} = 20$ ways that the rows could be chosen, but without loss of generality let them be the first three rows. (Change the order of the rows to make this true.) We will multiply whatever answer we get by 20 to get our final answer. Now consider the 3x3 that is next to the 3 boxes we have filled in. We must put one ball in each row (since there must be 2 balls in each row and we've already put one in each). We split into three cases: All three balls are in the same column. In this case, there are $3$ choices for which column that is. From here, the bottom half of the board is fixed. Two balls are in one column, and one is in the other. In this case, there are 3 ways to choose which column gets 2 balls and 2 ways to choose which one gets the other ball. Then, there are 3 ways to choose which row the lone ball is in. Now, what happens in the bottom half of the board? Well, the 3 boxes in the column with no balls in the top half must all be filled in, so there are no choices here. In the column with two balls already, we can choose any of the 3 boxes for the third ball. This forces the location for the last two balls. So we have $3 \\cdot 2 \\cdot 3 \\cdot 3 = 54$. All three balls are in different columns. Then there are 3 ways to choose which row the ball in column 2 goes and 2 ways to choose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the board. There is 1 ball in each of the columns 1, 2, and 3 now, so in the 3x3 where we still have choices, each row and column has one square that is not filled in. Similar to the upper 3x3, there are 6 ways to do this. So there are $3 \\cdot 2 \\cdot 6 = 36$ ways. Therefor there are $20(3+54+36) = 1860$ different shadings, and the solution is $860.",
"We start by showing that every group of $6$ rows can be grouped into $3$ complementary pairs. We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns $1$ and $2$ shaded. Note how if there is no complement to this, then all the other five rows must have at least one square in the first two columns shaded. That means that in total, the first two rows have $2+5=7$ squares shaded in- that is false since it should only be $6$. Thus, there exists another row that is complementary to the first. We remove those two and use a similar argument again to show that every group of $6$ rows can be grouped into $3$ complementary pairs. Now we proceed with three cases. There are $\\frac{\\binom42}2=3$ pairs of complementary pairs. The first case is that the three pairs are all different, meaning that every single possible pair of shaded squares is used once. This gives us $6!=720.$ Our second case is that two of the pairs are the same, and the third is different. We have $3$ to choose the pair that shows up twice and $2$ for the other, giving us $3\\cdot2\\cdot\\binom62\\binom42\\binom21=6\\cdot15\\cdot6\\cdot2=1080.$ Our final case is that all three pairs are the same. This is just $3\\cdot\\binom63=60.$ Our answer is thus $720+1080+60=1860,$ leaving us with a final answer of $860.",
"We draw a bijection between walking from $(0,0,0,0)$ to $(3,3,3,3)$ as follows: if in the $i$th row, the $j$th and $k$th columns are shaded, then the $(2i-1)$st step is in the direction corresponding to $j$, and the $(2i)$th step is in the direction corresponding to $k$ ($j < k$) here. We can now use the Principle of Inclusion-Exclusion based on the stipulation that $j\\ne k$ to solve the problem: $\\frac{1}{2^6}\\left(\\frac{12!}{3!^4}-4\\cdot{6 \\choose 1}\\frac{10!}{3!^3}+4\\cdot 3\\cdot{6 \\choose 2}\\frac{8!}{3!^2}-4\\cdot 3\\cdot 2\\cdot{6 \\choose 3}\\frac{6!}{3!}+4\\cdot 3\\cdot 2\\cdot 1\\cdot {6 \\choose 4}4!\\right) = 1860$ So that the answer is $860$. Solution 4 There are ${6\\choose3}$ to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns: One example of each case for the first two columns If column 1 and column 2 do not share any two filled squares on the same row, then there are ${6\\choose3}$ combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives ${6\\choose3}^2 = 400$ arrangements. If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row ($6 - 1 = 5$ places), share another empty square on a row, and have 2 squares each on different rows. This gives $6 \\cdot 5 \\cdot {4\\choose2} = 180$. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get ${4\\choose2} = 6$. We get $180 \\cdot 6 = 1080$. If column 1 and column 2 share 2 filled squares on the same row (${6\\choose2} = 15$ places), they must also share 2 empty squares on the same row (${4\\choose2} = 6$). The last two squares can be arranged in ${2\\choose1} = 2$ positions; this totals to $15 \\cdot 6 \\cdot 2 = 180$. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have ${2\\choose1} = 2$ places, giving $180 \\cdot 2 = 360$. If column 1 and column 2 share 3 filled squares on the same row (${6\\choose3} = 20$ places), then the squares on columns 3 and 4 are fixed. Thus, there are $400 + 1080 + 360 + 20 = 1860$ number of shadings, so the answer is $860$. Solution 5 Consider all possible shadings for a single row. There are ${4 \\choose 2}=6$ ways to do so, and denote these as $a=1+2$, $b=3+4$, $c=1+4$, $d=2+3$, $e=1+3$, and $f=2+4$ where $x+y$ indicates that columns $x$ and $y$ are shaded. From our condition on the columns, we have $a+c+e=a+d+f=b+d+e=b+c+f=3$ Summing the first two and the last two equations, we have $2a+c+d+e+f=6=2b+c+d+e+f$, from which we have $a=b$. Likewise, $c=d$ and $e=f$ since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for $c$/$d$ and $e$/$f$, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives $3{6 \\choose 3}=60$ solutions; the second gives $6\\cdot 6\\cdot5\\cdot{4 \\choose 2}=1080$ solutions, and the final case gives $6!=720$ solutions. In all, we have 1860 solutions, for an answer of $860$. Solution 6 Each shading can be brought, via row swapping operations, to a state with a $3\\times2$ shaded $L$ in the lower left hand corner. The number of such arrangements multiplied by ${5 \\choose 2}{3\\choose 2}$ will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form $\\{3,0,0\\},\\{0,1,2\\},\\{1,1,1\\}$. Form 1: The entire lower left $3\\times2$ rectangle is shaded, forcing the opposite $3\\times2$ rectangle to also be shaded; thus 1 arrangement Form 2: There is a column with nothing shaded in the bottom right $3\\times2$, so it must be completely shaded in the upper right $3\\times2$. Now consider the upper right half column that will have $1$ shade. There are $3$ ways of choosing this shade, and all else is determined from here; thus 3 arrangements Form 3: The upper right $3\\times3$ will have exactly $2$ shades per column and row. This is equivalent to the number of terms in a $3\\times3$ determinant, or $6$ arrangements Of the $3^2$ ways of choosing to complete the bottom half of the $4\\times6$, form 1 is achieved in exactly 1 way; form 2 is achieved in $3\\times2$ ways; and form $3$ in the remaining $2$ ways. Thus, the weighted total is $1+6\\times3+2\\times6=31$. Complete: $31\\times60=860\\mod{1000}$ Solution 7 Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares. There are ${6 \\choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a \"1-row\") within the first 3 columns and which rows have 2 (we'll call these \"2-rows\") within the first 3 columns. Next, we do some casework: If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only ${3 \\choose 3}{3 \\choose 0} \\times 1= 1$ valid shading in this case. If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are ${3 \\choose 2}{3 \\choose 1}\\times 2=18$ valid shadings in this case. If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are ${3 \\choose 1}{3 \\choose 2}\\times {4 \\choose 2}=54$ valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row). If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are ${3 \\choose 0}{3 \\choose 3}\\times{6 \\choose 3} = 20$ valid shadings in this case. In total, we have ${6\\choose3}(1+18+54+20)=20*93=1860$. Thus our answer is $860.",
"There are ${6\\choose3}$ to choose the arrangement of the shaded squares in each column. Examine the positioning of the shaded squares in the first two columns: One example of each case for the first two columns If column 1 and column 2 do not share any two filled squares on the same row, then there are ${6\\choose3}$ combinations for column 1, and then column 2 is fixed. Now, any row cannot have more than 2 shaded squares, so after we pick three more squares in the third column, the fourth column is also fixed. This gives ${6\\choose3}^2 = 400$ arrangements. If column 1 and column 2 share 1 filled square on the same row (6 places), then they each share 1 filled square on a row ($6 - 1 = 5$ places), share another empty square on a row, and have 2 squares each on different rows. This gives $6 \\cdot 5 \\cdot {4\\choose2} = 180$. Now, the third and fourths columns must also share a fixed shared shaded square in the row in which the first two columns both had spaces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get ${4\\choose2} = 6$. We get $180 \\cdot 6 = 1080$. If column 1 and column 2 share 2 filled squares on the same row (${6\\choose2} = 15$ places), they must also share 2 empty squares on the same row (${4\\choose2} = 6$). The last two squares can be arranged in ${2\\choose1} = 2$ positions; this totals to $15 \\cdot 6 \\cdot 2 = 180$. Now, the third and fourth columns have a fixed 2 filled squares in common rows and 2 empty squares in common rows. The remaining 2 squares have ${2\\choose1} = 2$ places, giving $180 \\cdot 2 = 360$. If column 1 and column 2 share 3 filled squares on the same row (${6\\choose3} = 20$ places), then the squares on columns 3 and 4 are fixed. Thus, there are $400 + 1080 + 360 + 20 = 1860$ number of shadings, so the answer is $860$. Solution 5 Consider all possible shadings for a single row. There are ${4 \\choose 2}=6$ ways to do so, and denote these as $a=1+2$, $b=3+4$, $c=1+4$, $d=2+3$, $e=1+3$, and $f=2+4$ where $x+y$ indicates that columns $x$ and $y$ are shaded. From our condition on the columns, we have $a+c+e=a+d+f=b+d+e=b+c+f=3$ Summing the first two and the last two equations, we have $2a+c+d+e+f=6=2b+c+d+e+f$, from which we have $a=b$. Likewise, $c=d$ and $e=f$ since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for $c$/$d$ and $e$/$f$, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives $3{6 \\choose 3}=60$ solutions; the second gives $6\\cdot 6\\cdot5\\cdot{4 \\choose 2}=1080$ solutions, and the final case gives $6!=720$ solutions. In all, we have 1860 solutions, for an answer of $860$. Solution 6 Each shading can be brought, via row swapping operations, to a state with a $3\\times2$ shaded $L$ in the lower left hand corner. The number of such arrangements multiplied by ${5 \\choose 2}{3\\choose 2}$ will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form $\\{3,0,0\\},\\{0,1,2\\},\\{1,1,1\\}$. Form 1: The entire lower left $3\\times2$ rectangle is shaded, forcing the opposite $3\\times2$ rectangle to also be shaded; thus 1 arrangement Form 2: There is a column with nothing shaded in the bottom right $3\\times2$, so it must be completely shaded in the upper right $3\\times2$. Now consider the upper right half column that will have $1$ shade. There are $3$ ways of choosing this shade, and all else is determined from here; thus 3 arrangements Form 3: The upper right $3\\times3$ will have exactly $2$ shades per column and row. This is equivalent to the number of terms in a $3\\times3$ determinant, or $6$ arrangements Of the $3^2$ ways of choosing to complete the bottom half of the $4\\times6$, form 1 is achieved in exactly 1 way; form 2 is achieved in $3\\times2$ ways; and form $3$ in the remaining $2$ ways. Thus, the weighted total is $1+6\\times3+2\\times6=31$. Complete: $31\\times60=860\\mod{1000}$ Solution 7 Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares. There are ${6 \\choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a \"1-row\") within the first 3 columns and which rows have 2 (we'll call these \"2-rows\") within the first 3 columns. Next, we do some casework: If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only ${3 \\choose 3}{3 \\choose 0} \\times 1= 1$ valid shading in this case. If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are ${3 \\choose 2}{3 \\choose 1}\\times 2=18$ valid shadings in this case. If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are ${3 \\choose 1}{3 \\choose 2}\\times {4 \\choose 2}=54$ valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row). If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are ${3 \\choose 0}{3 \\choose 3}\\times{6 \\choose 3} = 20$ valid shadings in this case. In total, we have ${6\\choose3}(1+18+54+20)=20*93=1860$. Thus our answer is $860.",
"Consider all possible shadings for a single row. There are ${4 \\choose 2}=6$ ways to do so, and denote these as $a=1+2$, $b=3+4$, $c=1+4$, $d=2+3$, $e=1+3$, and $f=2+4$ where $x+y$ indicates that columns $x$ and $y$ are shaded. From our condition on the columns, we have $a+c+e=a+d+f=b+d+e=b+c+f=3$ Summing the first two and the last two equations, we have $2a+c+d+e+f=6=2b+c+d+e+f$, from which we have $a=b$. Likewise, $c=d$ and $e=f$ since these pairs shade in complimentary columns. So the six rows are paired up into a row and its compliment. In all, we can have 3 a's and 3 b's and similar setups for $c$/$d$ and $e$/$f$, 2 a's, 2 b's, 1 c and 1 d and similar setups for all six arrangements, or one of each. This first case gives $3{6 \\choose 3}=60$ solutions; the second gives $6\\cdot 6\\cdot5\\cdot{4 \\choose 2}=1080$ solutions, and the final case gives $6!=720$ solutions. In all, we have 1860 solutions, for an answer of $860$. Solution 6 Each shading can be brought, via row swapping operations, to a state with a $3\\times2$ shaded $L$ in the lower left hand corner. The number of such arrangements multiplied by ${5 \\choose 2}{3\\choose 2}$ will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form $\\{3,0,0\\},\\{0,1,2\\},\\{1,1,1\\}$. Form 1: The entire lower left $3\\times2$ rectangle is shaded, forcing the opposite $3\\times2$ rectangle to also be shaded; thus 1 arrangement Form 2: There is a column with nothing shaded in the bottom right $3\\times2$, so it must be completely shaded in the upper right $3\\times2$. Now consider the upper right half column that will have $1$ shade. There are $3$ ways of choosing this shade, and all else is determined from here; thus 3 arrangements Form 3: The upper right $3\\times3$ will have exactly $2$ shades per column and row. This is equivalent to the number of terms in a $3\\times3$ determinant, or $6$ arrangements Of the $3^2$ ways of choosing to complete the bottom half of the $4\\times6$, form 1 is achieved in exactly 1 way; form 2 is achieved in $3\\times2$ ways; and form $3$ in the remaining $2$ ways. Thus, the weighted total is $1+6\\times3+2\\times6=31$. Complete: $31\\times60=860\\mod{1000}$ Solution 7 Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares. There are ${6 \\choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a \"1-row\") within the first 3 columns and which rows have 2 (we'll call these \"2-rows\") within the first 3 columns. Next, we do some casework: If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only ${3 \\choose 3}{3 \\choose 0} \\times 1= 1$ valid shading in this case. If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are ${3 \\choose 2}{3 \\choose 1}\\times 2=18$ valid shadings in this case. If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are ${3 \\choose 1}{3 \\choose 2}\\times {4 \\choose 2}=54$ valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row). If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are ${3 \\choose 0}{3 \\choose 3}\\times{6 \\choose 3} = 20$ valid shadings in this case. In total, we have ${6\\choose3}(1+18+54+20)=20*93=1860$. Thus our answer is $860.",
"Each shading can be brought, via row swapping operations, to a state with a $3\\times2$ shaded $L$ in the lower left hand corner. The number of such arrangements multiplied by ${5 \\choose 2}{3\\choose 2}$ will be the total. Consider rows 2 and 3 up from the bottom: they each have one of their allotted two squares shaded. Depending how the remaining three shades are distributed, the column totals of columns 2,3, and 4 from the left can be of the form $\\{3,0,0\\},\\{0,1,2\\},\\{1,1,1\\}$. Form 1: The entire lower left $3\\times2$ rectangle is shaded, forcing the opposite $3\\times2$ rectangle to also be shaded; thus 1 arrangement Form 2: There is a column with nothing shaded in the bottom right $3\\times2$, so it must be completely shaded in the upper right $3\\times2$. Now consider the upper right half column that will have $1$ shade. There are $3$ ways of choosing this shade, and all else is determined from here; thus 3 arrangements Form 3: The upper right $3\\times3$ will have exactly $2$ shades per column and row. This is equivalent to the number of terms in a $3\\times3$ determinant, or $6$ arrangements Of the $3^2$ ways of choosing to complete the bottom half of the $4\\times6$, form 1 is achieved in exactly 1 way; form 2 is achieved in $3\\times2$ ways; and form $3$ in the remaining $2$ ways. Thus, the weighted total is $1+6\\times3+2\\times6=31$. Complete: $31\\times60=860\\mod{1000}$ Solution 7 Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares. There are ${6 \\choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a \"1-row\") within the first 3 columns and which rows have 2 (we'll call these \"2-rows\") within the first 3 columns. Next, we do some casework: If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only ${3 \\choose 3}{3 \\choose 0} \\times 1= 1$ valid shading in this case. If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are ${3 \\choose 2}{3 \\choose 1}\\times 2=18$ valid shadings in this case. If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are ${3 \\choose 1}{3 \\choose 2}\\times {4 \\choose 2}=54$ valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row). If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are ${3 \\choose 0}{3 \\choose 3}\\times{6 \\choose 3} = 20$ valid shadings in this case. In total, we have ${6\\choose3}(1+18+54+20)=20*93=1860$. Thus our answer is $860.",
"Note that if we find a valid shading of the first 3 columns, the shading of the last column is determined. We also note that within the first 3 columns, there will be 3 rows with 1 shaded square and 3 rows with 2 shaded squares. There are ${6 \\choose 3}$ ways to choose which rows have 1 shaded square (which we'll call a \"1-row\") within the first 3 columns and which rows have 2 (we'll call these \"2-rows\") within the first 3 columns. Next, we do some casework: If all of the shaded squares in the first column are in a 1-row, then the squares in the second and third columns must be in the 2-rows. Thus there is only ${3 \\choose 3}{3 \\choose 0} \\times 1= 1$ valid shading in this case. If 2 of the shaded squares in the first column are in a 1-row, and the third shaded square is in a 2-row, then the other shaded square in that 2-row can either be in column 3 or column 4. Once we determine that, the other shaded squares are uniquely determined. Thus, there are ${3 \\choose 2}{3 \\choose 1}\\times 2=18$ valid shadings in this case. If 1 shaded square in the first column is in a 1-row, and the other 2 are in 2-rows, then the 2-row without a shaded square in the first column must have shaded squares in both column 2 and column 3. This leaves 4 possible squares in the second column to be shaded (since there can't be another shaded square in the occupied 1-row). Thus, there are ${3 \\choose 1}{3 \\choose 2}\\times {4 \\choose 2}=54$ valid shadings in this case. (We only need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row). If all of the shaded squares in the first column are in the 3 2-rows, then if we choose any 3 squares in the second column to be shaded, then the third column is uniquely determined to create a valid shading. Thus, there are ${3 \\choose 0}{3 \\choose 3}\\times{6 \\choose 3} = 20$ valid shadings in this case. In total, we have ${6\\choose3}(1+18+54+20)=20*93=1860$. Thus our answer is $860.",
"We can use generating functions. Suppose that the variables $a$, $b$, $c$, and $d$ represent shading a square that appears in the first, second, third, or fourth columns, respectively. Then if two squares in the row are shaded, then the row is represented by the generating function $ab+ac+ad+bc+bd+cd$, which we can write as $P(a,b,c,d)=(ab+cd)+(a+b)(c+d)$. Therefore, $P(a,b,c,d)^6$ represents all of the possible ways to color six rows such that each row has two shaded squares. We only want the possibilities when each column has three shaded squares, or rather, the coefficient of $a^3b^3c^3d^3$ in $P(a,b,c,d)^6$. By the Binomial Theorem, \\[P(a,b,c,d)^6=\\sum_{k=0}^6 \\binom{6}{k} (ab+cd)^k(a+b)^{6-k}(c+d)^{6-k}.\\tag{1}\\] If we expand $(ab+cd)^k$, then the powers of $a$ and $b$ are always equal. Therefore, to obtain terms of the form $a^3b^3c^3d^3$, the powers of $a$ and $b$ in $(a+b)^{6-k}$ must be equal. In particular, only the central term in the binomial expansion will contribute, and this implies that $k$ must be even. We can use the same logic for $c$ and $d$. Therefore, the coefficient of $a^3b^3c^3d^3$ in the following expression is the same as the coefficient of $a^3b^3c^3d^3$ in (1). \\[\\sum_{k=0}^3 \\binom{6}{2k} (ab+cd)^{2k}(ab)^{3-k}(cd)^{3-k}\\binom{6-2k}{3-k}^2.\\tag{2}\\] Now we notice that the only way to obtain terms of the form $a^3b^3c^3d^3$ is if we take the central term in the binomial expansion of $(ab+cd)^{2k}$. Therefore, the terms that contribute to the coefficient of $a^3b^3c^3d^3$ in (2) are \\[\\sum_{k=0}^3 \\binom{6}{2k}\\binom{2k}{k}\\binom{6-2k}{3-k}^2(abcd)^3.\\] This sum is $400+1080+360+20=1860$ so the answer is $860."
] |
2007-I-11
| 2,007
| 11
|
For each positive integer $p$ , let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$ . For example, $b(6) = 2$ and $b(23) = 5$ . If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000.
| 955
|
I
|
[
"$\\left(k- \\frac 12\\right)^2=k^2-k+\\frac 14$ and $\\left(k+ \\frac 12\\right)^2=k^2+k+ \\frac 14$. Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\\leq k^2+k$. There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$. $44<\\sqrt{2007}<45$, so all numbers $1$ to $44$ have their full range. Summing this up with the formula for the sum of the first $n$ squares ($\\frac{n(n+1)(2n+1)}{6}$), we get $\\sum_{k=1}^{44}2k^2=2\\frac{44(44+1)(2*44+1)}{6}=58740$. We need only consider the $740$ because we are working with modulo $1000$. Now consider the range of numbers such that $b(p)=45$. These numbers are $\\left\\lceil\\frac{44^2 + 45^2}{2}\\right\\rceil = 1981$ to $2007$. There are $2007 - 1981 + 1 = 27$ (1 to be inclusive) of them. $27*45=1215$, and $215+740= 955, the answer.",
"Let $p$ be in the range of $a^2 \\le p < (a+1)^2$. Then, we need to find the point where the value of $b(p)$ flips from $k$ to $k+1$. This will happen when $p$ exceeds $(a+\\frac{1}{2})^2$ or $a(a+1)+\\frac{1}{4}$. Thus, if $a^2 \\le p \\le a(a+1)$ then $b(p)=a$. For $a(a+1) < p < (a+1)^2$, then $b(p)=a+1$. There are $a+1$ terms in the first set of $p$, and $a$ terms in the second set. Thus, the sum of $b(p)$ from $a^2 \\le p <(a+1)^2$ is $2a(a+1)$ or $4\\cdot\\binom{a+1}{2}$. For the time being, consider that $S = \\sum_{p=1}^{44^2-1} b(p)$. Then, the sum of the values of $b(p)$ is $4\\binom{2}{2}+4\\binom{3}{2}+\\cdots +4\\binom{44}{2}=4\\left(\\binom{2}{2}+\\binom{3}{2}+\\cdots +\\binom{44}{2}\\right)$. We can collapse this to $4\\binom{45}{3}=56760$. Now, we have to consider $p$ from $44^2 \\le p < 2007$. Considering $p$ from just $44^2 \\le p \\le 1980$, we see that all of these values have $b(p)=44$. Because there are $45$ values of $p$ in that range, the sum of $b(p)$ in that range is $45\\cdot44=1980$. Adding this to $56760$ we get $58740$ or $740$ mod $1000$. Now, take the range $1980 < p \\le 2007$. There are $27$ values of $p$ in this range, and each has $b(p)=45$. Thus, that contributes $27*45=1215$ or $215$ to the sum. Finally, adding $740$ and $215$ we get $740+215=955. ~firebolt360"
] |
2007-I-12
| 2,007
| 12
|
In isosceles triangle $ABC$ , $A$ is located at the origin and $B$ is located at (20,0). Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$ . If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$ -axis, the area of the region common to the original and the rotated triangle is in the form $p\sqrt{2} + q\sqrt{3} + r\sqrt{6} + s$ , where $p,q,r,s$ are integers. Find $(p-q+r-s)/2$ .
| 875
|
I
|
[
"[asy] defaultpen(fontsize(12)+0.6); size(300); var theta=15; pair A=origin, B=(20,0), C=extension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp); draw(A--B--C--A); draw(A--Bp--Cp--A, royalblue); markscalefactor=0.1; draw(rightanglemark(Y,X,A)); dot(\"$A$\",A,dir(210)); dot(\"$B$\",B,dir(-20)); dot(\"$C$\",C,up); dot(\"$B'$\",Bp,dir(-10)); dot(\"$C'$\",Cp,dir(100)); MA(\"60^\\circ\",Bp,A,C,2); MA(\"15^\\circ\",B,A,Bp,5); MA(\"75^\\circ\",C,B,A,2); MA(\"30^\\circ\",A,C,B,4); MA(\"30^\\circ\",A,Cp,Bp,4); MA(\"45^\\circ\",A,extension(A,C,Bp,Cp),Bp,3); MA(\"15^\\circ\",C,Y,Cp,8); MA(\"15^\\circ\",C,A,Cp,9); [/asy] Solution 1 Let the new triangle be $\\triangle AB'C'$ ($A$, the origin, is a vertex of both triangles). Let $\\overline{B'C'}$ intersect with $\\overline{AC}$ at point $D$, $\\overline{BC}$ intersect with $\\overline{B'C'}$ at $E$, and $\\overline{BC}$ intersect with $\\overline{AB'}$ at $F$. The region common to both triangles is the quadrilateral $ADEF$. Notice that $[ADEF] = [\\triangle ADB'] - [\\triangle EFB']$, where we let $[\\ldots]$ denote area. To find $[\\triangle ADB']$: Since $\\angle B'AC'$ and $\\angle BAC$ both have measures $75^{\\circ}$, both of their complements are $15^{\\circ}$, and $\\angle DAB' = 90 - 2(15) = 60^{\\circ}$. We know that $\\angle DB'A = 75^{\\circ}$, so $\\angle ADB' = 180 - 60 - 75 = 45^{\\circ}$. Thus $\\triangle ADB'$ is a $45 - 60 - 75 \\triangle$. It can be solved by drawing an altitude splitting the $75^{\\circ}$ angle into $30^{\\circ}$ and $45^{\\circ}$ angles, forming a $30-60-90$ right triangle and a $45-45-90$ isosceles right triangle. Since we know that $AB' = 20$, the base of the $30-60-90$ triangle is $10$, the base of the $45-45-90$ is $10\\sqrt{3}$, and their common height is $10\\sqrt{3}$. Thus, the total area of $[\\triangle ADB'] = \\frac{1}{2}(10\\sqrt{3})(10\\sqrt{3} + 10) = 150 + 50\\sqrt{3}.",
"Let the new triangle be $\\triangle AB'C'$ ($A$, the origin, is a vertex of both triangles). Let $\\overline{B'C'}$ intersect with $\\overline{AC}$ at point $D$, $\\overline{BC}$ intersect with $\\overline{B'C'}$ at $E$, and $\\overline{BC}$ intersect with $\\overline{AB'}$ at $F$. The region common to both triangles is the quadrilateral $ADEF$. Notice that $[ADEF] = [\\triangle ADB'] - [\\triangle EFB']$, where we let $[\\ldots]$ denote area. To find $[\\triangle ADB']$: Since $\\angle B'AC'$ and $\\angle BAC$ both have measures $75^{\\circ}$, both of their complements are $15^{\\circ}$, and $\\angle DAB' = 90 - 2(15) = 60^{\\circ}$. We know that $\\angle DB'A = 75^{\\circ}$, so $\\angle ADB' = 180 - 60 - 75 = 45^{\\circ}$. Thus $\\triangle ADB'$ is a $45 - 60 - 75 \\triangle$. It can be solved by drawing an altitude splitting the $75^{\\circ}$ angle into $30^{\\circ}$ and $45^{\\circ}$ angles, forming a $30-60-90$ right triangle and a $45-45-90$ isosceles right triangle. Since we know that $AB' = 20$, the base of the $30-60-90$ triangle is $10$, the base of the $45-45-90$ is $10\\sqrt{3}$, and their common height is $10\\sqrt{3}$. Thus, the total area of $[\\triangle ADB'] = \\frac{1}{2}(10\\sqrt{3})(10\\sqrt{3} + 10) = 150 + 50\\sqrt{3}.",
"Redefine the points in the same manner as the last time ($\\triangle AB'C'$, intersect at $D$, $E$, and $F$). This time, notice that $[ADEF] = [\\triangle AB'C'] - ([\\triangle ADC'] + [\\triangle EFB'])$. The area of $[\\triangle AB'C'] = [\\triangle ABC]$. The altitude of $\\triangle ABC$ is clearly $10 \\tan 75 = 10 \\tan (30 + 45)$. The tangent addition rule yields $10(2 + \\sqrt{3})$ (see above). Thus, $[\\triangle ABC] = \\frac{1}{2} 20 \\cdot (20 + 10\\sqrt{3}) = 200 + 100\\sqrt{3}$. The area of $[\\triangle ADC']$ (with a side on the y-axis) can be found by splitting it into two triangles, $30-60-90$ and $15-75-90$ right triangles. $AC' = AC = \\frac{10}{\\sin 15}$. The sine subtraction rule shows that $\\frac{10}{\\sin 15} = \\frac{10}{\\frac{\\sqrt{6} - \\sqrt{2}}4} = \\frac{40}{\\sqrt{6} - \\sqrt{2}} = 10(\\sqrt{6} + \\sqrt{2})$. $AC'$, in terms of the height of $\\triangle ADC'$, is equal to $h(\\sqrt{3} + \\tan 75) = h(\\sqrt{3} + 2 + \\sqrt{3})$. \\begin{align*} [ADC'] &= \\frac 12 AC' \\cdot h \\\\ &= \\frac 12 (10\\sqrt{6} + 10\\sqrt{2})\\left(\\frac{10\\sqrt{6} + 10\\sqrt{2}}{2\\sqrt{3} + 2}\\right) \\\\ &= \\frac{(800 + 400\\sqrt{3})}{(2 + \\sqrt{3})}\\cdot\\frac{2 - \\sqrt{3}}{2-\\sqrt{3}} \\\\ &= \\frac{400\\sqrt{3} + 400}8 = 50\\sqrt{3} + 50 \\end{align*} The area of $[\\triangle EFB']$ was found in the previous solution to be $- 500\\sqrt{2} + 400\\sqrt{3} - 300\\sqrt{6} +750$. Therefore, $[ADEF]$ $= (200 + 100\\sqrt{3}) - \\left((50 + 50\\sqrt{3}) + (-500\\sqrt{2} + 400\\sqrt{3} - 300\\sqrt{6} +750)\\right)$ $= 500\\sqrt{2} - 350\\sqrt{3} + 300\\sqrt{6} - 600$, and our answer is $875.",
"Call the points of the intersections of the triangles $D$, $E$, and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\\overline{AD}$ bisects $\\angle EDE'$. Through HL congruency, we can find that $\\triangle AED$ is congruent to $\\triangle AE'D$. This divides the region $AEDF$ (which we are trying to solve for) into two congruent triangles and an isosceles right triangle. $AE = 20 \\cos 15 = 20 \\cos (45 - 30) = 20 \\cdot \\frac{\\sqrt{6} + \\sqrt{2}}{4} = 5\\sqrt{6} + 5\\sqrt{2}$ Since $FE' = AE' = AE$, we find that $[AE'F] = \\frac 12 (5\\sqrt{6} + 5\\sqrt{2})^2 = 100 + 50\\sqrt{3}$. Now, we need to find $[AED] = [AE'D]$. The acute angles of the triangles are $\\frac{15}{2}$ and $90 - \\frac{15}{2}$. By repeated application of the half-angle formula, we can find that $\\tan \\frac{15}{2} = \\sqrt{2} - \\sqrt{3} + \\sqrt{6} - 2$. The area of $[AED] = \\frac 12 \\left(20 \\cos 15\\right)^2 \\left(\\tan \\frac{15}{2}\\right)$. Thus, $[AED] + [AE'D] = 2\\left(\\frac 12((5\\sqrt{6} + 5\\sqrt{2})^2 \\cdot (\\sqrt{2} - \\sqrt{3} + \\sqrt{6} - 2))\\right)$, which eventually simplifies to $500\\sqrt{2} - 350\\sqrt{3} + 300\\sqrt{6} - 600$. Adding them together, we find that the solution is $[AEDF] = [AE'F] + [AED] + [AE'D]$ $= 100 + 50\\sqrt{3} + 500\\sqrt{2} - 350\\sqrt{3} + 300\\sqrt{6} - 600=$ $= 500\\sqrt{2} - 350\\sqrt{3} + 300\\sqrt{6} - 600$, and the answer is $875.",
"From the given information, calculate the coordinates of all the points (by finding the equations of the lines, equating). Then, use the shoelace method to calculate the area of the intersection. $\\overline{AC}$: $y = (\\tan 75) x = (2 + \\sqrt{3})x$ $\\overline{AB'}$: $y = (\\tan 15) x = (2 - \\sqrt{3})x$ $\\overline{BC}$: It passes thru $(20,0)$, and has a slope of $-\\tan75 = -(2 + \\sqrt{3})$. The equation of its line is $y = (2+\\sqrt{3})(20 - x)$. $\\overline{B'C'}$: $AC' = AC = \\frac{10}{\\cos 75} = 10\\sqrt{6} + 10\\sqrt{2}$, so it passes thru point $(0, 10\\sqrt{6} + 10\\sqrt{2})$. It has a slope of $-\\tan 60 = -\\sqrt{3}$. So the equation of its line is $y = -\\sqrt{3}x + 10(\\sqrt{6} + \\sqrt{2})$. Now, we can equate the equations to find the intersections of all the points. $A (0,0)$ $D$ is the intersection of $\\overline{BC},\\ \\overline{B'C'}$. $(2+\\sqrt{3})(20-x) = -\\sqrt{3}x + 10(\\sqrt{6} + \\sqrt{2})$. Therefore, $x = 5(4 + 2\\sqrt{3}-\\sqrt{6}-\\sqrt{2})$, $y = 5(3\\sqrt{6} + 5\\sqrt{2}-4\\sqrt{3}-6)$. $E$ is the intersection of $\\overline{AB'},\\ \\overline{BC}$. $(2 - \\sqrt{3})x =(2+\\sqrt{3})(20-x)$. Therefore, $x = 5(2+\\sqrt{3})$, $y = 5$. $F$ is the intersection of $\\overline{AC},\\ \\overline{B'C'}$. $(2+\\sqrt{3})x = -\\sqrt{3}x + 10(\\sqrt{6} + \\sqrt{2})$. Therefore, $x = 5\\sqrt{2}$, $y = 10\\sqrt{2}+ 5\\sqrt{6}$. We take these points and tie them together by shoelace, and the answer should come out to be $875."
] |
2007-I-13
| 2,007
| 13
|
A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length 4. A plane passes through the midpoints of $AE$ , $BC$ , and $CD$ . The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$ . Find $p$ .
| 80
|
I
|
[
"Solution 1 Note first that the intersection is a pentagon. Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,0,2\\sqrt{2})$. Using the coordinates of the three points of intersection $(-1,1,\\sqrt{2}),\\ (2,0,0),\\ (0,-2,0)$, it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$, and using the points we find that $2a = d \\Longrightarrow d = \\frac{a}{2}$, $-2b = d \\Longrightarrow d = \\frac{-b}{2}$, and $-a + b + \\sqrt{2}c = d \\Longrightarrow -\\frac{d}{2} - \\frac{d}{2} + \\sqrt{2}c = d \\Longrightarrow c = d\\sqrt{2}$. It is then $x - y + 2\\sqrt{2}z = 2$. [asy]import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2); draw(A--B--C--D--A--E--B--E--C--E--D); label(\"A\",A, SE); label(\"B\",B,(1,0,0)); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE); draw(P--X--Q--R--Y--cycle,linetype(\"6 6\")+linewidth(0.7)); [/asy] [asy] pointpen = black; pathpen = black+linewidth(0.7); pair P = (0, 2.5^.5), X = (3/2^.5,0), Y = (-3/2^.5,0), Q = (2^.5,-2.5^.5), R = (-2^.5,-2.5^.5); D(MP(\"P\",P,N)--MP(\"X\",X,NE)--MP(\"Q\",Q)--MP(\"R\",R)--MP(\"Y\",Y,NW)--cycle); D(X--Y,linetype(\"6 6\") + linewidth(0.7)+blue); D(P--(0,-P.y),linetype(\"6 6\") + linewidth(0.7) + red); MP(\"\\color{blue}{3\\sqrt{2}}\",(X+Y)/2); MP(\"2\\sqrt{2}\",(Q+R)/2); MP(\"\\color{red}{\\sqrt{\\frac{5}{2}}}\",(0,-P.y/2),E); MP(\"\\color{red}{\\sqrt{\\frac{5}{2}}}\",(0,2*P.y/5),E); [/asy] Write the equation of the lines and substitute to find that the other two points of intersection on $\\overline{BE}$, $\\overline{DE}$ are $\\left(\\frac{\\pm 3}{2},\\frac{\\pm 3}{2},\\frac{\\sqrt{2}}{2}\\right)$. To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula ($\\sqrt{a^2 + b^2 + c^2}$), it is possible to find that the area of the triangle is $\\frac{1}{2}bh \\Longrightarrow \\frac{1}{2} 3\\sqrt{2} \\cdot \\sqrt{\\frac 52} = \\frac{3\\sqrt{5}}{2}$. The trapezoid has area $\\frac{1}{2}h(b_1 + b_2) \\Longrightarrow \\frac 12\\sqrt{\\frac 52}\\left(2\\sqrt{2} + 3\\sqrt{2}\\right) = \\frac{5\\sqrt{5}}{2}$. In total, the area is $4\\sqrt{5} = \\sqrt{80}$, and the solution is $080.",
"Note first that the intersection is a pentagon. Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,0,2\\sqrt{2})$. Using the coordinates of the three points of intersection $(-1,1,\\sqrt{2}),\\ (2,0,0),\\ (0,-2,0)$, it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$, and using the points we find that $2a = d \\Longrightarrow d = \\frac{a}{2}$, $-2b = d \\Longrightarrow d = \\frac{-b}{2}$, and $-a + b + \\sqrt{2}c = d \\Longrightarrow -\\frac{d}{2} - \\frac{d}{2} + \\sqrt{2}c = d \\Longrightarrow c = d\\sqrt{2}$. It is then $x - y + 2\\sqrt{2}z = 2$. [asy]import three; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2); draw(A--B--C--D--A--E--B--E--C--E--D); label(\"A\",A, SE); label(\"B\",B,(1,0,0)); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE); draw(P--X--Q--R--Y--cycle,linetype(\"6 6\")+linewidth(0.7)); [/asy] [asy] pointpen = black; pathpen = black+linewidth(0.7); pair P = (0, 2.5^.5), X = (3/2^.5,0), Y = (-3/2^.5,0), Q = (2^.5,-2.5^.5), R = (-2^.5,-2.5^.5); D(MP(\"P\",P,N)--MP(\"X\",X,NE)--MP(\"Q\",Q)--MP(\"R\",R)--MP(\"Y\",Y,NW)--cycle); D(X--Y,linetype(\"6 6\") + linewidth(0.7)+blue); D(P--(0,-P.y),linetype(\"6 6\") + linewidth(0.7) + red); MP(\"\\color{blue}{3\\sqrt{2}}\",(X+Y)/2); MP(\"2\\sqrt{2}\",(Q+R)/2); MP(\"\\color{red}{\\sqrt{\\frac{5}{2}}}\",(0,-P.y/2),E); MP(\"\\color{red}{\\sqrt{\\frac{5}{2}}}\",(0,2*P.y/5),E); [/asy] Write the equation of the lines and substitute to find that the other two points of intersection on $\\overline{BE}$, $\\overline{DE}$ are $\\left(\\frac{\\pm 3}{2},\\frac{\\pm 3}{2},\\frac{\\sqrt{2}}{2}\\right)$. To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula ($\\sqrt{a^2 + b^2 + c^2}$), it is possible to find that the area of the triangle is $\\frac{1}{2}bh \\Longrightarrow \\frac{1}{2} 3\\sqrt{2} \\cdot \\sqrt{\\frac 52} = \\frac{3\\sqrt{5}}{2}$. The trapezoid has area $\\frac{1}{2}h(b_1 + b_2) \\Longrightarrow \\frac 12\\sqrt{\\frac 52}\\left(2\\sqrt{2} + 3\\sqrt{2}\\right) = \\frac{5\\sqrt{5}}{2}$. In total, the area is $4\\sqrt{5} = \\sqrt{80}$, and the solution is $080.",
"Use the same coordinate system as above, and let the plane determined by $\\triangle PQR$ intersect $\\overline{BE}$ at $X$ and $\\overline{DE}$ at $Y$. Then the line $\\overline{XY}$ is the intersection of the planes determined by $\\triangle PQR$ and $\\triangle BDE$. Note that the plane determined by $\\triangle BDE$ has the equation $x=y$, and $\\overline{PQ}$ can be described by $x=2(1-t)-t,\\ y=t,\\ z=t\\sqrt{2}$. It intersects the plane when $2(1-t)-t=t$, or $t=\\frac{1}{2}$. This intersection point has $z=\\frac{\\sqrt{2}}{2}$. Similarly, the intersection between $\\overline{PR}$ and $\\triangle BDE$ has $z=\\frac{\\sqrt{2}}{2}$. So $\\overline{XY}$ lies on the plane $z=\\frac{\\sqrt{2}}{2}$, from which we obtain $X=\\left( \\frac{3}{2},\\frac{3}{2},\\frac{\\sqrt{2}}{2}\\right)$ and $Y=\\left( -\\frac{3}{2},-\\frac{3}{2},\\frac{\\sqrt{2}}{2}\\right)$. The area of the pentagon $EXQRY$ can be computed in the same way as above. Solution 3 [asy]import three; import math; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2), H=(4,2,0), I=(-2,-4,0); draw(A--B--C--D--A--E--B--E--C--E--D); draw(B--H--Q, linetype(\"6 6\")+linewidth(0.7)+blue); draw(X--H, linetype(\"6 6\")+linewidth(0.7)+blue); draw(D--I--R, linetype(\"6 6\")+linewidth(0.7)+blue); draw(Y--I, linetype(\"6 6\")+linewidth(0.7)+blue); label(\"A\",A, SE); label(\"B\",B,NE); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE); label(\"H\",H,NE); label(\"I\",I,S); draw(P--X--Q--R--Y--cycle,linetype(\"6 6\")+linewidth(0.7)); [/asy] Extend $\\overline{RQ}$ and $\\overline{AB}$. The point of intersection is $H$. Connect $\\overline{PH}$. $\\overline{EB}$ intersects $\\overline{PH}$ at $X$. Do the same for $\\overline{QR}$ and $\\overline{AD}$, and let the intersections be $I$ and $Y$ Because $Q$ is the midpoint of $\\overline{BC}$, and $\\overline{AB}\\parallel\\overline{DC}$, so $\\triangle{RQC}\\cong\\triangle{HQB}$. $\\overline{BH}=2$. Because $\\overline{BH}=2$, we can use mass point geometry to get that $\\overline{PX}=\\overline{XH}$. $|\\triangle{XHQ}|=\\frac{\\overline{XH}}{\\overline{PH}}\\cdot\\frac{\\overline{QH}}{\\overline{HI}}\\cdot|\\triangle{PHI}|=\\frac{1}{6}\\cdot|\\triangle{PHI}|$ Using the same principle, we can get that $|\\triangle{IYR}|=\\frac{1}{6}|\\triangle{PHI}|$ Therefore, the area of $PYRQX$ is $\\frac{2}{3}\\cdot|\\triangle{PHI}|$ $\\overline{RQ}=2\\sqrt{2}$, so $\\overline{IH}=6\\sqrt{2}$. Using the law of cosines, $\\overline{PH}=\\sqrt{28}$. The area of $\\triangle{PHI}=\\frac{1}{2}\\cdot\\sqrt{28-18}\\cdot6\\sqrt{2}=6\\sqrt{5}$ Using this, we can get the area of $PYRQX = \\sqrt{80}$ so the answer is $080.",
"[asy]import three; import math; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2), H=(4,2,0), I=(-2,-4,0); draw(A--B--C--D--A--E--B--E--C--E--D); draw(B--H--Q, linetype(\"6 6\")+linewidth(0.7)+blue); draw(X--H, linetype(\"6 6\")+linewidth(0.7)+blue); draw(D--I--R, linetype(\"6 6\")+linewidth(0.7)+blue); draw(Y--I, linetype(\"6 6\")+linewidth(0.7)+blue); label(\"A\",A, SE); label(\"B\",B,NE); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE); label(\"H\",H,NE); label(\"I\",I,S); draw(P--X--Q--R--Y--cycle,linetype(\"6 6\")+linewidth(0.7)); [/asy] Extend $\\overline{RQ}$ and $\\overline{AB}$. The point of intersection is $H$. Connect $\\overline{PH}$. $\\overline{EB}$ intersects $\\overline{PH}$ at $X$. Do the same for $\\overline{QR}$ and $\\overline{AD}$, and let the intersections be $I$ and $Y$ Because $Q$ is the midpoint of $\\overline{BC}$, and $\\overline{AB}\\parallel\\overline{DC}$, so $\\triangle{RQC}\\cong\\triangle{HQB}$. $\\overline{BH}=2$. Because $\\overline{BH}=2$, we can use mass point geometry to get that $\\overline{PX}=\\overline{XH}$. $|\\triangle{XHQ}|=\\frac{\\overline{XH}}{\\overline{PH}}\\cdot\\frac{\\overline{QH}}{\\overline{HI}}\\cdot|\\triangle{PHI}|=\\frac{1}{6}\\cdot|\\triangle{PHI}|$ Using the same principle, we can get that $|\\triangle{IYR}|=\\frac{1}{6}|\\triangle{PHI}|$ Therefore, the area of $PYRQX$ is $\\frac{2}{3}\\cdot|\\triangle{PHI}|$ $\\overline{RQ}=2\\sqrt{2}$, so $\\overline{IH}=6\\sqrt{2}$. Using the law of cosines, $\\overline{PH}=\\sqrt{28}$. The area of $\\triangle{PHI}=\\frac{1}{2}\\cdot\\sqrt{28-18}\\cdot6\\sqrt{2}=6\\sqrt{5}$ Using this, we can get the area of $PYRQX = \\sqrt{80}$ so the answer is $080."
] |
2007-I-14
| 2,007
| 14
|
A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ , $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$
| 224
|
I
|
[
"Solution 1 Define a function $f(n)$ on the non-negative integers, as \\[f(n) = \\frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \\frac{a_n}{a_{n+1}}+\\frac{a_{n+1}}{a_{n}}\\] We want $\\left\\lfloor f(2006) \\right\\rfloor$. Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$. Dividing through by $a_{n}a_{n-1}$, we get \\[.\\phantom{------------} \\frac{a_{n+1}}{a_{n}} = \\frac{a_{n}}{a_{n-1}} + \\frac{2007}{a_{n}a_{n-1}} \\phantom{------------} (1)\\] and dividing through by $a_{n}a_{n+1}$, we get \\[.\\phantom{------------} \\frac{a_{n-1}}{a_{n}} = \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} \\phantom{------------} (2)\\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \\[\\frac{a_{n+1}}{a_{n}} + \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} = \\frac{a_{n}}{a_{n-1}} + \\frac{a_{n-1}}{a_{n}} + \\frac{2007}{a_{n}a_{n-1}}\\] i.e. \\[f(n)+ \\frac{2007}{a_{n}a_{n+1}} = f(n-1) + \\frac{2007}{a_{n}a_{n-1}}\\] Rearranging this to the form $f(n)-f(n-1)$ and summing over $n=1$ to $n=2006$, we notice that most of the terms on each side telescope. Without rearranging, you can see that most terms cancel against the corresponding term on the other side. We are left with \\[f(2006) + \\frac{2007}{a_{2006}a_{2007}} = f(0) + \\frac{2007}{a_{1}a_{0}}\\] We have $f(0) = 2$, and $2007/a_0a_1 = 2007/9 = 223$. So \\[f(2006) = 2 + 223 - \\frac{2007}{a_{2006}a_{2007}} = 224 + \\left( 1 - \\frac{2007}{a_{2006}a_{2007}}\\right)\\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$, this means that the sequence $(a_n)$ is increasing. Since $a_2=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$. It follows that $0<1-\\frac{2007}{a_{2006}a_{2007}}<1$ and so $\\left\\lfloor f(2006) \\right\\rfloor = 224. ~anellipticcurveoverq",
"Define a function $f(n)$ on the non-negative integers, as \\[f(n) = \\frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \\frac{a_n}{a_{n+1}}+\\frac{a_{n+1}}{a_{n}}\\] We want $\\left\\lfloor f(2006) \\right\\rfloor$. Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$. Dividing through by $a_{n}a_{n-1}$, we get \\[.\\phantom{------------} \\frac{a_{n+1}}{a_{n}} = \\frac{a_{n}}{a_{n-1}} + \\frac{2007}{a_{n}a_{n-1}} \\phantom{------------} (1)\\] and dividing through by $a_{n}a_{n+1}$, we get \\[.\\phantom{------------} \\frac{a_{n-1}}{a_{n}} = \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} \\phantom{------------} (2)\\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \\[\\frac{a_{n+1}}{a_{n}} + \\frac{a_{n}}{a_{n+1}} + \\frac{2007}{a_{n}a_{n+1}} = \\frac{a_{n}}{a_{n-1}} + \\frac{a_{n-1}}{a_{n}} + \\frac{2007}{a_{n}a_{n-1}}\\] i.e. \\[f(n)+ \\frac{2007}{a_{n}a_{n+1}} = f(n-1) + \\frac{2007}{a_{n}a_{n-1}}\\] Rearranging this to the form $f(n)-f(n-1)$ and summing over $n=1$ to $n=2006$, we notice that most of the terms on each side telescope. Without rearranging, you can see that most terms cancel against the corresponding term on the other side. We are left with \\[f(2006) + \\frac{2007}{a_{2006}a_{2007}} = f(0) + \\frac{2007}{a_{1}a_{0}}\\] We have $f(0) = 2$, and $2007/a_0a_1 = 2007/9 = 223$. So \\[f(2006) = 2 + 223 - \\frac{2007}{a_{2006}a_{2007}} = 224 + \\left( 1 - \\frac{2007}{a_{2006}a_{2007}}\\right)\\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$, this means that the sequence $(a_n)$ is increasing. Since $a_2=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$. It follows that $0<1-\\frac{2007}{a_{2006}a_{2007}}<1$ and so $\\left\\lfloor f(2006) \\right\\rfloor = 224. ~anellipticcurveoverq",
"We are given that $a_{n+1}a_{n-1}= a_{n}^{2}+2007$, $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$. Add these two equations to get $a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})$ $\\frac{a_{n+1}+a_{n-1}}{a_{n}}= \\frac{a_{n}+a_{n-2}}{a_{n-1}}$. This is an invariant. Defining $b_{i}= \\frac{a_{i}}{a_{i-1}}$ for each $i \\ge 2$, the above equation means $b_{n+1}+\\frac{1}{b_{n}}= b_{n}+\\frac{1}{b_{n-1}}$. We can thus calculate that $b_{2007}+\\frac{1}{b_{2006}}= b_{3}+\\frac{1}{b_{2}}= 225$. Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$, notice that $b_{2007}= \\frac{a_{2007}}{a_{2006}}= \\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \\frac{a_{2006}}{a_{2005}}= b_{2006}$. This means that $b_{2007}+\\frac{1}{b_{2007}}< b_{2007}+\\frac{1}{b_{2006}}= 225$. It is only a tiny bit less because all the $b_i$ are greater than $1$, so we conclude that the floor of $\\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\\frac{1}{b_{2007}}$ is $224. ~anellipticcurveoverq",
"The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant \\[\\left|\\begin{array}{cc}a_{n+1}&a_n\\\\a_n&a_{n-1}\\end{array}\\right|=2007.\\] Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\\alpha b_n+\\beta b_{n-1}$ for $n\\ge 2$. We wish to find $\\alpha$ and $\\beta$ such that $a_n=b_n$ for all $n\\ge 1$. To do this, we use the following matrix form of a linear recurrence relation \\[\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\] When we take determinants, this equation becomes \\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\text{det}\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\] We want \\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=2007\\] for all $n$. Therefore, we replace the two matrices by $2007$ to find that \\[2007=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\cdot 2007\\] \\[1=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)=-\\beta.\\] Therefore, $\\beta=-1$. Computing that $a_3=672$, and using the fact that $b_3=\\alpha b_2-b_1$, we conclude that $\\alpha=225$. Clearly, $a_1=b_1$, $a_2=b_2$, and $a_3=b_3$. We claim that $a_n=b_n$ for all $n\\ge 1$. We proceed by induction. If $a_k=b_k$ for all $k\\le n$, then clearly, \\[b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.\\] We also know by the definition of $b_{n+1}$ that \\[\\text{det}\\left(\\begin{array}{cc}b_{n+1}&b_n\\\\b_n&b_{n-1}\\end{array}\\right)=\\text{det}\\left(\\begin{array}{cc}\\alpha&\\beta\\\\1&0\\end{array}\\right)\\text{det}\\left(\\begin{array}{cc}b_{n}&b_{n-1}\\\\b_{n-1}&b_{n-2}\\end{array}\\right).\\] We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$. After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$. Thinking of this as a linear equation in the variable $b_{n+1}$, we already know that this has the solution $b_{n+1}=a_{n+1}$. Therefore, by induction, $a_n=b_n$ for all $n\\ge 1$. We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$. It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\\ge 3$. Now \\[\\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\\frac{a_{2007}}{a_{2006}}+\\frac{a_{2006}}{a_{2007}}=\\frac{225a_{2006}-a_{2005}}{a_{2006}}+\\frac{a_{2006}}{a_{2007}}.\\] \\[=225+\\frac{a_{2006}}{a_{2007}}-\\frac{a_{2005}}{a_{2006}}=225+\\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.\\] \\[=225-\\frac{2007}{a_{2005}a_{2006}}.\\] The sequence certainly grows fast enough such that $\\frac{2007}{a_{2005}a_{2006}}<1$. Therefore, the largest integer less than or equal to this value is $224. ~anellipticcurveoverq",
"This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to \\[a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)\\] where $k$ is a positive integer and $a_0 = a_1 = 3.$ Lemma 1 : For $n \\geq 1$, \\[a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)\\] We shall prove by induction. From (1), $a_2 = 3k + 3$. From the lemma, $a_2 = (k + 2) 3 - 3 = 3k + 3.$ Base case proven. Assume that the lemma is true for some $t \\geq 1$. Then, eliminating the $a_{t-1}$ using (1) and (2) gives \\[(k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)\\] It follows from (2) that \\[(k+2)a_{t+1} - a_t =\\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},\\] where the last line followed from (1) for case $n = t+1$. Lemma 2 : For $n \\geq 0,$ \\[a_{n+1} \\geq a_{n}.\\] Base case is obvious. Assume that $a_{t+1} \\geq a_{t}$ for some $t \\geq 0$. Then it follows that \\[a_{t+2} =\\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\\frac{a_{t+1}}{a_t} ) + 9k \\geq a_{t+1} + 9k > a_{t+1}.\\] This completes the induction. Lemma 3 : For $n \\geq 1,$ \\[a_n a_{n+1} > 9k\\] Using (1) and Lemma 2, for $n \\geq 1,$ \\[a_{n+1}a_n \\geq a_{n+1}a_{n-1} = a_n^2 + 9k > 9k\\] Finally, using (3), for $n \\geq 1,$ \\[\\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\\frac{9k}{a_n a_{n+1}}.\\] Using lemma 3, the largest integer less than or equal to this value would be $k + 1$. Solution 5 (pure algebra) We will try to manipulate $\\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\\frac{a_1^2+a_2^2}{a_1a_2}$. $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_0+\\frac{a_1^2}{a_0}}{a_1} = \\frac{a_0+\\frac{a_1^2+2007}{a_0}-\\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \\frac{a_0+a_2-\\frac{2007}{a_0}}{a_1} = \\frac{a_0a_2+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$, $= \\frac{a_1^2+2007+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2} = \\frac{a_1^2+a_2^2}{a_1a_2}+\\frac{2007}{a_1a_2}-\\frac{2007}{a_0a_1}$ We can keep on using this method to get that $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{2005}a_{2006}}+\\frac{2007}{a_{2005}a_{2006}}-\\ldots-\\frac{2007}{a_{0}a_{1}}$ This telescopes to $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{0}a_{1}}$ or $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \\frac{a_0^2+a_1^2}{a_0a_1}+\\frac{2007}{a_{0}a_{1}}-\\frac{2007}{a_{2006}a_{2007}}$ Finding the first few values, we notice that they increase rapidly, so $\\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\\frac{2007}{a_{2006}a_{2007}}$. The greatest number that does not exceed this is $224. ~anellipticcurveoverq",
"This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to \\[a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)\\] where $k$ is a positive integer and $a_0 = a_1 = 3.$ Lemma 1 : For $n \\geq 1$, \\[a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)\\] We shall prove by induction. From (1), $a_2 = 3k + 3$. From the lemma, $a_2 = (k + 2) 3 - 3 = 3k + 3.$ Base case proven. Assume that the lemma is true for some $t \\geq 1$. Then, eliminating the $a_{t-1}$ using (1) and (2) gives \\[(k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)\\] It follows from (2) that \\[(k+2)a_{t+1} - a_t =\\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},\\] where the last line followed from (1) for case $n = t+1$. Lemma 2 : For $n \\geq 0,$ \\[a_{n+1} \\geq a_{n}.\\] Base case is obvious. Assume that $a_{t+1} \\geq a_{t}$ for some $t \\geq 0$. Then it follows that \\[a_{t+2} =\\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\\frac{a_{t+1}}{a_t} ) + 9k \\geq a_{t+1} + 9k > a_{t+1}.\\] This completes the induction. Lemma 3 : For $n \\geq 1,$ \\[a_n a_{n+1} > 9k\\] Using (1) and Lemma 2, for $n \\geq 1,$ \\[a_{n+1}a_n \\geq a_{n+1}a_{n-1} = a_n^2 + 9k > 9k\\] Finally, using (3), for $n \\geq 1,$ \\[\\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\\frac{9k}{a_n a_{n+1}}.\\] Using lemma 3, the largest integer less than or equal to this value would be $k + 1$. Solution 5 (pure algebra) We will try to manipulate $\\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\\frac{a_1^2+a_2^2}{a_1a_2}$. $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_0+\\frac{a_1^2}{a_0}}{a_1} = \\frac{a_0+\\frac{a_1^2+2007}{a_0}-\\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \\frac{a_0+a_2-\\frac{2007}{a_0}}{a_1} = \\frac{a_0a_2+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$, $= \\frac{a_1^2+2007+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2} = \\frac{a_1^2+a_2^2}{a_1a_2}+\\frac{2007}{a_1a_2}-\\frac{2007}{a_0a_1}$ We can keep on using this method to get that $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{2005}a_{2006}}+\\frac{2007}{a_{2005}a_{2006}}-\\ldots-\\frac{2007}{a_{0}a_{1}}$ This telescopes to $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{0}a_{1}}$ or $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \\frac{a_0^2+a_1^2}{a_0a_1}+\\frac{2007}{a_{0}a_{1}}-\\frac{2007}{a_{2006}a_{2007}}$ Finding the first few values, we notice that they increase rapidly, so $\\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\\frac{2007}{a_{2006}a_{2007}}$. The greatest number that does not exceed this is $224. ~anellipticcurveoverq",
"We will try to manipulate $\\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\\frac{a_1^2+a_2^2}{a_1a_2}$. $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_0+\\frac{a_1^2}{a_0}}{a_1} = \\frac{a_0+\\frac{a_1^2+2007}{a_0}-\\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \\frac{a_0+a_2-\\frac{2007}{a_0}}{a_1} = \\frac{a_0a_2+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$, $= \\frac{a_1^2+2007+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2} = \\frac{a_1^2+a_2^2}{a_1a_2}+\\frac{2007}{a_1a_2}-\\frac{2007}{a_0a_1}$ We can keep on using this method to get that $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{2005}a_{2006}}+\\frac{2007}{a_{2005}a_{2006}}-\\ldots-\\frac{2007}{a_{0}a_{1}}$ This telescopes to $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{0}a_{1}}$ or $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \\frac{a_0^2+a_1^2}{a_0a_1}+\\frac{2007}{a_{0}a_{1}}-\\frac{2007}{a_{2006}a_{2007}}$ Finding the first few values, we notice that they increase rapidly, so $\\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\\frac{2007}{a_{2006}a_{2007}}$. The greatest number that does not exceed this is $224. ~anellipticcurveoverq",
"We will try to manipulate $\\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\\frac{a_1^2+a_2^2}{a_1a_2}$. $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_0+\\frac{a_1^2}{a_0}}{a_1} = \\frac{a_0+\\frac{a_1^2+2007}{a_0}-\\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \\frac{a_0+a_2-\\frac{2007}{a_0}}{a_1} = \\frac{a_0a_2+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$, $= \\frac{a_1^2+2007+a_2^2-\\frac{2007a_2}{a_0}}{a_1a_2} = \\frac{a_1^2+a_2^2}{a_1a_2}+\\frac{2007}{a_1a_2}-\\frac{2007}{a_0a_1}$ We can keep on using this method to get that $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{2005}a_{2006}}+\\frac{2007}{a_{2005}a_{2006}}-\\ldots-\\frac{2007}{a_{0}a_{1}}$ This telescopes to $\\frac{a_0^2+a_1^2}{a_0a_1} = \\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\\frac{2007}{a_{2006}a_{2007}}-\\frac{2007}{a_{0}a_{1}}$ or $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \\frac{a_0^2+a_1^2}{a_0a_1}+\\frac{2007}{a_{0}a_{1}}-\\frac{2007}{a_{2006}a_{2007}}$ Finding the first few values, we notice that they increase rapidly, so $\\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, $\\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\\frac{2007}{a_{2006}a_{2007}}$. The greatest number that does not exceed this is $224. ~anellipticcurveoverq",
"Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$, and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$, respectively. It will be clear why I decided to factor these expressions as I did momentarily. Next, let's see what the expression $\\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$. For $n = 1$, we get $\\frac{1 + 224^2}{224}$, the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\\frac{224^4 + 224^2\\cdot223\\cdot2 + 224^2 + 223^2}{224^3 + 224\\cdot223}$. It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$, while the second-largest term in the denominator is smaller than $224^2$. Thus, the floor of this expression will come out to be $224$ as well. Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$, the degrees of both the numerator and denominator increase by $2$, because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\\frac{a_{n}^2 + 2007}{a_{n-1}})^2$). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\\approx224:1$ ratio between the two. For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$, they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\\geq2$, whose $\\lim_{k\\to\\infty}=$ $224. ~anellipticcurveoverq",
"Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$, and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$, respectively. It will be clear why I decided to factor these expressions as I did momentarily. Next, let's see what the expression $\\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$. For $n = 1$, we get $\\frac{1 + 224^2}{224}$, the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\\frac{224^4 + 224^2\\cdot223\\cdot2 + 224^2 + 223^2}{224^3 + 224\\cdot223}$. It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$, while the second-largest term in the denominator is smaller than $224^2$. Thus, the floor of this expression will come out to be $224$ as well. Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$, the degrees of both the numerator and denominator increase by $2$, because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\\frac{a_{n}^2 + 2007}{a_{n-1}})^2$). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\\approx224:1$ ratio between the two. For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$, they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\\geq2$, whose $\\lim_{k\\to\\infty}=$ $224. ~anellipticcurveoverq"
] |
2007-I-15
| 2,007
| 15
|
Let $ABC$ be an equilateral triangle, and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ . Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ . The area of triangle $DEF$ is $14\sqrt{3}$ . The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime. Find $r$ .
| 989
|
I
|
[
"Denote the length of a side of the triangle $x$, and of $\\overline{AE}$ as $y$. The area of the entire equilateral triangle is $\\frac{x^2\\sqrt{3}}{4}$. Add up the areas of the triangles using the $\\frac{1}{2}ab\\sin C$ formula (notice that for the three outside triangles, $\\sin 60 = \\frac{\\sqrt{3}}{2}$): $\\frac{x^2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{4}(5 \\cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\\sqrt{3}$. This simplifies to $\\frac{x^2\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)$. Some terms will cancel out, leaving $y = \\frac{5}{3}x - 22$. $\\angle FEC$ is an exterior angle to $\\triangle AEF$, from which we find that $60 + \\angle CED = 60 + \\angle AFE$, so $\\angle CED = \\angle AFE$. Similarly, we find that $\\angle EDC = \\angle AEF$. Thus, $\\triangle AEF \\sim \\triangle CDE$. Setting up a ratio of sides, we get that $\\frac{5}{x-y} = \\frac{y}{2}$. Using the previous relationship between $x$ and $y$, we can solve for $x$. $xy - y^2 = 10$ $\\frac{5}{3}x^2 - 22x - \\left(\\frac{5}{3}x - 22\\right)^2 - 10 = 0$ $\\frac{5}{3}x^2 - \\frac{25}{9}x^2 - 22x + 2 \\cdot \\frac{5 \\cdot 22}{3}x - 22^2 - 10= 0$ $10x^2 - 462x + 66^2 + 90 = 0$ Use the quadratic formula, though we only need the root of the discriminant. This is $\\sqrt{(7 \\cdot 66)^2 - 4 \\cdot 10 \\cdot (66^2 + 90)} = \\sqrt{49 \\cdot 66^2 - 40 \\cdot 66^2 - 4 \\cdot 9 \\cdot 100}$$= \\sqrt{9 \\cdot 4 \\cdot 33^2 - 9 \\cdot 4 \\cdot 100} = 6\\sqrt{33^2 - 100}$. The answer is $989.",
"First of all, assume $EC=x,BD=m, ED=a, EF=b$, then we can find $BF=m-3, AE=2+m-x$ It is not hard to find $ab*sin60^{\\circ}*\\frac{1}{2}=14\\sqrt{3}, ab=56$, we apply LOC on $\\triangle{DEF}, \\triangle{BFD}$, getting that $(m-3)^2+m^2-m(m-3)=a^2+b^2-ab$, leads to $a^2+b^2=m^2-3m+65$ Apply LOC on $\\triangle{CED}, \\triangle{AEF}$ separately, getting $4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.$ Add those terms together and use the equality $a^2+b^2=m^2-3m+65$, we can find: $2x^2-(2m+1)x+2m-42=0$ According to basic angle chasing, $\\angle{A}=\\angle{C}; \\angle{AFE}=\\angle{CED}$, so $\\triangle{AFE}\\sim \\triangle{CED}$, the ratio makes $\\frac{5}{x}=\\frac{2+m-x}{2}$, getting that $x^2-(2+m)x+10=0$ Now we have two equations with $m$, and $x$ values for both equations must be the same, so we can solve for $x$ in two equations. $x=\\frac{2m+1 \\pm \\sqrt{4m^2+4m+1-16m+336}}{4}; x=\\frac{4+2m \\pm \\sqrt{4m^2+16m-144}}{4}$, then we can just use positive sign to solve, simplifies to $3+\\sqrt{4m^2+16m-144}=\\sqrt{4m^2-12m+337}$, getting $m=\\frac{211-3\\sqrt{989}}{10}$, since the triangle is equilateral, $AB=BC=2+m=\\frac{231-3\\sqrt{989}}{10}$, and the desired answer is $989 ~bluesoul"
] |
2007-II-1
| 2,007
| 1
|
A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in $2007$ . No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in $2007$ . A set of plates in which each possible sequence appears exactly once contains N license plates. Find N/10.
| 372
|
II
|
[
"There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice. If $0$ appears 0 or 1 times amongst the sequence, there are $\\frac{7!}{(7-5)!} = 2520$ sequences possible. If $0$ appears twice in the sequence, there are ${5\\choose2} = 10$ places to place the $0$s. There are $\\frac{6!}{(6-3)!} = 120$ ways to place the remaining three characters. In total, that gives us $10 \\cdot 120 = 1200$. Thus, $N = 2520 + 1200 = 3720$, and $\\frac{N}{10} = 372."
] |
2007-II-2
| 2,007
| 2
|
Find the number of ordered triples $(a,b,c)$ where $a$ , $b$ , and $c$ are positive integers , $a$ is a factor of $b$ , $a$ is a factor of $c$ , and $a+b+c=100$ .
| 200
|
II
|
[
"Denote $x = \\frac{b}{a}$ and $y = \\frac{c}{a}$. The last condition reduces to $a(1 + x + y) = 100$. Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$. Subtracting the one, we see that $x + y = \\{0,1,3,4,9,19,24,49,99\\}$. There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$. Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = 200."
] |
2007-II-4
| 2,007
| 4
|
The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$ .
| 450
|
II
|
[
"Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit. Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on): $100 = 300x + 200y$ $2(60) = 240x + 300y$ $3(50) = 150x + my$ Solve the system of equations with the first two equations to find that $(x,y) = \\left(\\frac{1}{7}, \\frac{2}{7}\\right)$. Substitute this into the third equation to find that $1050 = 150 + 2m$, so $m = 450."
] |
2007-II-5
| 2,007
| 5
|
The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant ?
| 888
|
II
|
[
"Solution 1 There are $223 \\cdot 9 = 2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),\\ (0,9)$. Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal $y = \\frac{223}{9}x$. This passes through 8 horizontal lines ($y = 1 \\ldots 8$) and 222 vertical lines ($x = 1 \\ldots 222$). Every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through $222 + 8 + 1 = 231$ squares. The number of non-diagonal squares is $2007 - 231 = 1776$. Divide this in 2 to get the number of squares in one of the triangles, with the answer being $\\frac{1776}2 = 888$. Solution 2 Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\\ (0,9)$. In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\\frac{2007 - 8(223)}{9} = 24 \\frac 79$, which means that $\\lfloor 24 \\frac 79\\rfloor = 24$ unit squares can fit in that row. In general, there are $\\sum_{i=0}^{8} \\lfloor \\frac{223i}{9} \\rfloor$ triangles. Since $\\lfloor \\frac{223}{9} \\rfloor = 24$, we see that there are more than $24(0 + 1 + \\ldots + 8) = 24(\\frac{8 \\times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\\lfloor \\frac{0}{9} \\rfloor = 0, \\lfloor \\frac{7}{9} \\rfloor = 0, \\lfloor \\frac{14}{9} \\rfloor = 1,$ $\\lfloor \\frac{21}{9} \\rfloor = 2, \\lfloor \\frac{28}{9} \\rfloor = 3, \\lfloor \\frac{35}{9} \\rfloor = 3,$ $\\lfloor \\frac{42}{9} \\rfloor = 4, \\lfloor \\frac{49}{9} \\rfloor = 5, \\lfloor \\frac{56}{9} \\rfloor = 6$. Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888$. Solution 3 From Pick's Theorem, $\\frac{2007}{2}=\\frac{233}{2}-\\frac{2}{2}+\\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$. Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points. Solution 4 We know that the number of squares intersected in an $m\\times{n}$ rectangle is $m + n -\\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is: $9 + 223 - 1 = 231$. Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line. So, $\\frac{2007 - 231}{2} = \\frac{1776}{2} = 888",
"There are $223 \\cdot 9 = 2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),\\ (0,9)$. Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal $y = \\frac{223}{9}x$. This passes through 8 horizontal lines ($y = 1 \\ldots 8$) and 222 vertical lines ($x = 1 \\ldots 222$). Every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through $222 + 8 + 1 = 231$ squares. The number of non-diagonal squares is $2007 - 231 = 1776$. Divide this in 2 to get the number of squares in one of the triangles, with the answer being $\\frac{1776}2 = 888$. Solution 2 Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\\ (0,9)$. In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\\frac{2007 - 8(223)}{9} = 24 \\frac 79$, which means that $\\lfloor 24 \\frac 79\\rfloor = 24$ unit squares can fit in that row. In general, there are $\\sum_{i=0}^{8} \\lfloor \\frac{223i}{9} \\rfloor$ triangles. Since $\\lfloor \\frac{223}{9} \\rfloor = 24$, we see that there are more than $24(0 + 1 + \\ldots + 8) = 24(\\frac{8 \\times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\\lfloor \\frac{0}{9} \\rfloor = 0, \\lfloor \\frac{7}{9} \\rfloor = 0, \\lfloor \\frac{14}{9} \\rfloor = 1,$ $\\lfloor \\frac{21}{9} \\rfloor = 2, \\lfloor \\frac{28}{9} \\rfloor = 3, \\lfloor \\frac{35}{9} \\rfloor = 3,$ $\\lfloor \\frac{42}{9} \\rfloor = 4, \\lfloor \\frac{49}{9} \\rfloor = 5, \\lfloor \\frac{56}{9} \\rfloor = 6$. Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888$. Solution 3 From Pick's Theorem, $\\frac{2007}{2}=\\frac{233}{2}-\\frac{2}{2}+\\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$. Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points. Solution 4 We know that the number of squares intersected in an $m\\times{n}$ rectangle is $m + n -\\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is: $9 + 223 - 1 = 231$. Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line. So, $\\frac{2007 - 231}{2} = \\frac{1776}{2} = 888",
"Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\\ (0,9)$. In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\\frac{2007 - 8(223)}{9} = 24 \\frac 79$, which means that $\\lfloor 24 \\frac 79\\rfloor = 24$ unit squares can fit in that row. In general, there are $\\sum_{i=0}^{8} \\lfloor \\frac{223i}{9} \\rfloor$ triangles. Since $\\lfloor \\frac{223}{9} \\rfloor = 24$, we see that there are more than $24(0 + 1 + \\ldots + 8) = 24(\\frac{8 \\times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\\lfloor \\frac{0}{9} \\rfloor = 0, \\lfloor \\frac{7}{9} \\rfloor = 0, \\lfloor \\frac{14}{9} \\rfloor = 1,$ $\\lfloor \\frac{21}{9} \\rfloor = 2, \\lfloor \\frac{28}{9} \\rfloor = 3, \\lfloor \\frac{35}{9} \\rfloor = 3,$ $\\lfloor \\frac{42}{9} \\rfloor = 4, \\lfloor \\frac{49}{9} \\rfloor = 5, \\lfloor \\frac{56}{9} \\rfloor = 6$. Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888$. Solution 3 From Pick's Theorem, $\\frac{2007}{2}=\\frac{233}{2}-\\frac{2}{2}+\\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$. Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points. Solution 4 We know that the number of squares intersected in an $m\\times{n}$ rectangle is $m + n -\\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is: $9 + 223 - 1 = 231$. Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line. So, $\\frac{2007 - 231}{2} = \\frac{1776}{2} = 888",
"From Pick's Theorem, $\\frac{2007}{2}=\\frac{233}{2}-\\frac{2}{2}+\\frac{2I}{2}$. In other words, $2I=1776$ and I is $888$. Do you see why we simply set $I$ as the answer as well? That's because every interior point, if moved down and left one (southwest direction), can have that point and the previous point create a unit square. For example, $(1, 1)$ moves to $(0, 0)$, so the square of points ${(0, 0), (1, 0), (1, 1), (0, 1)}$ is one example. This applies, of course, for $888$ points. Solution 4 We know that the number of squares intersected in an $m\\times{n}$ rectangle is $m + n -\\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is: $9 + 223 - 1 = 231$. Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line. So, $\\frac{2007 - 231}{2} = \\frac{1776}{2} = 888",
"We know that the number of squares intersected in an $m\\times{n}$ rectangle is $m + n -\\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is: $9 + 223 - 1 = 231$. Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below the line. So, $\\frac{2007 - 231}{2} = \\frac{1776}{2} = 888"
] |
2007-II-6
| 2,007
| 6
|
An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd , and $a_{i}>a_{i+1}$ if $a_{i}$ is even . How many four-digit parity-monotonic integers are there?
| 640
|
II
|
[
"Let's set up a table of values. Notice that 0 and 9 both cannot appear as any of $a_1,\\ a_2,\\ a_3$ because of the given conditions. (Also note that 0 cannot appear as 0 cannot be the first digit of an integer.) A clear pattern emerges. For example, for $3$ in the second column, we note that $3$ is less than $4,6,8$, but greater than $1$, so there are four possible places to align $3$ as the second digit. Digit 1st 2nd 3rd 4th 0 0 0 0 64 1 1 4 16 64 2 1 4 16 64 3 1 4 16 64 4 1 4 16 64 5 1 4 16 64 6 1 4 16 64 7 1 4 16 64 8 1 4 16 64 9 0 0 0 64 For any number from 1-8, there are exactly 4 numbers from 1-8 that are odd and less than the number or that are even and greater than the number (the same will happen for 0 and 9 in the last column). Thus, the answer is $4^{k-1} \\cdot 10 = 4^3\\cdot10 = 640$. -yeetdayeet",
"This problem can be solved via recursion since we are \"building a string\" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications($0$ can't be at the front and no digit is greater than $9$). There are $4$ options to add no matter what(try some examples if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $640."
] |
2007-II-7
| 2,007
| 7
|
Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$ Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$
| 553
|
II
|
[
"Solution 1 For $x = 1$, we see that $\\sqrt[3]{1} \\ldots \\sqrt[3]{7}$ all work, giving 7 integers. For $x=2$, we see that in $\\sqrt[3]{8} \\ldots \\sqrt[3]{26}$, all of the even numbers work, giving 10 integers. For $x = 3$, we get 13, and so on. We can predict that at $x = 22$ we get 70. To prove this, note that all of the numbers from $\\sqrt[3]{x^3} \\ldots \\sqrt[3]{(x+1)^3 - 1}$ divisible by $x$ work. Thus, $\\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \\frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4$ (the one to be inclusive) integers will fit the conditions. $3k + 4 = 70 \\Longrightarrow k = 22$. The maximum value of $n_i = (x + 1)^3 - 1$. Therefore, the solution is $\\frac{23^3 - 1}{22} = \\frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553$. Solution 2 Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$. The wording of this problem (which uses \"exactly\") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \\leq k^3+70k$. Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\\leq 70k$. Since $69k$, $3k^2+3k+1$, and $70k$ are all integers, this inequality is equivalent to $69k\\leq 3k^2+3k<70k$. Since $k$ is positive, we can divide the inequality by $k$ to get $69 \\leq 3k+3 < 70$. Clearly the only $k$ that satisfies is $k=22$. Then, $\\frac{n_{70}}{k}=k^2+69=484+69=553!)",
"For $x = 1$, we see that $\\sqrt[3]{1} \\ldots \\sqrt[3]{7}$ all work, giving 7 integers. For $x=2$, we see that in $\\sqrt[3]{8} \\ldots \\sqrt[3]{26}$, all of the even numbers work, giving 10 integers. For $x = 3$, we get 13, and so on. We can predict that at $x = 22$ we get 70. To prove this, note that all of the numbers from $\\sqrt[3]{x^3} \\ldots \\sqrt[3]{(x+1)^3 - 1}$ divisible by $x$ work. Thus, $\\frac{(x+1)^3 - 1 - x^3}{x} + 1 = \\frac{3x^2 + 3x + 1 - 1}{x} + 1 = 3x + 4$ (the one to be inclusive) integers will fit the conditions. $3k + 4 = 70 \\Longrightarrow k = 22$. The maximum value of $n_i = (x + 1)^3 - 1$. Therefore, the solution is $\\frac{23^3 - 1}{22} = \\frac{(23 - 1)(23^2 + 23 + 1)}{22} = 529 + 23 + 1 = 553$. Solution 2 Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$. The wording of this problem (which uses \"exactly\") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \\leq k^3+70k$. Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\\leq 70k$. Since $69k$, $3k^2+3k+1$, and $70k$ are all integers, this inequality is equivalent to $69k\\leq 3k^2+3k<70k$. Since $k$ is positive, we can divide the inequality by $k$ to get $69 \\leq 3k+3 < 70$. Clearly the only $k$ that satisfies is $k=22$. Then, $\\frac{n_{70}}{k}=k^2+69=484+69=553!)",
"Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$. The wording of this problem (which uses \"exactly\") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \\leq k^3+70k$. Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\\leq 70k$. Since $69k$, $3k^2+3k+1$, and $70k$ are all integers, this inequality is equivalent to $69k\\leq 3k^2+3k<70k$. Since $k$ is positive, we can divide the inequality by $k$ to get $69 \\leq 3k+3 < 70$. Clearly the only $k$ that satisfies is $k=22$. Then, $\\frac{n_{70}}{k}=k^2+69=484+69=553!)"
] |
2007-II-9
| 2,007
| 9
|
Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$
| 259
|
II
|
[
"Solution 1 Several Pythagorean triples exist amongst the numbers given. $BE = DF = \\sqrt{63^2 + 84^2} = 21\\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \\sqrt{63^2 + (448 - 2\\cdot84)^2} = 7\\sqrt{9^2 + 40^2} = 287$. Use the Two Tangent Theorem on $\\triangle BEF$. Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \\frac{287 - PQ}{2}$. By the Two Tangent theorem, note that $EP = EX = \\frac{287 - PQ}{2}$, making $BX = 105 - EX = 105 - \\left[\\frac{287 - PQ}{2}\\right]$. Also, $BX = BY$. $FY = 364 - BY = 364 - \\left[105 - \\left[\\frac{287 - PQ}{2}\\right]\\right]$. Finally, $FP = FY = 364 - \\left[105 - \\left[\\frac{287 - PQ}{2}\\right]\\right] = \\frac{805 - PQ}{2}$. Also, $FP = FQ + PQ = \\frac{287 - PQ}{2} + PQ$. Equating, we see that $\\frac{805 - PQ}{2} = \\frac{287 + PQ}{2}$, so $PQ = 259.",
"Several Pythagorean triples exist amongst the numbers given. $BE = DF = \\sqrt{63^2 + 84^2} = 21\\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \\sqrt{63^2 + (448 - 2\\cdot84)^2} = 7\\sqrt{9^2 + 40^2} = 287$. Use the Two Tangent Theorem on $\\triangle BEF$. Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \\frac{287 - PQ}{2}$. By the Two Tangent theorem, note that $EP = EX = \\frac{287 - PQ}{2}$, making $BX = 105 - EX = 105 - \\left[\\frac{287 - PQ}{2}\\right]$. Also, $BX = BY$. $FY = 364 - BY = 364 - \\left[105 - \\left[\\frac{287 - PQ}{2}\\right]\\right]$. Finally, $FP = FY = 364 - \\left[105 - \\left[\\frac{287 - PQ}{2}\\right]\\right] = \\frac{805 - PQ}{2}$. Also, $FP = FQ + PQ = \\frac{287 - PQ}{2} + PQ$. Equating, we see that $\\frac{805 - PQ}{2} = \\frac{287 + PQ}{2}$, so $PQ = 259.",
"By the Two Tangent Theorem, we have that $FY = PQ + QF$. Solve for $PQ = FY - QF$. Also, $QF = EP = EX$, so $PQ = FY - EX$. Since $BX = BY$, this can become $PQ = FY - EX + (BY - BX)$$= \\left(FY + BY\\right) - \\left(EX + BX\\right) = FB - EB$. Substituting in their values, the answer is $364 - 105 = 259$. Solution 3 Call the incenter of $\\triangle BEF$ $O_1$ and the incenter of $\\triangle DFE$ $O_2$. Draw triangles $\\triangle O_1PQ,\\triangle PQO_2$. Drawing $BE$, We find that $BE = \\sqrt {63^2 + 84^2} = 105$. Applying the same thing for $F$, we find that $FD = 105$ as well. Draw a line through $E,F$ parallel to the sides of the rectangle, to intersect the opposite side at $E_1,F_1$ respectively. Drawing $\\triangle EE_1F$ and $FF_1E$, we can find that $EF = \\sqrt {63^2 + 280^2} = 287$. We then use Heron's formula to get: \\[[BEF] = [DEF] = 11 466\\]. So the inradius of the triangle-type things is $\\frac {637}{21}$. Now, we just have to find $O_1Q = O_2P$, which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find $PQ$.",
"Call the incenter of $\\triangle BEF$ $O_1$ and the incenter of $\\triangle DFE$ $O_2$. Draw triangles $\\triangle O_1PQ,\\triangle PQO_2$. Drawing $BE$, We find that $BE = \\sqrt {63^2 + 84^2} = 105$. Applying the same thing for $F$, we find that $FD = 105$ as well. Draw a line through $E,F$ parallel to the sides of the rectangle, to intersect the opposite side at $E_1,F_1$ respectively. Drawing $\\triangle EE_1F$ and $FF_1E$, we can find that $EF = \\sqrt {63^2 + 280^2} = 287$. We then use Heron's formula to get: \\[[BEF] = [DEF] = 11 466\\]. So the inradius of the triangle-type things is $\\frac {637}{21}$. Now, we just have to find $O_1Q = O_2P$, which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find $PQ$.",
"Why not first divide everything by its greatest common factor, $7$? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by $7$. From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means: $A = rs$ indicating $26(9)=r(54)$ so $r = 13/3$. Now, we can start applying the equivalent tangents. Calling them $a$, $b$, and $c$ (with $c$ being the longest and a being the shortest), $a+b+c$ is the semi perimeter or $54$. And since the longest side (which has $b+c$) is $52$, $a=2$. Note that the distance $PQ$ we desired to find is just $c - a$. What is $b$ then? $b = 13$. And $c$ is $39$. Therefore the answer is $37$... $NOT.$ Multiply by $7$ back again (I hope you remembered to write this in $huge$ letters on top of the scrap paper!), we actually get $259$.",
"Scaling everything by 7, we have that $AE = 12, AB = 9, BF = 52$. Note that if the perpendicular of $F$ dropped down to $ED$ is $X$, then $EX = 52-12 = 40$. But $FX = 9$ and so we have a $9-40-41$ right triangle with $EFX$ meaning $EF = 41$. Now, by symmetry, we know that $EP = QF = a$ meaning $PF = 41-a$. If the tangent of the circle inscribed in $BEF$ is tangent to $BE$ at $Y$, then if $BY = b$ we have a system of equations. $a+b = 15, b+41-a = 52$. We can then solve for $a$, and since $PQ = 41-2a$, the rest follows."
] |
2007-II-10
| 2,007
| 10
|
Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ , $n$ , and $r$ are positive integers , $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$ )
| 710
|
II
|
[
"Use casework: $B$ has 6 elements: Probability: $\\frac{1}{2^6} = \\frac{1}{64}$ $A$ must have either 0 or 6 elements, probability: $\\frac{2}{2^6} = \\frac{2}{64}$. $B$ has 5 elements: Probability: ${6\\choose5}/64 = \\frac{6}{64}$ $A$ must have either 0, 6, or 1, 5 elements. The total probability is $\\frac{2}{64} + \\frac{2}{64} = \\frac{4}{64}$. $B$ has 4 elements: Probability: ${6\\choose4}/64 = \\frac{15}{64}$ $A$ must have either 0, 6; 1, 5; or 2,4 elements. If there are 1 or 5 elements, the set which contains 5 elements must have four emcompassing $B$ and a fifth element out of the remaining $2$ numbers. The total probability is $\\frac{2}{64}\\left({2\\choose0} + {2\\choose1} + {2\\choose2}\\right) = \\frac{2}{64} + \\frac{4}{64} + \\frac{2}{64} = \\frac{8}{64}$. We could just continue our casework. In general, the probability of picking B with $n$ elements is $\\frac{{6\\choose n}}{64}$. Since the sum of the elements in the $k$th row of Pascal's Triangle is $2^k$, the probability of obtaining $A$ or $S-A$ which encompasses $B$ is $\\frac{2^{7-n}}{64}$. In addition, we must count for when $B$ is the empty set (probability: $\\frac{1}{64}$), of which all sets of $A$ will work (probability: $1$). Thus, the solution we are looking for is $\\left(\\sum_{i=1}^6 \\frac{{6\\choose i}}{64} \\cdot \\frac{2^{7-i}}{64}\\right) + \\frac{1}{64} \\cdot \\frac{64}{64}$ $=\\frac{(1)(64)+(6)(64)+(15)(32)+(20)(16)+(15)(8)+(6)(4)+(1)(2)}{(64)(64)}$ $=\\frac{1394}{2^{12}}$ $=\\frac{697}{2^{11}}$. The answer is $697 + 2 + 11 = 710$.",
"we need $B$ to be a subset of $A$ or $S-A$ we can divide each element of $S$ into 4 categories: it is in $A$ and $B$ it is in $A$ but not in $B$ it is not in $A$ but is in $B$ or it is not in $A$ and not in $B$ these can be denoted as $+A+B$, $+A-B$,$-A+B$, and $-A-B$ we note that if all of the elements are in $+A+B$, $+A-B$ or $-A-B$ we have that $B$ is a subset of $A$ which can happen in $\\dfrac{3^6}{4^6}$ ways similarly if the elements are in $+A-B$,$-A+B$, or $-A-B$ we have that $B$ is a subset of $S-A$ which can happen in $\\dfrac{3^6}{4^6}$ ways as well but we need to make sure we don't over-count ways that are in both sets these are when $+A-B$ or $-A-B$ which can happen in $\\dfrac{2^6}{4^6}$ ways so our probability is $\\dfrac{2\\cdot 3^6-2^6}{4^6}= \\dfrac{3^6-2^5}{2^{11}}=\\dfrac{697}{2^{11}}$. so the final answer is $697 + 2 + 11 = 710$.",
"$B$ must be in $A$ or $B$ must be in $S-A$. This is equivalent to saying that $B$ must be in $A$ or $B$ is disjoint from $A$. The probability of this is the sum of the probabilities of each event individually minus the probability of each event occurring simultaneously. There are $\\binom{6}{x}$ ways to choose $A$, where $x$ is the number of elements in $A$. From those $x$ elements, there are ${2^x}$ ways to choose $B$. Thus, the probability that $B$ is in $A$ is the sum of all the values $\\binom{6}{x}({2^x})$ for values of $x$ ranging from $0$ to $6$. For the second probability, the ways to choose $A$ stays the same but the ways to choose $B$ is now ${2^{6-x}}$. We see that these two summations are simply from the Binomial Theorem and that each of them is ${(2+1)^6}$. We subtract the case where both of them are true. This only happens when $B$ is the null set. $A$ can be any subset of $S$, so there are ${2^6}$ possibilities. Our final sum of possibilities is $2\\cdot 3^6-2^6$. We have ${2^6}$ total possibilities for both $A$ and $B$, so there are ${2^{12}}$ total possibilities. This reduces down to $\\dfrac{2\\cdot 3^6-2^6}{4^6}= \\dfrac{3^6-2^5}{2^{11}}=\\dfrac{697}{2^{11}}$. The answer is thus $697 + 2 + 11 = 710$.",
"Let $|S|$ denote the number of elements in a general set $S$. We use complementary counting. There is a total of $2^6$ elements in $P$, so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$. Note that the number of $x$-element subset of $S$ is $\\binom{6}{x}$. In general, for $0 \\le |A| \\le 6$, in order for $B$ to be in neither $A$ nor $S-A$, $B$ must have at least one element from both $A$ and $S-A$. In other words, $B$ must contain any subset of $A$ and $S-A$ except for the empty set $\\{\\}$. This can be done in $\\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)$ ways. As $|A|$ ranges from $0$ to $6$, we can calculate the total number of unsuccessful outcomes to be \\[\\sum_{|A| = 0}^{6} \\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.\\] So our desired answer is \\[1 - \\dfrac{2702}{2^{12}} = \\dfrac{697}{2^{11}} \\Rightarrow 710.\\] -MP8148",
"To begin with, we note that there are $2^6$ subsets of $S$(which we can assume is $\\{1,2,3,4,5,6\\}$), including the null set. This gives a total of $(2^6)^2 = 2^{12}$ total possibilities for A and B. Case 1: B is contained in A. If B has $0$ elements, which occurs in $\\binom{6}{0}$ ways, A can be anything, giving us $\\binom{6}{0} \\cdot 2^6$. If B has $1$ element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us $\\binom{6}{1} \\cdot 2^5$. Summing, we end up with the binomial expansion of $(2 + 1)^6 = 3^6$. Case 2: B is contained in S-A. By symmetry, this case is the same as Case 1, once again giving us $3^6$ possibilities. Case 3: B is contained in both. We claim here that B can only be the null set. For contradiction, assume that there exists some element $x$ in B which satisfies this restriction. Then, A must contain $x$ as well, but we also know that $S-A$ contains $x$, contradiction. Hence, B is the null set, whereas A can be anything. This gives us $2^6$ possibilities. Since we have overcounted Case 3 in both of the other two cases, our final count is $2 \\cdot 3^6 - 2^6$. This gives us the probability $\\frac{2 \\cdot 3^6 - 2^6}{2^{12}}$. Upon simplifying, we end up with $\\frac{697}{2^{11}}$, giving the desired answer of $710. - Spacesam ~clarifications by LeonidasTheConquerer"
] |
2007-II-11
| 2,007
| 11
|
Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface . The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$ . It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi+b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime . Find $a+b+c.$
| 179
|
II
|
[
"If it weren’t for the small tube, the larger tube would travel $144\\pi$. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube. Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the diagram has a length of $72 + 24 = 96$. The horizontal line divides the radius of the larger circle into $72 - 24 = 48$ on the top half, which indicates that the right triangle has leg of 48 and hypotenuse of 96, a $30-60-90 \\triangle$. Find the length of the purple arc in the diagram (the distance the tube rolled, but not the horizontal distance). The sixty degree central angle indicates to take $\\frac{60}{360} = \\frac 16$ of the circumference of the larger circle (twice), while the $180 - 2(30) = 120^{\\circ}$ central angle in the smaller circle indicates to take $\\frac{120}{360} = \\frac 13$ of the circumference. This adds up to $2 \\cdot \\frac 16 144\\pi + \\frac 13 48\\pi = 64\\pi$. The actual horizontal distance it takes can be found by using the $30-60-90 \\triangle$s. The missing leg is equal in length to $48\\sqrt{3}$. Thus, the total horizontal distance covered is $96\\sqrt{3}$. Thus, we get $144\\pi - 64\\pi + 96\\sqrt{3} = 80\\pi + 96\\sqrt{3}$, and our answer is $179."
] |
2007-II-12
| 2,007
| 12
|
The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that $\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$ find $\log_{3}(x_{14}).$
| 91
|
II
|
[
"Suppose that $x_0 = a$, and that the common ratio between the terms is $r$. The first conditions tells us that $\\log_3 a + \\log_3 ar + \\ldots + \\log_3 ar^7 = 308$. Using the rules of logarithms, we can simplify that to $\\log_3 a^8r^{1 + 2 + \\ldots + 7} = 308$. Thus, $a^8r^{28} = 3^{308}$. Since all of the terms of the geometric sequence are integral powers of $3$, we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$. We find that $8x + 28y = 308$. The possible positive integral pairs of $(x,y)$ are $(35,1),\\ (28,3),\\ (21,5),\\ (14,7),\\ (7,9),\\ (0,11)$. The second condition tells us that $56 \\le \\log_3 (a + ar + \\ldots ar^7) \\le 57$. Using the sum formula for a geometric series and substituting $x$ and $y$, this simplifies to $3^{56} \\le 3^x \\frac{3^{8y} - 1}{3^y-1} \\le 3^{57}$. The fractional part $\\approx \\frac{3^{8y}}{3^y} = 3^{7y}$. Thus, we need $\\approx 56 \\le x + 7y \\le 57$. Checking the pairs above, only $(21,5)$ is close. Our solution is therefore $\\log_3 (ar^{14}) = \\log_3 3^x + \\log_3 3^{14y} = x + 14y = 091.",
"All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$s and $0$s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\\dots x_0 = 3^{308}$, and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\\dots +(56-7r) = 308$, and $r = 5$. Thus $\\log_3 (x_{14}) = 091.",
"Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$, $x_1$, $x_2$..., as $3^n$, $3^{n+m}$, and $3^{n+2m}$... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$. Simplify to get $2n+7m=77$. (1) Now, rewrite the second given equation as $3^{56} \\leq \\left( \\sum_{n=0}^{7}x_{n} \\right) \\leq 3^{57}$. Obviously, $x_7$, aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\\leq56$. Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$. Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$. (2) Use equations 1 and 2 to get $m=5$ and $n=21$. $\\log_{3}{x_{14}} = \\log_{3}{3^{21+14*5}} = 21+14*5 = 091 -jackshi2006",
"Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$. As stated above, we then get that $a+a+n+...+a+7n=308$, from which we simplify to $2a+7n=77$. From here, we just go brute force using the second statement (that $3^{56}\\leq 3^{a}+...+3^{a+7n}\\leq 3^{57}$). Rearranging the equation from earlier, we get \\[n=11-\\frac{2a}{7}\\] from which it is clear that $a$ is a multiple of $7$. Testing the first few values of $a$, we get: Case 1 ($a=7,n=9$) The sequence is then $3^{7}+...+3^{70}$, which breaks the upper bound. Case 2 ($a=14,n=7$) The sequence is then $3^{14}+...+3^{63}$, which also breaks the upper bound. Case 3 ($a=21,n=5$) This is the first reasonable one, giving $3^{21}+...+3^{56}$. It seems like this would break the upper bound, but from some testing we get: \\[3^{21}+...+3^{56}<3^{57}\\] \\[1+...+3^{35}<3^{36}\\] \\[1+...+3^{30}<2*3^{35}\\] \\[3^{5}+...+3^{30}<2*3^{35}-1\\] \\[1+...+3^{25}<2*3^{30}-3^{-5}\\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \\[1<2*3^{5}\\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$, so $x_14=3^{a+14n}\\rightarrow3^{91}$. We then get the requested answer, $\\log_3(3^{91})=091 ~ amcrunner",
"Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$. As stated above, we then get that $a+a+n+...+a+7n=308$, from which we simplify to $2a+7n=77$. From here, we just go brute force using the second statement (that $3^{56}\\leq 3^{a}+...+3^{a+7n}\\leq 3^{57}$). Rearranging the equation from earlier, we get \\[n=11-\\frac{2a}{7}\\] from which it is clear that $a$ is a multiple of $7$. Testing the first few values of $a$, we get: Case 1 ($a=7,n=9$) The sequence is then $3^{7}+...+3^{70}$, which breaks the upper bound. Case 2 ($a=14,n=7$) The sequence is then $3^{14}+...+3^{63}$, which also breaks the upper bound. Case 3 ($a=21,n=5$) This is the first reasonable one, giving $3^{21}+...+3^{56}$. It seems like this would break the upper bound, but from some testing we get: \\[3^{21}+...+3^{56}<3^{57}\\] \\[1+...+3^{35}<3^{36}\\] \\[1+...+3^{30}<2*3^{35}\\] \\[3^{5}+...+3^{30}<2*3^{35}-1\\] \\[1+...+3^{25}<2*3^{30}-3^{-5}\\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \\[1<2*3^{5}\\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$, so $x_14=3^{a+14n}\\rightarrow3^{91}$. We then get the requested answer, $\\log_3(3^{91})=091 ~ amcrunner"
] |
2007-II-13
| 2,007
| 13
|
A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$ th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3$ ? [asy] for (int i=0; i<12; ++i){ for (int j=0; j<i; ++j){ //dot((-j+i/2,-i)); draw((-j+i/2,-i)--(-j+i/2+1,-i)--(-j+i/2+1,-i+1)--(-j+i/2,-i+1)--cycle); } } [/asy]
| 640
|
II
|
[
"Label each of the bottom squares as $x_0, x_1 \\ldots x_9, x_{10}$. Through induction, we can find that the top square is equal to ${10\\choose0}x_0 + {10\\choose1}x_1 + {10\\choose2}x_2 + \\ldots {10\\choose10}x_{10}$. (This also makes sense based on a combinatorial argument: the number of ways a number can \"travel\" to the top position going only up is equal to the number of times it will be counted in the final sum.) Examine the equation $\\mod 3$. All of the coefficients from $x_2 \\ldots x_8$ will be multiples of $3$ (since the numerator will have a $9$). Thus, the expression boils down to $x_0 + 10x_1 + 10x_9 + x_{10} \\equiv 0 \\mod 3$. Reduce to find that $x_0 + x_1 + x_9 + x_{10} \\equiv 0 \\mod 3$. Out of $x_0,\\ x_1,\\ x_9,\\ x_{10}$, either all are equal to $0$, or three of them are equal to $1$. This gives ${4\\choose0} + {4\\choose3} = 1 + 4 = 5$ possible combinations of numbers that work. The seven terms from $x_2 \\ldots x_8$ can assume either $0$ or $1$, giving us $2^7$ possibilities. The answer is therefore $5 \\cdot 2^7 = 640.",
"If we label the bottom row of values $a_1,a_2,…,a_11$ then the sum of the bottom row is $a_1+…+a_11$, the sum of the second row is $a_1+2(a_2+…+a_10)+a_11$ and this pattern continues until the top element is $(a_1+a_11)+2(a_2+a_10)+…+4(a_5+a_7)+5a_6$. Then, the terms with coefficient 0 mod 3 can be anything, and we do casework on the other solutions mod 3. We get 640. ~joeythetoey"
] |
2007-II-14
| 2,007
| 14
|
Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$ , $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$
| 676
|
II
|
[
"Let $r$ be a root of $f(x)$. Then we have $f(r)f(2r^2)=f(2r^3+r)$; since $r$ is a root, we have $f(r)=0$; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $|2r^3+r|>r$, so $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots. Note that $f(x)$ is not constant. We then find two complex roots: $r = \\pm i$. We find that $f(i)f(-2) = f(-i)$, and that $f(-i)f(-2) = f(i)$. This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \\Longrightarrow f(i)f(-i)(f(-2)^2 - 1) = 0$. Thus, $\\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial. (Note: This requires the assumption that $f(-2)\\neq1$. Clearly, $f(-2)\\neq-1$, because that would imply the existence of a real root.) The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$. Substituting into the given expression, we have \\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\\] \\[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\\] Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$, or $g(x)$ satisfies the same constraints as $f(x)$. Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$. Since $f(2)+f(3)=125=5^n+10^n$ for some $n$, we have $n=2$; so $f(5) = 676.",
"Let $r$ be a root of $f(x).$ This means that $f(r)f(2r^2)=f(2r^3+r).$ In other words, $2r^3+r$ is a root of $f(x)$ too. Since $f(x)$ can't have infinitely many roots, \\[Q(x)=P(P(\\dotsb P(P(r)) \\dotsb))\\] is cyclic, where $P(x)=2x^3+x.$ Now, we will do casework. Case 1: $\\deg f\\geq1$ Subcase 1: $|r|>1$ This means that \\[|2r^3+r|\\geq|2r^3|-|r|=|r|(2|r|^2-1)>|r|(2\\cdot1^2-1)=|r|.\\] It follows that $|2r^3+r|>|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true. Subcase 2: $|r|<1$ This means that $|2r^3+r|\\geq2|r^3|-|r|=|r|(|2r^2|-1)<|r|.$ It follows that $|2r^3+r|<|r|$ for all $r.$ This implies that $Q(r)$ can't be cyclic. Thus, it is impossible for $|r|>1$ to be true. Subcase 3: $|r|=1.$ Since $|r|$ is not greater than or less than 1, $|r|=1.$ This means that all the roots of the polynomial have a magnitude of $1.$ More specifically, $|2r^3+r|$ has a magnitude of one. Since this would mean an equality condition from the triangle inequality, $2r^3$ and $r$ are collinear with the origin in the complex plane. In other words, $\\frac{2r^3}{r}=\\pm c\\Leftrightarrow cr=2r^3\\Leftrightarrow 2r^2=c\\Leftrightarrow r=\\pm\\sqrt{\\pm\\frac{c}{2}},$ for some real constant $c.$ Now, from $|r|=1,$ we find that $\\left|\\pm\\sqrt{\\pm\\frac{c}{2}}\\right|=1\\Leftrightarrow \\sqrt{\\pm\\frac{c}{2}}=1\\Leftrightarrow \\pm\\frac{c}{2}=1\\Leftrightarrow c=\\pm2.$ Putting this back into the equation, we find that $r=1,-1,i,-i.$ Now, this means that $2r^3+r=3,-3,i,-i.$ $3$ and $-3$ obviously doesn't have a magnitude of $1.$ Thus, $i,-i$ are the only possible roots of the polynomial. Since roots come in conjugate pairs, $f(x)=[(x-i)(x+i)]^n=(x^2+1)^n,$ works for all constants $n\\neq0.$ Case 2: $\\deg f=0.$ This means that $f(x)=c,$ for some constant $c.$ In other words, $c^2=c.$ We can easily find that this means that $c=0,1.$ Combining all the cases, we conclude that $f(x)=(x^2+1)^n,0,1$ are the only polynomials that satisfy this equation. Now, we can test! $f(x)=0,1$ obviously don't satisfy $f(2)+f(3)=125.$ Thus, $f(x)=(x^2+1)^n.$ Substituting, we find that $5^n+10^n=125\\Leftrightarrow n=2.$ We conclude that $f(5)=(5^2+1)^2=26^2=676 ~ pinkpig"
] |
2007-II-15
| 2,007
| 15
|
Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ , $\omega_{B}$ to $BC$ and $BA$ , $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 389
|
II
|
[
"[asy] defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label(\"$A$\",A,N); label(\"$B$\",B,0.15*(B-P)); label(\"$C$\",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot(\"$O_A$\",X,right,p); dot(\"$O_B$\",Y,left+up,p); dot(\"$O_C$\",Z,down,p); dot(\"$O$\",Zp,dir(-45),p+red); dot(\"$I$\",P,left+up,p); [/asy] First, apply Heron's formula to find that $[ABC] = \\sqrt{21 \\cdot 8 \\cdot 7 \\cdot 6} = 84$. The semiperimeter is $21$, so the inradius is $\\frac{A}{s} = \\frac{84}{21} = 4$. Now consider the incenter $I$ of $\\triangle ABC$. Let the radius of one of the small circles be $r$. Let the centers of the three little circles tangent to the sides of $\\triangle ABC$ be $O_A$, $O_B$, and $O_C$. Let the center of the circle tangent to those three circles be $O$. The homothety $\\mathcal{H}\\left(I, \\frac{4-r}{4}\\right)$ maps $\\triangle ABC$ to $\\triangle XYZ$; since $OO_A = OO_B = OO_C = 2r$, $O$ is the circumcenter of $\\triangle XYZ$ and $\\mathcal{H}$ therefore maps the circumcenter of $\\triangle ABC$ to $O$. Thus, $2r = R \\cdot \\frac{4 - r}{4}$, where $R$ is the circumradius of $\\triangle ABC$. Substituting $R = \\frac{abc}{4[ABC]} = \\frac{65}{8}$, $r = \\frac{260}{129}$ and the answer is $389. https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png",
"[asy] defaultpen(fontsize(12)+0.8); size(350); pair A,B,C,X,Y,Z,P,Q,R,Zp; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); Zp=circumcenter(X,Y,Z); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r), gray); draw(A--P--B^^P--C^^X--Y--Z--X, royalblue); draw(X--foot(X,A,C)^^Z--foot(Z,A,C),royalblue); draw(CR(Zp,r), gray); draw(incircle(A,B,C)^^incircle(X,Y,Z)^^P--foot(P,A,C), heavygreen+0.6); draw(rightanglemark(X,foot(X,A,C),C,10),linewidth(0.6)); draw(rightanglemark(Z,foot(Z,A,C),A,10),linewidth(0.6)); label(\"$A$\",A,N); label(\"$B$\",B,0.15*(B-P)); label(\"$C$\",C,0.1*(C-P)); pen p=fontsize(10)+linewidth(3); dot(\"$O_A$\",X,right,p); dot(\"$O_B$\",Y,left+up,p); dot(\"$O_C$\",Z,down,p); dot(\"$O$\",Zp,dir(-45),p+red); dot(\"$I$\",P,left+up,p); [/asy] First, apply Heron's formula to find that $[ABC] = \\sqrt{21 \\cdot 8 \\cdot 7 \\cdot 6} = 84$. The semiperimeter is $21$, so the inradius is $\\frac{A}{s} = \\frac{84}{21} = 4$. Now consider the incenter $I$ of $\\triangle ABC$. Let the radius of one of the small circles be $r$. Let the centers of the three little circles tangent to the sides of $\\triangle ABC$ be $O_A$, $O_B$, and $O_C$. Let the center of the circle tangent to those three circles be $O$. The homothety $\\mathcal{H}\\left(I, \\frac{4-r}{4}\\right)$ maps $\\triangle ABC$ to $\\triangle XYZ$; since $OO_A = OO_B = OO_C = 2r$, $O$ is the circumcenter of $\\triangle XYZ$ and $\\mathcal{H}$ therefore maps the circumcenter of $\\triangle ABC$ to $O$. Thus, $2r = R \\cdot \\frac{4 - r}{4}$, where $R$ is the circumradius of $\\triangle ABC$. Substituting $R = \\frac{abc}{4[ABC]} = \\frac{65}{8}$, $r = \\frac{260}{129}$ and the answer is $389. https://latex.artofproblemsolving.com/9/4/7/947b7f06d947dbf8bc5d8f61cdd193c330377372.png",
"Consider a 13-14-15 triangle. $A=84.$ (By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.) The inradius is $r=\\frac{A}{s}=\\frac{84}{21}=4$, where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$ The circumradius is $R=\\frac{abc}{4rs}=\\frac{13\\cdot 14\\cdot 15}{4\\cdot 4\\cdot 21}=\\frac{65}{8},$ where $a,$ $b,$ and $c$ are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, $v$. Cut and combine the triangles, as shown. Then solve for $4u$: $\\frac{65}{8}v=8u$ $v=\\frac{64}{65}u$ $u+v=1$ $u+\\frac{64}{65}u=1$ $\\frac{129}{65}u=1$ $4u=\\frac{260}{129}$ The solution is $260+129=389.",
"Let $A'$, $B'$, $C'$, and $O$ be the centers of circles $\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$, $\\omega$, respectively, and let $x$ be their radius. Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$. Since $OA'=OB'=OC'=2x$, $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$. Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$. Then the inradius of $A'B'C'=r-x$, so now we compute r. Computing the inradius by $A=rs$, we find that the inradius of $ABC$ is $4$. Additionally, using the circumradius formula $R=\\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\\frac{65}{8}$. Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$. \\[\\frac{\\frac{65}{8}}{4}=\\frac{2x}{4-x}\\] Solving, we get $x=\\frac{260}{129}$, so our answer is $260+129=389.",
"Let $A'$, $B'$, $C'$, and $O$ be the centers of circles $\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$, $\\omega$, respectively, and let $x$ be their radius. Now, triangles $ABC$ and $A'B'C'$ are similar by parallel sides, so we can find ratios of two quantities in each triangle and set them equal to solve for $x$. Since $OA'=OB'=OC'=2x$, $O$ is the circumcenter of triangle $A'B'C'$ and its circumradius is $2x$. Let $I$ denote the incenter of triangle $ABC$ and $r$ the inradius of $ABC$. Then the inradius of $A'B'C'=r-x$, so now we compute r. Computing the inradius by $A=rs$, we find that the inradius of $ABC$ is $4$. Additionally, using the circumradius formula $R=\\frac{abc}{4K}$ where $K$ is the area of $ABC$ and $R$ is the circumradius, we find $R=\\frac{65}{8}$. Now we can equate the ratio of circumradius to inradius in triangles $ABC$ and $A'B'C'$. \\[\\frac{\\frac{65}{8}}{4}=\\frac{2x}{4-x}\\] Solving, we get $x=\\frac{260}{129}$, so our answer is $260+129=389.",
"According to the diagram, it is easily to see that there is a small triangle made by the center of three circles which aren't in the middle. The circumradius of them is$2r$. Now denoting $AB=13;BC=14;AC=15$, and centers of circles tangent to $AB,AC;AC,BC;AB,BC$ are relatively $M,N,O$ with $OJ,NK$ both perpendicular to $BC$. It is easy to know that $tanB=\\frac{12}{5}$, so $tan\\angle OBJ=\\frac{2}{3}$ according to half angle formula. Similarly, we can find $tan\\angle NCK=\\frac{1}{2}$. So we can see that $JK=ON=14-\\frac{7x}{2}$. Obviously, $\\frac{2x}{14-\\frac{7x}{2}}=\\frac{65}{112}$ . After solving, we get $x=\\frac{260}{129}$, so our answer is $260+129=389. ~bluesoul",
"[asy] defaultpen(fontsize(12)+0.8); size(300); pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label(\"$A$\",A,N); label(\"$B$\",B,0.15*(B-P)); label(\"$C$\",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot(\"$O_A$\",X,right,p); dot(\"$O_B$\",Y,left+up,p); dot(\"$O_C$\",Z,right+up,p); dot(\"$O$\",circumcenter(X,Y,Z),right+down,p); dot(\"$I$\",P,left+up,p); dot(\"$H$\",(7,0),down,p); [/asy] Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$, respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$. Since $AB||O_AO_B$, $BC||O_BO_C$, and $CA||O_CO_A$, we know that \\[\\angle BAI = \\angle O_BO_AI, \\angle CBI = \\angle O_CO_BI, \\angle ACI = \\angle O_AO_CI \\text{ and }\\angle CAI = \\angle O_CO_AI, \\angle ABI = \\angle O_AO_BI, \\angle BCI = \\angle O_BO_CI.\\] That means $\\angle ABC = \\angle O_AO_BO_C$, $\\angle BAC = \\angle O_BO_AO_C$, and $\\angle ACB = \\angle O_AO_CO_B$. Thus, $\\triangle ABC \\sim \\triangle O_AO_BO_C$. We also know that we are scaling each side of $\\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear; same apply with $B$ and $C$). Now, let the homothety $\\mathcal{H} (I, x)$ map $\\triangle ABC$ to $\\triangle O_AO_BO_C$. To start off, we know the circumradius of $O_AO_BO_C$ is $O$, since $OO_A = OO_B = OO_C = 2r$. Since $O_AO_B = 13x$, $O_BO_C = 14x$, $O_CO_A = 15x$, we can get an relationship involving $x$ and $r$ via another way to find the circumradius: \\[[\\triangle O_AO_BO_C ] =\\frac{abc}{4R} \\Longrightarrow 84x^2 =\\frac{13x\\cdot 14x\\cdot 15x}{4\\cdot 2r} \\Longrightarrow r=\\frac{65x}{16}\\] Take notice of the inradius of $ABC$. We get the inradius to be $[\\triangle ABC ] = sr_0 \\Longrightarrow r_0=4$. Let the tangency point of the incircle and side $BC$ be $H$. We know $IH = 4$. We also know that we can cut off the part of $IH$ that is outside of $\\triangle O_AO_BO_C$ to get the inradius of $\\triangle O_AO_BO_C$. To part that is outside $\\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\\triangle O_AO_BO_C$ is just $4-r$. We can calculate that incradius in another way, though. We know that the inradius of $\\triangle ABC$ is $4$, which means the inradius of $\\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio). Thus, we have $4x = 4-r = 4-\\dfrac{65x}{16} \\Longrightarrow x = \\dfrac{64}{129} \\Longrightarrow r = \\dfrac{260}{129}$. That gives $389 ~sml1809"
] |
2008-I-1
| 2,008
| 1
|
Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance?
| 252
|
I
|
[
"Solution 1 Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance. Thus, $\\dfrac{3k}{5k + 20} = \\dfrac{29}{50}$, solving gives $k = 116$. Thus, the number of people that like to dance is $2k + 20 = 252",
"Say that there were $3k$ girls and $2k$ boys at the party originally. $2k$ like to dance. Then, there are $3k$ girls and $2k + 20$ boys, and $2k + 20$ like to dance. Thus, $\\dfrac{3k}{5k + 20} = \\dfrac{29}{50}$, solving gives $k = 116$. Thus, the number of people that like to dance is $2k + 20 = 252",
"Let the number of girls be $g$. Let the number of total people originally be $t$. We know that $\\frac{g}{t}=\\frac{3}{5}$ from the problem. We also know that $\\frac{g}{t+20}=\\frac{29}{50}$ from the problem. We now have a system and we can solve. The first equation becomes: $3t=5g$. The second equation becomes: $50g=29t+580$ Now we can sub in $30t=50g$ by multiplying the first equation by $10$. We can plug this into our second equation. $30t=29t+580$ $t=580$ We know that there were originally $580$ people. Of those, $\\frac{2}{5}*580=232$ like to dance. We also know that with these people, $20$ boys joined, all of whom like to dance. We just simply need to add $20$ to get $232+20=252",
"Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \\[0.4p+20 = 0.42(p+20)\\] Solving for p gives us $p=580$, so the solution is $0.4p+20 = 252",
"Assume all the boys like to dance and none of the girls like to dance. We then proceed like the previous solutions. ~Arcticturn",
"Assume all the boys like to dance and none of the girls like to dance. We then proceed like the previous solutions. ~Arcticturn"
] |
2008-I-2
| 2,008
| 2
|
Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$ , and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$ .
| 25
|
I
|
[
"Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$. This implies that vertex G must be located outside of square $AIME$. [asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label(\"\\(G\\)\",G,N); label(\"\\(A\\)\",A,NW); label(\"\\(I\\)\",I,NE); label(\"\\(M\\)\",M,NE); label(\"\\(E\\)\",E,NW); label(\"\\(10\\)\",(M+E)/2,S); [/asy] Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\\triangle GXY \\sim \\triangle GEM$. Let the height of $GXY$ be $h$. By the similarity, $\\dfrac{h}{6} = \\dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = 025."
] |
2008-I-3
| 2,008
| 3
|
Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
| 314
|
I
|
[
"Let the biking rate be $b$, swimming rate be $s$, jogging rate be $j$, all in km/h. We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$. Subtracting the second from twice the first gives $4j + 5s = 57$. Mod 4, we need $s\\equiv1\\pmod{4}$. Thus, $(j,s) = (13,1),(8,5),(3,9)$. $(13,1)$ and $(3,9)$ give non-integral $b$, but $(8,5)$ gives $b = 15$. Thus, our answer is $15^{2} + 8^{2} + 5^{2} = 314.",
"Let $b$, $j$, and $s$ be the biking, jogging, and swimming rates of the two people. Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$. Subtracting gives us that $2b - j - s = 17$. Adding three times this to the first equation gives that $8b + s = 125\\implies b\\le 15$. Adding four times the previous equation to the first given one gives us that $10b - j = 142\\implies b > 14\\implies b\\ge 15$. This gives us that $b = 15$, and then $j = 8$ and $s = 5$. Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = 314.",
"Creating two systems, we get $2x+3y+4z=74$, and $2y+3z+4x=91$. Subtracting the two expressions we get $y+z-2x=-17$. Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly: When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$, $4z \\equiv 2 \\pmod{6} = 5$, we get that $x=15$ and $y=8$, which works because of integer values. Therefore, $225+64+25=314",
"Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\\implies j=8$. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$, we get $b=15$. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=314 hours is also pretty insane, so none of the swimming speeds make any sense.",
"Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\\implies j=8$. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$, we get $b=15$. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=314 hours is also pretty insane, so none of the swimming speeds make any sense."
] |
2008-I-4
| 2,008
| 4
|
There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$ . Find $x + y$ .
| 80
|
I
|
[
"Completing the square, $y^2 = x^2 + 84x + 2008 = (x+42)^2 + 244$. Thus $244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)$ by difference of squares. Since $244$ is even, one of the factors is even. A parity check shows that if one of them is even, then both must be even. Since $244 = 2^2 \\cdot 61$, the factors must be $2$ and $122$. Since $x,y > 0$, we have $y - x - 42 = 2$ and $y + x + 42 = 122$; the latter equation implies that $x + y = 080 is the unique solution.",
"We complete the square like in the first solution: $y^2 = (x+42)^2 + 244$. Since consecutive squares differ by the consecutive odd numbers, we note that $y$ and $x+42$ must differ by an even number. We can use casework with the even numbers, starting with $y-(x+42)=2$. \\begin{align*}2(x+42)+1+2(x+42)+3&=244\\\\ \\Leftrightarrow x&=18\\end{align*} Thus, $y=62$ and the answer is $080.",
"We see that $y^2 \\equiv x^2 + 4 \\pmod{6}$. By quadratic residues, we find that either $x \\equiv 0, 3 \\pmod{6}$. Also, $y^2 \\equiv (x+42)^2 + 244 \\equiv (x+2)^2 \\pmod{4}$, so $x \\equiv 0, 2 \\mod{4}$. Combining, we see that $x \\equiv 0 \\mod{6}$. Testing $x = 6$ and other multiples of $6$, we quickly find that $x = 18, y = 62$ is the solution. $18+62=080",
"We solve for x: $x^2 + 84x + 2008-y^2 = 0$ $x=\\dfrac{-84+\\sqrt{84^2-4\\cdot 2008+4y^2}}{2}=-42+\\sqrt{y^2-244}$ So $y^2-244$ is a perfect square. Since 244 is even, the difference $\\sqrt{y^2-244} -y^2$ is even, so we try $y^2-244=(y-2)^2$: $-244=-4y+4$, $y=62$. Plugging into our equation, we find that $x=18$, and $(x,y)=(18,62)$ indeed satisfies the original equation. $x+y=080",
"Let $y=x+d$ for some $d>0$, substitute into the original equation to get $84x + 2008 = 2xd + d^2$. All terms except for the last one are even, hence $d^2$ must be even, hence let $d=2e$. We obtain $21x + 502 = xe + e^2$. Rearrange to $502-e^2 = x(e-21)$. Obviously for $0<e<21$ the right hand side is negative and the left hand side is positive. Hence $e\\geq 21$. Let $e=21+f$, then $f\\geq 0$. We have $502 - (21+f)^2 = xf$. Left hand side simplifies to $61 - 42f + f^2$. As $x$ must be an integer, $f$ must divide the left hand side. But $61$ is a prime, which only leaves two options: $f=1$ and $f=61$. Option $f=61$ gives us a negative $x$. Option $f=1$ gives us $x=61/f - 42 + f = 18$, and $y = x + d= x + 2e = x + 2(21+f) = 18 + 44 = 62$, hence $x+y=080.",
"First complete the square to get $y^2 = (x+42)^2 + 244$. Remember that squares are the sums of consecutive odd integers, so when the difference between the two squares is 244, the two squares must be an even number of odd integers apart. However, there is only one distinct solution, as the problem states, and very quickly you will realize that only two odd integers work. When there are four, then the numbers are not odd, and when it is any other even integer it does not divide. So we need two consecutive odd integers that sum to 244. Easily we find 121 and 123. 121 is the 61st odd integer and 123 is the 62nd odd integer, so $(x+42)^2$ is the sum of the first 60 odd integers, or $(60)^2$, while $y^2$ is $62^2$ for the same reasons. That way we get $x=12$, $y=62$, hence $x+y=080. -jackshi2006"
] |
2008-I-5
| 2,008
| 5
|
A right circular cone has base radius $r$ and height $h$ . The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .
| 14
|
I
|
[
"The path is a circle with radius equal to the slant height of the cone, which is $\\sqrt {r^{2} + h^{2}}$. Thus, the length of the path is $2\\pi\\sqrt {r^{2} + h^{2}}$. Also, the length of the path is 17 times the circumference of the base, which is $34r\\pi$. Setting these equal gives $\\sqrt {r^{2} + h^{2}} = 17r$, or $h^{2} = 288r^{2}$. Thus, $\\dfrac{h^{2}}{r^{2}} = 288$, and $\\dfrac{h}{r} = 12\\sqrt {2}$, giving an answer of $12 + 2 = 014."
] |
2008-I-6
| 2,008
| 6
|
A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$ ?
| 17
|
I
|
[
"Let the $k$th number in the $n$th row be $a(n,k)$. Writing out some numbers, we find that $a(n,k) = 2^{n-1}(n+2k-2)$.[1] We wish to find all $(n,k)$ such that $67| a(n,k) = 2^{n-1} (n+2k-2)$. Since $2^{n-1}$ and $67$ are relatively prime, it follows that $67|n+2k-2$. Since every row has one less element than the previous row, $1 \\le k \\le 51-n$ (the first row has $50$ elements, the second $49$, and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence $n+2k-2 \\le n + 2(51-n) - 2 = 100 - n \\le 100,$ it follows that $67| n - 2k + 2$ implies that $n-2k+2 = 67$ itself. Now, note that we need $n$ to be odd, and also that $n+2k-2 = 67 \\le 100-n \\Longrightarrow n \\le 33$. We can check that all rows with odd $n$ satisfying $1 \\le n \\le 33$ indeed contains one entry that is a multiple of $67$, and so the answer is $\\frac{33+1}{2} = 017, and by our inductive hypothesis, \\begin{align*}a(n,k) &= a(n-1,k)+a(n-1,k+1)\\\\ &= 2^{n-2}(n-1+2k-2)+2^{n-2}(n-1+2(k+1)-2)\\\\&=2^{n-2}(2n+4k-4)\\\\&=2^{n-1}(n+2k-2)\\end{align*} thereby completing our induction.",
"The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$). The second row will be $2, 4, 6, \\cdots , 98$, the third row will be $3, 5, \\cdots, 97$, and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$. However, with each row the row will have one less element, and the $99-67+1 = 33$rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\\frac{33+1}{2} = 017."
] |
2008-I-7
| 2,008
| 7
|
Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$ . For example, $S_4$ is the set ${400,401,402,\ldots,499}$ . How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square?
| 708
|
I
|
[
"The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$, which means that all squares above $50^2 = 2500$ are more than $100$ apart. Then the first $26$ sets ($S_0,\\cdots S_{25}$) each have at least one perfect square because the differences between consecutive squares in them are all less than $100$. Also, since $316$ is the largest $x$ such that $x^2 < 100000$ ($100000$ is the upper bound which all numbers in $S_{999}$ must be less than), there are $316 - 50 = 266$ other sets after $S_{25}$ that have a perfect square. There are $1000 - 266 - 26 = 708 sets without a perfect square."
] |
2008-I-8
| 2,008
| 8
|
Find the positive integer $n$ such that \[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\]
| 47
|
I
|
[
"Since we are dealing with acute angles, $\\tan(\\arctan{a}) = a$. Note that $\\tan(\\arctan{a} + \\arctan{b}) = \\dfrac{a + b}{1 - ab}$, by tangent addition. Thus, $\\arctan{a} + \\arctan{b} = \\arctan{\\dfrac{a + b}{1 - ab}}$. Applying this to the first two terms, we get $\\arctan{\\dfrac{1}{3}} + \\arctan{\\dfrac{1}{4}} = \\arctan{\\dfrac{7}{11}}$. Now, $\\arctan{\\dfrac{7}{11}} + \\arctan{\\dfrac{1}{5}} = \\arctan{\\dfrac{23}{24}}$. We now have $\\arctan{\\dfrac{23}{24}} + \\arctan{\\dfrac{1}{n}} = \\dfrac{\\pi}{4} = \\arctan{1}$. Thus, $\\dfrac{\\dfrac{23}{24} + \\dfrac{1}{n}}{1 - \\dfrac{23}{24n}} = 1$; and simplifying, $23n + 24 = 24n - 23 \\Longrightarrow n = 47.",
"From the expansion of $e^{iA}e^{iB}e^{iC}e^{iD}$, we can see that \\[\\cos(A + B + C + D) = \\cos A \\cos B \\cos C \\cos D - \\tfrac{1}{4} \\sum_{\\rm sym} \\sin A \\sin B \\cos C \\cos D + \\sin A \\sin B \\sin C \\sin D,\\] and \\[\\sin(A + B + C + D) = \\sum_{\\rm cyc}\\sin A \\cos B \\cos C \\cos D - \\sum_{\\rm cyc} \\sin A \\sin B \\sin C \\cos D .\\] If we divide both of these by $\\cos A \\cos B \\cos C \\cos D$, then we have \\[\\tan(A + B + C + D) = \\frac {1 - \\sum \\tan A \\tan B + \\tan A \\tan B \\tan C \\tan D}{\\sum \\tan A - \\sum \\tan A \\tan B \\tan C},\\] which makes for more direct, less error-prone computations. Substitution gives the desired answer.",
"From the expansion of $e^{iA}e^{iB}e^{iC}e^{iD}$, we can see that \\[\\cos(A + B + C + D) = \\cos A \\cos B \\cos C \\cos D - \\tfrac{1}{4} \\sum_{\\rm sym} \\sin A \\sin B \\cos C \\cos D + \\sin A \\sin B \\sin C \\sin D,\\] and \\[\\sin(A + B + C + D) = \\sum_{\\rm cyc}\\sin A \\cos B \\cos C \\cos D - \\sum_{\\rm cyc} \\sin A \\sin B \\sin C \\cos D .\\] If we divide both of these by $\\cos A \\cos B \\cos C \\cos D$, then we have \\[\\tan(A + B + C + D) = \\frac {1 - \\sum \\tan A \\tan B + \\tan A \\tan B \\tan C \\tan D}{\\sum \\tan A - \\sum \\tan A \\tan B \\tan C},\\] which makes for more direct, less error-prone computations. Substitution gives the desired answer.",
"Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\\arctan\\frac{1}{n}$, is the argument of $n+i$. The sum of these angles is then just the argument of the product \\[(3+i)(4+i)(5+i)(n+i)\\] and expansion give us $(48n-46)+(48+46n)i$. Since the argument of this complex number is $\\frac{\\pi}{4}$, its real and imaginary parts must be equal; then, we can we set them equal to get \\[48n - 46 = 48 + 46n.\\] Therefore, $n=47.",
"You could always just bash out $\\sin(a+b+c)$ (where a,b,c,n are the angles of the triangles respectively) using the sum identities again and again until you get a pretty ugly radical and use a triangle to get $\\cos(a+b+c)$ and from there you use a sum identity again to get $\\sin(a+b+c+n)$ and using what we found earlier you can find $\\tan(n)$ by division that gets us $\\frac{23}{24}$ ~YBSuburbanTea"
] |
2008-I-9
| 2,008
| 9
|
Ten identical crates each of dimensions $3$ ft $\times$ $4$ ft $\times$ $6$ ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41$ ft tall, where $m$ and $n$ are relatively prime positive integers. Find $m$ .
| 190
|
I
|
[
"Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following: \\begin{align*}3a + 4b + 6c &= 41\\\\ a + b + c &= 10\\end{align*} Subtracting 3 times the second from the first gives $b + 3c = 11$, or $(b,c) = (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous. For $(5,2,3),(3,5,2)$, we see that there are $2\\cdot\\dfrac{10!}{5!2!3!} = 10\\cdot9\\cdot8\\cdot7$ ways to stack the crates. For $(1,8,1)$, there are $\\dfrac{10!}{8!1!1!} = 90$. Also, there are $3^{10}$ total ways to stack the crates to any height. Thus, our probability is $\\dfrac{10\\cdot9\\cdot8\\cdot7 + 90}{3^{10}} = \\dfrac{10\\cdot8\\cdot7 + 10}{3^{8}} = \\dfrac{570}{3^8} = \\dfrac{190}{3^{7}}$. Our answer is the numerator, $190.",
"Let's make two observations. We are trying to find the number of ways we can add $3\\text{s}, 4\\text{s}$, and $6\\text{s}$ to get $41$, and the total number of (non-distinct) sums possible is $3^{10}$. Then we just use casework to easily and directly solve for the number of ways to get $41$. To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\\text{s}, 1\\text{s}$, and $3\\text{s}$. Case I: $0$ zeroes, $0$ threes, $11$ ones Impossible, because there are only ten available spots. Case II: $1$ zero, $1$ three, $8$ ones This is just $\\frac{10!}{8!}$, so there are $90$ possibilities. Case III: $3$ zeroes, $2$ threes, $5$ ones This is just $\\frac{10!}{3!2!5!}$. This gives $2520$ possibilities. Case IV: 5 zeroes, 3 threes, and 2 ones. This is the same as case $3$, so also $2520$ possibilities. $90+2520+2520=5130$ $5130$ has three powers of $3$, so $5130$ divided by $27$ is $190. -jackshi2006",
"Let's make two observations. We are trying to find the number of ways we can add $3\\text{s}, 4\\text{s}$, and $6\\text{s}$ to get $41$, and the total number of (non-distinct) sums possible is $3^{10}$. Then we just use casework to easily and directly solve for the number of ways to get $41$. To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\\text{s}, 1\\text{s}$, and $3\\text{s}$. Case I: $0$ zeroes, $0$ threes, $11$ ones Impossible, because there are only ten available spots. Case II: $1$ zero, $1$ three, $8$ ones This is just $\\frac{10!}{8!}$, so there are $90$ possibilities. Case III: $3$ zeroes, $2$ threes, $5$ ones This is just $\\frac{10!}{3!2!5!}$. This gives $2520$ possibilities. Case IV: 5 zeroes, 3 threes, and 2 ones. This is the same as case $3$, so also $2520$ possibilities. $90+2520+2520=5130$ $5130$ has three powers of $3$, so $5130$ divided by $27$ is $190. -jackshi2006",
"Note we are placing 10 crates where each \"height\" is 3, 4, 6 and we want all the heights to sum to 41. We can model this as the generating function \\[\\left(x^3+x^4+x^6\\right)^{10}\\] where we want the coefficient of $x^{41}.$ First off, factor this to get \\[{x^{30} \\left(1 + x + x^3\\right)^{10}}\\] and then see that we want the coefficient of $x^{11}$ in $\\left(1+x+x^3\\right)^{10}.$ From multinomial theorem, this expansion is \\[\\sum_{a+b+c=10}\\binom{10}{a,b,c}1^ax^bx^{3c}\\] If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \\le 10$(from the multinomial expansion). This has the solutions $(b, c) = \\{8, 1\\}, \\{5, 2\\}, \\{2, 3\\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is \\[\\frac{\\binom{10}{8,1,1} + \\binom{10}{5,2,3} + \\binom{10}{2,3,5}}{3^{10}} = \\frac{190}{3^7} \\rightarrow 190\\]",
"Note we are placing 10 crates where each \"height\" is 3, 4, 6 and we want all the heights to sum to 41. We can model this as the generating function \\[\\left(x^3+x^4+x^6\\right)^{10}\\] where we want the coefficient of $x^{41}.$ First off, factor this to get \\[{x^{30} \\left(1 + x + x^3\\right)^{10}}\\] and then see that we want the coefficient of $x^{11}$ in $\\left(1+x+x^3\\right)^{10}.$ From multinomial theorem, this expansion is \\[\\sum_{a+b+c=10}\\binom{10}{a,b,c}1^ax^bx^{3c}\\] If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \\le 10$(from the multinomial expansion). This has the solutions $(b, c) = \\{8, 1\\}, \\{5, 2\\}, \\{2, 3\\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is \\[\\frac{\\binom{10}{8,1,1} + \\binom{10}{5,2,3} + \\binom{10}{2,3,5}}{3^{10}} = \\frac{190}{3^7} \\rightarrow 190\\]"
] |
2008-I-10
| 2,008
| 10
|
Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$ . The diagonals have length $10\sqrt {21}$ , and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$ , respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$ . The distance $EF$ can be expressed in the form $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ .
| 32
|
I
|
[
"[asy] size(300); defaultpen(1); pair A=(0,0), D=(4,0), B= A+2 expi(1/3*pi), C= D+2expi(2/3*pi), E=(-4/3,0), F=(3,0); draw(F--C--B--A); draw(E--A--D--C); draw(A--C,dashed); draw(circle(A,abs(C-A)),dotted); label(\"\\(A\\)\",A,S); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NE); label(\"\\(D\\)\",D,SE); label(\"\\(E\\)\",E,N); label(\"\\(F\\)\",F,S); clip(currentpicture,(-1.5,-1)--(5,-1)--(5,3)--(-1.5,3)--cycle); [/asy] Key observation. $AD = 20\\sqrt{7}$. Proof 1. By the triangle inequality, we can immediately see that $AD \\geq 20\\sqrt{7}$. However, notice that $10\\sqrt{21} = 20\\sqrt{7}\\cdot\\sin\\frac{\\pi}{3}$, so by the law of sines, when $AD = 20\\sqrt{7}$, $\\angle ACD$ is right and the circle centered at $A$ with radius $10\\sqrt{21}$, which we will call $\\omega$, is tangent to $\\overline{CD}$. Thus, if $AD$ were increased, $\\overline{CD}$ would have to be moved even farther outwards from $A$ to maintain the angle of $\\frac{\\pi}{3}$ and $\\omega$ could not touch it, a contradiction. Proof 2. Again, use the triangle inequality to obtain $AD \\geq 20\\sqrt{7}$. Let $x = AD$ and $y = CD$. By the law of cosines on $\\triangle ADC$, $2100 = x^2+y^2-xy \\iff y^2-xy+(x^2-2100) = 0$. Viewing this as a quadratic in $y$, the discriminant $\\Delta$ must satisfy $\\Delta = x^2-4(x^2-2100) = 8400-3x^2 \\geq 0 \\iff x \\leq 20\\sqrt{7}$. Combining these two inequalities yields the desired conclusion. This observation tells us that $E$, $A$, and $D$ are collinear, in that order. Then, $\\triangle ADC$ and $\\triangle ACF$ are $30-60-90$ triangles. Hence $AF = 15\\sqrt {7}$, and $EF = EA + AF = 10\\sqrt {7} + 15\\sqrt {7} = 25\\sqrt {7}$. Finally, the answer is $25+7=032.",
"Extend $\\overline {AB}$ through $B$, to meet $\\overline {DC}$ (extended through $C$) at $G$. $ADG$ is an equilateral triangle because of the angle conditions on the base. If $\\overline {GC} = x$ then $\\overline {CD} = 40\\sqrt{7}-x$, because $\\overline{AD}$ and therefore $\\overline{GD}$ $= 40\\sqrt{7}$. By simple angle chasing, $CFD$ is a 30-60-90 triangle and thus $\\overline{FD} = \\frac{40\\sqrt{7}-x}{2}$, and $\\overline{CF} = \\frac{40\\sqrt{21} - \\sqrt{3}x}{2}$ Similarly $CAF$ is a 30-60-90 triangle and thus $\\overline{CF} = \\frac{10\\sqrt{21}}{2} = 5\\sqrt{21}$. Equating and solving for $x$, $x = 30\\sqrt{7}$ and thus $\\overline{FD} = \\frac{40\\sqrt{7}-x}{2} = 5\\sqrt{7}$. $\\overline{ED}-\\overline{FD} = \\overline{EF}$ $30\\sqrt{7} - 5\\sqrt{7} = 25\\sqrt{7}$ and $25 + 7 = 032. ~polya_mouse. (how is this a P10 what)"
] |
2008-I-11
| 2,008
| 11
|
Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ , $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have length 14?
| 172
|
I
|
[
"Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$. If a sequence ends in an $A$, then it must have been formed by appending two $A$s to the end of a string of length $n-2$. If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a string of length $n-1$ ending in an $A$, or by appending two $B$s to a string of length $n-2$ ending in a $B$. Thus, we have the recursions \\begin{align*} a_n &= a_{n-2} + b_{n-2}\\\\ b_n &= a_{n-1} + b_{n-2} \\end{align*} By counting, we find that $a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0$. \\[\\begin{array}{|r||r|r|||r||r|r|} \\hline n & a_n & b_n & n & a_n & b_n\\\\ \\hline 1&0&1& 8&6&10\\\\ 2&1&0& 9&11&11\\\\ 3&1&2& 10&16&21\\\\ 4&1&1& 11&22&27\\\\ 5&3&3& 12&37&43\\\\ 6&2&4& 13&49&64\\\\ 7&6&5& 14&80&92\\\\ \\hline \\end{array}\\] Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = 172.",
"Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$. Additionally, let $t_n$ denote the total number of sequences of length $n$. Then, $t_n=a_n+b_n$, as the total amount of sequences of length $n$ consists of the sequences of length $n$ ending in $A$ and the sequences of length $n$ ending in $B$. \\begin{align*} a_n &= a_{n-2} + b_{n-2}\\\\ b_n &= a_{n-1} + b_{n-2} \\end{align*} The recursion for $a_n$ tells us that $a_n=a_{n-2}+b_{n-2}$. However, this is also the definition for $t_{n-2}$. Therefore, $a_n=t_{n-2}$. We also know from our recursion for $b_n$ that $b_n=a_{n-1}+b_{n-2}$. Substituting for $a_n$ and $b_n$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+a_{n-1}+b_{n-2}$. Furthermore, note that since $a_n=t_{n-2}$, $a_{n-1}=t_{n-3}$. Furthermore, using our definition for $t_{n-2}$, we can rewrite $b_{n-2}$ as $t_{n-2}-a_{n-2}$. Substituting for $a_{n-1}$ and $b_{n-2}$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}$. Finally, note that since $a_n=t_{n-2}$, $a_{n-2}=t_{n-4}$. Substituting for $a_{n-2}$ into our recursion for $t_n$ gives us $t_n=2t_{n-2}+t_{n-3}-t_{n-4}$. We now have a recursion only in terms of $t$. By counting, we find that $t_1=1$, $t_2=1$, $t_3=3$, and $t_4=2$. \\[\\begin{array}{|r|r||r|r|} \\hline n & t_n & n & t_n\\\\ \\hline 1&1&8&16\\\\ 2&1&9&22\\\\ 3&3&10&37\\\\ 4&2&11&49\\\\ 5&6&12&80\\\\ 6&6&13&113\\\\ 7&11&14&172\\\\ \\hline \\end{array}\\] Therefore, the number of such sequences of length 14 is $172.",
"We replace \"14\" with \"$2n$\". We first note that we must have an even number of chunks of $B$'s, because of parity issues. We then note that every chunk of $B$'s except the last one must end in the sequence $BAA$, since we need at least two $A$'s to separate it from the next chunk of $B$'s. The last chunk of $B$'s must, of course, end with a $B$. Thus our sequence must look like this : \\[\\quad A\\text{'s} \\quad.",
"There must be an even amount of runs of consecutive $B$s due to parity. Thus, we can split this sequence into the following cases: $A$, $BAAB$, $AABAAB$, $BAABAA$, $AABAABAA$, $BAABAABAAB$, $AABAABAABAAB$, $BAABAABAABAA$, and $AABAABAABAABAA$, in which the amount of letters in one run does not necessarily represent the amount of letters there can be. For the first case and the last case, there is only one possible sequence of letters. For all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two $A$s and make the sequence $BAAAAB$. There are three \"slots\" in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars, the number of ways for the second case is $\\binom{7}{2}=21$. Applying this logic to all of the other cases gives us $\\binom{7}{3}$, $\\binom{7}{3}$, $\\binom{7}{4}$, $\\binom{8}{6}$, $\\binom{8}{1}$, and $\\binom{8}{1}$. Adding 1+$\\binom{7}{2}$+$\\binom{7}{3}$+$\\binom{7}{3}$+$\\binom{7}{4}$+$\\binom{8}{6}$+$\\binom{8}{1}$+$\\binom{8}{1}$ gives us the answer $172."
] |
2008-I-12
| 2,008
| 12
|
On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling 52 kilometers per hour will be four car lengths behind the back of the car in front of it.) A photoelectric eye by the side of the road counts the number of cars that pass in one hour. Assuming that each car is 4 meters long and that the cars can travel at any speed, let $M$ be the maximum whole number of cars that can pass the photoelectric eye in one hour. Find the quotient when $M$ is divided by 10.
| 375
|
I
|
[
"Let $n$ be the number of car lengths that separates each car (it is easy to see that this should be the same between each pair of consecutive cars.) Then their speed is at most $15n$. Let a unit be the distance between the cars (front to front). Then the length of each unit is $4(n + 1)$. To maximize, in a unit, the CAR comes first, THEN the empty space. So at time zero, the car is right at the eye. Hence, we count the number of units that pass the eye in an hour: $\\frac {15,000n\\frac{\\text{meters}}{\\text{hour}}}{4(n + 1)\\frac{\\text{meters}}{\\text{unit}}} = \\frac {15,000n}{4(n + 1)}\\frac{\\text{units}}{\\text{hour}}$. We wish to maximize this. Observe that as $n$ gets larger, the $+ 1$ gets less and less significant, so we take the limit as $n$ approaches infinity $\\lim_{n\\rightarrow \\infty}\\frac {15,000n}{4(n + 1)} = \\lim_{n\\rightarrow \\infty}\\frac {15,000}{4} = 3750$ Now, as the speeds are clearly finite, we can never actually reach $3750$ full UNITs. However, we only need to find the number of CARS. We can increase their speed so that the camera stops (one hour goes by) after the car part of the $3750$th unit has passed, but not all of the space behind it. Hence, $3750$ cars is possible, and the answer is $375.",
"Disclaimer: This is for the people who may not understand calculus, and is also how I did it. First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we start with one car in front of the photoelectric eye. We first set the speed of the cars as $15k$. Then, the distance between them is $\\frac{4}{1000} \\times k\\text{km}$. Therefore, it takes the car closest to the eye not on the eye $\\frac{\\frac{k}{250}}{15k}$ hours to get to the eye. There is one hour, so the amount of cars that can pass is $\\frac{1}{\\frac{\\frac{k}{250}}{15k}}$, or $3750$ cars. When divided by ten, you get the quotient of $375",
"Call the speed of each car $s$, so that the distance between each car will be $\\lfloor a/15 \\rfloor$. To maximize the number of cars that pass, let the first car back end line up with the eye. Notice that every $4/1000+ (4/1000)(\\lfloor a/15 \\rfloor)$ kilometers (distance from front of one car to the front of the consecutive car), one car will be captured. The time it takes to travel this distance is $\\frac{4/1000+(4/1000)(\\lfloor a/15 \\rfloor)}{a}$, and the number of these intervals in an hour is $\\frac{1}{\\frac{4/1000+(4/1000)(\\lfloor a/15 \\rfloor)}{a}}=\\frac{250a}{1+\\lfloor a/15 \\rfloor}$. Now, let $a=15n$ and we can proceed with Solution 1."
] |
2008-I-13
| 2,008
| 13
|
Let $p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3$ . Suppose that $p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1) = p(1,1) = p(1, - 1) = p(2,2) = 0$ . There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$ , $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$ . Find $a + b + c$ .
| 40
|
I
|
[
"Solution 1 \\begin{align*} p(0,0) &= a_0 \\\\ &= 0 \\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\\\ &= a_1 + a_3 + a_6 \\\\ &= 0 \\\\ p(-1,0) &= -a_1 + a_3 - a_6 \\\\ &= 0 \\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\\\ &= a_4 + a_7 + a_8 \\\\ &= 0 \\\\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2 \\\\ &= -a_4 - a_7 + a_8 \\\\ &= 0 \\end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, \\begin{align*} p(2,2) &= 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 \\\\ &= -6 a_1 - 6 a_2 - 4 a_4 \\\\ &= 0 \\end{align*} So, $3a_1 + 3a_2 + 2a_4 = 0$, or equivalently $a_4 = -\\frac{3(a_1 + a_2)}{2}$. Substituting these equations into the original polynomial $p$, we find that at $\\left(\\frac{a}{c}, \\frac{b}{c}\\right)$, \\[a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \\iff\\] \\[a_1x + a_2y - \\frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \\frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 = 0 \\iff\\] \\[a_1x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) + a_2y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0\\]. The remaining coefficients $a_1$ and $a_2$ are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible $p$, we must have $x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) = y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0$. As the answer format implies that the $x$-coordinate of the root is non-integral, $x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) = 0 \\iff x + 1 - \\frac{3}{2}y = 0 \\iff y = \\frac{2}{3}(x + 1)\\ (1)$. The format also implies that $y$ is positive, so $y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0 \\iff y^2 - 1 - \\frac{3}{2}x(x - 1) = 0\\ (2)$. Substituting $(1)$ into $(2)$ and reducing to a quadratic yields $(19x - 5)(x - 2) = 0$, in which the only non-integral root is $x = \\frac{5}{19}$, so $y = \\frac{16}{19}$. The answer is $5 + 16 + 19 = 040\", (2,2), SE, fontsize(8)); dot((5/19,16/19), green); [/asy]",
"\\begin{align*} p(0,0) &= a_0 \\\\ &= 0 \\\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 \\\\ &= a_1 + a_3 + a_6 \\\\ &= 0 \\\\ p(-1,0) &= -a_1 + a_3 - a_6 \\\\ &= 0 \\end{align*} Adding the above two equations gives $a_3 = 0$, and so we can deduce that $a_6 = -a_1$. Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. Now, \\begin{align*} p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\\\ &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\\\ &= a_4 + a_7 + a_8 \\\\ &= 0 \\\\ p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2 \\\\ &= -a_4 - a_7 + a_8 \\\\ &= 0 \\end{align*} Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally, \\begin{align*} p(2,2) &= 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 \\\\ &= -6 a_1 - 6 a_2 - 4 a_4 \\\\ &= 0 \\end{align*} So, $3a_1 + 3a_2 + 2a_4 = 0$, or equivalently $a_4 = -\\frac{3(a_1 + a_2)}{2}$. Substituting these equations into the original polynomial $p$, we find that at $\\left(\\frac{a}{c}, \\frac{b}{c}\\right)$, \\[a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \\iff\\] \\[a_1x + a_2y - \\frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \\frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 = 0 \\iff\\] \\[a_1x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) + a_2y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0\\]. The remaining coefficients $a_1$ and $a_2$ are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible $p$, we must have $x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) = y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0$. As the answer format implies that the $x$-coordinate of the root is non-integral, $x(x - 1)\\left(x + 1 - \\frac{3}{2}y\\right) = 0 \\iff x + 1 - \\frac{3}{2}y = 0 \\iff y = \\frac{2}{3}(x + 1)\\ (1)$. The format also implies that $y$ is positive, so $y\\left(y^2 - 1 - \\frac{3}{2}x(x - 1)\\right) = 0 \\iff y^2 - 1 - \\frac{3}{2}x(x - 1) = 0\\ (2)$. Substituting $(1)$ into $(2)$ and reducing to a quadratic yields $(19x - 5)(x - 2) = 0$, in which the only non-integral root is $x = \\frac{5}{19}$, so $y = \\frac{16}{19}$. The answer is $5 + 16 + 19 = 040\", (2,2), SE, fontsize(8)); dot((5/19,16/19), green); [/asy]",
"Consider the cross section of $z = p(x, y)$ on the plane $z = 0$. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of $p(x, y)$ (same degree of $x$ and $y$ in terms) and they include the eight given points. One simple way to do this would be to use the equations $x = 0$, $x = 1$, and $y = \\frac{2}{3}x + \\frac{2}{3}$, giving us $p_1(x, y) = x\\left(x - 1\\right)\\left( \\frac{2}{3}x - y + \\frac{2}{3}\\right) = \\frac{2}{3}x + xy + \\frac{2}{3}x^3-x^2y$. Another way to do this would to use the line $y = x$ and the ellipse, $x^2 + xy + y^2 = 1$. This would give $p_2(x, y) = \\left(x - y\\right)\\left(x^2 + xy + y^2 - 1\\right) = -x + y + x^3 - y^3$. (But (Another way would be to use the hyperbola from Solution 1. Interesting that different curves both work.) At this point, we consider that $p_1$ and $p_2$ both must have $\\left(\\frac{a}{c}, \\frac{b}{c}\\right)$ as a zero. A quick graph of the 4 lines and the ellipse used to create $p_1$ and $p_2$ gives nine intersection points. Eight of them are the given ones, and the ninth is $\\left(\\frac{5}{19}, \\frac{16}{19}\\right)$. The last intersection point can be found by finding the intersection points of $y = \\frac{2}{3}x + \\frac{2}{3}$ and $x^2 + xy + y^2 = 1$. Finally, just add the values of $a$, $b$, and $c$ to get $5 + 16 + 19 = 040\", (2,2), SE, fontsize(8)); dot((5/19,16/19), green); [/asy]",
"We can plug in the values to obtain \\[p(0,0)=0\\Longrightarrow a_0=0\\] \\[p(1,0)=0\\Longrightarrow a_1+a_3+a_6=0\\] \\[p(0,1)=0\\Longrightarrow a_2+a_5+a_9=0\\] \\[p(-1,0)=0\\Longrightarrow a_1-a_3+a_6=0\\] \\[p(0,-1)=0\\Longrightarrow a_2-a-5+a_9=0\\] \\[p(1,1)=0\\Longrightarrow a_4+a_7+a_8=0\\] \\[p(1,-1)=0\\Longrightarrow a_4+a_7-a_8=0\\] \\[p(2,2)=0\\Longrightarrow2a_4=3a_6+3a_9\\Leftrightarrow2a_4+3a_1+3a_2=0.\\] Now, this means that \\[p(x,y)=a_1x+a_2y+a_4xy-a_1x^3+a_7x^2y+a_8xy^2-a_2y^3.\\] After some simplifying, we obtain \\[p(x,y)=a_1(x-x^3)+a_2(y-y^3)+a_4xy(1-x).\\] Since $p(x,y)=0$, $3a_1+3a_2+2a_4=0,$ and we suspect that: \\[x-x^3=y-y^3\\] and \\[\\frac{x-x^3}{xy(1-x)}=\\frac{3}{2}\\Leftrightarrow\\frac{1+x}{y}=\\frac{3}{2} \\Leftrightarrow\\frac{2}{3}+\\frac{2x}{3}=y\\]. Plugging this into the first equation, and factoring, and cancelling $(x+1)$, and simplifying, we get $19x^2 -43x+10=0$, so we find that $(x,y)=\\left(\\frac{5}{19},\\frac{16}{19}\\right)\\Longrightarrow a+b+c=5+16+19=40 ~~pinkpig",
"We can plug in the values to obtain \\[p(0,0)=0\\Longrightarrow a_0=0\\] \\[p(1,0)=0\\Longrightarrow a_1+a_3+a_6=0\\] \\[p(0,1)=0\\Longrightarrow a_2+a_5+a_9=0\\] \\[p(-1,0)=0\\Longrightarrow a_1-a_3+a_6=0\\] \\[p(0,-1)=0\\Longrightarrow a_2-a-5+a_9=0\\] \\[p(1,1)=0\\Longrightarrow a_4+a_7+a_8=0\\] \\[p(1,-1)=0\\Longrightarrow a_4+a_7-a_8=0\\] \\[p(2,2)=0\\Longrightarrow2a_4=3a_6+3a_9\\Leftrightarrow2a_4+3a_1+3a_2=0.\\] Now, this means that \\[p(x,y)=a_1x+a_2y+a_4xy-a_1x^3+a_7x^2y+a_8xy^2-a_2y^3.\\] After some simplifying, we obtain \\[p(x,y)=a_1(x-x^3)+a_2(y-y^3)+a_4xy(1-x).\\] Since $p(x,y)=0$, $3a_1+3a_2+2a_4=0,$ and we suspect that: \\[x-x^3=y-y^3\\] and \\[\\frac{x-x^3}{xy(1-x)}=\\frac{3}{2}\\Leftrightarrow\\frac{1+x}{y}=\\frac{3}{2} \\Leftrightarrow\\frac{2}{3}+\\frac{2x}{3}=y\\]. Plugging this into the first equation, and factoring, and cancelling $(x+1)$, and simplifying, we get $19x^2 -43x+10=0$, so we find that $(x,y)=\\left(\\frac{5}{19},\\frac{16}{19}\\right)\\Longrightarrow a+b+c=5+16+19=40 ~~pinkpig"
] |
2008-I-14
| 2,008
| 14
|
Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $\overline{AB} = 18$ , and let $m$ denote the maximum possible length of segment $BP$ . Find $m^{2}$ .
| 432
|
I
|
[
"Solution 1 [asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NW); label(\"\\(O\\)\",O,NW); label(\"\\(P\\)\",P,SE); label(\"\\(T\\)\",T,SE); label(\"\\(9\\)\",(O+A)/2,N); label(\"\\(9\\)\",(O+B)/2,N); label(\"\\(x-9\\)\",(C+A)/2,N); [/asy] Let $x = OC$. Since $OT, AP \\perp TC$, it follows easily that $\\triangle APC \\sim \\triangle OTC$. Thus $\\frac{AP}{OT} = \\frac{CA}{CO} \\Longrightarrow AP = \\frac{9(x-9)}{x}$. By the Law of Cosines on $\\triangle BAP$, \\begin{align*}BP^2 = AB^2 + AP^2 - 2 \\cdot AB \\cdot AP \\cdot \\cos \\angle BAP \\end{align*} where $\\cos \\angle BAP = \\cos (180 - \\angle TOA) = - \\frac{OT}{OC} = - \\frac{9}{x}$, so: \\begin{align*}BP^2 &= 18^2 + \\frac{9^2(x-9)^2}{x^2} + 2(18) \\cdot \\frac{9(x-9)}{x} \\cdot \\frac 9x = 405 + 729\\left(\\frac{2x - 27}{x^2}\\right)\\end{align*} Let $k = \\frac{2x-27}{x^2} \\Longrightarrow kx^2 - 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \\ge 0 \\Longleftrightarrow k \\le \\frac{1}{27}$. Thus, \\[BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = 432",
"[asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NW); label(\"\\(O\\)\",O,NW); label(\"\\(P\\)\",P,SE); label(\"\\(T\\)\",T,SE); label(\"\\(9\\)\",(O+A)/2,N); label(\"\\(9\\)\",(O+B)/2,N); label(\"\\(x-9\\)\",(C+A)/2,N); [/asy] Let $x = OC$. Since $OT, AP \\perp TC$, it follows easily that $\\triangle APC \\sim \\triangle OTC$. Thus $\\frac{AP}{OT} = \\frac{CA}{CO} \\Longrightarrow AP = \\frac{9(x-9)}{x}$. By the Law of Cosines on $\\triangle BAP$, \\begin{align*}BP^2 = AB^2 + AP^2 - 2 \\cdot AB \\cdot AP \\cdot \\cos \\angle BAP \\end{align*} where $\\cos \\angle BAP = \\cos (180 - \\angle TOA) = - \\frac{OT}{OC} = - \\frac{9}{x}$, so: \\begin{align*}BP^2 &= 18^2 + \\frac{9^2(x-9)^2}{x^2} + 2(18) \\cdot \\frac{9(x-9)}{x} \\cdot \\frac 9x = 405 + 729\\left(\\frac{2x - 27}{x^2}\\right)\\end{align*} Let $k = \\frac{2x-27}{x^2} \\Longrightarrow kx^2 - 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \\ge 0 \\Longleftrightarrow k \\le \\frac{1}{27}$. Thus, \\[BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = 432",
"Proceed as follows for Solution 1. Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$. Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$. Solving, we obtain $x=27$ as the critical value. Hence, $k$ has the maximum value of $(2*27-27)/27^2=1/27$. Since $BP^2=405+729k$, the maximum value of $\\overline {BP}$ occurs at $k=1/27$, so $BP^2$ has a maximum value of $405+729/27=432",
"Proceed as follows for Solution 1. Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$. Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$. Solving, we obtain $x=27$ as the critical value. Hence, $k$ has the maximum value of $(2*27-27)/27^2=1/27$. Since $BP^2=405+729k$, the maximum value of $\\overline {BP}$ occurs at $k=1/27$, so $BP^2$ has a maximum value of $405+729/27=432",
"[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"\\(B\\)\",B,NW); label(\"\\(A\\)\",A,NE); label(\"\\(O\\)\",O,N); label(\"\\(P\\)\",P,S); label(\"\\(T\\)\",T,S); label(\"\\(Q\\)\",Q,S); label(\"\\(C\\)\",C,E); label(\"\\(\\theta\\)\",C + (-1.7,-0.2), NW); label(\"\\(9\\)\", (B+O)/2, N); label(\"\\(9\\)\", (O+A)/2, N); label(\"\\(9\\)\", (O+T)/2,W); [/asy] From the diagram, we see that $BQ = OT + BO \\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = 432",
"[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"\\(B\\)\",B,NW); label(\"\\(A\\)\",A,NE); label(\"\\(\\omega\\)\",O,N); label(\"\\(P\\)\",P,S); label(\"\\(T\\)\",T,S); label(\"\\(Q\\)\",Q,S); label(\"\\(C\\)\",C,E); label(\"\\(9\\)\", (B+O)/2, N); label(\"\\(9\\)\", (O+A)/2, N); label(\"\\(9\\)\", (O+T)/2,W); [/asy] (Diagram credit goes to Solution 2) We let $AC=x$. From similar triangles, we have that $PC=\\frac{x\\sqrt{x^2+18x}}{x+9}$ (Use Pythagorean on $\\triangle\\omega TC$ and then using $\\triangle\\omega CT\\sim\\triangle ACP$). Similarly, $TP=QT=\\frac{9\\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem again and $\\triangle CAP\\sim\\triangle CBQ$, $BQ=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem $\\bold{again}$, $BP=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2+(\\frac{18\\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\\sqrt{405+\\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\\sqrt{432}$, so the answer is $(\\sqrt{432})^2=432",
"[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"\\(B\\)\",B,NW); label(\"\\(A\\)\",A,NE); label(\"\\(\\omega\\)\",O,N); label(\"\\(P\\)\",P,S); label(\"\\(T\\)\",T,S); label(\"\\(Q\\)\",Q,S); label(\"\\(C\\)\",C,E); label(\"\\(9\\)\", (B+O)/2, N); label(\"\\(9\\)\", (O+A)/2, N); label(\"\\(9\\)\", (O+T)/2,W); [/asy] (Diagram credit goes to Solution 2) We let $AC=x$. From similar triangles, we have that $PC=\\frac{x\\sqrt{x^2+18x}}{x+9}$ (Use Pythagorean on $\\triangle\\omega TC$ and then using $\\triangle\\omega CT\\sim\\triangle ACP$). Similarly, $TP=QT=\\frac{9\\sqrt{x^2+18x}}{x+9}$. Using the Pythagorean Theorem again and $\\triangle CAP\\sim\\triangle CBQ$, $BQ=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2}$. Using the Pythagorean Theorem $\\bold{again}$, $BP=\\sqrt{(x+18)^2-(\\frac{(x+18)\\sqrt{x^2+18x}}{x+9})^2+(\\frac{18\\sqrt{x^2+18x}}{x+9})^2}$. After a large bashful simplification, $BP=\\sqrt{405+\\frac{1458x-6561}{x^2+18x+81}}$. The fraction is equivalent to $729\\frac{2x-9}{(x+9)^2}$. Taking the derivative of the fraction and solving for x, we get that $x=18$. Plugging $x=18$ back into the expression for $BP$ yields $\\sqrt{432}$, so the answer is $(\\sqrt{432})^2=432",
"[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"\\(B\\)\",B,NW); label(\"\\(A\\)\",A,NE); label(\"\\(\\omega\\)\",O,N); label(\"\\(P\\)\",P,S); label(\"\\(T\\)\",T,S); label(\"\\(Q\\)\",Q,S); label(\"\\(C\\)\",C,E); label(\"\\(9\\)\", (B+O)/2, N); label(\"\\(9\\)\", (O+A)/2, N); label(\"\\(9\\)\", (O+T)/2,W); [/asy] (Diagram credit goes to Solution 2) Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\\frac{9x}{x+9}$. Now, apply law of cosines on $\\triangle PAB$. \\[BP^2=\\left(\\frac{9x^2}{x+9}\\right)^2+18^2-2(18)\\left(\\frac{9x}{x+9}\\right)\\cos(\\angle PAB).\\] We can see that $\\cos(\\angle PAB)=\\cos(180^{\\circ}-\\angle PAC)=\\cos(\\angle PAC -90^{\\circ})=-\\sin(\\angle PCA)$. We can find $-\\sin(\\angle PCA)=-\\frac{9}{x+9}$. Plugging this into our equation, we get: \\[BP^2=\\left(\\frac{9x^2}{x+9}\\right)^2+18^2-2(18)\\left(\\frac{9x}{x+9}\\right)\\left(-\\frac{9}{x+9}\\right).\\] Eventually, \\[BP^2 = 81\\left(\\frac{x^2+36x}{(x+9)^2}+4\\right).\\] We want to maximize $\\frac{x^2+36x}{(x+9)^2}$. There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is $\\frac{4}{3}$. Finally, evaluating $BP^2$ for this value $81\\left(\\frac{4}{3}+4\\right) = 432",
"Let $h$ be the distance from $A$ to $CT$. Observe that $h$ takes any value from $0$ to $r$, where $r$ is the radius of the circle. Let $Q$ be the foot of the altitude from $B$ to $CT$. It is clear that $T$ is the midpoint of $PQ$, and so the length $OT$ is the average of $AP$ and $BQ$. It follows thus that $BQ = 2r - h$. We compute $PT = \\sqrt{r^2 - (r - h)^2} = \\sqrt{h(2r - h)},$ and so $BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)$. This is $\\frac{1}{3}(6r - 3h)(2r + 3h) \\le \\frac{1}{3} \\cdot \\left( \\frac{8r}{2} \\right)^2$. Equality is attained, so thus we extract the answer of $\\frac{16 \\cdot 9^2}{3} = 27 \\cdot 16 = 432",
"Let $h$ be the distance from $A$ to $CT$. Observe that $h$ takes any value from $0$ to $r$, where $r$ is the radius of the circle. Let $Q$ be the foot of the altitude from $B$ to $CT$. It is clear that $T$ is the midpoint of $PQ$, and so the length $OT$ is the average of $AP$ and $BQ$. It follows thus that $BQ = 2r - h$. We compute $PT = \\sqrt{r^2 - (r - h)^2} = \\sqrt{h(2r - h)},$ and so $BP^2 = PQ^2 + BQ^2 = 4PT^2 + BQ^2 = 4h(2r - h) + (2r-h)^2 = (2r-h)(2r + 3h)$. This is $\\frac{1}{3}(6r - 3h)(2r + 3h) \\le \\frac{1}{3} \\cdot \\left( \\frac{8r}{2} \\right)^2$. Equality is attained, so thus we extract the answer of $\\frac{16 \\cdot 9^2}{3} = 27 \\cdot 16 = 432"
] |
2008-II-1
| 2,008
| 1
|
Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$ , where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$ .
| 100
|
II
|
[
"Rewriting this sequence with more terms, we have \\begin{align*} N &= 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + 95^2 - 94^2 - 93^2 + 92^2 + 91^2 + \\ldots - 10^2 - 9^2 + 8^2 + 7^2 - 6^2 - 5^2 + 4^2 + 3^2 - 2^2 - 1^2 \\mbox{, and reordering, we get}\\\\ N &= (100^2 - 98^2) + (99^2 - 97^2) + (96^2 - 94^2) + (95^2 - 93^2) + (92^2 - 90^2) + \\ldots + (8^2 - 6^2) + (7^2 - 5^2) +(4^2 - 2^2) + (3^2 - 1^2) \\mbox{.} \\end{align*} Factoring this expression yields \\begin{align*} N &= (100 - 98)(100 + 98) + (99 - 97)(99 + 97) + (96 - 94)(96 + 94) + (95 - 93)(95 + 93) + (92 - 90)(90 + 92) + \\ldots + (8 - 6)(8 + 6) + (7 - 5)(7 + 5) + (4 - 2)(4 + 2) + (3 - 1)(3 + 1) \\mbox{, leading to}\\\\ N &= 2(100 + 98) + 2(99 + 97) + 2(96 + 94) + 2(95 + 93) + 2(92 + 90) + \\ldots + 2(8 + 6) + 2(7 + 5) + 2(4 + 2) + 2(3 + 1) \\mbox{.} \\end{align*} Next, we get \\begin{align*} N &= 2(100 + 98 + 99 + 97 + 96 + 94 + 95 + 93 + 92 + 90 + \\ldots + 8 + 6 + 7 + 5 + 4 + 2 + 3 + 1 \\mbox{, and rearranging terms yields}\\\\ N &= 2(100 + 99 + 98 + 97 + 96 + \\ldots + 5 + 4 + 3 + 2 + 1) \\mbox{.} \\end{align*} Then, \\begin{align*} N &= 2\\left(\\frac{(100)(101)}{2}\\right) \\mbox{, and simplifying, we get}\\\\ N &= (100)(101) \\mbox{, so}\\\\ N &= 10100 \\mbox{.} \\end{align*} Dividing $10100$ by $1000$ yields a remainder of $100.",
"Since we want the remainder when $N$ is divided by $1000$, we may ignore the $100^2$ term. Then, applying the difference of squares factorization to consecutive terms, \\begin{align*} N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \\cdots + (3-2)(3+2) - 1 \\\\ &= \\underbrace{197 - 193}_4 + \\underbrace{189 - 185}_4 + \\cdots + \\underbrace{5 - 1}_4 \\\\ &= 4 \\cdot \\left(\\frac{197-5}{8}+1\\right) = 100 \\end{align*}",
"By observation, we realize that the sequence \\[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2\\] alternates every 4 terms. Simplifying, we get \\[(a+3)^2 + (a+2)^2 - (a+1)^2 - a^2 = 8a + 12\\], turning $N$ into a arithmetic sequence with 25 terms, them being $1, 5, 9, \\dots ,97$, as the series $8a + 12$ alternates every 4 terms. Applying the sum of arithmetic sequence formula, we get \\begin{align*} N &= \\frac{25\\cdot((8\\cdot1 + 12) + (8\\cdot97 + 12))}{2} \\\\ &= \\frac{25\\cdot(20 + 788)}{2} = 10100 \\end{align*} So the answer would be \\[\\frac{10100}{1000} = 100\\]. - erdaifuu",
"We can remove the $100^2$ since $100^2 \\equiv 0 \\pmod{1000}$ and use difference of squares to factor out the rest. This gives \\[(1)(99+98) + (-1)(96+97) + ... +(1)(3+2) +(-1)(1+0)\\] Writing this another way, we get \\[(99(2) - 1) - (97(2)- 1) + (95(2) - 1) - ... +(3(2)-1) - (1(2) -1)\\] We know that the last one is negative because all the numbers before multiplying that are in the form $4a - 1$(eg. $99 = 25(4) - 1$) are positive. Let $x = 99$. This makes the expression \\[(2x-1) - (2(x-2) - 1) + (2(x-4) - 1) - ... + (2(x-96) - 1) - (2(x-98) - 1)\\] This simplifies to \\[(2x-1) - (2x-5) + (2x-9) - ... +(2x - 193) - (2x-197)\\] Since the first one is positive and the last one is negative, that means there are an even number of terms and using the associative property and the distributive property, all of the $2x$ terms cancel out. A consequence of this is that all of the positive integers turn negative and all the negative ones turn positive(eg. $-(2x-5) = -2x + 5$). We are left with the sequence \\[-1+5-9+...-193+197\\] We can notice the property that the number of terms in the sequence to a positive number n is equal to $(n+3)/4$, as well as the fact that every pair sums up to 4. Therefore the total number of terms is $(197+3) / 4 = 50$. Therefore, there are 25 pairs each summing up to 4, leaving us with $25(4) = 100. ~idk12345678",
"We simply take the outer pairs and from there, we use the inside terms. That is, $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \\cdots + 4^2 + 3^2 - 2^2 - 1^2$ becomes $N = (100^2 - 1^2) + (99^2 - 2^2) - (98^2 - 3^2) - (97^2 - 4^2) + (96^2 - 5^2) ...$ and thus is reduced to $N = 101(99 + 97 - 95 - 93 + 81 + 89 - ... - 7 - 5 + 3 + 1)$ since the entire sequence has 50 terms. Therefore, the last two terms must be positive as the above inner sequence repeats as follows : Positive - positive - negative - negative. It follows that $N = 101[8(12) + 4] = 101[100] = 10100$. So the requested answer is $100. ~elpianista227"
] |
2008-II-2
| 2,008
| 2
|
Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $50$ -mile mark at exactly the same time. How many minutes has it taken them?
| 620
|
II
|
[
"Let Rudolf bike at a rate $r$, so Jennifer bikes at the rate $\\dfrac 34r$. Let the time both take be $t$. Then Rudolf stops $49$ times (because the rest after he reaches the finish does not count), losing a total of $49 \\cdot 5 = 245$ minutes, while Jennifer stops $24$ times, losing a total of $24 \\cdot 5 = 120$ minutes. The time Rudolf and Jennifer actually take biking is then $t - 245,\\, t-120$ respectively. Using the formula $r = \\frac dt$, since both Jennifer and Rudolf bike $50$ miles, \\begin{align}r &= \\frac{50}{t-245}\\\\ \\frac{3}{4}r &= \\frac{50}{t-120} \\end{align} Substituting equation $(1)$ into equation $(2)$ and simplifying, we find \\begin{align*}50 \\cdot \\frac{3}{4(t-245)} &= 50 \\cdot \\frac{1}{t-120}\\\\ \\frac{1}{3}t &= \\frac{245 \\cdot 4}{3} - 120\\\\ t &= 620\\ \\text{minutes} \\end{align*}",
"Let the total time that Jennifer and Rudolph bike, including rests, to be $t$ minutes. Furthermore, let Rudolph's biking rate be $r$ so Jennifer's biking rate is $\\frac{3}{4}r$. Note that Rudolf takes 49 breaks, taking $49\\cdot 5$ minutes, and Jennifer takes 24 breaks, taking $24\\cdot 5$ minutes. Since they both reach the 50 mile mark, then by $d=rt$, the rate times time taken for Rudolph and Jennifer must equal. Hence, we disregard the breaks from the total time taken and get the equation \\[r(t-49\\cdot 5)=\\frac{3}{4}r(t-24\\cdot 5),\\]yielding $t=620."
] |
2008-II-3
| 2,008
| 3
|
A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic cm of the remaining block of cheese after ten slices have been cut off?
| 729
|
II
|
[
"Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$, so that the desired volume is $abc$. Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$. By AM-GM, $\\frac{a+b+c}{3} = 9 \\ge \\sqrt[3]{abc} \\Longrightarrow abc \\le 729 cm edge.",
"A more intuitive way to solve it is by seeing that to keep the volume of the rectangular cheese the greatest, we must slice the cheese off to decrease the greatest length of the cheese (this is easy to check). Here are the ten slices: ${10, 13, 14} \\rightarrow {10, 13, 13} \\rightarrow {10, 12, 13} \\rightarrow {10, 12, 12} \\rightarrow {10, 11, 12} \\rightarrow {10, 11, 11} \\rightarrow {10, 10, 11} \\rightarrow {10, 10, 10} \\rightarrow {9, 10, 10} \\rightarrow {9, 9, 10} \\rightarrow {9, 9, 9}.$ So the greatest possible volume is thus $9 \\times 9 \\times 9 = 729"
] |
2008-II-4
| 2,008
| 4
|
There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ integers $a_k$ ( $1\le k\le r$ ) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$ .
| 21
|
II
|
[
"In base $3$, we find that $\\overline{2008}_{10} = \\overline{2202101}_{3}$. In other words, $2008 = 2 \\cdot 3^{6} + 2 \\cdot 3^{5} + 2 \\cdot 3^3 + 1 \\cdot 3^2 + 1 \\cdot 3^0$ In order to rewrite as a sum of perfect powers of $3$, we can use the fact that $2 \\cdot 3^k = 3^{k+1} - 3^k$: $2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0$ The answer is $7+5+4+3+2+0 = 021"
] |
2008-II-5
| 2,008
| 5
|
In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$ , let $BC = 1000$ and $AD = 2008$ . Let $\angle A = 37^\circ$ , $\angle D = 53^\circ$ , and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively. Find the length $MN$ .
| 504
|
II
|
[
"Solution 1 Extend $\\overline{AB}$ and $\\overline{CD}$ to meet at a point $E$. Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$. [asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label(\"\\(A\\)\",A,SW); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NE); label(\"\\(D\\)\",D,SE); label(\"\\(E\\)\",E,NE); label(\"\\(M\\)\",M[0],SW); label(\"\\(N\\)\",N,S); label(\"\\(1004\\)\",(N+D)/2,S); label(\"\\(500\\)\",(M[0]+C)/2,S); [/asy] As $\\angle AED = 90^{\\circ}$, note that the midpoint of $\\overline{AD}$, $N$, is the center of the circumcircle of $\\triangle AED$. We can do the same with the circumcircle about $\\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that \\[NE = ND = \\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500.\\] Thus $MN = NE - ME = 504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"Extend $\\overline{AB}$ and $\\overline{CD}$ to meet at a point $E$. Then $\\angle AED = 180 - 53 - 37 = 90^{\\circ}$. [asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); draw(A--B--C--D--cycle); draw(B--E--C,dashed); draw(M[0]--N); draw(N--E,dashed); draw(rightanglemark(D,E,A,2)); picture p = new picture; draw(p,Circle(N,r),dashed+linewidth(0.5)); clip(p,A--D--D+(0,20)--A+(0,20)--cycle); add(p); label(\"\\(A\\)\",A,SW); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NE); label(\"\\(D\\)\",D,SE); label(\"\\(E\\)\",E,NE); label(\"\\(M\\)\",M[0],SW); label(\"\\(N\\)\",N,S); label(\"\\(1004\\)\",(N+D)/2,S); label(\"\\(500\\)\",(M[0]+C)/2,S); [/asy] As $\\angle AED = 90^{\\circ}$, note that the midpoint of $\\overline{AD}$, $N$, is the center of the circumcircle of $\\triangle AED$. We can do the same with the circumcircle about $\\triangle BEC$ and $M$ (or we could apply the homothety to find $ME$ in terms of $NE$). It follows that \\[NE = ND = \\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500.\\] Thus $MN = NE - ME = 504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"\\(A\\)\",A,SW); label(\"\\(B\\)\",B,NW); label(\"\\(C\\)\",C,NE); label(\"\\(D\\)\",D,NE); label(\"\\(F\\)\",F,S); label(\"\\(G\\)\",G,SW); label(\"\\(M\\)\",M[0],SW); label(\"\\(N\\)\",N,S); label(\"\\(H\\)\",H,S); label(\"\\(x\\)\",(N+H)/2+(0,1),S); label(\"\\(h\\)\",(B+F)/2,W); label(\"\\(h\\)\",(C+G)/2,W); label(\"\\(1000\\)\",(B+C)/2,NE); label(\"\\(504-x\\)\",(G+D)/2,S); label(\"\\(504+x\\)\",(A+F)/2,S); label(\"\\(h\\)\",(M[0]+H)/2,(1,0)); [/asy] Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so \\[\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.\\] By the Pythagorean Theorem on $\\triangle MHN$, \\[MN^{2} = x^2 + h^2 = 504^2,\\] so $MN = 504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since $\\angle{A} = 37$ and $\\angle{D} = 53$, you can solve the equation [by cross-multiplication]: \\begin{align*}\\tan{37}\\times (1008-x) &= \\tan{53} \\times x\\\\ \\frac{(1008-x)}{x} &= \\frac{\\tan{53}}{\\tan{37}} = \\frac{\\sin{53}}{\\cos{53}} \\times\\frac{\\cos{37}}{\\sin{37}}\\end{align*} However, we know that $\\cos{90-x} = \\sin{x}$ and $\\sin{90-x} = \\cos{x}$ are co-functions. Applying this, \\begin{align*}\\frac{(1008-x)}{x} &= \\frac{\\sin^2{53}}{\\cos^2{53}} \\\\ x\\sin^2{53} &= 1008\\cos^2{53} - x\\cos^2{53}\\\\ x(\\sin^2{53} + \\cos^2{53}) &= 1008\\cos^2{53}\\\\ x = 1008\\cos^2{53} &\\Longrightarrow 1008-x = 1008\\sin^2{53} \\end{align*} Now, if we can find $1004 - (EA + 500)$, and the height of the trapezoid, we can create a right triangle and use the Pythagorean Theorem to find $MN$. The leg of the right triangle along the horizontal is: \\[1004 - 1008\\sin^2{53} - 500 = 504 - 1008\\sin^2{53}.\\] Now to find the other leg of the right triangle (also the height of the trapezoid), we can simplify the following expression: \\begin{align*}\\tan{37} \\times 1008 \\sin^2{53} = \\tan{37} \\times 1008 \\cos^2{37} = 1008\\cos{37}\\sin{37} = 504\\sin74\\end{align*} Now we used Pythagorean Theorem and get that $MN$ is equal to: \\begin{align*}&\\sqrt{(1008\\sin^2{53} + 500 -1004)^2 + (504\\sin{74})^2} = 504\\sqrt{1-2\\sin^2{53} + \\sin^2{74}} \\end{align*} However, $1-2\\sin^2{53} = \\cos^2{106}$ and $\\sin^2{74} = \\sin^2{106}$ so now we end up with: \\[504\\sqrt{\\cos^2{106} + \\sin^2{106}} =504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"Plot the trapezoid such that $B=\\left(1000\\cos 37^\\circ, 0\\right)$, $C=\\left(0, 1000\\sin 37^\\circ\\right)$, $A=\\left(2008\\cos 37^\\circ, 0\\right)$, and $D=\\left(0, 2008\\sin 37^\\circ\\right)$. The midpoints of the requested sides are $\\left(500\\cos 37^\\circ, 500\\sin 37^\\circ\\right)$ and $\\left(1004\\cos 37^\\circ, 1004\\sin 37^\\circ\\right)$. To find the distance from $M$ to $N$, we simply apply the distance formula and the Pythagorean identity $\\sin^2 x + \\cos^2 x = 1$ to get $MN=504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"Similar to solution 1; Notice that it forms a right triangle. Remembering that the median to the hypotenuse is simply half the length of the hypotenuse, we quickly see that the length is $\\frac{2008}{2}-\\frac{1000}{2}=504$. Solution 6 Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $504. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.)",
"Let the height be h. Note that if $\\overline{NH} = x$ then if we draw perpendiculars like in solution 1, $\\overline{FN} = 500 - x, \\overline{AF} = 504 + x, \\overline{HG} = 500, \\overline{GD} = 504 - x.$ Note that we wish to find $\\overline{MN} = \\sqrt{x^2 + h^2}.$ Let's find $\\tan(53)$ in two ways. Finding $\\tan(53)$ from $\\triangle BAF$ yields $\\tan(53) = \\frac{504+x}{h}.$ Finding it from $\\triangle CDG$ yields $\\frac{h}{504-x}.$ Setting these equal yields \\[\\frac{504+x}{h}=\\frac{h}{504-x} \\rightarrow h^2 = 504^2-x^2 \\rightarrow \\sqrt{x^2+h^2} = \\sqrt{504^2} = 504\\]",
"Rotate trapezoid $MNCD$ 180 degrees around point $N$ so that $AN$ coincides with $ND$. Let the image of trapezoid $MNCD$ be $ANM'C'$. Since angles are preserved during rotations, $\\angle BAC' = 37^{\\circ} + 53 ^{\\circ} =90 ^{\\circ}$. Since $BM=CM=C'M'$ and $BM || C'M'$, $BMM'C'$ is a parallelogram. Thus, $MM'=BC'$. Let the point where $BC'$ intersects $AD$ be $E$. Since $BMNE$ is a parallelogram, $AE=AN-BM=1004-500-504$. Since $BE=EC$ and $\\angle BAC= 90^{\\circ}$, $AE$ is a median to the hypotenuse of $BAC'$. Therefore, $BC'=2 AE= 1008$, and $BE=MN=504",
"Draw line $ME$ from point $M$ parallel to $CD$ that intersects $AD$. Draw $MF$ from point $M$ parallel to $BA$ that intersects $AD$. Note that triangle $EMF$ is a right triangle because $\\angle BAC' = 90^{\\circ}$. $EN$ has length $DN - CM = 1004 - 500 = 504$ because $CDEM$ is a parallelogram. Similarly, $FN$ has length $AN - BM = 504$ so N is the midpoint of the hypotenuse of right triangle $EMF$. The midpoint of the hypotenuse in a right triangle is equidistant from all three vertices, so $MN = 504.",
"Draw line $ME$ from point $M$ parallel to $CD$ that intersects $AD$. Draw $MF$ from point $M$ parallel to $BA$ that intersects $AD$. Note that triangle $EMF$ is a right triangle because $\\angle BAC' = 90^{\\circ}$. $EN$ has length $DN - CM = 1004 - 500 = 504$ because $CDEM$ is a parallelogram. Similarly, $FN$ has length $AN - BM = 504$ so N is the midpoint of the hypotenuse of right triangle $EMF$. The midpoint of the hypotenuse in a right triangle is equidistant from all three vertices, so $MN = 504."
] |
2008-II-6
| 2,008
| 6
|
The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$ .
| 561
|
II
|
[
"Rearranging the definitions, we have \\[\\frac{a_n}{a_{n-1}} = \\frac{a_{n-1}}{a_{n-2}} + 1,\\quad \\frac{b_n}{b_{n-1}} = \\frac{b_{n-1}}{b_{n-2}} + 1\\] from which it follows that $\\frac{a_n}{a_{n-1}} = 1+ \\frac{a_{n-1}}{a_{n-2}} = \\cdots = (n-1) + \\frac{a_{1}}{a_0} = n$ and $\\frac{b_n}{b_{n-1}} = (n-1) + \\frac{b_{1}}{b_0} = n+2$. These recursions, $a_{n} = na_{n-1}$ and $b_{n} = (n+2)b_{n-1}$, respectively, correspond to the explicit functions $a_n = n!$ and $b_n = \\frac{(n+2)!}{2}$ (after applying our initial conditions). It follows that $\\frac{b_{32}}{a_{32}} = \\frac{\\frac{34!}{2}}{32!} = \\frac{34 \\cdot 33}{2} = 561 corresponds to the triangular numbers."
] |
2008-II-7
| 2,008
| 7
|
Let $r$ , $s$ , and $t$ be the three roots of the equation \[8x^3 + 1001x + 2008 = 0.\] Find $(r + s)^3 + (s + t)^3 + (t + r)^3$ .
| 753
|
II
|
[
"By Vieta's formulas, we have $r + s + t = 0$ so $t = -r - s.$ Substituting this into our problem statement, our desired quantity is \\[(r + s)^3 - r^3 - s^3 = 3r^2s + 3rs^2 = 3rs(r + s).\\] Also by Vieta's formulas we have \\[rst = -rs(r + s) = -\\dfrac{2008}{8} = -251\\] so negating both sides and multiplying through by 3 gives our answer of $753",
"By Vieta's formulas, we have $r+s+t = 0$, and so the desired answer is $(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)$. Additionally, using the factorization \\[r^3 + s^3 + t^3 - 3rst = (r+s+t)(r^2 + s^2 + t^2 - rs - st - tr) = 0\\] we have that $r^3 + s^3 + t^3 = 3rst$. By Vieta's again, $rst = \\frac{-2008}8 = -251 \\Longrightarrow -(r^3 + s^3 + t^3) = -3rst = 753",
"Vieta's formulas gives $r + s + t = 0$. Since $r$ is a root of the polynomial, $8r^3 + 1001r + 2008 = 0\\Longleftrightarrow - 8r^3 = 1001r + 2008$, and the same can be done with $s,\\ t$. Therefore, we have \\begin{align*}8\\{(r + s)^3 + (s + t)^3 + (t + r)^3\\} &= - 8(r^3 + s^3 + t^3)\\\\ &= 1001(r + s + t) + 2008\\cdot 3 = 3\\cdot 2008\\end{align*}yielding the answer $753. Also, Newton's Sums yields an answer through the application. http://www.artofproblemsolving.com/Wiki/index.php/Newton's_Sums",
"Expanding, you get: \\[r^3 + 3r^2s + 3s^2r +s^3 +\\] \\[s^3 + 3s^2t + 3t^2s +t^3 +\\] \\[r^3 + 3r^2t + 3t^2r +t^3\\] \\[= 2r^3 + 2s^3 + 2t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r\\] This looks similar to $(r+s+t)^3 = r^3 + s^3 + t^3 + 3r^2s + 3s^2r + 3s^2t + 3t^2s + 3r^2t + 3t^2r + 6rst$ Substituting: \\[(r+s+t)^3 - 6rst + r^3+s^3+t^3 = (r + s)^3 + (s + t)^3 + (t + r)^3\\] Since $r+s+t = 0$, \\[(r+s)^3 + (s+t)^3 + (t+r)^3 = (0-t)^3 + (0-r)^3 + (0-s)^3 = -(r^3 + s^3 + t^3)\\] Substituting, we get \\[(r+s+t)^3 - 6rst + r^3+s^3+t^3 = -(r^3 + s^3 + t^3)\\] or, \\[0^3 - 6rst + r^3+s^3+t^3 = -(r^3 + s^3 + t^3) \\implies 2(r^3 + s^3 + t^3) = 6rst\\] We are trying to find $-(r^3 + s^3 + t^3)$. Substituting: \\[-(r^3 + s^3 + t^3) = -3srt = \\frac{-2008*3}{8} = 753.\\]",
"Write $(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)$ and let $f(x)=8x^3+1001x+2008$. Then \\[f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.\\] Solving for $r^3+s^3+t^3$ and negating the result yields the answer $753",
"Here by Vieta's formulas: $r+s+t = 0$ --(1) $rst = \\frac{-2008}{8} = -251$ --(2) By the factorisation formula: Let $a = r+s$, $b = s+t$, $c = t+r$, $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$ (By (1)) So \\[a^3+b^3+c^3 = 3abc = 3(r+s)(s+t)(t+r) = 3(-t)(-r)(-s) = 3[-(-251)] = 753.\\]",
"Let's construct a polynomial with the roots $(r+s), (s+t),$ and $(t+r)$. sum of the roots: $=2(r+s+t)=2\\cdot0=0$ pairwise product of the roots: $(r+s)(s+t)+(s+t)(t+r)+(t+r)(r+s)=r^2+s^2+t^2+3(rs+st+tr)$ $=(r+s+t)^2+rs+st+tr=0+\\frac{1001}{8}$ product of the roots: $(r+s)(s+t)(t+r)=r^2t+r^2s+s^2r+s^2t+t^2r+t^2s+3rst$ $=(rs+st+tr)(r+s+t)-3rst+2rst=-rst=-\\frac{2008}{8}$ thus, the polynomial we get is $x^3+\\frac{1001}{8}x+-\\frac{2008}{8}=0$ as $(r+s), (s+t),$ and $(t+r)$ are roots of this polynomial, we know that (using power reduction) $(r+s)^3+\\frac{1001}{8}(r+s)-\\frac{2008}{8}=0$ $(s+t)^3+\\frac{1001}{8}(s+t)-\\frac{2008}{8}=0$ $(t+r)^3+\\frac{1001}{8}(t+r)-\\frac{2008}{8}=0$ adding all of the equations up, we see that $(r+s)^3+(s+t)^3+(t+r)^3=3\\cdot\\frac{2008}{8}-\\frac{1001}{8}(2r+2s+2t)=251(3)+0=753",
"We want to find what is $-(r^3+s^3+t^3)$ which reminds us of Newton sum. So we can see that $8S_3+0\\cdot S_2+1001\\cdot S_1+3\\cdot 2008=0$ Notice that $S_1=0$ so it is just $S_3=-\\frac{2008\\cdot 3}{8}=-753$, the desired answer is $753 ~bluesoul",
"This solution uses Vietas, as with everyone else's solution. Expanding the expression we get \\[(r+s)^3+(s+t)^3+(t+r)^3 = r^3+3r^2s+3rs^2+\\dots +3s^2t+3ts^2+t^3\\] Seeing the cubes, we try to find a $(r+s+t)^3$ and upon doing so, we get \\[(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-6rst+(r^3+s^3+t^3)\\] Recall that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Thus, we get \\[(r+s)^3+(s+t)^3+(t+r)^3=(r+s+t)^3-3rst+(r+s+t)(r^2+s^2+t^2-rs-st-tr)\\] Plugging in $(r+s+t)=0$ we get \\[(r+s)^3+(s+t)^3+(t+r)^3=0-3rst+0=-3\\cdot -251=753\\] ~firebolt360",
"$8x^3+1001x+2008=0$ We want to find $(r+s)^3+(s+t)^3+(t+r)^3.$ Let's call this result n. From vieta's formulas, we find that $r+s+t=-0/8=0$, $rs+st+tr=1001/8$, and $rst=-2008/8=-251.$ Expanding and rearranging gives us $n=(r+s)^3+(s+t)^3+(t+r)^3=r^3+3r^2s+3rs^2+s^3+s^3+3s^2t+3st^2+t^3+t^3+3t^2r+3tr^2+r^3=2r^3+2s^3+3r^2s+3rs^2+3s^2t+3st^2+3t^2r+3tr^2=3(r^3+s^3+t^3-(r^2s+rs^2+s^2t+st^2+t^2r+tr^2))-(r^3+s^3+t^3)=3((r+s+t)(r^2+s^2+t^2))-((r+s+t)^3-3(r^2s+rs^2+s^2t+st^2+t^2r+tr^2)-6rst)=3(r+s+t)((r+s+t)^2-2(rs+st+tr))-((r+s+t)^3-3(r^2s+rs^2+s^2t+st^2+t^2r+tr^2)-6rst)$ Let $k=r^2s+rs^2+s^2t+st^2+t^2r+tr^2$ $k=r^2s+rs^2+s^2t+st^2+t^2r+tr^2=(r+s+t)(r^2+s^2+t^2)-(r^3+s^3+t^3)=(r+s+t)((r+s+t)^2-2(rs+st+tr))-((r+s+t)^3-3k-6rst)=(0)(0^2-2(1001/8))-(0^3-3k-6(-251))=0-(0-3k+1506)=3k-1506$ Solving gives us $k=753$ $n=3(r+s+t)((r+s+t)^2-2(rs+st+tr))-((r+s+t)^3-3(r^2s+rs^2+s^2t+st^2+t^2r+tr^2)-6rst)=3(0)(0^2-2(1001/8))-(0^3-3k-6(-251))=0-(0-3(753)+1506)=753$ Therefore, the answer is $753.$ Also note that this is the only solution that still would have worked effectively if $r+s+t$ was nonzero."
] |
2008-II-8
| 2,008
| 8
|
Let $a = \pi/2008$ . Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer.
| 251
|
II
|
[
"Solution 1 By the product-to-sum identities, we have that $2\\cos a \\sin b = \\sin (a+b) - \\sin (a-b)$. Therefore, this reduces to a telescope series: \\begin{align*} \\sum_{k=1}^{n} 2\\cos(k^2a)\\sin(ka) &= \\sum_{k=1}^{n} [\\sin(k(k+1)a) - \\sin((k-1)ka)]\\\\ &= -\\sin(0) + \\sin(2a)- \\sin(2a) + \\sin(6a) - \\cdots - \\sin((n-1)na) + \\sin(n(n+1)a)\\\\ &= -\\sin(0) + \\sin(n(n+1)a) = \\sin(n(n+1)a) \\end{align*} Thus, we need $\\sin \\left(\\frac{n(n+1)\\pi}{2008}\\right)$ to be an integer; this can be only $\\{-1,0,1\\}$, which occur when $2 \\cdot \\frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \\cdot 251 | n(n+1) \\Longrightarrow 251 | n, n+1$. We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$, so 1004 doesn't divide $n(n+1) = 250 \\cdot 251$. Therefore, it is clear that $n = 251. It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.",
"By the product-to-sum identities, we have that $2\\cos a \\sin b = \\sin (a+b) - \\sin (a-b)$. Therefore, this reduces to a telescope series: \\begin{align*} \\sum_{k=1}^{n} 2\\cos(k^2a)\\sin(ka) &= \\sum_{k=1}^{n} [\\sin(k(k+1)a) - \\sin((k-1)ka)]\\\\ &= -\\sin(0) + \\sin(2a)- \\sin(2a) + \\sin(6a) - \\cdots - \\sin((n-1)na) + \\sin(n(n+1)a)\\\\ &= -\\sin(0) + \\sin(n(n+1)a) = \\sin(n(n+1)a) \\end{align*} Thus, we need $\\sin \\left(\\frac{n(n+1)\\pi}{2008}\\right)$ to be an integer; this can be only $\\{-1,0,1\\}$, which occur when $2 \\cdot \\frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \\cdot 251 | n(n+1) \\Longrightarrow 251 | n, n+1$. We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$, so 1004 doesn't divide $n(n+1) = 250 \\cdot 251$. Therefore, it is clear that $n = 251. It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.",
"We proceed with complex trigonometry. We know that for all $\\theta$, we have $\\cos \\theta = \\dfrac{1}{2} \\left( z + \\dfrac{1}{z} \\right)$ and $\\sin \\theta = \\dfrac{1}{2i} \\left( z - \\dfrac{1}{z} \\right)$ for some complex number $z$ on the unit circle. Similarly, we have $\\cos n \\theta = \\dfrac{1}{2} \\left( z^n + \\dfrac{1}{z^n} \\right)$ and $\\sin n \\theta = \\dfrac{1}{2i} \\left(z^n - \\dfrac{1}{z^n} \\right)$. Thus, we have $\\cos n^2 a \\sin n a = \\dfrac{1}{4i} \\left( z^{n^2} + \\dfrac{1}{z^{n^2}} \\right) \\left( z^{n} - \\dfrac{1}{z^n} \\right)$ $= \\dfrac{1}{4i} \\left( z^{n^2 + n} - \\dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \\dfrac{1}{z^{n^2 - n}} \\right)$ $= \\dfrac{1}{2} \\left( \\dfrac{1}{2i} \\left(z^{n^2 + n} - \\dfrac{1}{z^{n^2 + n}} \\right) - \\dfrac{1}{2i} \\left(z^{n^2 - n} - \\dfrac{1}{z^{n^2 - n}} \\right) \\right)$ $= \\dfrac{1}{2} \\left( \\sin ((n^2 + n)a) - \\sin ((n^2 - n)a) \\right)$ $= \\dfrac{1}{2} \\left( \\sin(((n+1)^2 - (n+1))a) - \\sin((n^2 - n)a) \\right)$ which clearly telescopes! Since the $2$ outside the brackets cancels with the $\\dfrac{1}{2}$ inside, we see that the sum up to $n$ terms is $\\sin ((2^2 - 2)a) - \\sin ((1^2 - 1)a) + \\sin ((3^3 - 3)a) - \\sin ((2^2 - 2)a) \\cdots + \\sin (((n+1)^2 - (n+1))a) - \\sin ((n^2 - n)a)$ $= \\sin (((n+1)^2 - (n+1))a) - \\sin(0)$ $= \\sin ((n^2 + n)a) - 0$ $= \\sin \\left( \\dfrac{n(n+1) \\pi}{2008} \\right)$. This expression takes on an integer value iff $\\dfrac{2n(n+1)}{2008} = \\dfrac{n(n+1)}{1004}$ is an integer; that is, $1004 \\mid n(n+1)$. Clearly, $1004 = 2^2 \\cdot 251$, implying that $251 \\mid n(n+1)$. Since we want the smallest possible value of $n$, we see that we must have ${n,n+1} = 251$. If $n+1 = 251 \\rightarrow n=250$, then we have $n(n+1) = 250(251)$, which is clearly not divisible by $1004$. However, if $n = 251$, then $1004 \\mid n(n+1)$, so our answer is $251. It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above."
] |
2008-II-9
| 2,008
| 9
|
A particle is located on the coordinate plane at $(5,0)$ . Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$ -direction. Given that the particle's position after $150$ moves is $(p,q)$ , find the greatest integer less than or equal to $|p| + |q|$ .
| 19
|
II
|
[
"Solution 1 Let $P(x, y)$ be the position of the particle on the $xy$-plane, $r$ be the length $OP$ where $O$ is the origin, and $\\theta$ be the inclination of OP to the x-axis. If $(x', y')$ is the position of the particle after a move from $P$, then we have two equations for $x'$ and $y'$: \\[x'=r\\cos(\\pi/4+\\theta)+10 = \\frac{\\sqrt{2}(x - y)}{2} + 10\\] \\[y' = r\\sin(\\pi/4+\\theta) = \\frac{\\sqrt{2}(x + y)}{2}.\\] Let $(x_n, y_n)$ be the position of the particle after the nth move, where $x_0 = 5$ and $y_0 = 0$. Then $x_{n+1} + y_{n+1} = \\sqrt{2}x_n+10$, $x_{n+1} - y_{n+1} = -\\sqrt{2}y_n+10$. This implies $x_{n+2} = -y_n + 5\\sqrt{2}+ 10$, $y_{n+2}=x_n + 5\\sqrt{2}$. Substituting $x_0 = 5$ and $y_0 = 0$, we have $x_8 = 5$ and $y_8 = 0$ again for the first time. Thus, $p = x_{150} = x_6 = -5\\sqrt{2}$ and $q = y_{150} = y_6 = 5 + 5\\sqrt{2}$. Hence, the final answer is $5\\sqrt {2} + 5(\\sqrt {2} + 1) \\approx 19.1 \\Longrightarrow 019.",
"Let $P(x, y)$ be the position of the particle on the $xy$-plane, $r$ be the length $OP$ where $O$ is the origin, and $\\theta$ be the inclination of OP to the x-axis. If $(x', y')$ is the position of the particle after a move from $P$, then we have two equations for $x'$ and $y'$: \\[x'=r\\cos(\\pi/4+\\theta)+10 = \\frac{\\sqrt{2}(x - y)}{2} + 10\\] \\[y' = r\\sin(\\pi/4+\\theta) = \\frac{\\sqrt{2}(x + y)}{2}.\\] Let $(x_n, y_n)$ be the position of the particle after the nth move, where $x_0 = 5$ and $y_0 = 0$. Then $x_{n+1} + y_{n+1} = \\sqrt{2}x_n+10$, $x_{n+1} - y_{n+1} = -\\sqrt{2}y_n+10$. This implies $x_{n+2} = -y_n + 5\\sqrt{2}+ 10$, $y_{n+2}=x_n + 5\\sqrt{2}$. Substituting $x_0 = 5$ and $y_0 = 0$, we have $x_8 = 5$ and $y_8 = 0$ again for the first time. Thus, $p = x_{150} = x_6 = -5\\sqrt{2}$ and $q = y_{150} = y_6 = 5 + 5\\sqrt{2}$. Hence, the final answer is $5\\sqrt {2} + 5(\\sqrt {2} + 1) \\approx 19.1 \\Longrightarrow 019.",
"Let the particle's position be represented by a complex number. Recall that multiplying a number by cis$\\left( \\theta \\right)$ rotates the object in the complex plane by $\\theta$ counterclockwise. In this case, we use $a = cis(\\frac{\\pi}{4})$. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to $a_{150} = (((5a + 10)a + 10)a + 10 \\ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \\ldots + 10$ where a is cis$\\left( \\theta \\right)$. By De-Moivre's theorem, $\\left(cis( \\theta \\right)^n )$=cis$\\left(n \\theta \\right)$. Therefore, $10(a^{150} + \\ldots + 1)= 10(1 + a + \\ldots + a^6) = - 10(a^7) = - 10(\\frac{ \\sqrt {2} }{2} - \\frac{i\\sqrt {2}} {2})$ Furthermore, $5a^{150} = - 5i$. Thus, the final answer is $5\\sqrt {2} + 5(\\sqrt {2} + 1) \\approx 19.1 \\Longrightarrow 019.",
"As before, consider $z$ as a complex number. Consider the transformation $z \\to (z-\\omega)e^{i\\theta} + \\omega$. This is a clockwise rotation of $z$ by $\\theta$ radians about the points $\\omega$. Let $f(z)$ denote one move of $z$. Then Therefore, $z$ rotates along a circle with center $\\omega = 5+(5+5\\sqrt2)i$. Since $8 \\cdot \\frac{\\pi}{4} = 2\\pi$, $f^9(z) = f(z) \\implies f^{150}(z) = f^6(z) \\implies p+q = 019.",
"Let $T:\\begin{pmatrix}x\\\\y\\end{pmatrix}\\rightarrow R(\\frac{\\pi}{4})\\begin{pmatrix}x\\\\y\\end{pmatrix}+\\begin{pmatrix}10\\\\0\\end{pmatrix}$. We assume that the rotation matrix $R(\\frac{\\pi}{4}) = R$ here. Then we have $T^{150}\\begin{pmatrix}5\\\\0\\end{pmatrix}=R(R(...R(R\\begin{pmatrix}5\\\\0\\end{pmatrix}+\\begin{pmatrix}10\\\\0\\end{pmatrix})+\\begin{pmatrix}10\\\\0\\end{pmatrix}...)+\\begin{pmatrix}10\\\\0\\end{pmatrix})+\\begin{pmatrix}10\\\\0\\end{pmatrix}$ This simplifies to $R^{150}\\begin{pmatrix}5\\\\0\\end{pmatrix}+(I+R^2+R^3+...+R^{149})\\begin{pmatrix}10\\\\0\\end{pmatrix}$ Since $R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O$, so we have $R^6\\begin{pmatrix}5\\\\0\\end{pmatrix}+(-R^6-R^7)\\begin{pmatrix}10\\\\0\\end{pmatrix}$, giving $p=-5\\sqrt{2}, q=5\\sqrt{2}+5$. The answer is yet $\\lfloor10\\sqrt{2}+5\\rfloor=019."
] |
2008-II-11
| 2,008
| 11
|
In triangle $ABC$ , $AB = AC = 100$ , and $BC = 56$ . Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$ . Circle $Q$ is externally tangent to circle $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$ . No point of circle $Q$ lies outside of $\bigtriangleup\overline{ABC}$ . The radius of circle $Q$ can be expressed in the form $m - n\sqrt{k}$ ,where $m$ , $n$ , and $k$ are positive integers and $k$ is the product of distinct primes. Find $m +nk$ .
| 254
|
II
|
[
"[asy] size(200); pathpen=black;pointpen=black;pen f=fontsize(9); real r=44-6*35^.5; pair A=(0,96),B=(-28,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); path PC=CR(P,16),QC=CR(Q,r); D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); D(PC); D(QC); MP(\"A\",A,N,f);MP(\"B\",B,f);MP(\"C\",C,f);MP(\"X\",X,f);MP(\"Y\",Y,f);D(MP(\"P\",P,NW,f));D(MP(\"Q\",Q,NW,f)); [/asy] Let $X$ and $Y$ be the feet of the perpendiculars from $P$ and $Q$ to $BC$, respectively. Let the radius of $\\odot Q$ be $r$. We know that $PQ = r + 16$. From $Q$ draw segment $\\overline{QM} \\parallel \\overline{BC}$ such that $M$ is on $PX$. Clearly, $QM = XY$ and $PM = 16-r$. Also, we know $QPM$ is a right triangle. To find $XC$, consider the right triangle $PCX$. Since $\\odot P$ is tangent to $\\overline{AC},\\overline{BC}$, then $PC$ bisects $\\angle ACB$. Let $\\angle ACB = 2\\theta$; then $\\angle PCX = \\angle QBX = \\theta$. Dropping the altitude from $A$ to $BC$, we recognize the $7 - 24 - 25$ right triangle, except scaled by $4$. So we get that $\\tan(2\\theta) = 24/7$. From the half-angle identity, we find that $\\tan(\\theta) = \\frac {3}{4}$. Therefore, $XC = \\frac {64}{3}$. By similar reasoning in triangle $QBY$, we see that $BY = \\frac {4r}{3}$. We conclude that $XY = 56 - \\frac {4r + 64}{3} = \\frac {104 - 4r}{3}$. So our right triangle $QPM$ has sides $r + 16$, $r - 16$, and $\\frac {104 - 4r}{3}$. By the Pythagorean Theorem, simplification, and the quadratic formula, we can get $r = 44 - 6\\sqrt {35}$, for a final answer of $254.",
"First let $\\theta = \\angle{PCB}$ ; now connect the points as shown in the first solution's diagram ; realise that $\\tan\\theta = r/x = 16/y = r + 16/(x+y)$ where $x = BY$ and $y = CX$ (the 2 tangents) ; then we have that $QM = 64r = 56 - x - y \\implies (x+y) = 56 - 64r$ hence $r/x = 16+r/(56-64r)$ ; now drop altitude $AY$ to solve for $\\tan{2\\theta}$ ; now since we know $\\tan{2\\theta}$ we know $\\tan \\theta = r/x$ in terms of $r$ hence solve the resulting equation in $r$.",
"Refer to the above diagram. Let the larger circle have center $O_1$, the smaller have center $O_2$, and the incenter be $I$. We can easily calculate that the area of $\\triangle ABC = 2688$, and $s = 128$ and $R = 21$, where $R$ is the inradius. Now, Line $\\overline{AI}$ is the perpendicular bisector of $\\overline{BC}$, as $\\triangle ABC$ is isosceles. Letting the point of intersection be $X$, we get that $BX = 28$ and $IX = 21$, and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\\overline{AB}$ and $\\overline{BC}$. By Pythagoras, $BI = 35$, and we notice that $\\triangle BIX$ is a 3-4-5 right triangle. Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\\overline{BC}$, we find that $BY = \\frac{4r}{3}$. Similarly, we find that the distance between the projection from $O_1$ onto $\\overline{BC}$, $W$, and $C$, is $\\frac{64}{3}$. From there, we let the projection of $O_2$ onto $\\overline{O_1W}$ be $Z$, and we have $O_2Z = 28 - \\frac{4r}{3} + \\frac{20}{3}$, $O_1Z = 16 - r$, and $O_1O_2 = 16 + r$. We finish with Pythagoras on $\\triangle O_1O_2Z$, whence we get the desired answer of $254. - Spacesam",
"Refer to the above diagram. Let the larger circle have center $O_1$, the smaller have center $O_2$, and the incenter be $I$. We can easily calculate that the area of $\\triangle ABC = 2688$, and $s = 128$ and $R = 21$, where $R$ is the inradius. Now, Line $\\overline{AI}$ is the perpendicular bisector of $\\overline{BC}$, as $\\triangle ABC$ is isosceles. Letting the point of intersection be $X$, we get that $BX = 28$ and $IX = 21$, and $B, O_2, I$ are collinear as $O_2$ is equidistant from $\\overline{AB}$ and $\\overline{BC}$. By Pythagoras, $BI = 35$, and we notice that $\\triangle BIX$ is a 3-4-5 right triangle. Letting $r$ be the desired radius and letting $Y$ be the projection of $O_2$ onto $\\overline{BC}$, we find that $BY = \\frac{4r}{3}$. Similarly, we find that the distance between the projection from $O_1$ onto $\\overline{BC}$, $W$, and $C$, is $\\frac{64}{3}$. From there, we let the projection of $O_2$ onto $\\overline{O_1W}$ be $Z$, and we have $O_2Z = 28 - \\frac{4r}{3} + \\frac{20}{3}$, $O_1Z = 16 - r$, and $O_1O_2 = 16 + r$. We finish with Pythagoras on $\\triangle O_1O_2Z$, whence we get the desired answer of $254. - Spacesam",
"Let the incenter be O and the altitude from A to $\\overline{BC}$ be T. Note that by AA, $\\triangle BQY \\sim \\triangle OBT$ and $\\triangle PXC \\sim \\triangle OTC.$ Note that from $A = rs$, the inradius of the big triangle is $21$ Using ravi substitution(or noticing that $\\overline{AT}$ is an altitude), we then have that $TB = TC = 28.$ From similar triangles, we can now find $\\overline{BY}.$ We have \\[\\frac{\\overline{BY}}{QY} = \\frac{7}{{21}} \\rightarrow \\overline{BY} = \\frac{4}{3} r\\] Now, note that as in solution 1, drawing the perpendicular from Q to $\\overline{PX}$(call it Z) yields $\\overline{PZ} = 16 - r, \\overline{ZX} = r.$ Then, from this, \\[\\overline{QZ} = \\overline{YX} = \\sqrt{(\\overline{PQ})^2 - (\\overline{PZ})^2} = \\sqrt{(16+r)^2-(16-r)^2} = 8\\sqrt{r}\\] Using similar similarity as was done to find $\\overline{BY}$ we have $\\frac{\\overline{PX}}{\\overline{XC}} = \\frac{\\overline{OT}}{\\overline{TC}} \\rightarrow \\frac{16}{\\overline{XC}} = \\frac{21}{28} \\rightarrow \\overline{XC} = \\frac{64}{3}$. Now adding all these up and equating them to $\\overline{BC}$ yields \\[\\frac{4}{3}r + 8\\sqrt{r}+ \\frac{16}{3} = 56 \\rightarrow r = 44 - 6\\sqrt{35} \\rightarrow 44 + 6\\cdot 35 = 254\\]"
] |
2008-II-12
| 2,008
| 12
|
There are two distinguishable flagpoles, and there are $19$ flags, of which $10$ are identical blue flags, and $9$ are identical green flags. Let $N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when $N$ is divided by $1000$ .
| 310
|
II
|
[
"Solution 1 The well known problem of ordering $x$ elements of a string of $y$ elements such that none of the $x$ elements are next to each other has ${y-x+1\\choose x}$ solutions. (1) We generalize for $a$ blues and $b$ greens. Consider a string of $a+b$ elements such that we want to choose the greens such that none of them are next to each other. We would also like to choose a place where we can divide this string into two strings such that the left one represents the first pole, and the right one represents the second pole, in order up the pole according to position on the string. However, this method does not account for the fact that the first pole could end with a green, and the second pole could start with a green, since the original string assumed that no greens could be consecutive. We solve this problem by introducing an extra blue, such that we choose our divider by choosing one of these $a+1$ blues, and not including that one as a flag on either pole. From (1), we now have ${a+2\\choose b}$ ways to order the string such that no greens are next to each other, and $a+1$ ways to choose the extra blue that will divide the string into the two poles: or $(a+1){a+2\\choose b}$ orderings in total. However, we have overcounted the solutions where either pole has no flags, we have to count these separately. This is the same as choosing our extra blue as one of the two ends, and ordering the other $a$ blues and $b$ greens such that no greens are next to each other: for a total of $2{a+1\\choose b}$ such orderings. Thus, we have $(a+1){a+2\\choose b}-2{a+1\\choose b}$ orderings that satisfy the conditions in the problem: plugging in $a=10$ and $b=9$, we get $2310 \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"The well known problem of ordering $x$ elements of a string of $y$ elements such that none of the $x$ elements are next to each other has ${y-x+1\\choose x}$ solutions. (1) We generalize for $a$ blues and $b$ greens. Consider a string of $a+b$ elements such that we want to choose the greens such that none of them are next to each other. We would also like to choose a place where we can divide this string into two strings such that the left one represents the first pole, and the right one represents the second pole, in order up the pole according to position on the string. However, this method does not account for the fact that the first pole could end with a green, and the second pole could start with a green, since the original string assumed that no greens could be consecutive. We solve this problem by introducing an extra blue, such that we choose our divider by choosing one of these $a+1$ blues, and not including that one as a flag on either pole. From (1), we now have ${a+2\\choose b}$ ways to order the string such that no greens are next to each other, and $a+1$ ways to choose the extra blue that will divide the string into the two poles: or $(a+1){a+2\\choose b}$ orderings in total. However, we have overcounted the solutions where either pole has no flags, we have to count these separately. This is the same as choosing our extra blue as one of the two ends, and ordering the other $a$ blues and $b$ greens such that no greens are next to each other: for a total of $2{a+1\\choose b}$ such orderings. Thus, we have $(a+1){a+2\\choose b}-2{a+1\\choose b}$ orderings that satisfy the conditions in the problem: plugging in $a=10$ and $b=9$, we get $2310 \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Split the problem into two cases: Case 1 - Both poles have blue flags: There are 9 ways to place the 10 blue flags on the poles. In each of these configurations, there are 12 spots that a green flag could go. (either in between two blues or at the tops or bottoms of the poles) Then, since there are 9 green flags, all of which must be used, there are ${12\\choose9}$ possiblities for each of the 9 ways to place the blue flags. Total: ${12\\choose9}*9$ possibilities. Case 2 - One pole has no blue flags: Since each pole is non empty, the pole without a blue flag must have one green flag. The other pole has 10 blue flags and, by the argument used in case 1, there are ${11\\choose8}$ possibilities, and since the poles are distinguishable, there are a total of $2*{11\\choose8}$ possiblities for this case. Finally, we have $9*{12\\choose9}+2*{11\\choose8}=2310 \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Call the two flagpoles Flagpole A and Flagpole B. Case 1: Flag distribution: 1|18 B|variable: $\\dbinom{10}{9}=10$ ways G|variable: $\\dbinom{11}{8}=165$ ways Case 2: Flag distribution: 2|17 BB|variable: $\\dbinom{9}{9}=1$ way BG|variable (two ways to arrange the first sequence): $\\dbinom{10}{8} \\times 2 = 90$ ways GG|variable: can't happen Case 3: Flag distribution: 3|16 BBB|variable: can't happen BBG|variable (3 legal ways to arrange): $\\dbinom{9}{8} \\times 3 = 27$ ways GBG|variable (1 legal way to arrange): $\\dbinom{10}{7} = 120$ ways GGG|variable: clearly can't happen Case 4: Flag distribution: 4|15 BBBB|variable: can't happen BBBG|variable (4 legal ways to arrange): $\\dbinom{8}{8} \\times 4 = 4$ ways GBBG|variable (3 legal ways to arrange): $\\dbinom{9}{7} \\times 3 = 108$ ways GGBG|variable: can't happen GGGG|variable: can't happen Case 5: Flag distribution: 5|14 BBBBB|variable: can't happen BBBBG|variable: can't happen BBGBG|variable (6 legal ways to arrange): $\\dbinom{8}{7} \\times 6 = 48$ ways GBGBG|variable (1 legal way to arrange): $\\dbinom{9}{6} = 84$ ways GGGGB|variable: can't happen GGGGG|variable: can't happen Case 6: Flag distribution: 6|13 BBBBBB|variable: can't happen BBBBBG|variable: can't happen BBBBGG|variable (10 legal ways to arrange): $\\dbinom{7}{7} \\times 10 = 10$ ways BBBGGG|variable (4 legal ways to arrange): $\\dbinom{8}{6} \\times 4 = 112$ ways BBGGGG|variable: can't happen BGGGGG|variable: can't happen GGGGGG|variable: can't happen Case 7: Flag distribution: 7|12 BBBBBBB|variable: can't happen BBBBBBG|variable: can't happen BBBBBGG|variable: can't happen BBBBGGG|variable (10 legal ways to arrange): $\\dbinom{7}{6} \\times 10 = 70$ ways BBBGGGG|variable (1 legal way to arrange): $\\dbinom{8}{5} = 56$ ways rest can't happen Case 8: Flag distribution: 8|11 BBBBBBBB|variable: can't happen BBBBBBBG|variable: can't happen BBBBBBGG|variable: can't happen BBBBBGGG|variable (20 legal ways to arrange): $\\dbinom{6}{6} \\times 20 = 20$ ways BBBBGGGG|variable: (5 legal ways to arrange): $\\dbinom{7}{5} \\times 5 = 105$ ways others can't happen Case 9: Flag distribution: 9|10 BBBBBGGGG|variable (15 legal ways to arrange): $\\dbinom{6}{5} \\times 15 = 90$ ways BBBBGGGGG|variable (1 legal way to arrange): $\\dbinom{7}{4} = 35$ ways others can't happen So our total number of ways is $2$ times $10+165+1+90+27+120+4+108+48+84+10+112+70+56+20+105+90+35$ (since the two flagpoles are distinguishable) which is $2310$ ways. We have to find the last 3 digits, so our final answer is $310$. Solution by fidgetboss_4000 Note: Do not attempt unless you are good at bashing. Solution 4 (Vandermonde's) Let the number of blue flags on the first flagpole be $b$ and define $g$ similarly. Now note the bound $g \\le b+1.$ Now we cannot have any two consecutive greens. This condition is the same as \"there are $b+1$ spaces between each blue flag. How many ways can we put the $g$ green flags in the $b+1$ spaces?\" Therefore given $b$ blue flags on the first flagpole and $g$ green flags on the first flagpole we have a total of $\\binom{b+1}{g}.$ Similarly there are $\\binom{11-b}{9-g}$ ways for the second flagpole. Summing over all possible possibilities we see the sum is $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Swapping the sum we have $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g} = \\sum_{b=0}^{10} \\sum_{g = 0}^{b+1}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Then applying Vandermonde's yields $\\sum_{b=0}^{10}\\binom{12}{9}$ so the sum is $11 \\cdot\\binom{12}{9}.$ However, this answer overcounts. We cannot have no flags on a pole, so we subtract $2 \\cdot\\binom{11}{9}$ since we can have empty flags for the first or second flag. Therefore the answer is $11 \\cdot\\binom{12}{9} - 2 \\cdot\\binom{11}{9} \\pmod{1000} \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Call the two flagpoles Flagpole A and Flagpole B. Case 1: Flag distribution: 1|18 B|variable: $\\dbinom{10}{9}=10$ ways G|variable: $\\dbinom{11}{8}=165$ ways Case 2: Flag distribution: 2|17 BB|variable: $\\dbinom{9}{9}=1$ way BG|variable (two ways to arrange the first sequence): $\\dbinom{10}{8} \\times 2 = 90$ ways GG|variable: can't happen Case 3: Flag distribution: 3|16 BBB|variable: can't happen BBG|variable (3 legal ways to arrange): $\\dbinom{9}{8} \\times 3 = 27$ ways GBG|variable (1 legal way to arrange): $\\dbinom{10}{7} = 120$ ways GGG|variable: clearly can't happen Case 4: Flag distribution: 4|15 BBBB|variable: can't happen BBBG|variable (4 legal ways to arrange): $\\dbinom{8}{8} \\times 4 = 4$ ways GBBG|variable (3 legal ways to arrange): $\\dbinom{9}{7} \\times 3 = 108$ ways GGBG|variable: can't happen GGGG|variable: can't happen Case 5: Flag distribution: 5|14 BBBBB|variable: can't happen BBBBG|variable: can't happen BBGBG|variable (6 legal ways to arrange): $\\dbinom{8}{7} \\times 6 = 48$ ways GBGBG|variable (1 legal way to arrange): $\\dbinom{9}{6} = 84$ ways GGGGB|variable: can't happen GGGGG|variable: can't happen Case 6: Flag distribution: 6|13 BBBBBB|variable: can't happen BBBBBG|variable: can't happen BBBBGG|variable (10 legal ways to arrange): $\\dbinom{7}{7} \\times 10 = 10$ ways BBBGGG|variable (4 legal ways to arrange): $\\dbinom{8}{6} \\times 4 = 112$ ways BBGGGG|variable: can't happen BGGGGG|variable: can't happen GGGGGG|variable: can't happen Case 7: Flag distribution: 7|12 BBBBBBB|variable: can't happen BBBBBBG|variable: can't happen BBBBBGG|variable: can't happen BBBBGGG|variable (10 legal ways to arrange): $\\dbinom{7}{6} \\times 10 = 70$ ways BBBGGGG|variable (1 legal way to arrange): $\\dbinom{8}{5} = 56$ ways rest can't happen Case 8: Flag distribution: 8|11 BBBBBBBB|variable: can't happen BBBBBBBG|variable: can't happen BBBBBBGG|variable: can't happen BBBBBGGG|variable (20 legal ways to arrange): $\\dbinom{6}{6} \\times 20 = 20$ ways BBBBGGGG|variable: (5 legal ways to arrange): $\\dbinom{7}{5} \\times 5 = 105$ ways others can't happen Case 9: Flag distribution: 9|10 BBBBBGGGG|variable (15 legal ways to arrange): $\\dbinom{6}{5} \\times 15 = 90$ ways BBBBGGGGG|variable (1 legal way to arrange): $\\dbinom{7}{4} = 35$ ways others can't happen So our total number of ways is $2$ times $10+165+1+90+27+120+4+108+48+84+10+112+70+56+20+105+90+35$ (since the two flagpoles are distinguishable) which is $2310$ ways. We have to find the last 3 digits, so our final answer is $310$. Solution by fidgetboss_4000 Note: Do not attempt unless you are good at bashing. Solution 4 (Vandermonde's) Let the number of blue flags on the first flagpole be $b$ and define $g$ similarly. Now note the bound $g \\le b+1.$ Now we cannot have any two consecutive greens. This condition is the same as \"there are $b+1$ spaces between each blue flag. How many ways can we put the $g$ green flags in the $b+1$ spaces?\" Therefore given $b$ blue flags on the first flagpole and $g$ green flags on the first flagpole we have a total of $\\binom{b+1}{g}.$ Similarly there are $\\binom{11-b}{9-g}$ ways for the second flagpole. Summing over all possible possibilities we see the sum is $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Swapping the sum we have $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g} = \\sum_{b=0}^{10} \\sum_{g = 0}^{b+1}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Then applying Vandermonde's yields $\\sum_{b=0}^{10}\\binom{12}{9}$ so the sum is $11 \\cdot\\binom{12}{9}.$ However, this answer overcounts. We cannot have no flags on a pole, so we subtract $2 \\cdot\\binom{11}{9}$ since we can have empty flags for the first or second flag. Therefore the answer is $11 \\cdot\\binom{12}{9} - 2 \\cdot\\binom{11}{9} \\pmod{1000} \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Let the number of blue flags on the first flagpole be $b$ and define $g$ similarly. Now note the bound $g \\le b+1.$ Now we cannot have any two consecutive greens. This condition is the same as \"there are $b+1$ spaces between each blue flag. How many ways can we put the $g$ green flags in the $b+1$ spaces?\" Therefore given $b$ blue flags on the first flagpole and $g$ green flags on the first flagpole we have a total of $\\binom{b+1}{g}.$ Similarly there are $\\binom{11-b}{9-g}$ ways for the second flagpole. Summing over all possible possibilities we see the sum is $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Swapping the sum we have $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g} = \\sum_{b=0}^{10} \\sum_{g = 0}^{b+1}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Then applying Vandermonde's yields $\\sum_{b=0}^{10}\\binom{12}{9}$ so the sum is $11 \\cdot\\binom{12}{9}.$ However, this answer overcounts. We cannot have no flags on a pole, so we subtract $2 \\cdot\\binom{11}{9}$ since we can have empty flags for the first or second flag. Therefore the answer is $11 \\cdot\\binom{12}{9} - 2 \\cdot\\binom{11}{9} \\pmod{1000} \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Let the number of blue flags on the first flagpole be $b$ and define $g$ similarly. Now note the bound $g \\le b+1.$ Now we cannot have any two consecutive greens. This condition is the same as \"there are $b+1$ spaces between each blue flag. How many ways can we put the $g$ green flags in the $b+1$ spaces?\" Therefore given $b$ blue flags on the first flagpole and $g$ green flags on the first flagpole we have a total of $\\binom{b+1}{g}.$ Similarly there are $\\binom{11-b}{9-g}$ ways for the second flagpole. Summing over all possible possibilities we see the sum is $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Swapping the sum we have $\\sum_{g = 0}^{b+1} \\sum_{b=0}^{10}\\binom{b+1}{g}\\binom{11-b}{9-g} = \\sum_{b=0}^{10} \\sum_{g = 0}^{b+1}\\binom{b+1}{g}\\binom{11-b}{9-g}.$ Then applying Vandermonde's yields $\\sum_{b=0}^{10}\\binom{12}{9}$ so the sum is $11 \\cdot\\binom{12}{9}.$ However, this answer overcounts. We cannot have no flags on a pole, so we subtract $2 \\cdot\\binom{11}{9}$ since we can have empty flags for the first or second flag. Therefore the answer is $11 \\cdot\\binom{12}{9} - 2 \\cdot\\binom{11}{9} \\pmod{1000} \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"Consider this as arranging the flags on one big flagpole, then splitting the flagpole into two. Example: Start with big flagpole $BGBGBGGBBGBGBGBGBGB$, and then split it into $BGBGBG$ and $GBBGBGBGBGBGB$. We will split this problem into two cases: Case 1: The split is not two greens. Basically if arranging 10 blue flags and 9 green flags does not give any consecutive green flags, then we can split it into two flagpoles wherever. For example, \\[BGBBGBGBGBGBGBGBGBG\\] can become \\[BGB|BGBGBGBGBGBGBGBG\\] or \\[BGBBG|BGBGBGBGBGBGBG\\]. By stars and bars, there are $\\binom{11}{2}$ ways to arrange the flags, then $18$ ways to split the big flagpole, so total $\\binom{11}{2}\\cdot 18$ ways. Case 2: The split is two greens. Here the big flagpole has two consecutive green flags, so we are forced to split the two green flags. There are $\\binom{11}{3}\\cdot 8$ ways for this case, since we first place 8 “$G$”s among 10 blue flags, then choose one to become $GG$. (Look at the first example) Thus the total number of ways is \\[\\binom{11}{2}\\cdot 18 + \\binom{11}{3}\\cdot 8 = 2310 \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky",
"We split the 9 green flags over the two poles: $0$ and $9$, $1$ and $8$, $2$ and $7$, and so on, respectively. For the first case, $0$ and $9$, 8 of the 10 blue flags must be placed between the 9 green flags on the same pole, and the 9th blue flag must be placed on the other pole so that there is at least one flag on each pole. This leaves $\\textbf{11}$ choices for the last blue flag. Moving on, notice that now for every other case there will always be 3 blue flags left to place. Since we want to distribute 3 indistinguishable flags to 11 distinguishable spots, we use stars and bars. There are $\\binom{13}{10} = \\textbf{286}$ ways to distribute the flags for the other cases. Then the total number of ways is $2 * (11 + 4 * 286) = 2310 \\equiv 310, since the poles are distinguishable. ~ AXcatterwocky"
] |
2008-II-13
| 2,008
| 13
|
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$ . Then the area of $S$ has the form $a\pi + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ .
| 29
|
II
|
[
"If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution.",
"If a point $z = r\\text{cis}\\,\\theta$ is in $R$, then the point $\\frac{1}{z} = \\frac{1}{r} \\text{cis}\\, \\left(-\\theta\\right)$ is in $S$ (where cis denotes $\\text{cis}\\, \\theta = \\cos \\theta + i \\sin \\theta$). Since $R$ is symmetric every $60^{\\circ}$ about the origin, it suffices to consider the area of the result of the transformation when $-30 \\le \\theta \\le 30$, and then to multiply by $6$ to account for the entire area. We note that if the region $S_2 = \\left\\lbrace\\frac{1}{z}|z \\in R_2\\right\\rbrace$, where $R_2$ is the region (in green below) outside the circle of radius $1/\\sqrt{3}$ centered at the origin, then $S_2$ is simply the region inside a circle of radius $\\sqrt{3}$ centered at the origin. It now suffices to find what happens to the mapping of the region $R-R_2$ (in blue below). The equation of the hexagon side in that region is $x = r \\cos \\theta = \\frac{1}{2}$, which is transformed to $\\frac{1}{r} \\cos (-\\theta) = \\frac{1}{r} \\cos \\theta =$2 . Let $r\\text{cis}\\,\\theta = a+bi$ where $a,b \\in \\mathbb{R}$; then $r = \\sqrt{a^2 + b^2}, \\cos \\theta = \\frac{a}{\\sqrt{a^2 + b^2}}$, so the equation becomes $a^2 - 2a + b^2 = 0 \\Longrightarrow (a-1)^2 + b^2 = 1$. Hence the side is sent to an arc of the unit circle centered at $(1,0)$, after considering the restriction that the side of the hexagon is a segment of length $1/\\sqrt{3}$. Including $S_2$, we find that $S$ is the union of six unit circles centered at $\\text{cis}\\, \\frac{k\\pi}{6}$, $k = 0,1,2,3,4,5$, as shown below. [asy] unitsize(1.5cm); defaultpen(linewidth(0.7)); picture p; real max = .5 + 1/3^.5; pen d = linetype(\"4 4\"); fill(1.5*expi(-pi/6)--arc((0,0),1,-30,30)--1.5*expi(pi/6)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);for(int i = 0; i < 6; ++i) add(rotate(i*60)*p); draw((0,max)--(0,-max),d,Arrows(4));draw((max,0)--(-max,0),d,Arrows(4)); draw(Circle((0,0),1),d); draw(expi(pi/6)--1.5*expi(pi/6),EndArrow(4)); draw(expi(-pi/6)--1.5*expi(-pi/6),EndArrow(4)); label(\"$1/\\sqrt{3}$\",(0,-0.5),W,fontsize(8)); [/asy] $\\Longrightarrow$ [asy]unitsize(4.5cm); defaultpen(linewidth(0.7)); picture p; fill((0,0)--arc((0,0),1,-30,30)--cycle,rgb(0.5,1,0.5));fill(arc((0,0),1,-30,30)--arc(1/3^.5,1/3^.5,60,-60)--cycle,rgb(0.5,0.5,1)); draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); draw(arc(1/3^.5*expi(pi/3),1/3^.5,120,359.99),linetype(\"4 4\")); draw(expi(pi/2)--1/3^.5*expi(pi/3)--expi(pi/6),linetype(\"4 4\")); draw(Circle((0,0),1),linetype(\"4 4\")); label(\"$\\sqrt{3}$\",(0,-0.5),W,fontsize(8)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy] The area of the regular hexagon is $6 \\cdot \\left( \\frac{\\left(\\sqrt{3}\\right)^2 \\sqrt{3}}{4} \\right) = \\frac{9}{2}\\sqrt{3}$. The total area of the six $120^{\\circ}$ sectors is $6\\left(\\frac{1}{3}\\pi - \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\sqrt{3}\\right) = 2\\pi - \\frac{3}{2}\\sqrt{3}$. Their sum is $2\\pi + \\sqrt{27}$, and $a+b = 029. - Th3Numb3rThr33",
"We can describe the line parallel to the imaginary axis $x=\\frac{1}{2}$ using polar coordinates as $r(\\theta)=\\dfrac{1}{2\\cos{\\theta}},$ which rearranges to $z=\\left(\\dfrac{1}{2\\cos{\\theta}}\\right)(cis{\\theta})\\implies \\frac{1}{z}=2\\cos{\\theta}cis(-\\theta).$ Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}r^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}(2\\cos\\theta)^2 d\\theta.$ Thankfully, this is a routine computation: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}2(\\cos\\theta)^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta$ $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta = 12\\left[\\frac{1}{2}\\sin{2\\theta}+\\theta\\right]_0^{\\frac{\\pi}{6}}=12\\left(\\frac{\\sqrt{3}}{4}+\\frac{\\pi}{6}\\right)=2\\pi+3\\sqrt{3}=2\\pi + \\sqrt{27}$ $a+b = 029.",
"We can describe the line parallel to the imaginary axis $x=\\frac{1}{2}$ using polar coordinates as $r(\\theta)=\\dfrac{1}{2\\cos{\\theta}},$ which rearranges to $z=\\left(\\dfrac{1}{2\\cos{\\theta}}\\right)(cis{\\theta})\\implies \\frac{1}{z}=2\\cos{\\theta}cis(-\\theta).$ Denote the area of $S$ as $[S].$ Now, dividing the hexagon to 12 equal parts, we have the following integral: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}r^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}\\frac{1}{2}(2\\cos\\theta)^2 d\\theta.$ Thankfully, this is a routine computation: $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}2(\\cos\\theta)^2 d\\theta = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta$ $[S] = 12\\int_{0}^{\\frac{\\pi}{6}}(\\cos{2\\theta}+1)d\\theta = 12\\left[\\frac{1}{2}\\sin{2\\theta}+\\theta\\right]_0^{\\frac{\\pi}{6}}=12\\left(\\frac{\\sqrt{3}}{4}+\\frac{\\pi}{6}\\right)=2\\pi+3\\sqrt{3}=2\\pi + \\sqrt{27}$ $a+b = 029."
] |
2008-II-14
| 2,008
| 14
|
Let $a$ and $b$ be positive real numbers with $a \ge b$ . Let $\rho$ be the maximum possible value of $\dfrac{a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a-x)^2 + (b-y)^2\] has a solution $(x,y)$ satisfying $0 \le x < a$ and $0 \le y < b$ . Then $\rho^2$ can be expressed as a fraction $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 7
|
II
|
[
"Solution 1 Notice that the given equation implies $a^2 + y^2 = b^2 + x^2 = 2(ax + by)$ We have $2by \\ge y^2$, so $2ax \\le a^2 \\implies x \\le \\frac {a}{2}$. Then, notice $b^2 + x^2 = a^2 + y^2 \\ge a^2$, so $b^2 \\ge \\frac {3}{4}a^2 \\implies \\rho^2 \\le \\frac {4}{3}$. The solution $(a, b, x, y) = \\left(1, \\frac {\\sqrt {3}}{2}, \\frac {1}{2}, 0\\right)$ satisfies the equation, so $\\rho^2 = \\frac {4}{3}$, and the answer is $3 + 4 = 007",
"Notice that the given equation implies $a^2 + y^2 = b^2 + x^2 = 2(ax + by)$ We have $2by \\ge y^2$, so $2ax \\le a^2 \\implies x \\le \\frac {a}{2}$. Then, notice $b^2 + x^2 = a^2 + y^2 \\ge a^2$, so $b^2 \\ge \\frac {3}{4}a^2 \\implies \\rho^2 \\le \\frac {4}{3}$. The solution $(a, b, x, y) = \\left(1, \\frac {\\sqrt {3}}{2}, \\frac {1}{2}, 0\\right)$ satisfies the equation, so $\\rho^2 = \\frac {4}{3}$, and the answer is $3 + 4 = 007",
"Consider the points $(a,y)$ and $(x,b)$. They form an equilateral triangle with the origin. We let the side length be $1$, so $a = \\cos{\\theta}$ and $b = \\sin{\\left(\\theta + \\frac {\\pi}{3}\\right)}$. Thus $f(\\theta) = \\frac {a}{b} = \\frac {\\cos{\\theta}}{\\sin{\\left(\\theta + \\frac {\\pi}{3}\\right)}}$ and we need to maximize this for $0 \\le \\theta \\le \\frac {\\pi}{6}$. Taking the derivative shows that $-f'(\\theta) = \\frac {\\cos{\\frac {\\pi}{3}}}{\\sin^2{\\left(\\theta + \\frac {\\pi}{3}\\right)}} \\ge 0$, so the maximum is at the endpoint $\\theta = 0$. We then get $\\rho = \\frac {\\cos{0}}{\\sin{\\frac {\\pi}{3}}} = \\frac {2}{\\sqrt {3}}$ Then, $\\rho^2 = \\frac {4}{3}$, and the answer is $3+4=007",
"Consider a cyclic quadrilateral $ABCD$ with $\\angle B = \\angle D = 90^{\\circ}$, and $AB = y, BC = a, CD = b, AD = x$. Then \\[AC^2 = a^2 + y^2 = b^2 + x^2\\] From Ptolemy's Theorem, $ax + by = AC(BD)$, so \\[AC^2 = (a - x)^2 + (b - y)^2 = a^2 + y^2 + b^2 + x^2 - 2(ax + by) = 2AC^2 - 2AC*BD\\] Simplifying, we have $BD = AC/2$. Note the circumcircle of $ABCD$ has radius $r = AC/2$, so $BD = r$ and has an arc of $60^{\\circ}$, so $\\angle C = 30^{\\circ}$. Let $\\angle BDC = \\theta$. $\\frac ab = \\frac{BC}{CD} = \\frac{\\sin \\theta}{\\sin(150^{\\circ} - \\theta)}$, where both $\\theta$ and $150^{\\circ} - \\theta$ are $\\leq 90^{\\circ}$ since triangle $BCD$ must be acute. Since $\\sin$ is an increasing function over $(0, 90^{\\circ})$, $\\frac{\\sin \\theta}{\\sin(150^{\\circ} - \\theta)}$ is also increasing function over $(60^{\\circ}, 90^{\\circ})$. $\\frac ab$ maximizes at $\\theta = 90^{\\circ} \\Longrightarrow \\frac ab$ maximizes at $\\frac 2{\\sqrt {3}}$. This squared is $(\\frac 2{\\sqrt {3}})^2 = \\frac4{3}$, and $4 + 3 = 007",
"The problem is looking for an intersection in the said range between parabola $P$: $y = \\tfrac{(x-a)^2 + b^2-a^2}{2b}$ and the hyperbola $H$: $y^2 = x^2 + b^2 - a^2$. The vertex of $P$ is below the x-axis and it's x-coordinate is a, which is to the right of the vertex of the $H$, which is $\\sqrt{a^2 - b^2}$. So for the intersection to exist with $x<a$ and $y \\geq 0$, $P$ needs to cross x-axis between $\\sqrt{a^2 - b^2}$, and $a$, meaning, \\[(\\sqrt{a^2 - b^2}-a)^2 + b^2-a^2 \\geq 0\\] Divide both side by $b^2$, \\[(\\sqrt{\\rho^2 - 1}-\\rho)^2 + 1-\\rho^2 \\geq 0\\] which can be easily solved by moving $1-\\rho^2$ to RHS and taking square roots. Final answer $\\rho^2 \\leq \\frac{4}{3}$ $007",
"The given system is equivalent to the points $(a,y)$ and $(x,b)$ forming an equilateral triangle with the origin. WLOG let this triangle have side length $1$, so $x=\\sqrt{1-a^2}$. The condition $x<a$ implies $(x,b)$ lies to the left of $(a,y)$, so $(x,b)$ is the top vertex. Now we can compute (by complex numbers, or the sine angle addition identity) that $b = \\frac{\\sqrt{3}}{2}a + \\frac{1}{2}\\sqrt{1-a^2}$, so $\\frac{a}{b} = \\frac{a}{\\frac{\\sqrt{3}}{2}a + \\frac{1}{2}\\sqrt{1-a^2}} = \\frac{1}{\\frac{\\sqrt{3}}{2} + \\frac{1}{2a}\\sqrt{1-a^2}}$. Minimizing this is equivalent to minimizing the denominator, which happens when $\\sqrt{1-a^2} = 0$ and thus $a=1$, resulting in $\\rho = \\frac{2}{\\sqrt{3}}$, so $\\rho^2 = \\frac{4}{3}$ and the answer is $007",
"Notice that by Pythagorean theorem, if we take a triangle with vertices $(0,0),$ $(a,y),$ and $(x,b)$ forming an equilateral triangle. Now, take a rectangle with vertices $(0,0), (a,0), (0,b), (a,b).$ Notice that $(a,y)$ and $(x,b)$ are on the sides. Let $\\alpha$ be the angle formed by the points $(0,b), (0,0), (x,b).$ Then, we have that \\[\\cos \\alpha = \\frac{b}{s},\\] where $s$ is the side of the equilateral triangle. Also, we have that $30^{\\circ}-\\alpha$ is the angle formed by the points $(a,0), (0,0), (a,y),$ and so \\[\\cos (30^{\\circ}-\\alpha) = \\frac{a}{s}.\\] Thus, we have that \\[\\frac{a}{b} = \\frac{\\cos (30^{\\circ}-\\alpha)}{\\cos \\alpha}.\\] We see that this expression is maximized when $\\alpha$ is maximized (at least when $\\alpha$ is in the interval $(0,90^{\\circ}),$ which it is). Then, $\\alpha \\ge 30^{\\circ},$ so ew have that the maximum of $\\frac{a}{b}$ is \\[\\frac{\\cos 0}{\\cos 30^{\\circ}} = \\frac{2}{\\sqrt{3}},\\] and so our answer is $4+3 = 7.$",
"Notice that by Pythagorean theorem, if we take a triangle with vertices $(0,0),$ $(a,y),$ and $(x,b)$ forming an equilateral triangle. Now, take a rectangle with vertices $(0,0), (a,0), (0,b), (a,b).$ Notice that $(a,y)$ and $(x,b)$ are on the sides. Let $\\alpha$ be the angle formed by the points $(0,b), (0,0), (x,b).$ Then, we have that \\[\\cos \\alpha = \\frac{b}{s},\\] where $s$ is the side of the equilateral triangle. Also, we have that $30^{\\circ}-\\alpha$ is the angle formed by the points $(a,0), (0,0), (a,y),$ and so \\[\\cos (30^{\\circ}-\\alpha) = \\frac{a}{s}.\\] Thus, we have that \\[\\frac{a}{b} = \\frac{\\cos (30^{\\circ}-\\alpha)}{\\cos \\alpha}.\\] We see that this expression is maximized when $\\alpha$ is maximized (at least when $\\alpha$ is in the interval $(0,90^{\\circ}),$ which it is). Then, $\\alpha \\ge 30^{\\circ},$ so ew have that the maximum of $\\frac{a}{b}$ is \\[\\frac{\\cos 0}{\\cos 30^{\\circ}} = \\frac{2}{\\sqrt{3}},\\] and so our answer is $4+3 = 7.$",
"As the previous solutions pointed out, we have that the triangle with vertices $(0,0),$ $(a,y),$ and $(x,b)$ is equilateral. Then, note that that equilateral triangle is inscribed in a rectangle with vertices $(0,0),$ $(a,0),$ $(a,b),$ and $(0,b).$ Thus, our objective is to find the most \"oblique\" rectangle that has an equilateral triangle inscribed in it. This is when a side of the equilateral triangle coincides with a side of the rectangle, and therefore an axis. We see this because if we rotate it so that it is no longer the case, the longer side decreases in length while the shorter side increases in length. Rotating it to coincides with the other axis, it is just the original rectangle reflected over $y=x.$ WLOG let the side length of the equilateral triangle be $1.$ Then, by our above argument, \\[(a,b)=\\left(1,\\dfrac{\\sqrt{3}}{2}\\right)\\] is optimal, so our answer is \\[\\left(\\dfrac{2}{\\sqrt{3}}\\right)^2=\\dfrac{4}{3}\\implies 007.\\]",
"As the previous solutions pointed out, we have that the triangle with vertices $(0,0),$ $(a,y),$ and $(x,b)$ is equilateral. Then, note that that equilateral triangle is inscribed in a rectangle with vertices $(0,0),$ $(a,0),$ $(a,b),$ and $(0,b).$ Thus, our objective is to find the most \"oblique\" rectangle that has an equilateral triangle inscribed in it. This is when a side of the equilateral triangle coincides with a side of the rectangle, and therefore an axis. We see this because if we rotate it so that it is no longer the case, the longer side decreases in length while the shorter side increases in length. Rotating it to coincides with the other axis, it is just the original rectangle reflected over $y=x.$ WLOG let the side length of the equilateral triangle be $1.$ Then, by our above argument, \\[(a,b)=\\left(1,\\dfrac{\\sqrt{3}}{2}\\right)\\] is optimal, so our answer is \\[\\left(\\dfrac{2}{\\sqrt{3}}\\right)^2=\\dfrac{4}{3}\\implies 007.\\]"
] |
2009-I-1
| 2,009
| 1
|
Call a $3$ -digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.
| 840
|
I
|
[
"Assume that the largest geometric number starts with a $9$. We know that the common ratio must be a rational of the form $k/3$ for some integer $k$, because a whole number should be attained for the 3rd term as well. When $k = 1$, the number is $931$. When $k = 2$, the number is $964$. When $k = 3$, we get $999$, but the integers must be distinct. By the same logic, the smallest geometric number is $124$. The largest geometric number is $964$ and the smallest is $124$. Thus the difference is $964 - 124 = 840.",
"Consider the three-digit number $abc$. If its digits form a geometric progression, we must have that ${a \\over b} = {b \\over c}$, that is, $b^2 = ac$. The minimum and maximum geometric numbers occur when $a$ is minimized and maximized, respectively. The minimum occurs when $a = 1$; letting $b = 2$ and $c = 4$ achieves this, so the smallest possible geometric number is 124. For the maximum, we have that $b^2 = 9c$; $b$ is maximized when $9c$ is the greatest possible perfect square; this happens when $c = 4$, yielding $b = 6$. Thus, the largest possible geometric number is 964. Our answer is thus $964 - 124 = 840.",
"The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $840"
] |
2009-I-2
| 2,009
| 2
|
There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that \[\frac {z}{z + n} = 4i.\] Find $n$ .
| 697
|
I
|
[
"Let $z = a + 164i$. Then \\[\\frac {a + 164i}{a + 164i + n} = 4i\\] and \\[a + 164i = \\left (4i \\right ) \\left (a + n + 164i \\right ) = 4i \\left (a + n \\right ) - 656.\\] By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation, we conclude that \\[a = -656.\\] By equating the imaginary terms on each side of the equation, we conclude that \\[164i = 4i \\left (a + n \\right ) = 4i \\left (-656 + n \\right ).\\] We now have an equation for $n$: \\[4i \\left (-656 + n \\right ) = 164i,\\] and this equation shows that $n = 697",
"\\[\\frac {z}{z+n}=4i\\] \\[1-\\frac {n}{z+n}=4i\\] \\[1-4i=\\frac {n}{z+n}\\] \\[\\frac {1}{1-4i}=\\frac {z+n}{n}\\] \\[\\frac {1+4i}{17}=\\frac {z}{n}+1\\] Since their imaginary part has to be equal, \\[\\frac {4i}{17}=\\frac {164i}{n}\\] \\[n=\\frac {(164)(17)}{4}=697\\] \\[n = 697.\\]",
"Below is an image of the complex plane. Let $\\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$. [asy] unitsize(1cm); xaxis(\"Re\",Arrows); yaxis(\"Im\",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label(\"Im$(z)=164$\",F(-330),N); pair z = (-656,164); dot(Label(\"$z$\",align=N),z); dot(Label(\"$z+n$\",align=N),z+(697,0)); draw(Label(\"$4x$\"),z--(0,0)); draw(Label(\"$x$\"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$. Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\\angle\\theta_1 \\cdot r_2\\angle\\theta_2 = r_1r_2\\angle(\\theta_1+\\theta_2)$ and $\\frac{z_1}{z_2} = \\frac{r_1\\angle\\theta_1}{r_2\\angle\\theta_2} = \\frac{r_1}{r_2}\\angle(\\theta_1-\\theta_2)$, where $r$ is the magnitude and $\\theta$ is the phase, and $z_n=r_n\\angle\\theta_n$. Since $4i$ has magnitude $4$ and phase $90^\\circ$ (since the positive imaginary axis points in a direction $90^\\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$. This means that $z$, the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$, since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\\frac{x \\cdot 4x}{2}$, or using the hypotenuse and its corresponding altitude, as $\\frac{164n}{2}$, so $\\frac{x \\cdot 4x}{2} = \\frac{164n}{2} \\implies x^2 = 41n$. By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \\implies 17x^2 = n^2$. Substituting out $x^2$ using the earlier equation, we get $17\\cdot41n = n^2 \\implies n = 697. ~emerald_block",
"Below is an image of the complex plane. Let $\\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$. [asy] unitsize(1cm); xaxis(\"Re\",Arrows); yaxis(\"Im\",Arrows); real f(real x) {return 164;} pair F(real x) {return (x,f(x));} draw(graph(f,-700,100),red,Arrows); label(\"Im$(z)=164$\",F(-330),N); pair z = (-656,164); dot(Label(\"$z$\",align=N),z); dot(Label(\"$z+n$\",align=N),z+(697,0)); draw(Label(\"$4x$\"),z--(0,0)); draw(Label(\"$x$\"),(0,0)--z+(697,0)); markscalefactor=2; draw(rightanglemark(z,(0,0),z+(697,0))); [/asy] $z$ must lie on the line $\\operatorname{Im}(z)=164$. $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$. Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\\angle\\theta_1 \\cdot r_2\\angle\\theta_2 = r_1r_2\\angle(\\theta_1+\\theta_2)$ and $\\frac{z_1}{z_2} = \\frac{r_1\\angle\\theta_1}{r_2\\angle\\theta_2} = \\frac{r_1}{r_2}\\angle(\\theta_1-\\theta_2)$, where $r$ is the magnitude and $\\theta$ is the phase, and $z_n=r_n\\angle\\theta_n$. Since $4i$ has magnitude $4$ and phase $90^\\circ$ (since the positive imaginary axis points in a direction $90^\\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$. We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$. Additionally, $z$, the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$. This means that $z$, the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$, since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\\frac{x \\cdot 4x}{2}$, or using the hypotenuse and its corresponding altitude, as $\\frac{164n}{2}$, so $\\frac{x \\cdot 4x}{2} = \\frac{164n}{2} \\implies x^2 = 41n$. By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \\implies 17x^2 = n^2$. Substituting out $x^2$ using the earlier equation, we get $17\\cdot41n = n^2 \\implies n = 697. ~emerald_block",
"Taking the reciprocal of our equation gives us $1 + \\frac{n}{z} = \\frac{1}{4i}.$ Therefore, \\[\\frac{n}{z} = \\frac{1-4i}{4i} = \\frac{17}{-16+4i}.\\] Since $z$ has an imaginary part of $164$, we must multiply both sides of our RHS fraction by $\\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \\[\\frac{n}{z} = \\frac{697}{-656 + 164i}.\\] Therefore, we can conclude the real part of $z$ is $-656$ and $n = 697 (it wasn't necessary to find the real part) ~Maximilian113",
"Taking the reciprocal of our equation gives us $1 + \\frac{n}{z} = \\frac{1}{4i}.$ Therefore, \\[\\frac{n}{z} = \\frac{1-4i}{4i} = \\frac{17}{-16+4i}.\\] Since $z$ has an imaginary part of $164$, we must multiply both sides of our RHS fraction by $\\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \\[\\frac{n}{z} = \\frac{697}{-656 + 164i}.\\] Therefore, we can conclude the real part of $z$ is $-656$ and $n = 697 (it wasn't necessary to find the real part) ~Maximilian113"
] |
2009-I-3
| 2,009
| 3
|
A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 11
|
I
|
[
"The probability of three heads and five tails is $\\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\\binom {8}{3}p^5(1-p)^3$. \\begin{align*} 25\\binom {8}{3}p^3(1-p)^5&=\\binom {8}{3}p^5(1-p)^3 \\\\ 25(1-p)^2&=p^2 \\\\ 25p^2-50p+25&=p^2 \\\\ 24p^2-50p+25&=0 \\\\ p&=\\frac {5}{6}\\end{align*} Therefore, the answer is $5+6=011.",
"We start as shown above. However, when we get to $25(1-p)^2=p^2$, we square root both sides to get $5(1-p)=p$. We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$, so they are both positive. Now, we have: \\begin{align*} 5(1-p)&=p \\\\ 5-5p&=p \\\\ 5&=6p \\\\ p&=\\frac {5}{6}\\end{align*} Now, we get $5+6=011. ~Jerry_Guo",
"Rewrite it as : $(P)^3$$(1-P)^5=\\frac {1}{25}$ $(P)^5$$(1-P)^3$ This can be simplified as $24P^2 -50P + 25 = 0$ This can be factored into $(4P-5)(6P-5)$ This yields two solutions: $\\frac54$ (ignored because it would result in $1-p<0$ ) or $\\frac56$ Therefore, the answer is $5+6$ = $011"
] |
2009-I-4
| 2,009
| 4
|
In parallelogram $ABCD$ , point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$ . Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$ . Find $\frac {AC}{AP}$ .
| 177
|
I
|
[
"One of the ways to solve this problem is to make this parallelogram a straight line. So the whole length of the line is $APC$($AMC$ or $ANC$), and $ABC$ is $1000x+2009x=3009x.$ $AP$($AM$ or $AN$) is $17x.$ So the answer is $3009x/17x = 177",
"Draw a diagram with all the given points and lines involved. Construct parallel lines $\\overline{DF_2F_1}$ and $\\overline{BB_1B_2}$ to $\\overline{MN}$, where for the lines the endpoints are on $\\overline{AM}$ and $\\overline{AN}$, respectively, and each point refers to an intersection. Also, draw the median of quadrilateral $BB_2DF_1$ $\\overline{E_1E_2E_3}$ where the points are in order from top to bottom. Clearly, by similar triangles, $BB_2 = \\frac {1000}{17}MN$ and $DF_1 = \\frac {2009}{17}MN$. It is not difficult to see that $E_2$ is the center of quadrilateral $ABCD$ and thus the midpoint of $\\overline{AC}$ as well as the midpoint of $\\overline{B_1}{F_2}$ (all of this is easily proven with symmetry). From more triangle similarity, $E_1E_3 = \\frac12\\cdot\\frac {3009}{17}MN\\implies AE_2 = \\frac12\\cdot\\frac {3009}{17}AP\\implies AC = 2\\cdot\\frac12\\cdot\\frac {3009}{17}AP$ $= 177.",
"Using vectors, note that $\\overrightarrow{AM}=\\frac{17}{1000}\\overrightarrow{AB}$ and $\\overrightarrow{AN}=\\frac{17}{2009}\\overrightarrow{AD}$. Note that $\\overrightarrow{AP}=\\frac{x\\overrightarrow{AM}+y\\overrightarrow{AN}}{x+y}$ for some positive x and y, but at the same time is a scalar multiple of $\\overrightarrow{AB}+\\overrightarrow{AD}$. So, writing the equation $\\overrightarrow{AP}=\\frac{x\\overrightarrow{AM}+y\\overrightarrow{AN}}{x+y}$ in terms of $\\overrightarrow{AB}$ and $\\overrightarrow{AD}$, we have $\\overrightarrow{AP}=\\frac{\\frac{17x}{1000}\\overrightarrow{AB}+\\frac{17y}{2009}\\overrightarrow{AD}}{x+y}$. But the coefficients of the two vectors must be equal because, as already stated, $\\overrightarrow{AP}$ is a scalar multiple of $\\overrightarrow{AB}+\\overrightarrow{AD}$. We then see that $\\frac{x}{x+y}=\\frac{1000}{3009}$ and $\\frac{y}{x+y}=\\frac{2009}{3009}$. Finally, we have $\\overrightarrow{AP}=\\frac{17}{3009}(\\overrightarrow{AB}+\\overrightarrow{AD})$ and, simplifying, $\\overrightarrow{AB}+\\overrightarrow{AD}=177\\overrightarrow{AP}$ and the desired quantity is $177$.",
"We approach the problem using mass points on triangle $ABD$ as displayed below. [asy] pair A=(0,0),B=(50,0),D=(10,40),C=B+D,M=(8,0),NN=(2,8); draw(A--B--C--D--cycle); draw(B--D^^A--C^^M--NN); pair O=extension(A,C,B,D); pair P=extension(A,C,M,NN); dot(A);dot(B);dot(C);dot(D);dot(O);dot(M);dot(NN);dot(P); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,NE); label(\"$D$\",D,NW); label(\"$M$\",M,S); label(\"$N$\",NN,NW); label(\"$P$\",P,NNE); label(\"$O$\",O,N); [/asy] But as $MN$ does not protrude from a vertex, we will have to \"split the mass\" at point $A$. First, we know that $DO$ is congruent to $BO$ because diagonals of parallelograms bisect each other. Therefore, we can assign equal masses to points $B$ and $D$. In this case, we assign $B$ and $D$ a mass of 17 each. Now we split the mass at $A$, so we balance segments $AB$ and $AD$ separately, and then the mass of $A$ is the sum of those masses. A mass of 983 is required to balance segment $AB$, while a mass of 1992 is required to balance segment $AD$. Therefore, $A$ has a mass of $1992+983=2975$. Also, $O$ has a mass of 34. Therefore, $\\frac{AO}{AP}=\\frac{2975+34}{34}=\\frac{3009}{34}$, so $\\frac{AC}{AP}=\\frac{2 (3009)}{34}=177$.",
"Assume, for the ease of computation, that $AM=AN=17$, $AB=1000$, and $AD=2009$. Now, let line $MN$ intersect line $CD$ at point $X$ and let $Y$ be a point such that $XY\\parallel AD$ and $AY\\parallel DX$. As a result, $ADXY$ is a parallelogram. By construction, $\\triangle MAN\\sim \\triangle MYX$ so \\[\\frac{MY}{MA}=\\frac{YX}{AN}=\\frac{AD}{AN}=\\frac{2009}{17}\\implies MY=2009\\] and $AY=DX=2009-17$. Also, because $AM\\parallel XC$, we have $\\triangle PAM\\sim \\triangle PCX$ so \\[\\frac{PC}{PA}=\\frac{CX}{AM}=\\frac{DX+CD}{AM}=\\frac{2009-17+1000}{17}=176.\\] Hence, $\\frac{AC}{AP}=\\frac{PC}{PA}+1=177",
"Assign $A = (0,0)$. Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \\times 2009.$ Now, we can assign \\[A=(0,0)\\] \\[B=(1000, 0)\\] \\[C=(1000,-2009)\\] \\[D=(0,-2009).\\] Then, since $\\frac{AM}{AB} = \\frac{17}{1000}$ and $AB = 1000$, we can place $M$ at $(17, 0).$ Similarly, place $AN$ at $(0, 17).$ Then, the equation of line $MN$ is $y=x-17,$ and the equation of $AC$ is $y=\\frac{-2009}{1000}x.$ Solve to find point $P$ at $\\left( \\frac{1000}{177}, \\frac{-2009}{177} \\right)$. We can calculate vectors to represent the distances: \\[\\overrightarrow{AC}= <1000, -2009>\\] \\[\\overrightarrow{AP}= \\frac{1}{177}<1000, -2009>.\\] In this way, we can see that \\[AC:AP = 177:1,\\] and our answer is $177 as a rectangle, either by projecting the plane onto another (tilted) plane or removing the restriction that the axes have to be perpendicular.",
"Assign $A = (0,0)$. Since there are no constraints in the problem against this, assume $ABCD$ to be a rectangle with dimensions $1000 \\times 2009.$ Now, we can assign \\[A=(0,0)\\] \\[B=(1000, 0)\\] \\[C=(1000,-2009)\\] \\[D=(0,-2009).\\] Then, since $\\frac{AM}{AB} = \\frac{17}{1000}$ and $AB = 1000$, we can place $M$ at $(17, 0).$ Similarly, place $AN$ at $(0, 17).$ Then, the equation of line $MN$ is $y=x-17,$ and the equation of $AC$ is $y=\\frac{-2009}{1000}x.$ Solve to find point $P$ at $\\left( \\frac{1000}{177}, \\frac{-2009}{177} \\right)$. We can calculate vectors to represent the distances: \\[\\overrightarrow{AC}= <1000, -2009>\\] \\[\\overrightarrow{AP}= \\frac{1}{177}<1000, -2009>.\\] In this way, we can see that \\[AC:AP = 177:1,\\] and our answer is $177 as a rectangle, either by projecting the plane onto another (tilted) plane or removing the restriction that the axes have to be perpendicular.",
"Let $ABCD$ be a square with side length $1$ where $D = (0,0)$, $A = (0,1)$, $B = (1,1)$, and $C = (1,0)$. Draw $M$ on $AB$ such that $AM$ has length $\\frac{17}{1000}$ and $AN$ has length $\\frac{17}{2009}$. Draw $AC$ with $P$ as the intersection point of $AC$ and $MN$. $N$ has coordinates $(0, 1-\\frac{17}{2009}) = (0, \\frac{1992}{2009})$ Extend lines $MN$ and $CD$ such that their intersection point is point $E$, which lies on $y=0$. Line $ME$ has slope $\\frac{AN}{AM} = \\frac{1000}{2009}$. With the y-intercept $N$ it has the equation $y = \\frac{1000}{2009}x + \\frac{1992}{2009}$. Solving for the $x$ coordinate on point $E$ $(y = 0)$, $x = -\\frac{1992}{1000}$. $ED$ has length $\\frac{1992}{1000}$ and $EC$ has length $\\frac{1992}{1000} + 1 = \\frac{2992}{1000}$. Triangles $APM$ and $CPE$ are similar (AA). $\\frac{AP}{PC} = \\frac{AM}{EC} = \\frac{2992}{17}$. $PC = \\frac{2992}{17}AP$ $\\frac{AC}{AP} = \\frac{PC+AP}{AP} = \\frac{3009}{17} = 177$. ~unhappyfarmer",
"Let $ABCD$ be a square with side length $1$ where $D = (0,0)$, $A = (0,1)$, $B = (1,1)$, and $C = (1,0)$. Draw $M$ on $AB$ such that $AM$ has length $\\frac{17}{1000}$ and $AN$ has length $\\frac{17}{2009}$. Draw $AC$ with $P$ as the intersection point of $AC$ and $MN$. $N$ has coordinates $(0, 1-\\frac{17}{2009}) = (0, \\frac{1992}{2009})$ Extend lines $MN$ and $CD$ such that their intersection point is point $E$, which lies on $y=0$. Line $ME$ has slope $\\frac{AN}{AM} = \\frac{1000}{2009}$. With the y-intercept $N$ it has the equation $y = \\frac{1000}{2009}x + \\frac{1992}{2009}$. Solving for the $x$ coordinate on point $E$ $(y = 0)$, $x = -\\frac{1992}{1000}$. $ED$ has length $\\frac{1992}{1000}$ and $EC$ has length $\\frac{1992}{1000} + 1 = \\frac{2992}{1000}$. Triangles $APM$ and $CPE$ are similar (AA). $\\frac{AP}{PC} = \\frac{AM}{EC} = \\frac{2992}{17}$. $PC = \\frac{2992}{17}AP$ $\\frac{AC}{AP} = \\frac{PC+AP}{AP} = \\frac{3009}{17} = 177$. ~unhappyfarmer"
] |
2009-I-5
| 2,009
| 5
|
Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ .
| 72
|
I
|
[
"[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label(\"$A$\",A,(-1,-1));label(\"$B$\",B,(1,-1));label(\"$C$\",C,(1,1)); label(\"$K$\",K,(0,2));label(\"$L$\",L,(0,-2));label(\"$M$\",M,(-1,1)); label(\"$P$\",P,(1,1)); label(\"$180$\",(A+M)/2,(-1,0));label(\"$180$\",(P+C)/2,(-1,0));label(\"$225$\",(A+K)/2,(0,2));label(\"$225$\",(K+C)/2,(0,2)); label(\"$300$\",(B+C)/2,(1,1)); [/asy] Since $K$ is the midpoint of $\\overline{PM}$ and $\\overline{AC}$, quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\\bigtriangleup{AMB}$ is similar to $\\bigtriangleup{LPB}$ Thus, \\[\\frac {AM}{LP}=\\frac {AB}{LB}=\\frac {AL+LB}{LB}=\\frac {AL}{LB}+1\\] Now let's apply the angle bisector theorem. \\[\\frac {AL}{LB}=\\frac {AC}{BC}=\\frac {450}{300}=\\frac {3}{2}\\] \\[\\frac {AM}{LP}=\\frac {AL}{LB}+1=\\frac {5}{2}\\] \\[\\frac {180}{LP}=\\frac {5}{2}\\] \\[LP=072\\]",
"Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \\[\\frac{BL}{CB}=\\frac{AL}{CA}\\implies\\frac{BL}{300}=\\frac{AL}{450}\\implies 3BL=2AL\\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$. Since $K$ is the midpoint of $A$ and $C$, the weight of $A$ is equal to the weight of $C$, which equals $2$. Also, since the weight of $L$ is $5$ and $C$ is $2$, we can weight $P$ as $7$. By the definition of mass points, \\[\\frac{LP}{CP}=\\frac{2}{5}\\implies LP=\\frac{2}{5}CP\\] By vertical angles, angle $MKA =$ angle $PKC$. Also, it is given that $AK=CK$ and $PK=MK$. By the SAS congruence, $\\triangle MKA$ = $\\triangle PKC$. So, $MA$ = $CP$ = $180$. Since $LP=\\frac{2}{5}CP$, $LP = \\frac{2}{5}(180) = 072",
"Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \\[\\frac{BL}{CB}=\\frac{AL}{CA}\\implies\\frac{BL}{300}=\\frac{AL}{450}\\implies 3BL=2AL\\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$. Since $K$ is the midpoint of $A$ and $C$, the weight of $A$ is equal to the weight of $C$, which equals $2$. Also, since the weight of $L$ is $5$ and $C$ is $2$, we can weight $P$ as $7$. By the definition of mass points, \\[\\frac{LP}{CP}=\\frac{2}{5}\\implies LP=\\frac{2}{5}CP\\] By vertical angles, angle $MKA =$ angle $PKC$. Also, it is given that $AK=CK$ and $PK=MK$. By the SAS congruence, $\\triangle MKA$ = $\\triangle PKC$. So, $MA$ = $CP$ = $180$. Since $LP=\\frac{2}{5}CP$, $LP = \\frac{2}{5}(180) = 072",
"Using the diagram from solution $1$, we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$. Applying the angle bisector theorem to $\\triangle CKB$, we get that $\\frac{KP}{PB} = \\frac{225}{300} = \\frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$. Now, apply law of cosines on $\\triangle CKP$ and $\\triangle CPB.$ If we let $\\angle KCP = \\angle PCB = \\alpha$, then the law of cosines gives the following system of equations: \\[9x^2 = 225^2 + 180^2 - 2\\cdot 225 \\cdot 180 \\cdot \\cos \\alpha\\] \\[16x^2 = 180^2 + 300^2 - 2 \\cdot 180 \\cdot 300 \\cdot \\cos \\alpha.\\] Bashing those out, we get that $x = 15 \\sqrt{13}$ and $\\cos \\alpha = \\frac{7}{10}.$ Since $\\cos \\alpha = \\frac{7}{10}$, we can use the double angle formula to calculate that $\\cos \\left(2 \\alpha \\right) = -\\frac{1}{50}.$ Now, apply Law of Cosines on $\\triangle ABC$ to find $AB$. We get: \\[AB^2 = 450^2 + 300^2 - 2 \\cdot 450 \\cdot 300 \\cdot \\left(- \\frac{1}{50} \\right).\\] Bashing gives $AB = 30 \\sqrt{331}.$ From the angle bisector theorem on $\\triangle ABC$, we know that $\\frac{AL}{BL} = \\frac{450}{300} = \\frac{3}{2}.$ So, $AL = 18 \\sqrt{331}$ and $BL = 12 \\sqrt{331}.$ Now, we apply Law of Cosines on $\\triangle ALC$ and $\\triangle BLC$ in order to solve for the length of $LC$. We get the following system: \\[(18 \\sqrt{331})^2 = 450^2 + LC^2 - 2 \\cdot 450 \\cdot LC \\cdot \\frac{7}{10}\\] \\[(12 \\sqrt{331})^2 = LC^2 + 300^2 - 2 \\cdot 300 \\cdot LC \\cdot \\frac{7}{10}\\] The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$. The only value that satisfies both equations is $LC = 252$, and since $LP = LC - PC$, we have \\[LC = 252 - 180 = 072.\\]",
"Using the diagram from solution $1$, we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$. Applying the angle bisector theorem to $\\triangle CKB$, we get that $\\frac{KP}{PB} = \\frac{225}{300} = \\frac{3}{4}.$ So, we can let $MK = KP = 3x$ and $BP = 4x$. Now, apply law of cosines on $\\triangle CKP$ and $\\triangle CPB.$ If we let $\\angle KCP = \\angle PCB = \\alpha$, then the law of cosines gives the following system of equations: \\[9x^2 = 225^2 + 180^2 - 2\\cdot 225 \\cdot 180 \\cdot \\cos \\alpha\\] \\[16x^2 = 180^2 + 300^2 - 2 \\cdot 180 \\cdot 300 \\cdot \\cos \\alpha.\\] Bashing those out, we get that $x = 15 \\sqrt{13}$ and $\\cos \\alpha = \\frac{7}{10}.$ Since $\\cos \\alpha = \\frac{7}{10}$, we can use the double angle formula to calculate that $\\cos \\left(2 \\alpha \\right) = -\\frac{1}{50}.$ Now, apply Law of Cosines on $\\triangle ABC$ to find $AB$. We get: \\[AB^2 = 450^2 + 300^2 - 2 \\cdot 450 \\cdot 300 \\cdot \\left(- \\frac{1}{50} \\right).\\] Bashing gives $AB = 30 \\sqrt{331}.$ From the angle bisector theorem on $\\triangle ABC$, we know that $\\frac{AL}{BL} = \\frac{450}{300} = \\frac{3}{2}.$ So, $AL = 18 \\sqrt{331}$ and $BL = 12 \\sqrt{331}.$ Now, we apply Law of Cosines on $\\triangle ALC$ and $\\triangle BLC$ in order to solve for the length of $LC$. We get the following system: \\[(18 \\sqrt{331})^2 = 450^2 + LC^2 - 2 \\cdot 450 \\cdot LC \\cdot \\frac{7}{10}\\] \\[(12 \\sqrt{331})^2 = LC^2 + 300^2 - 2 \\cdot 300 \\cdot LC \\cdot \\frac{7}{10}\\] The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$. The only value that satisfies both equations is $LC = 252$, and since $LP = LC - PC$, we have \\[LC = 252 - 180 = 072.\\]",
"Note that we are given that $\\overline{MK} = \\overline{KP}$, that $\\overline{AK} = \\overline{CK}.$ Note then that $\\angle MKA = \\angle CKB$ by vertical angles. From this, we have $\\triangle MKA \\cong PKC.$ This means that $\\overline{CP}$ is 180. Applying angle bisector theorem on $\\triangle ACB$ gives $\\frac{\\overline{AL}}{\\overline{LB}} = \\frac{450}{300} = \\frac{3}{2}.$ Applying it on $\\triangle KCB$ yields \\[\\frac{\\overline{KP}}{\\overline{PB}} = \\frac{225}{300} = \\frac{3}{4}\\] Now we can proceed with area ratios. Suppose the area of $\\triangle ACB = A.$ This means that \\[[\\triangle AKL] = \\left(\\frac{225}{225+225}\\right)\\left(\\frac{3}{5}\\right)A = \\frac{3}{10}A\\] Continuing on $\\triangle LPB$ we have \\[[\\triangle LPB] = \\left(\\frac{2}{2+3}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{4}{4+3}\\right) = \\frac{4}{35}A\\] Since $\\overline{AK}=\\overline{KC}$ $[\\triangle KPL] = [\\triangle AKB]-[\\triangle AKL] - [\\triangle LPB] = \\frac{1}{2}A - \\frac{3}{10}A - \\frac{4}{35}A = \\frac{3}{35}A.$ Area ratios on $\\triangle KCP$ yield $[\\triangle KCP] = \\left(\\frac{1}{2}\\right)\\left(\\frac{3}{3+4}\\right) = \\frac{3}{14}.$ Now, suppose $\\overline{LP} = x.$ We have that the ratio of areas of $\\triangle LKP$ and $\\triangle PKC$ is $\\frac{x}{180}$ and is also $\\frac{\\frac{3}{35}}{\\frac{3}{14}}$ and equating these gives \\[x = 72\\]",
"Note that we are given that $\\overline{MK} = \\overline{KP}$, that $\\overline{AK} = \\overline{CK}.$ Note then that $\\angle MKA = \\angle CKB$ by vertical angles. From this, we have $\\triangle MKA \\cong PKC.$ This means that $\\overline{CP}$ is 180. Applying angle bisector theorem on $\\triangle ACB$ gives $\\frac{\\overline{AL}}{\\overline{LB}} = \\frac{450}{300} = \\frac{3}{2}.$ Applying it on $\\triangle KCB$ yields \\[\\frac{\\overline{KP}}{\\overline{PB}} = \\frac{225}{300} = \\frac{3}{4}\\] Now we can proceed with area ratios. Suppose the area of $\\triangle ACB = A.$ This means that \\[[\\triangle AKL] = \\left(\\frac{225}{225+225}\\right)\\left(\\frac{3}{5}\\right)A = \\frac{3}{10}A\\] Continuing on $\\triangle LPB$ we have \\[[\\triangle LPB] = \\left(\\frac{2}{2+3}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{4}{4+3}\\right) = \\frac{4}{35}A\\] Since $\\overline{AK}=\\overline{KC}$ $[\\triangle KPL] = [\\triangle AKB]-[\\triangle AKL] - [\\triangle LPB] = \\frac{1}{2}A - \\frac{3}{10}A - \\frac{4}{35}A = \\frac{3}{35}A.$ Area ratios on $\\triangle KCP$ yield $[\\triangle KCP] = \\left(\\frac{1}{2}\\right)\\left(\\frac{3}{3+4}\\right) = \\frac{3}{14}.$ Now, suppose $\\overline{LP} = x.$ We have that the ratio of areas of $\\triangle LKP$ and $\\triangle PKC$ is $\\frac{x}{180}$ and is also $\\frac{\\frac{3}{35}}{\\frac{3}{14}}$ and equating these gives \\[x = 72\\]"
] |
2009-I-6
| 2,009
| 6
|
How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ ?
| 412
|
I
|
[
"First, $x$ must be less than $5$, since otherwise $x^{\\lfloor x\\rfloor}$ would be at least $3125$ which is greater than $1000$. Because ${\\lfloor x\\rfloor}$ must be an integer, let’s do case work based on ${\\lfloor x\\rfloor}$: For ${\\lfloor x\\rfloor}=0$, $N=1$ as long as $x \\neq 0$. This gives us $1$ value of $N$. For ${\\lfloor x\\rfloor}=1$, $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$ Therefore, $N=1$. However, we got $N=1$ in case 1 so it got counted twice. For ${\\lfloor x\\rfloor}=2$, $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$ This gives us $3^2-2^2=5$ $N$'s For ${\\lfloor x\\rfloor}=3$, $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$ This gives us $4^3-3^3=37$ $N$'s For ${\\lfloor x\\rfloor}=4$, $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$ This gives us $5^4-4^4=369$ $N$'s Since $x$ must be less than $5$, we can stop here and the answer is $1+5+37+369= 412 to get the same answer.",
"For a positive integer $k$, we find the number of positive integers $N$ such that $x^{\\lfloor x\\rfloor}=N$ has a solution with ${\\lfloor x\\rfloor}=k$. Then $x=\\sqrt[k]{N}$, and because $k \\le x < k+1$, we have $k^k \\le x^k < (k+1)^k$, and because $(k+1)^k$ is an integer, we get $k^k \\le x^k \\le (k+1)^k-1$. The number of possible values of $x^k$ is equal to the number of integers between $k^k$ and $(k+1)^k-1$ inclusive, which is equal to the larger number minus the smaller number plus one or $((k+1)^k-1)-(k^k)+1$, and this is equal to $(k+1)^k-k^k$. If $k>4$, the value of $x^k$ exceeds $1000$, so we only need to consider $k \\le 4$. The requested number of values of $N$ is the same as the number of values of $x^k$, which is $\\sum^{4}_{k=1} [(k+1)^k-k^k]=2-1+9-4+64-27+625-256=412."
] |
2009-I-7
| 2,009
| 7
|
The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$ . Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$ .
| 41
|
I
|
[
"The best way to solve this problem is to get the iterated part out of the exponent: \\[5^{(a_{n + 1} - a_n)} = \\frac {1}{n + \\frac {2}{3}} + 1\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {n + \\frac {5}{3}}{n + \\frac {2}{3}}\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {3n + 5}{3n + 2}\\] \\[a_{n + 1} - a_n = \\log_5{\\left(\\frac {3n + 5}{3n + 2}\\right)}\\] \\[a_{n + 1} - a_n = \\log_5{(3n + 5)} - \\log_5{(3n + 2)}\\] Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: \\[\\log_5{5},\\log_5{8}, \\log_5{11}, \\log_5{14}.\\] We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \\log_5{5} = \\log_5{(3(1) + 2)}$, we can easily use induction to show that $a_n = \\log_5{(3n + 2)}$. So now we only need to find the next value of $n$ that makes $\\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$. We test $25$: \\[3n + 2 = 25\\] \\[3n = 23\\] This has no integral solutions, so we try $125$: \\[3n + 2 = 125\\] \\[3n = 123\\] \\[n = 041\\]",
"We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$, we get: \\[5^{a_n-a_1}=\\dfrac{n+\\tfrac23}{1+\\tfrac23}\\] This simplifies to \\[5^{a_n}=3n+2.\\] We can now test powers of $5$. $5$ - that gives us $n=1$, which is useless. $25$ - that gives a non-integer $n$. $125$ - that gives $n=41. -integralarefun",
"We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$, we get: \\[5^{a_n-a_1}=\\dfrac{n+\\tfrac23}{1+\\tfrac23}\\] This simplifies to \\[5^{a_n}=3n+2.\\] We can now test powers of $5$. $5$ - that gives us $n=1$, which is useless. $25$ - that gives a non-integer $n$. $125$ - that gives $n=41. -integralarefun"
] |
2009-I-8
| 2,009
| 8
|
Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$ . Consider all possible positive differences of pairs of elements of $S$ . Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$ .
| 398
|
I
|
[
"When computing $N$, the number $2^x$ will be added $x$ times (for terms $2^x-2^0$, $2^x-2^1$, ..., $2^x - 2^{x-1}$), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\\cdot 2^{10} + 8\\cdot 2^9 + 6\\cdot 2^8 + \\cdots - 8\\cdot 2^1 - 10\\cdot 2^0$. Evaluating $N \\bmod {1000}$ yields: \\begin{align*} N & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\\\ & = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\\\ & = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\\\ & \\equiv (0 + 230) + (0 + 80) + (500 + 12) + 480 + 96 \\\\ & \\equiv 398 \\end{align*}",
"When computing $N$, the number $2^x$ will be added $x$ times (for terms $2^x-2^0$, $2^x-2^1$, ..., $2^x - 2^{x-1}$), and subtracted $10-x$ times. Hence $N$ can be computed as $N=10\\cdot 2^{10} + 8\\cdot 2^9 + 6\\cdot 2^8 + \\cdots - 8\\cdot 2^1 - 10\\cdot 2^0$. Evaluating $N \\bmod {1000}$ yields: \\begin{align*} N & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\\\ & = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\\\ & = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\\\ & \\equiv (0 + 230) + (0 + 80) + (500 + 12) + 480 + 96 \\\\ & \\equiv 398 \\end{align*}",
"This solution can be generalized to apply when $10$ is replaced by other positive integers. Extending from Solution 2, we get that the sum $N$ of all possible differences of pairs of elements in $S$ when $S = \\{2^0,2^1,2^2,\\ldots,2^{n}\\}$ is equal to $\\sum_{x=0}^{n} (2x-n) 2^x$. Let $A = \\sum_{x=0}^{n} x2^x$, $B=\\sum_{x=0}^{n} 2^x$. Then $N=2A - nB$. For $n = 10$, $B = 2^{11}-1 = 2047 \\equiv 47 \\pmod{1000}$ by the geometric sequence formula. $2A = \\sum_{x=1}^{n+1} (x-1)2^x$, so $2A - A = A = n2^{n+1} - \\sum_{x=1}^{n} 2^x$. Hence, for $n = 10$, $A = 10 \\cdot 2^{11} - 2^{11} + 2 = 9 \\cdot 2^{11} + 2 \\equiv 48 \\cdot 9 + 2 =$ $434 \\pmod{1000}$, by the geometric sequence formula and the fact that $2^{10} = 1024 \\equiv 24 \\pmod{1000}$. Thus, for $n = 10$, $N = 2A - 10B \\equiv 2\\cdot 434 - 10\\cdot 47 = 868 - 470 = 398.",
"Consider the unique differences $2^{a + n} - 2^a$. Simple casework yields a sum of $\\sum_{n = 1}^{10}(2^n - 1)(2^{11 - n} - 1) = \\sum_{n = 1}^{10}2^{11} + 1 - 2^n - 2^{11 - n} = 10\\cdot2^{11} + 10 - 2(2 + 2^2 + \\cdots + 2^{10})$ $= 10\\cdot2^{11} + 10 - 2^2(2^{10} - 1)\\equiv480 + 10 - 4\\cdot23\\equiv398. This method generalizes nicely as well.",
"Find the positive differences in all $55$ pairs and you will get $398. (This is not recommended unless you can't find any other solutions to this problem)",
"Find the positive differences in all $55$ pairs and you will get $398. (This is not recommended unless you can't find any other solutions to this problem)"
] |
2009-I-9
| 2,009
| 9
|
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $\text{\textdollar}1$ to $\text{\textdollar}9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$ . Find the total number of possible guesses for all three prizes consistent with the hint.
| 420
|
I
|
[
"[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$, that could have been on that day.] Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is \\[1131331.\\] Since the strings have seven digits and three threes, there are $\\binom{7}{3}=35$ arrangements of all such strings. In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups. Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to \\[x+y+z=7, x,y,z>0.\\] This gives us \\[\\binom{6}{2}=15\\] ways by stars and bars. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$. Thus, each arrangement has \\[\\binom{6}{2}-3=12\\] ways per arrangement, and there are $12\\times35=420 ways."
] |
2009-I-10
| 2,009
| 10
|
The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$ . Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $N \cdot (5!)^3$ . Find $N$ .
| 346
|
I
|
[
"Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es. Pretend the table only seats $3$ \"people\", with $1$ \"person\" from each planet. Counting clockwise, only the arrangement M, V, E satisfies the given constraints. Therefore, in the actual problem, the members must sit in cycles of M, V, E, but not necessarily with one M, one V, and one E in each cycle(for example, MMVVVE, MVVVEEE, MMMVVVEE all count as cycles). These cycles of MVE must start at seat $1$, since an M is at seat $1$. We simply count the number of arrangements through casework. 1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE 2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using stars and bars, we get $\\binom{4}{1}=4$ ways for the members of each planet. Therefore, there are $4^3=64$ ways in total. 3. Three cycles - 2 Ms, Vs, Es left, so $\\binom{4}{2}=6$, making there $6^3=216$ ways total. 4. Four cycles - 1 M, V, E left, each M can go to any of the four MVE cycles and likewise for V and E, $4^3=64$ ways total 5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just $1$ way. Combining all these cases, we get $1+1+64+64+216= 346",
"The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \\choose k-1}$ choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have $N = 1^3+4^3+6^3+4^3+1^3 = 346 - aops5234"
] |
2009-I-11
| 2,009
| 11
|
Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer.
| 600
|
I
|
[
"Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). $\\det \\left(\\begin{array}{c} P \\\\ Q\\end{array}\\right)=\\det \\left(\\begin{array}{cc}x_1 &y_1\\\\x_2&y_2\\end{array}\\right).$ Since the triangle has half the area of the parallelogram, we just need the determinant to be even. The determinant is \\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\\] Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=600 such triangles.",
"Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). $\\det \\left(\\begin{array}{c} P \\\\ Q\\end{array}\\right)=\\det \\left(\\begin{array}{cc}x_1 &y_1\\\\x_2&y_2\\end{array}\\right).$ Since the triangle has half the area of the parallelogram, we just need the determinant to be even. The determinant is \\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\\] Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=600 such triangles.",
"As in Solution 1, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. Let the line $41x+y=2009$ intersect the x-axis at $X$ and the y-axis at $Y$. $X$ has coordinates $(49,0)$, and $Y$ has coordinates $(0,2009)$. As such, there are exactly $50$ lattice points on this line that can be used for $P$ and $Q$. WLOG, let the x-coordinate of $P$ be less than the x-coordinate of $Q$. Note that $[OPQ]=[OYX]-[OYP]-[OXQ]$. We know that $OY=2009$ and $OX= 49$; as such, $[OYX]=\\frac{1}{2} \\cdot OY \\cdot OX = \\frac{1}{2} \\cdot 2009 \\cdot 49$. In addition, $[OYP]=\\frac{1}{2} \\cdot 2009 \\cdot x_1$ and $[OXQ]=\\frac{1}{2} \\cdot 49 \\cdot y_2$. Since $2009 \\cdot 49$ is odd, the total area of $OYX$ is not an integer; rather, it is of the form $k + \\frac{1}{2}$ where $k$ is an integer. To ensure $[OPQ]$ has an integral value, exactly one of $[OPY]$ and $[OQX]$ must have an integral value as well (the other must be of the form $k + \\frac{1}{2}$ where $k$ is an integer). Returning to $41x+y=2009$, we notice that integer pairs of $x$ and $y$ that satisfy the equation always have different parities. To satisfy exactly one of $[OPY]$ and $[OQX]$ having an integral area, we must have $x_1$ and $y_2$ having different parities. This is because having an even number for $x_1$ or $y_2$ makes the area of the triangle an integer. We can therefore deduce that $x_1$ and $x_2$ have the same parity. Out of the $50$ usable lattice points for $P$ and $Q$, $25$ have even x-coordinates and $25$ have odd x-coordinates. Since we must pick two points with even x-coordinates or two points with odd x-coordinates, our desired answer is $\\binom{25}{2}+\\binom{25}{2}=300+300=600.",
"As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either on the nonnegative x-axis on the nonnegative y-axis in the first quadrant We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$. Using the point-to-line distance formula, we can calculate the height of $\\triangle OPQ$ from vertex $O$ (the origin) to be: $\\dfrac{|41(0) + 1(0) - 2009|}{\\sqrt{41^2 + 1^2}} = \\dfrac{2009}{\\sqrt{1682}} = \\dfrac{2009}{29\\sqrt2}$ Let $b$ be the base of the triangle that is part of the line $41x+y=2009$. The area is calculated as: $\\dfrac{1}{2}\\times b \\times \\dfrac{2009}{29\\sqrt2} = \\dfrac{2009}{58\\sqrt2}\\times b$ Let the numerical area of the triangle be $k$. So, $k = \\dfrac{2009}{58\\sqrt2}\\times b$ We know that $k$ is an integer. So, $b = 58\\sqrt2 \\times z$, where $z$ is also an integer. We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. Changing the y-coordinates to be in terms of x, we get: $P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$. The distance between them equals $b$. Using the distance formula, we get $PQ = b = \\sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\\sqrt2 \\times |x_2 - x_1| = 58\\sqrt2\\times z$ $(*)$ WLOG, we can assume that $x_2 > x_1$. Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = 600 triangles.",
"As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either on the nonnegative x-axis on the nonnegative y-axis in the first quadrant We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$. Using the point-to-line distance formula, we can calculate the height of $\\triangle OPQ$ from vertex $O$ (the origin) to be: $\\dfrac{|41(0) + 1(0) - 2009|}{\\sqrt{41^2 + 1^2}} = \\dfrac{2009}{\\sqrt{1682}} = \\dfrac{2009}{29\\sqrt2}$ Let $b$ be the base of the triangle that is part of the line $41x+y=2009$. The area is calculated as: $\\dfrac{1}{2}\\times b \\times \\dfrac{2009}{29\\sqrt2} = \\dfrac{2009}{58\\sqrt2}\\times b$ Let the numerical area of the triangle be $k$. So, $k = \\dfrac{2009}{58\\sqrt2}\\times b$ We know that $k$ is an integer. So, $b = 58\\sqrt2 \\times z$, where $z$ is also an integer. We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. Changing the y-coordinates to be in terms of x, we get: $P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$. The distance between them equals $b$. Using the distance formula, we get $PQ = b = \\sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\\sqrt2 \\times |x_2 - x_1| = 58\\sqrt2\\times z$ $(*)$ WLOG, we can assume that $x_2 > x_1$. Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get $29\\sqrt2\\times (x_2-x_1) = 58\\sqrt2\\times z$. Dividing both sides by $29\\sqrt2$, we get $x_2-x_1 = 2z$ As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well. There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2. There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4. ... Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48. Summing them up, we get that there are $2+4+\\dots + 48 = 600 triangles.",
"We present a non-analytic solution; consider the lattice points on the line $41x+y=2009$. The line has intercepts $(0, 2009)$ and $(49, 0)$, so the lattice points for $x=0, 1, \\ldots, 49$ divide the line into $49$ equal segments. Call the area of the large triangle $A$. Any triangle formed with the origin having a base of one of these segments has area $A/49$ (call this value $B$) because the height is the same as that of large triangle, and the bases are in the ratio $1:49$. A segment comprised of $n$ small segments (all adjacent to each other) will have area $nB$. Rewriting in terms of the original area, $A=(\\frac{1}{2})(49)(2009)$, $B=\\frac{2009}{2}$, and $nB=n(\\frac{2009}{2})$. It is clear that in order to have a nonnegative integer for $nB$ as desired, $n$ must be even. This is equivalent to finding the number of ways to choose two distinct $x$-values $x_1$ and $x_2$ ($0 \\leq x_1, x_2 \\leq 49$) such that their positive difference ($n$) is even. Follow one of the previous methods above to choose these pairs and arrive at the answer of 600."
] |
2009-I-12
| 2,009
| 12
|
In right $\triangle ABC$ with hypotenuse $\overline{AB}$ , $AC = 12$ , $BC = 35$ , and $\overline{CD}$ is the altitude to $\overline{AB}$ . Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$ . The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 11
|
I
|
[
"First, note that $AB=37$; let the tangents from $I$ to $\\omega$ have length $x$. Then the perimeter of $\\triangle ABI$ is equal to \\[2(x+AD+DB)=2(x+37).\\] It remains to compute $\\dfrac{2(x+37)}{37}=2+\\dfrac{2}{37}x$. Observe $CD=\\dfrac{12\\cdot 35}{37}=\\dfrac{420}{37}$, so the radius of $\\omega$ is $\\dfrac{210}{37}$. We may also compute $AD=\\dfrac{12^{2}}{37}$ and $DB=\\dfrac{35^{2}}{37}$ by similar triangles. Let $O$ be the center of $\\omega$; notice that \\[\\tan(\\angle DAO)=\\dfrac{DO}{AD}=\\dfrac{210/37}{144/37}=\\dfrac{35}{24}\\] so it follows \\[\\sin(\\angle DAO)=\\dfrac{35}{\\sqrt{35^{2}+24^{2}}}=\\dfrac{35}{\\sqrt{1801}}\\] while $\\cos(\\angle DAO)=\\dfrac{24}{\\sqrt{1801}}$. By the double-angle formula $\\sin(2\\theta)=2\\sin\\theta\\cos\\theta$, it turns out that \\[\\sin(\\angle BAI)=\\sin(2\\angle DAO)=\\dfrac{2\\cdot 35\\cdot 24}{1801}=\\dfrac{1680}{1801}\\] Using the area formula $\\dfrac{1}{2}ab\\sin(C)$ in $\\triangle ABI$, \\[[ABI]=\\left(\\dfrac{1}{2}\\right)\\left(\\dfrac{144}{37}+x\\right)(37)\\left(\\dfrac{1680}{1801}\\right)=\\left(\\dfrac{840}{1801}\\right)(144+37x).\\] But also, using $rs$, \\[[ABI]=\\left(\\dfrac{210}{37}\\right)(37+x).\\] Now we can get \\[\\dfrac{[ABI]}{210}=\\dfrac{4(144+37x)}{1801}=\\dfrac{37+x}{37}\\] so multiplying everything by $37\\cdot 1801=66637$ lets us solve for $x$: \\[21312+5476x=66637+1801x.\\] We have $x=\\dfrac{66637-21312}{5476-1801}=\\dfrac{45325}{3675}=\\dfrac{37}{3}$, and now \\[2+\\dfrac{2}{37}x=2+\\dfrac{2}{3}=\\dfrac{8}{3}\\] giving the answer, $011.",
"Let $O$ be center of the circle and $P$,$Q$ be the two points of tangent such that $P$ is on $BI$ and $Q$ is on $AI$. We know that $AD:CD = CD:BD = 12:35$. Since the ratios between corresponding lengths of two similar diagrams are equal, we can let $AD = 144, CD = 420$ and $BD = 1225$. Hence $AQ = 144, BP = 1225, AB = 1369$ and the radius $r = OD = 210$. Since we have $\\tan OAB = \\frac {35}{24}$ and $\\tan OBA = \\frac{6}{35}$ , we have $\\sin {(OAB + OBA)} = \\frac {1369}{\\sqrt {(1801*1261)}},$$\\cos {(OAB + OBA)} = \\frac {630}{\\sqrt {(1801*1261)}}$. Hence $\\sin I = \\sin {(2OAB + 2OBA)} = \\frac {2*1369*630}{1801*1261}$. let $IP = IQ = x$ , then we have Area$(IBC)$ = $(2x + 1225*2 + 144*2)*\\frac {210}{2}$ = $(x + 144)(x + 1225)* \\sin {\\frac {I}{2}}$. Then we get $x + 1369 = \\frac {3*1369*(x + 144)(x + 1225)}{1801*1261}$. Now the equation looks very complex but we can take a guess here. Assume that $x$ is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of $\\frac {m}{n}$) that can be expressed as $\\frac {a}{b}$ such that $(a,b) = 1$. Look at both sides; we can know that $a$ has to be a multiple of $1369$ and not of $3$ and it's reasonable to think that $b$ is divisible by $3$ so that we can cancel out the $3$ on the right side of the equation. Let's see if $x = \\frac {1369}{3}$ fits. Since $\\frac {1369}{3} + 1369 = \\frac {4*1369}{3}$, and $\\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \\frac {3*1369* \\frac {1801}{3} * \\frac {1261*4}{3}} {1801*1261} = \\frac {4*1369}{3}$. Amazingly it fits! Since we know that $3*1369*144*1225 - 1369*1801*1261 < 0$, the other solution of this equation is negative which can be ignored. Hence $x = 1369/3$. Hence the perimeter is $1225*2 + 144*2 + \\frac {1369}{3} *2 = 1369* \\frac {8}{3}$, and $BC$ is $1369$. Hence $\\frac {m}{n} = \\frac {8}{3}$, $m + n = 11$.",
"As in Solution $1$, let $P$ and $Q$ be the intersections of $\\omega$ with $BI$ and $AI$ respectively. First, by pythagorean theorem, $AB = \\sqrt{12^2+35^2} = 37$. Now the area of $ABC$ is $1/2*12*35 = 1/2*37*CD$, so $CD=\\frac{420}{37}$ and the inradius of $\\triangle ABI$ is $r=\\frac{210}{37}$. Now from $\\triangle CDB \\sim \\triangle ACB$ we find that $\\frac{BC}{BD} = \\frac{AB}{BC}$ so $BD = BC^2/AB = 35^2/37$ and similarly, $AD = 12^2/37$. Note $IP=IQ=x$, $BP=BD$, and $AQ=AD$. So we have $AI = 144/37+x$, $BI = 1225/37+x$. Now we can compute the area of $\\triangle ABI$ in two ways: by heron's formula and by inradius times semiperimeter, which yields $rs=210/37(37+x) = \\sqrt{(37+x)(37-144/37)(37-1225/37)x}$ $210/37(37+x) = 12*35/37 \\sqrt{x(37+x)}$ $37+x = 2 \\sqrt{x(x+37)}$ $x^2+74x+1369 = 4x^2 + 148x$ $3x^2 + 74x - 1369 = 0$ The quadratic formula now yields $x=37/3$. Plugging this back in, the perimeter of $ABI$ is $2s=2(37+x)=2(37+37/3) = 37(8/3)$ so the ratio of the perimeter to $AB$ is $8/3$ and our answer is $8+3=011.",
"As in Solution $2$, let $P$ and $Q$ be the intersections of $\\omega$ with $BI$ and $AI$ respectively. Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point. Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse. Let $x = \\overline{AD} = \\overline{AQ}$. Let $y = \\overline{BD} = \\overline{BP}$. Let $z = \\overline{PI} = \\overline{QI}$. The semi-perimeter of $ABI$ is $x + y + z$. Since the lengths of the sides of $ABI$ are $x + y$, $y + z$ and $x + z$, the square of its area by Heron's formula is $(x+y+z)xyz$. The radius $r$ of $\\omega$ is $\\overline{CD}/2$. Therefore $r^2 = xy/4$. As $\\omega$ is the in-circle of $ABI$, the area of $ABI$ is also $r(x+y+z)$, and so the square area is $r^2(x+y+z)^2$. Therefore \\[(x+y+z)xyz = r^2(x+y+z)^2 = \\frac{xy(x+y+z)^2}{4}\\] Dividing both sides by $xy(x+y+z)/4$ we get: \\[4z = (x+y+z),\\] and so $z = (x+y)/3$. The semi-perimeter of $ABI$ is therefore $\\frac{4}{3}(x+y)$ and the whole perimeter is $\\frac{8}{3}(x+y)$. Now $x + y = \\overline{AB}$, so the ratio of the perimeter of $ABI$ to the hypotenuse $\\overline{AB}$ is $8/3$ and our answer is $8+3=011",
"[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,O,X; C=origin; A=(0,12); B=(18,0); D=foot(C,A,B); O = (C+D)/2; real r = length(D-C)/2; path c = CR(O, r); pair OA = (O+A)/2; real rA = length(A-O)/2; pair Ap = OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot(\"$A$\", A, N); dot(\"$B$\", B, SE); dot(\"$C$\", C, SW); dot(\"$D$\", D, 0.2*(D-C)); dot(\"$I$\", X, 0.5*(X-C)); dot(\"$P$\", Ap, 0.3*(Ap-O)); dot(\"$Q$\", Bp, 0.3*(Bp-O)); dot(\"$O$\", O, W); label(\"$\\beta$\",B,10*dir(157)); label(\"$\\alpha$\",A,5*dir(-55)); label(\"$\\theta$\",X,5*dir(55)); [/asy] Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\\alpha$, $\\beta$, and $\\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\\alpha + \\beta + \\theta = 90^\\circ$, so \\[\\theta = 90^\\circ - (\\alpha+\\beta).\\] The perimeter of $\\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired ratio is then \\[\\rho = 2\\left(1+\\frac z{37}\\right)\\] We need to find $z$. In $\\triangle OPI$, $z=r\\cot\\theta = r\\tan (\\alpha+\\beta)$. We get \\[\\tan\\alpha = \\frac{OD}{AD} = \\frac 12 \\frac{CD}{AD} = \\frac 12 \\tan A = \\frac 12 \\frac{BC}{AC} = \\frac{35}{24}.\\] Similarly, $\\tan\\beta = \\tfrac 6{35}$. Then \\[z = r\\cdot \\tan (\\alpha+\\beta) = r\\cdot \\frac{\\tan\\alpha + \\tan\\beta}{1-\\tan\\alpha\\tan\\beta}= \\frac{37^2\\cdot r}{18\\cdot 35}\\] Computing $[ABC]$ in two ways we get $CD = \\tfrac{12\\cdot 35}{37}$, so $r=\\tfrac{6\\cdot 35}{37}$. Using this value of $r$ we get $z=\\tfrac {37}3$. Thus \\[\\rho = 2\\left(1+\\frac 1{3}\\right) = \\frac 8{3},\\] and $8+3=011.",
"This solution is not a real solution and is solving the problem with a ruler and compass. Draw $AC = 4.8, BC = 14, AB = 14.8$. Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$. We then find the ratio to be around $\\frac{39.45}{14.8}$. Using long division, we find that this ratio is approximately 2.666, which you should recognize as $\\frac{8}{3}$. Since this seems reasonable, we find that the answer is $11 ~ilp",
"Denoting three tangents has length $h_1,h_2,h_3$ while $h_1,h_3$ lies on $AB$ with $h_1>h_3$.The area of $ABC$ is $1/2*12*35 = 1/2*37*CD$, so $CD=\\frac{420}{37}$ and the inradius of $\\triangle ABI$ is $r=\\frac{210}{37}$.As we know that the diameter of the circle is the height of $\\triangle ACB$ from $C$ to $AB$. Assume that $\\tan\\alpha=\\frac{h_1}{r}$ and $\\tan\\beta=\\frac{h_3}{r}$ and $\\tan\\omega=\\frac{h_2}{r}$. But we know that $\\tan(\\alpha+\\beta)=-\\tan(180-\\alpha-\\beta)=-\\tan\\omega$ According to the basic computation, we can get that $\\tan(\\alpha)=\\frac{35}{6}$; $\\tan(\\beta)=\\frac{24}{35}$ So we know that $\\tan(\\omega)=\\frac{1369}{630}$ according to the tangent addition formula. Hence, it is not hard to find that the length of $h_2$ is $\\frac{37}{3}$. According to basic addition and division, we get the answer is $\\frac{8}{3}$ which leads to $8+3=11 ~bluesoul",
"Notice $CD$ is the altitude of $\\triangle ABC$, and so $CD = \\frac{AC \\cdot BC}{AB} = \\frac{210}{37}.$ This means in the inradius of the circle is $r=\\frac{CD}{2}=\\frac{210}{37}.$ Next, by $\\triangle ABC \\sim \\triangle ACD \\sim \\triangle CBD$, we have $AD=CD \\cdot \\frac{12}{35} = \\frac{144}{37}$ and $BD = CD \\cdot \\frac{35}{12} = \\frac{1225}{37}$. Lastly, denote the tangent point of $AI, BI$ with the circle as $P, Q$ respectively. Then by the properties of inscribed circles, we have that $AP=AD=\\frac{144}{37}$ and $BQ=BD=\\frac{1225}{37}$, and $PI=QI=x$. We want to make an algebraic expression utilizing what we know about ABI. We use the inradius formula: $r = \\frac{[\\triangle ABI]}{s}$. We have $s=37+x$. With Heron's (very basic calculations) we get $[\\triangle ABI] = \\sqrt{(x+37)(x)(\\frac{1225}{37})(\\frac{144}{37})} = \\frac{420}{37} \\sqrt{x(37+x)}.$ Using the inradius formula, we get \\[\\frac{210}{37} = \\frac{\\frac{420}{37} \\sqrt{x(37+x)}}{37+x}\\] which simplifies to $x=\\frac{37}{3}$. The answer is then $\\frac{8}{3}. 11~ brocolimanx",
"Notice $CD$ is the altitude of $\\triangle ABC$, and so $CD = \\frac{AC \\cdot BC}{AB} = \\frac{210}{37}.$ This means in the inradius of the circle is $r=\\frac{CD}{2}=\\frac{210}{37}.$ Next, by $\\triangle ABC \\sim \\triangle ACD \\sim \\triangle CBD$, we have $AD=CD \\cdot \\frac{12}{35} = \\frac{144}{37}$ and $BD = CD \\cdot \\frac{35}{12} = \\frac{1225}{37}$. Lastly, denote the tangent point of $AI, BI$ with the circle as $P, Q$ respectively. Then by the properties of inscribed circles, we have that $AP=AD=\\frac{144}{37}$ and $BQ=BD=\\frac{1225}{37}$, and $PI=QI=x$. We want to make an algebraic expression utilizing what we know about ABI. We use the inradius formula: $r = \\frac{[\\triangle ABI]}{s}$. We have $s=37+x$. With Heron's (very basic calculations) we get $[\\triangle ABI] = \\sqrt{(x+37)(x)(\\frac{1225}{37})(\\frac{144}{37})} = \\frac{420}{37} \\sqrt{x(37+x)}.$ Using the inradius formula, we get \\[\\frac{210}{37} = \\frac{\\frac{420}{37} \\sqrt{x(37+x)}}{37+x}\\] which simplifies to $x=\\frac{37}{3}$. The answer is then $\\frac{8}{3}. 11~ brocolimanx"
] |
2009-I-13
| 2,009
| 13
|
The terms of the sequence $\{a_i\}$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$ .
| 90
|
I
|
[
"Our expression is \\[a_{n + 2} = \\frac {a_n + 2009} {1 + a_{n + 1}}.\\] Manipulate this to obtain: \\[a_{n + 2}a_{n + 1}+a_{n+2}= a_n + 2009.\\] Our goal is to cancel terms. If we substitute in $n+1$ for $n,$ we get: \\[a_{n+3}a_{n+2}+a_{n+3}=a_{n+1}+2009.\\] Subtracting these two equations and manipulating the expression yields: \\[(a_{n+2}+1)(a_{n+3}-a_{n+1})=a_{n+2}-a_n.\\] Notice we have the form $a_{k+2}-a_k$ on both sides. Let $b_n=a_{n+2}-a_n.$ Then: \\[b_{n+1}(a_{n+2}+1)=b_n.\\] Notice that since $a_n$ is always an integer, $a_{n+2}+1$ and $b_n$ must also always be an integer. It is also clear that $b_n$ is a multiple of $b_{n+1},$ implying a decreasing sequence. However, if the decreasing factor is nonzero, we will eventually have a $b_k$ that is not an integer, contradicting our conditions for $b_n$. Thus, we need either $a_{n+2}+1=0 \\Rightarrow a_{n+2}=-1$ (impossible as $a_n$ for all indices must be positive integers) or $b_n=0 \\Rightarrow a_{n+2}=a_n.$ Given this, we want to find the minimum of $a_1+a_2.$ We have, from the problem: \\[\\frac{a_1+2009}{1+a_2}=a_3=a_1 \\Rightarrow a_1 a_2 = 2009.\\] By AM-GM, to minimize this, we have to make $a_1$ and $a_2$ factors as close as possible. Hence, the smallest possible sum is $41+49=90.$ ~mathboy282",
"This question is guessable but let's prove our answer \\[a_{n + 2} = \\frac {a_n + 2009} {1 + a_{n + 1}}\\] \\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\\] \\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\\] lets put $n+1$ into $n$ now \\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\\] and set them equal now \\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\\] \\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\\] let's rewrite it \\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\\] Let's make it look nice and let $b_n=a_{n + 2}-a_n$ \\[(b_{n+1})(a_{n + 2}+1)= b_n\\] Since $b_n$ and $b_{n+1}$ are integers, we can see $b_n$ is divisible by $b_{n+1}$ But we can't have an infinite sequence of proper factors, unless $b_n=0$ Thus, $a_{n + 2}-a_n=0$ \\[a_{n + 2}=a_n\\] So now, we know $a_3=a_1$ \\[a_{3} = \\frac {a_1 + 2009} {1 + a_{2}}\\] \\[a_{1} = \\frac {a_1 + 2009} {1 + a_{2}}\\] \\[a_{1}+a_{1}a_{2} = a_1 + 2009\\] \\[a_{1}a_{2} = 2009\\] To minimize $a_{1}+a_{2}$, we need $41$ and $49$ Thus, our answer $= 41+49=090",
"If $a_{n} \\ne \\frac {2009}{a_{n+1}}$, then either \\[a_{n} = \\frac {a_{n}}{1} < \\frac {a_{n} + 2009}{1 + a_{n+1}} < \\frac {2009}{a_{n+1}}\\] or \\[\\frac {2009}{a_{n+1}} < \\frac {2009 + a_{n}}{a_{n+1} + 1} < \\frac {a_{n}}{1} = a_{n}\\] All the integers between $a_{n}$ and $\\frac {2009}{a_{n+1}}$ would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term. So $a_{n} = \\frac {2009}{a_{n+1}}$, which $a_{n} \\cdot a_{n+1} = 2009$. When $n = 1$, $a_{1} \\cdot a_{2} = 2009$. The smallest sum of two factors which have a product of $2009$ is $41 + 49=090",
"Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \\begin{align*} a_{1} &= a \\\\ a_{2} &= b \\\\ a_{3} &=\\frac{a+2009}{1+b} \\\\ a_{4} &=\\frac{(b+1)(b+2009)}{a+b+2010} \\\\ \\end{align*} The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$ Solving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41$ or vice versa for an answer of $90 This solution is VERY non rigorous and not recommended.",
"Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \\begin{align*} a_{1} &= a \\\\ a_{2} &= b \\\\ a_{3} &=\\frac{a+2009}{1+b} \\\\ a_{4} &=\\frac{(b+1)(b+2009)}{a+b+2010} \\\\ \\end{align*} The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$ Solving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41$ or vice versa for an answer of $90 This solution is VERY non rigorous and not recommended."
] |
2009-I-14
| 2,009
| 14
|
For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ .
| 905
|
I
|
[
"Because the order of the $a$'s doesn't matter, we simply need to find the number of $1$s $2$s $3$s and $4$s that minimize $S_2$. So let $w, x, y,$ and $z$ represent the number of $1$s, $2$s, $3$s, and $4$s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a$s, we know that $w + x + y + z = 350$. We also know that $w + 2x + 3y + 4z = 513$ and $w + 16x + 81y + 256z = 4745$. We can now solve these down to two variables: \\[w = 350 - x - y - z\\] Substituting this into the second and third equations, we get \\[x + 2y + 3z = 163\\] and \\[15x + 80y + 255z = 4395.\\] The second of these can be reduced to \\[3x + 16y + 51z = 879.\\] Now we substitute $x$ from the first new equation into the other new equation. \\[x = 163 - 2y - 3z\\] \\[3(163 - 2y - 3z) + 16y + 51z = 879\\] \\[489 + 10y + 42z = 879\\] \\[5y + 21z = 195\\] Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$. If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$, $x = 112$, $y = 18$, and $z = 5$, so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = 905."
] |
2009-I-15
| 2,009
| 15
|
In triangle $ABC$ , $AB = 10$ , $BC = 14$ , and $CA = 16$ . Let $D$ be a point in the interior of $\overline{BC}$ . Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$ , respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$ . The maximum possible area of $\triangle BPC$ can be expressed in the form $a - b\sqrt {c}$ , where $a$ , $b$ , and $c$ are positive integers and $c$ is not divisible by the square of any prime. Find $a + b + c$ .
| 150
|
I
|
[
"First, by the Law of Cosines, we have \\[\\cos BAC = \\frac {16^2 + 10^2 - 14^2}{2\\cdot 10 \\cdot 16} = \\frac {256+100-196}{320} = \\frac {1}{2},\\] so $\\angle BAC = 60^\\circ$. Let $O_1$ and $O_2$ be the circumcenters of triangles $BI_BD$ and $CI_CD$, respectively. We first compute \\[\\angle BO_1D = \\angle BO_1I_B + \\angle I_BO_1D = 2\\angle BDI_B + 2\\angle I_BBD.\\] Because $\\angle BDI_B$ and $\\angle I_BBD$ are half of $\\angle BDA$ and $\\angle ABD$, respectively, the above expression can be simplified to \\[\\angle BO_1D = \\angle BO_1I_B + \\angle I_BO_1D = 2\\angle BDI_B + 2\\angle I_BBD = \\angle ABD + \\angle BDA.\\] Similarly, $\\angle CO_2D = \\angle ACD + \\angle CDA$. As a result \\begin{align*}\\angle CPB &= \\angle CPD + \\angle BPD \\\\&= \\frac {1}{2} \\cdot \\angle CO_2D + \\frac {1}{2} \\cdot \\angle BO_1D \\\\&= \\frac {1}{2}(\\angle ABD + \\angle BDA + \\angle ACD + \\angle CDA) \\\\&= \\frac {1}{2} (2 \\cdot 180^\\circ - \\angle BAC) \\\\&= \\frac {1}{2} \\cdot 300^\\circ = 150^\\circ.\\end{align*} Therefore $\\angle CPB$ is constant ($150^\\circ$). Also, $P$ is $B$ or $C$ when $D$ is $B$ or $C$. Let point $L$ be on the same side of $\\overline{BC}$ as $A$ with $LC = LB = BC = 14$; $P$ is on the circle with $L$ as the center and $\\overline{LC}$ as the radius, which is $14$. The shortest (and only) distance from $L$ to $\\overline{BC}$ is $7\\sqrt {3}$. When the area of $\\triangle BPC$ is the maximum, the distance from $P$ to $\\overline{BC}$ has to be the greatest. In this case, it's $14 - 7\\sqrt {3}$. The maximum area of $\\triangle BPC$ is \\[\\frac {1}{2} \\cdot 14 \\cdot (14 - 7\\sqrt {3}) = 98 - 49 \\sqrt {3}\\] and the requested answer is $98 + 49 + 3 = 150.",
"From Law of Cosines on $\\triangle{ABC}$, \\[\\cos{A}=\\frac{16^2+10^2-14^2}{2\\cdot 10\\cdot 16}=\\frac{1}{2}\\implies\\angle{A}=60^\\circ.\\]Now, \\[\\angle{CI_CD}+\\angle{BI_BD}=180^\\circ+\\frac{\\angle{A}}{2}=210^\\circ.\\]Since $CI_CDP$ and $BI_BDP$ are cyclic quadrilaterals, it follows that \\[\\angle{BPC}=\\angle{CPD}+\\angle{DPB}=(180^\\circ-\\angle{CI_CD})+(180^\\circ-\\angle{BI_BD})=360^\\circ-210^\\circ=150^\\circ.\\]Next, applying Law of Cosines on $\\triangle{CPB}$, \\begin{align*} & BC^2=14^2=PC^2+PB^2+2\\cdot PB\\cdot PC\\cdot\\frac{\\sqrt{3}}{2} \\\\ & \\implies \\frac{PC^2+PB^2-196}{PC\\cdot PB}=-\\sqrt{3} \\\\ & \\implies \\frac{PC}{PB}+\\frac{PB}{PC}-\\frac{196}{PC\\cdot PB}=-\\sqrt{3} \\\\ & \\implies PC\\cdot PB = 196\\left(\\frac{1}{\\frac{PC}{PB}+\\frac{PB}{PC}+\\sqrt{3}}\\right). \\end{align*} By AM-GM, $\\frac{PC}{PB}+\\frac{PB}{PC}\\geq{2}$, so \\[PB\\cdot PC\\leq 196\\left(\\frac{1}{2+\\sqrt{3}}\\right)=196(2-\\sqrt{3}).\\]Finally, \\[[\\triangle{BPC}]=\\frac12 \\cdot PB\\cdot PC\\cdot\\sin{150^\\circ}=\\frac14 \\cdot PB\\cdot PC,\\]and the maximum area would be $49(2-\\sqrt{3})=98-49\\sqrt{3},$ so the answer is $150.",
"Proceed as in Solution 2 until you find $\\angle CPB = 150$. The locus of points $P$ that give $\\angle CPB = 150$ is a fixed arc from $B$ to $C$ ($P$ will move along this arc as $D$ moves along $BC$) and we want to maximise the area of [$\\triangle BPC$]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\\triangle BPC$ isosceles). We know that $BC=14$ and $\\angle CPB = 150$, so $\\angle PBC = \\angle PCB = 15$. Let $O$ be the foot of the perpendicular from $P$ to line $BC$. Then the area of [$\\triangle BPC$] is the same as $7OP$ because base $BC$ has length $14$. We can split $\\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$, with $BO=CO=7$ and $OP=7 \\tan 15=7(2-\\sqrt{3})=14-7\\sqrt3$. Then, the area of [$\\triangle BPC$] is equal to $7 \\cdot OP=98-49\\sqrt{3}$, and so the answer is $98+49+3=150."
] |
2009-II-1
| 2,009
| 1
|
Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.
| 114
|
II
|
[
"Let $x$ be the amount of paint in a strip. The total amount of paint is $130+164+188=482$. After painting the wall, the total amount of paint is $482-4x$. Because the total amount in each color is the same after finishing, we have \\[\\frac{482-4x}{3} = 130-x\\] \\[482-4x=390-3x\\] \\[x=92\\] Our answer is $482-4\\cdot 92 = 482 - 368 = 114. ~ carolynlikesmath",
"Let the stripes be $b, r, w,$ and $p$, respectively. Let the red part of the pink be $\\frac{r_p}{p}$ and the white part be $\\frac{w_p}{p}$ for $\\frac{r_p+w_p}{p}=p$. Since the stripes are of equal size, we have $b=r=w=p$. Since the amounts of paint end equal, we have $130-b=164-r-\\frac{r_p}{p}=188-w-\\frac{w_p}{p}$. Thus, we know that \\[130-p=164-p-\\frac{r_p}{p}=188-p-\\frac{w_p}{p}\\] \\[130=164-\\frac{r_p}{p}=188-\\frac{w_p}{p}\\] \\[r_p=34p, w_p=58p\\] \\[\\frac{r_p+w_p}{p}=92=p=b.\\] Each paint must end with $130-92=38$ oz left, for a total of $3 \\cdot 38 = 114 oz.",
"Let the stripes be $b, r, w,$ and $p$, respectively. Let the red part of the pink be $\\frac{r_p}{p}$ and the white part be $\\frac{w_p}{p}$ for $\\frac{r_p+w_p}{p}=p$. Since the stripes are of equal size, we have $b=r=w=p$. Since the amounts of paint end equal, we have $130-b=164-r-\\frac{r_p}{p}=188-w-\\frac{w_p}{p}$. Thus, we know that \\[130-p=164-p-\\frac{r_p}{p}=188-p-\\frac{w_p}{p}\\] \\[130=164-\\frac{r_p}{p}=188-\\frac{w_p}{p}\\] \\[r_p=34p, w_p=58p\\] \\[\\frac{r_p+w_p}{p}=92=p=b.\\] Each paint must end with $130-92=38$ oz left, for a total of $3 \\cdot 38 = 114 oz.",
"After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$. The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$. The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = 114.",
"We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are $188 - r$ and $164 - r$ respectively. The pink stripe is also r ounces of paint, but let there be $k$ ounces of red paint in the mixture and $r - k$ ounces of white paint. We now have two equations: $164 - r - k = 188 - r - (r-k)$ and $164 - r - k = 130 - r$. Solving yields k = 34 and r = 92. We now see that there will be $130 - 92 = 38$ ounces of paint left in each can. $38 * 3 = 114",
"Let the amount of paint each stripe painted used be $x$. Also, let the amount of paint of each color left be $y$. 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, $164 + 188 = 352$ and obtain the following equations : $352 - 3x = 2y$ and $130 - x = y$. Solve to obtain $x = 92$. Therefore $y$ is $130 - 92 = 38$, with three cans of equal amount of paint, the answer is $38 * 3 = 114.",
"Let $x$ be the number of ounces of paint needed for a single stripe. We know that in the end, the total amount of red and white paint will equal double the amount of blue paint. After painting, the amount of red and white paint remaining is equal to $164+188-2x$, and then minus another $x$ for the pink stripe. The amount of blue paint remaining is equal to $130-x$. So, we get the equation $2\\cdot(130-x)=164+188-3x$. Simplifying, we get $x=38$ and $3x=114. ~LegionOfAvatars",
"Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$. The answer is $130+164+188-4(92)=114"
] |
2009-II-2
| 2,009
| 2
|
Suppose that $a$ , $b$ , and $c$ are positive real numbers such that $a^{\log_3 7} = 27$ , $b^{\log_7 11} = 49$ , and $c^{\log_{11}25} = \sqrt{11}$ . Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\]
| 469
|
II
|
[
"First, we have: \\[x^{(\\log_y z)^2} = x^{\\left( (\\log_y z)^2 \\right) } = x^{(\\log_y z) \\cdot (\\log_y z) } = \\left( x^{\\log_y z} \\right)^{\\log_y z}\\] Now, let $x=y^w$, then we have: \\[x^{\\log_y z} = \\left( y^w \\right)^{\\log_y z} = y^{w\\log_y z} = y^{\\log_y (z^w)} = z^w\\] This is all we need to evaluate the given formula. Note that in our case we have $27=3^3$, $49=7^2$, and $\\sqrt{11}=11^{1/2}$. We can now compute: \\[a^{(\\log_3 7)^2} = \\left( a^{\\log_3 7} \\right)^{\\log_3 7} = 27^{\\log_3 7} = (3^3)^{\\log_3 7} = 7^3 = 343\\] Similarly, we get \\[b^{(\\log_7 11)^2} = (7^2)^{\\log_7 11} = 11^2 = 121\\] and \\[c^{(\\log_{11} 25)^2} = (11^{1/2})^{\\log_{11} 25} = 25^{1/2} = 5\\] and therefore the answer is $343+121+5 = 469.",
"We know from the first three equations that $\\log_a27 = \\log_37$, $\\log_b49 = \\log_711$, and $\\log_c\\sqrt{11} = \\log_{11}25$. Substituting, we find \\[a^{(\\log_a27)(\\log_37)} + b^{(\\log_b49)(\\log_711)} + c^{(\\log_c\\sqrt {11})(\\log_{11}25)}.\\] We know that $x^{\\log_xy} =y$, so we find \\[27^{\\log_37} + 49^{\\log_711} + \\sqrt {11}^{\\log_{11}25}\\] \\[(3^{\\log_37})^3 + (7^{\\log_711})^2 + ({11^{\\log_{11}25}})^{1/2}.\\] The $3$ and the $\\log_37$ cancel to make $7$, and we can do this for the other two terms. Thus, our answer is \\[7^3 + 11^2 + 25^{1/2}\\] \\[= 343 + 121 + 5\\] \\[= 469.\\]",
"First, let us take the log base 3 of the first expression. We get $\\log_3{a^{\\log_3{7}}} = 3$. Simplifying, we get \\[(\\log_3{7})(\\log_3{a}) = 3\\]. So, \\[\\log_3 a = \\frac{3}{\\log_3{7}}\\], and \\[a = 3^\\frac{3}{log_3{7}}\\]. We can repeat the same process for the other equations, giving us \\[b = 7^\\frac{2}{\\log_7{11}}\\], and \\[c = (\\sqrt{11})^\\frac{1}{\\log_ {11}{25}}\\]. Raising $a$ to the power of $(\\log_3{7})^2$, we get \\[3^{3\\log_3{7}} = 3^{\\log_3{343}} = 343\\]. Repeating a similar process for the other ones(you have to turn the square root to a fractional power for c), we get $343+121+5 = 469 ~idk12345678"
] |
2009-II-3
| 2,009
| 3
|
In rectangle $ABCD$ , $AB=100$ . Let $E$ be the midpoint of $\overline{AD}$ . Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$ .
| 141
|
II
|
[
"Solution 1 [asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label(\"\\(E\\)\",E,N); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,SW); label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",D,NE); label(\"\\(F\\)\",F,W); label(\"\\(100\\)\",Q,W); [/asy] From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\\triangle FBA \\sim \\triangle BCA$, and $\\triangle FBA \\sim \\triangle ABE$, so $\\triangle ABE \\sim \\triangle BCA$. This gives $\\frac {AE}{AB}= \\frac {AB}{BC}$. $AE=\\frac{AD}{2}$ and $BC=AD$, so $\\frac {AD}{2AB}= \\frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \\sqrt{2}$, or $100 \\sqrt{2}$, so the answer is $141.",
"[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label(\"\\(E\\)\",E,N); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,SW); label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",D,NE); label(\"\\(F\\)\",F,W); label(\"\\(100\\)\",Q,W); [/asy] From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\\triangle FBA \\sim \\triangle BCA$, and $\\triangle FBA \\sim \\triangle ABE$, so $\\triangle ABE \\sim \\triangle BCA$. This gives $\\frac {AE}{AB}= \\frac {AB}{BC}$. $AE=\\frac{AD}{2}$ and $BC=AD$, so $\\frac {AD}{2AB}= \\frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \\sqrt{2}$, or $100 \\sqrt{2}$, so the answer is $141.",
"Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\\overline{AC}$ is $x$ and the slope of $\\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is, \\[x\\cdot-\\frac{x}{2}=-1,\\] which implies that $-x^2=-2$ or $x=\\sqrt 2$. Therefore $AD=100\\sqrt 2\\approx 141.42$ so $\\lfloor AD\\rfloor=141.",
"Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let the positive y-axis be in the direction of ray $BA$. Thus, the vector $BE=(x,100)$ and the vector $AC=(2x,-100)$ are perpendicular and thus have a dot product of 0. Thus, calculating the dot product: \\[x\\cdot2x+(100)\\cdot(-100)=2x^2-10000=0\\] \\[2x^2-10000=0\\rightarrow x^2=5000\\] Substituting AD/2 for x: \\[(AD/2)^2=5000\\rightarrow AD^2=20000\\] \\[AD=100\\sqrt2\\] Solution 4 [asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); draw(C--X--E); label(\"\\(E\\)\",E,N); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,SW); label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",D,NE); label(\"\\(X\\)\",X,S); label(\"\\(100\\)\",Q,W); [/asy] Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \\parallel AC$, $\\angle BEX=90^\\circ$ by the problem statement, so $\\triangle BEX$ is right. Letting $AE=y$, we have $BE=\\sqrt{100^2+y^2}$ and $AC=EX=\\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\\sqrt{100^2+y^2}^2+\\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have \\[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\\] \\[2\\cdot 100^2 = 4y^2\\] \\[\\frac{100^2}{2}=y^2\\] \\[\\frac{100}{\\sqrt{2}}=y\\] \\[\\frac{100\\sqrt{2}}{2}=y\\] \\[100\\sqrt{2}=2y=AD\\], so the answer is $141.",
"[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5),X=(21,0); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); draw(C--X--E); label(\"\\(E\\)\",E,N); label(\"\\(A\\)\",A,NW); label(\"\\(B\\)\",B,SW); label(\"\\(C\\)\",C,SE); label(\"\\(D\\)\",D,NE); label(\"\\(X\\)\",X,S); label(\"\\(100\\)\",Q,W); [/asy] Draw $CX$ and $EX$ to form a parallelogram $AEXC$. Since $EX \\parallel AC$, $\\angle BEX=90^\\circ$ by the problem statement, so $\\triangle BEX$ is right. Letting $AE=y$, we have $BE=\\sqrt{100^2+y^2}$ and $AC=EX=\\sqrt{100^2+(2y)^2}$. Since $CX=EA$, $\\sqrt{100^2+y^2}^2+\\sqrt{100^2+(2y)^2}=(3y)^2$. Solving this, we have \\[100^2+ 100^2 + y^2 + 4y^2 = 9y^2\\] \\[2\\cdot 100^2 = 4y^2\\] \\[\\frac{100^2}{2}=y^2\\] \\[\\frac{100}{\\sqrt{2}}=y\\] \\[\\frac{100\\sqrt{2}}{2}=y\\] \\[100\\sqrt{2}=2y=AD\\], so the answer is $141."
] |
2009-II-4
| 2,009
| 4
|
A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ .
| 89
|
II
|
[
"Resolving the ambiguity The problem statement is confusing, as it can be interpreted in two ways: Either as \"there is a $k>1$ such that the child in $k$-th place had eaten $n+2-2k$ grapes\", or \"for all $k$, the child in $k$-th place had eaten $n+2-2k$ grapes\". The second meaning was apparently the intended one. Hence we will restate the problem statement in this way: A group of $c$ children held a grape-eating contest. When the contest was over, the following was true: There was a $n$ such that for each $k$ between $1$ and $c$, inclusive, the child in $k$-th place had eaten exactly $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$. Find the smallest possible value of $n$. Solution 1 The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$-st place), difference $d=-2$, and number of terms $c$. We can easily compute that this sum is equal to $c(n-c+1)$. Hence we have the equation $2009=c(n-c+1)$, and we are looking for a solution $(n,c)$, where both $n$ and $c$ are positive integers, $n\\geq 2(c-1)$, and $n$ is minimized. (The condition $n\\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.) The prime factorization of $2009$ is $2009=7^2 \\cdot 41$. Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$: $c=1$, $n-c+1=2009$, then $n=2009$ is a solution. $c=7$, $n-c+1=7\\cdot 41=287$, then $n=293$ is a solution. $c=41$, $n-c+1=7\\cdot 7=49$, then $n=89$ is a solution. In each of the other three cases, $n$ will obviously be less than $2(c-1)$, hence these are not valid solutions. The smallest valid solution is therefore $c=41$, $n=089.) ~MC413551",
"The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$-st place), difference $d=-2$, and number of terms $c$. We can easily compute that this sum is equal to $c(n-c+1)$. Hence we have the equation $2009=c(n-c+1)$, and we are looking for a solution $(n,c)$, where both $n$ and $c$ are positive integers, $n\\geq 2(c-1)$, and $n$ is minimized. (The condition $n\\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.) The prime factorization of $2009$ is $2009=7^2 \\cdot 41$. Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$: $c=1$, $n-c+1=2009$, then $n=2009$ is a solution. $c=7$, $n-c+1=7\\cdot 41=287$, then $n=293$ is a solution. $c=41$, $n-c+1=7\\cdot 7=49$, then $n=89$ is a solution. In each of the other three cases, $n$ will obviously be less than $2(c-1)$, hence these are not valid solutions. The smallest valid solution is therefore $c=41$, $n=089.) ~MC413551",
"If the first child ate $n=2m$ grapes, then the maximum number of grapes eaten by all the children together is $2m + (2m-2) + (2m-4) + \\cdots + 4 + 2 = m(m+1)$. Similarly, if the first child ate $2m-1$ grapes, the maximum total number of grapes eaten is $(2m-1)+(2m-3)+\\cdots+3+1 = m^2$. For $m=44$ the value $m(m+1)=44\\cdot 45 =1980$ is less than $2009$. Hence $n$ must be at least $2\\cdot 44+1=89$. For $n=89$, the maximum possible sum is $45^2=2025$. And we can easily see that $2009 = 2025 - 16 = 2025 - (1+3+5+7)$, hence $2009$ grapes can indeed be achieved for $n=89$ by dropping the last four children. Hence we found a solution with $n=89$ and $45-4=41$ kids, and we also showed that no smaller solution exists. Therefore the answer is $089.) ~MC413551",
"If the winner ate n grapes, then 2nd place ate $n+2-4=n-2$ grapes, 3rd place ate $n+2-6=n-4$ grapes, 4th place ate $n-6$ grapes, and so on. Our sum can be written as $n+(n-2)+(n-4)+(n-6)\\dots$. If there are x places, we can express this sum as $(x+1)n-x(x+1)$, as there are $(x+1)$ occurrences of n, and $(2+4+6+\\dots)$ is equal to $x(x+1)$. This can be factored as $(x+1)(n-x)=2009$. Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If $x+1=41$, then $x=40$ and $n=40+49=089.) ~MC413551",
"If the winner ate n grapes, then 2nd place ate $n+2-4=n-2$ grapes, 3rd place ate $n+2-6=n-4$ grapes, 4th place ate $n-6$ grapes, and so on. Our sum can be written as $n+(n-2)+(n-4)+(n-6)\\dots$. If there are x places, we can express this sum as $(x+1)n-x(x+1)$, as there are $(x+1)$ occurrences of n, and $(2+4+6+\\dots)$ is equal to $x(x+1)$. This can be factored as $(x+1)(n-x)=2009$. Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If $x+1=41$, then $x=40$ and $n=40+49=089.) ~MC413551"
] |
2009-II-5
| 2,009
| 5
|
Equilateral triangle $T$ is inscribed in circle $A$ , which has radius $10$ . Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$ . Circles $C$ and $D$ , both with radius $2$ , are internally tangent to circle $A$ at the other two vertices of $T$ . Circles $B$ , $C$ , and $D$ are all externally tangent to circle $E$ , which has radius $\dfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90); pair Ep=(0,4-27/5); pair[] dotted={A,B,C,D,Ep}; draw(Circle(A,10)); draw(Circle(B,3)); draw(Circle(C,2)); draw(Circle(D,2)); draw(Circle(Ep,27/5)); dot(dotted); label("$E$",Ep,E); label("$A$",A,W); label("$B$",B,W); label("$C$",C,W); label("$D$",D,E); [/asy]
| 32
|
II
|
[
"Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC = 8$, and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$, we obtain $(6+p)^2 =p^2 + 64 - 2(8)(p) \\cos 60$. The $2$ and the $\\cos 60$ terms cancel out: $p^2 + 12p +36 = p^2 + 64 - 8p$ $12p+ 36 = 64 - 8p$ $p =\\frac {28}{20} = \\frac {7}{5}$. The radius of circle $E$ is $4 + \\frac {7}{5} = \\frac {27}{5}$, so the answer is $27 + 5 = 032.",
"Draw $\\overline{CD}$ from circle center $C$ to center $D$. Let its midpoint be point $F$. Draw $\\overline{BF}$ perpendicular to line segment $CD$ and intersecting $CD$ at point $F$. Let $G$ be the common external tangent point of circle $B$ and circle $E$. $\\overline{GA} = 4$ Circle centers $A$ and $E$ lie on $\\overline{BF}$. Draw $\\overline{AC}$ from circle center $A$ to center $C$. $\\overline{AC}$ has length $8$. Triangle $ACF$ is a right triangle with $\\angle ACF = 30^{\\circ}$. (If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle $ACF$ is similar to one of the triangles). $\\overline{AF} = 4$ and $\\overline{CF} = 4\\sqrt{3}$. Let the radius of circle $E$ have length $r$. Draw $\\overline{EC}$ from circle center $E$ to center $C$. $\\overline{EC}$ has length $r + 2$. $\\overline{EF}$ has length $GA + AF - GE = 8 - r$. Since $CEF$ is a right triangle, by the Pythagorean theorem $EF^2 + CF^2 = EC^2$. $(8 - r)^2 + (4\\sqrt{3})^2 = (r + 2)^2$. Solving, $r = \\frac {27}{5}$ and the answer is $27 + 5 = 032. -unhappyfarmer",
"Draw $\\overline{CD}$ from circle center $C$ to center $D$. Let its midpoint be point $F$. Draw $\\overline{BF}$ perpendicular to line segment $CD$ and intersecting $CD$ at point $F$. Let $G$ be the common external tangent point of circle $B$ and circle $E$. $\\overline{GA} = 4$ Circle centers $A$ and $E$ lie on $\\overline{BF}$. Draw $\\overline{AC}$ from circle center $A$ to center $C$. $\\overline{AC}$ has length $8$. Triangle $ACF$ is a right triangle with $\\angle ACF = 30^{\\circ}$. (If you take the original equilateral triangle with centroid A and draw the medians, dividing the triangle into 6 30 degree triangles, you find that triangle $ACF$ is similar to one of the triangles). $\\overline{AF} = 4$ and $\\overline{CF} = 4\\sqrt{3}$. Let the radius of circle $E$ have length $r$. Draw $\\overline{EC}$ from circle center $E$ to center $C$. $\\overline{EC}$ has length $r + 2$. $\\overline{EF}$ has length $GA + AF - GE = 8 - r$. Since $CEF$ is a right triangle, by the Pythagorean theorem $EF^2 + CF^2 = EC^2$. $(8 - r)^2 + (4\\sqrt{3})^2 = (r + 2)^2$. Solving, $r = \\frac {27}{5}$ and the answer is $27 + 5 = 032. -unhappyfarmer"
] |
2009-II-6
| 2,009
| 6
|
Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$ .
| 750
|
II
|
[
"We can use complementary counting. We can choose a five-element subset in ${14\\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$s and 5 $B$s, thereby showing that there are ${10\\choose 5}$ such sets. Given a five-element subset $S$ of $\\{1,2,\\dots,14\\}$ in which no two numbers are consecutive, we can start by writing down a string of length 14, in which the $i$-th character is $A$ if $i\\in S$ and $B$ otherwise. Now we got a string with 5 $A$s and 9 $B$s. As no two numbers were consecutive, we know that in our string no two $A$s are consecutive. We can now remove exactly one $B$ from between each pair of $A$s to get a string with 5 $A$s and 5 $B$s. And clearly this is a bijection, as from each string with 5 $A$s and 5 $B$s we can reconstruct one original set by reversing the construction. Hence we have $m = {14\\choose 5} - {10\\choose 5} = 2002 - 252 = 1750$, and the answer is $1750 \\bmod 1000 = 750.",
"Let $A$ be the number of ways in which $5$ distinct numbers can be selected from the set of the first $14$ natural numbers, and let $B$ be the number of ways in which $5$ distinct numbers, no two of which are consecutive, can be selected from the same set. Then $m = A -B$. Because $A = \\dbinom{14}{5}$, the problem is reduced to finding $B$. Consider the natural numbers $1 \\leq a_1 < a_2 < a_3 < a_4 < a_5 \\leq 14$. If no two of them are consecutive, the numbers $b_1 = a_1, b_2 = a_2 - 1$, $b_3 = a_3 - 2$, $b_4 = a_4 - 3$, and $b_5 = a_5 - 4$ are distinct numbers from the interval $[1, 10]$. Conversely, if $b_1 < b_2 < b_3 < b_4 < b_5$ are distinct natural numbers from the interval $[1, 10]$, then $a_1 = b_1$, $a_2 = b_2 + 1$, $a_3 = b_3 + 2$, $a_4 = b_4 + 3$, and $a_5 = b_5 + 4$ are from the interval $[1, 14]$, and no two of them are consecutive. Therefore counting $B$ is the same as counting the number of ways of choosing $5$ distinct numbers from the set of the first $10$ natural numbers. Thus $B = \\dbinom{10}{5}$. Hence $m = A - B = \\dbinom{14}{5} - \\dbinom{10}{5} = 2002 - 252 = 1750$ and the answer is $750$.",
"We will approach this problem using complementary counting. First it is obvious, there are $\\binom{14}{5}$ total sets. To find the number of sets with NO consecutive elements, we do a little experimentation. Consider the following: Start of with any of the $14$ numbers, say WLOG, $1$. Then we cannot have $2$. So again WLOG, we may pick $3$. Then we cannot have $4$, so again WLOG, we may pick $5$. Then not $6$, but $7$, then not $8$, but $9$. Now we have the set ${1,3,5,7,9}$, and we had to remove $4$ digits from the $14$ total amount, so there wasn't any consecutive numbers. So we have that the number of non-consecutive cardinality $5$ sets, is $\\binom{14-4}{5}=\\binom{10}{5}$. Now you already read the solutions above, and if you are here, you are either confused, or looking for more. So now the answer is simply $1750$, which is $750 and doing the steps I used for my solution).",
"We will proceed by complementary counting. We can easily see that there are $\\binom{14}{5}$ ways to select $5$ element subsets from the original set. Now, we must find the number of subsets part of that group that do NOT have any consecutive integers. We can think of this situation as stars and bars. We have $9$ stars and $5$ bars, where the stars represent the integers NOT in the subset while the bars ARE the integers in the chosen subset. We want the amount of combinations where there is at least one star between each bar (meaning that none of the integers in the subset are consecutive). To do this, we remove $4$ stars from the total $9$ stars ($4$ because that is the amount that is between each bar). Now, we have $5$ stars and $5$ bars with no restrictions, so we calculate this as $\\binom{10}{5}$, which is $252$. Hence, $\\binom{14}{5} - \\binom{10}{5} = 2002 - 252 = 1750$, which becomes $750.",
"We will proceed by complementary counting. We can easily see that there are $\\binom{14}{5}$ ways to select $5$ element subsets from the original set. Now, we must find the number of subsets part of that group that do NOT have any consecutive integers. We can think of this situation as stars and bars. We have $9$ stars and $5$ bars, where the stars represent the integers NOT in the subset while the bars ARE the integers in the chosen subset. We want the amount of combinations where there is at least one star between each bar (meaning that none of the integers in the subset are consecutive). To do this, we remove $4$ stars from the total $9$ stars ($4$ because that is the amount that is between each bar). Now, we have $5$ stars and $5$ bars with no restrictions, so we calculate this as $\\binom{10}{5}$, which is $252$. Hence, $\\binom{14}{5} - \\binom{10}{5} = 2002 - 252 = 1750$, which becomes $750."
] |
2009-II-7
| 2,009
| 7
|
Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ .
| 401
|
II
|
[
"First, note that $(2n)!! = 2^n \\cdot n!$, and that $(2n)!! \\cdot (2n-1)!! = (2n)!$. We can now take the fraction $\\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$. We get that this fraction is equal to $\\dfrac{(2i)!}{(2i)!!^2} = \\dfrac{(2i)!}{2^{2i}(i!)^2}$. Now we can recognize that $\\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \\choose i}$, hence this fraction is $\\dfrac{{2i\\choose i}}{2^{2i}}$, and our sum turns into $S=\\sum_{i=1}^{2009} \\dfrac{{2i\\choose i}}{2^{2i}}$. Let $c = \\sum_{i=1}^{2009} {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i}$. Obviously $c$ is an integer, and $S$ can be written as $\\dfrac{c}{2^{2\\cdot 2009}}$. Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\\leq 2\\cdot 2009$. In other words, we just showed that $b=1$. To determine $a$, we need to determine the largest power of $2$ that divides $c$. Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$. We can now return to the observation that $(2i)! = (2i)!! \\cdot (2i-1)!! = 2^i \\cdot i! \\cdot (2i-1)!!$. Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$. It immediately follows that $p\\left( {2i\\choose i} \\right) = p((2i)!) - 2p(i!) = i - p(i!)$, and hence $p\\left( {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i} \\right) = 2\\cdot 2009 - i - p(i!)$. Obviously, for $i\\in\\{1,2,\\dots,2009\\}$ the function $f(i)=2\\cdot 2009 - i - p(i!)$ is is a strictly decreasing function. Therefore $p(c) = p\\left( {2\\cdot 2009\\choose 2009} \\right) = 2009 - p(2009!)$. We can now compute $p(2009!) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\dfrac{2009}{2^k} \\right\\rfloor = 1004 + 502 + \\cdots + 3 + 1 = 2001$. Hence $p(c)=2009-2001=8$. And thus we have $a=2\\cdot 2009 - p(c) = 4010$, and the answer is $\\dfrac{ab}{10} = \\dfrac{4010\\cdot 1}{10} = 401. From which the conclusion follows. - (OmicronGamma)",
"Using the steps of the previous solution we get $c = \\sum_{i=1}^{2009} {2i\\choose i} \\cdot 2^{2\\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$) you realize that you can \"thin-slice\" the problem and simply look at the cases where $i=2009, 2008$(they're nearly identical in nature but one has $4$ with it) since $\\dbinom{2i}{I}$ hardly contains any powers of $2$ or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of $2$ in $\\dbinom{4018}{2009}$ and $\\dbinom{4016}{2008}$ and you get the minimum power of $2$ in either expression is $8$ so the answer is $\\frac{4010}{10} \\implies 401.",
"We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$. Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \\cdot 2^a$. Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \\cdot 2)! = 4018!$. This is simply $\\sum_{n=1}^{11} \\left \\lfloor{\\frac{4018}{2^n}}\\right \\rfloor = 4010$. Therefore, our answer is $401.",
"We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$. Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \\cdot 2^a$. Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \\cdot 2)! = 4018!$. This is simply $\\sum_{n=1}^{11} \\left \\lfloor{\\frac{4018}{2^n}}\\right \\rfloor = 4010$. Therefore, our answer is $401.",
"Using the initial steps from Solution 1, $S=\\sum_{i=1}^{2009} \\dfrac{{2i\\choose i}}{2^{2i}}$. Clearly $b = 1$ as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find $a$. Note that $2^a$ would be the largest denominator in all the summands, so when they are summed it is the common denominator. Taking the p-adic valuations of each term, the powers of 2 in the denominator for $i$ is $(2i) - v_2(\\binom{2i}{i}).$ We can use Kummers theorem to see that $v_2(\\binom{2i}{i})$ is the number of digits carried over when $i$ is added to $i$ in base $2$. This is simply the number of $1$'s in the binary representation of $i$. Looking at the binary representations of some of the larger $i$ we see $2009 = 11111011001_2$ having eight $1$'s. So the power of two is $2 \\cdot 2009 - 8 = 4010$. Experimenting with $2008, 2007, 2006$ we see that the power of two are all $< 4010$, and under $2005$ the power of two $2i - v_2(\\binom{2i}{i}) < 2i < 4010$. Therefore $a = 4010$ and $\\frac{ab}{10} = 401 ~Aaryabhatta1"
] |
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