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1990-I-13
| 1,990 | 13 |
Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$ . Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit?
| 184 | null |
[
"Lemma: For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9 (Not-so-rigorous) Proof: One can prove by contradiction that there must be at least either one or two n-digit power of 9 for all n. If there is exactly 1 n-digit power of 9, then such a number $m$ cannot begin with 9 since that would result in $\\frac{m}{9}$ also being an n-digits, hence a contradiction. Therefore, this single n-digit power of 9 must not begin with 9. In the case that there are 2 n-digit powers of 9, we have already discovered that one of them must begin with 9, let it be $9^k = m$. The integer $9^{k - 1} = \\frac{m}{9}$ must then be an n-digit number that begins with $1$. Using this lemma, out of the 4001 powers of 9 in T, exactly 3816 must not begin with 9 (one for each digit). Therefore, the answer is $4000 - 3816 = 184 numbers have 9 as their leftmost digits. ~Edited by Mathandski (with help from a math god)",
"Let's divide all elements of $T$ into sections. Each section ranges from $10^{i}$ to $10^{i+1}$ And, each section must have 1 or 2 elements. So, let's consider both cases. If a section has 1 element, we claim that the number doesn't have 9 as the leftmost digit. Let this element be $9^{k}$ and the section ranges from $10^{i}$ to $10^{i+1}$. To the contrary, let's assume the number ($9^{k}$) does have 9 as the leftmost digit. Thus, $9 \\cdot 10^i \\leq 9^k$. But, if you divide both sides by 9, you get $10^i \\leq 9^{k-1}$, and because $9^{k-1} < 9^{k} < 10^{i+1}$, so we have another number ($9^{k-1}$) in the same section ($10^i \\leq 9^{k-1} < 10^{i+1}$). Which is a contradiction to our assumption that the section only has 1 element. So in this case, the number doesn't have 9 as the leftmost digit. If a section has 2 elements, we claim one has to have a 9 as the leftmost digit, one doesn't. Let the elements be $9^{k}$ and $9^{k+1}$, and the section ranges from $10^{i}$ to $10^{i+1}$. So We know $10^{i} \\leq 9^{k} < 9^{k+1} < 10^{i+1}$. From $10^{i} \\leq 9^{k}$, we know $9 \\cdot 10^{i} \\leq 9^{k+1}$, and since $9^{k+1} < 10^{i+1}$. The number($9^{k+1}$)'s leftmost digit must be 9, and the other number($9^{k}$)'s leftmost digit is 1. There are total 4001 elements in $T$ and 3817 sections that have 1 or 2 elements. And, no matter how many elements a section has, each section contains exactly one element that doesn't begin with 9. We can take 4001 elements, subtract 3817 elements that don't have 9 as the leftmost digit, and get $184 numbers that have 9 as the leftmost digit. - AlexLikeMath",
"We know that $9$ is very close to $10$. But we also know that for all $k \\in \\mathbb{Z}^+$, $10^k$ has $k+1$ digits. Hence $10^{4000}$ has $4001$ digits. Now, notice that if a power of $9^n$ has $9$ as the leftmost digit, then $9^{n-1}$ has $1$ as the leftmost digit and will preserve the same number of digits (just due to division). Hence we find that consecutive powers of $9$ are \"supposed\" to increase by one digit, however, when consecutive powers of $9$ have leading digit $1$ then leading digit $9$ they \"lose\" this one digit. Therefore, it suffices to find the number of times that $9^k$ has \"lost\" a digit since the number of times that this occurs shows us the number of pairs where the leading digit of $9^k$ is $1$ and the leading digit of $9^{k+1}$ is $9$. Hence, we find that $9^{4000}$ is \"supposed\" to have $4001$ digits, but only has $3817$ digits. Thus, the number of times $9^k$ has lost a digit is $4001-3817 = 184$. Thus, $184 as their leading digit. ~qwertysri987"
] |
1990-I-14
| 1,990 | 14 |
The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$ . Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$ . If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overline{DP}$ , we obtain a triangular pyramid, all four of whose faces are isosceles triangles. Find the volume of this pyramid. AIME 1990 Problem 14.png
| 594 | null |
[
"Solution 1(Synthetic) [asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, (99/133)^.5); Pa=(P.x,P.y,0); draw(P--Pa--A); draw(C--Pa--D); draw((C+D)/2--A--C--D--P--C--P--A--D); label(\"A\", A, NE); label(\"P\", P, N); label(\"C\", C); label(\"D\", D, S); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, N, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); [/asy] Our triangular pyramid has base $12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \\frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. [asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P$'$\", Pa, SW); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, NW, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"$\\frac{\\sqrt{939}}{2}$\", (C+P)/2 ,NW); [/asy] To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem, \\begin{align*}h^2 &= PA^2 - R^2 \\\\ &= \\left(\\frac{\\sqrt{939}}{2}\\right)^2 - \\left(\\frac{13^2\\sqrt{3}}{2\\sqrt{133}}\\right)^2\\\\ &= \\frac{939 \\cdot 133 - 13^4 \\cdot 3}{4 \\cdot 133} = \\frac{13068 \\cdot 3}{4 \\cdot 133} = \\frac{99^2}{133}\\\\ h &= \\frac{99}{\\sqrt{133}} \\end{align*} Finally, we substitute $h$ into the volume equation to find $V = 6\\sqrt{133}\\left(\\frac{99}{\\sqrt{133}}\\right) = 594.",
"[asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, (99/133)^.5); Pa=(P.x,P.y,0); draw(P--Pa--A); draw(C--Pa--D); draw((C+D)/2--A--C--D--P--C--P--A--D); label(\"A\", A, NE); label(\"P\", P, N); label(\"C\", C); label(\"D\", D, S); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, N, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); [/asy] Our triangular pyramid has base $12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \\frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. [asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P$'$\", Pa, SW); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, NW, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"$\\frac{\\sqrt{939}}{2}$\", (C+P)/2 ,NW); [/asy] To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem, \\begin{align*}h^2 &= PA^2 - R^2 \\\\ &= \\left(\\frac{\\sqrt{939}}{2}\\right)^2 - \\left(\\frac{13^2\\sqrt{3}}{2\\sqrt{133}}\\right)^2\\\\ &= \\frac{939 \\cdot 133 - 13^4 \\cdot 3}{4 \\cdot 133} = \\frac{13068 \\cdot 3}{4 \\cdot 133} = \\frac{99^2}{133}\\\\ h &= \\frac{99}{\\sqrt{133}} \\end{align*} Finally, we substitute $h$ into the volume equation to find $V = 6\\sqrt{133}\\left(\\frac{99}{\\sqrt{133}}\\right) = 594.",
"[asy] import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0, 399^(0.5), 0); triple D=(108^(0.5), 0, 0); triple C=(-108^(0.5), 0, 0); triple Pa; pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, (99/133)^.5); Pa=(P.x,P.y,0); draw(P--Pa--A); draw(C--Pa--D); draw((C+D)/2--A--C--D--P--C--P--A--D); label(\"A\", A, NE); label(\"P\", P, N); label(\"C\", C); label(\"D\", D, S); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, N, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); [/asy] Our triangular pyramid has base $12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$. The area of this isosceles triangle is easy to find by $[ACD] = \\frac{1}{2}bh$, where we can find $h_{ACD}$ to be $\\sqrt{399}$ by the Pythagorean Theorem. Thus $A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$. [asy] size(280); import three; pointpen = black; pathpen = black+linewidth(0.7); pen small = fontsize(9); real h=169/2*(3/133)^.5; currentprojection = perspective(20,-20,12); triple O=(0,0,0); triple A=(0,399^.5,0); triple D=(108^.5,0,0); triple C=(-108^.5,0,0); pair Ci=circumcenter((A.x,A.y),(C.x,C.y),(D.x,D.y)); triple P=(Ci.x, Ci.y, 99/133^.5); triple Pa=(P.x,P.y,0); draw(A--C--D--P--C--P--A--D); draw(P--Pa--A); draw(C--Pa--D); draw(circle(Pa, h)); label(\"A\", A, NE); label(\"C\", C, NW); label(\"D\", D, S); label(\"P\",P , N); label(\"P$'$\", Pa, SW); label(\"$13\\sqrt{3}$\", (A+D)/2, E, small); label(\"$13\\sqrt{3}$\", (A+C)/2, NW, small); label(\"$12\\sqrt{3}$\", (C+D)/2, SW, small); label(\"h\", (P + Pa)/2, W); label(\"$\\frac{\\sqrt{939}}{2}$\", (C+P)/2 ,NW); [/asy] To find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$, so we need to find the height of the tetrahedron. By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$. If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$, then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$. From here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem, \\begin{align*}h^2 &= PA^2 - R^2 \\\\ &= \\left(\\frac{\\sqrt{939}}{2}\\right)^2 - \\left(\\frac{13^2\\sqrt{3}}{2\\sqrt{133}}\\right)^2\\\\ &= \\frac{939 \\cdot 133 - 13^4 \\cdot 3}{4 \\cdot 133} = \\frac{13068 \\cdot 3}{4 \\cdot 133} = \\frac{99^2}{133}\\\\ h &= \\frac{99}{\\sqrt{133}} \\end{align*} Finally, we substitute $h$ into the volume equation to find $V = 6\\sqrt{133}\\left(\\frac{99}{\\sqrt{133}}\\right) = 594.",
"Let $\\triangle{ABC}$ (or the triangle with sides $12\\sqrt {3}$, $13\\sqrt {3}$, $13\\sqrt {3}$) be the base of our tetrahedron. We set points $C$ and $D$ as $(6\\sqrt {3}, 0, 0)$ and $( - 6\\sqrt {3}, 0, 0)$, respectively. Using Pythagoras, we find $A$ as $(0, \\sqrt {399}, 0)$. We know that the vertex of the tetrahedron ($P$) has to be of the form $(x, y, z)$, where $z$ is the altitude of the tetrahedron. Since the distance from $P$ to points $A$, $B$, and $C$ is $\\frac {\\sqrt {939}}{2}$, we can write three equations using the distance formula: \\begin{align*} x^{2} + (y - \\sqrt {399})^{2} + z^{2} &= \\frac {939}{4}\\\\ (x - 6\\sqrt {3})^{2} + y^{2} + z^{2} &= \\frac {939}{4}\\\\ (x + 6\\sqrt {3})^{2} + y^{2} + z^{2} &= \\frac {939}{4} \\end{align*} Subtracting the last two equations, we get $x = 0$. Solving for $y,z$ with a bit of effort, we eventually get $x = 0$, $y = \\frac {291}{2\\sqrt {399}}$, $z = \\frac {99}{\\sqrt {133}}$. Since the area of a triangle is $\\frac {1}{2}\\cdot bh$, we have the base area as $18\\sqrt {133}$. Thus, the volume is $V = \\frac {1}{3}\\cdot18\\sqrt {133}\\cdot\\frac {99}{\\sqrt {133}} = 6\\cdot99 = 594$. Solution 3 (Law of Cosines) Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\\overline{CD}$. We find the side lengths of $\\triangle XMP$. $MP = \\frac{13\\sqrt3}{2}$. $PX$ is half of $AC$, which is $\\frac{\\sqrt{3\\cdot13^2+3\\cdot12^2}}{2} = \\frac{\\sqrt{939}}{2}$. To find $MX$, consider right triangle $XMD$; since $XD=13\\sqrt3$ and $MD=6\\sqrt3$, we have $MX=\\sqrt{3\\cdot13^2-3\\cdot6^2} = \\sqrt{399}$. Let $\\theta=\\angle XPM$. For calculating trig, let us double all sides of $\\triangle XMP$. By Law of Cosines, $\\cos\\theta = \\frac{3\\cdot13^2+939-4\\cdot399}{2\\cdot13\\cdot3\\sqrt{313}}=\\frac{-150}{6\\cdot13\\sqrt{313}}=-\\frac{25}{13\\sqrt{313}}$. Hence, \\[\\sin\\theta = \\sqrt{1-\\cos^2\\theta}\\] \\[=\\sqrt{1-\\frac{625}{13^2\\cdot313}}\\] \\[=\\frac{\\sqrt{13^2\\cdot313-625}}{13\\sqrt{313}}\\] \\[=\\frac{\\sqrt{3\\cdot132^2}}{13\\sqrt{313}}\\] \\[=\\frac{132\\sqrt3}{13\\sqrt{313}}\\] Thus, the height of the pyramid is $PX\\sin\\theta=\\frac{\\sqrt{939}}{2}\\cdot\\frac{132\\sqrt3}{13\\sqrt{313}}=\\frac{66\\cdot3}{13}$. Since $[CPD]=3\\sqrt{3}\\cdot13\\sqrt{3}=9\\cdot13$, the volume of the pyramid is $\\frac13 \\cdot 9\\cdot13\\cdot \\frac{66\\cdot3}{13}=594.",
"Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\\overline{CD}$. We find the side lengths of $\\triangle XMP$. $MP = \\frac{13\\sqrt3}{2}$. $PX$ is half of $AC$, which is $\\frac{\\sqrt{3\\cdot13^2+3\\cdot12^2}}{2} = \\frac{\\sqrt{939}}{2}$. To find $MX$, consider right triangle $XMD$; since $XD=13\\sqrt3$ and $MD=6\\sqrt3$, we have $MX=\\sqrt{3\\cdot13^2-3\\cdot6^2} = \\sqrt{399}$. Let $\\theta=\\angle XPM$. For calculating trig, let us double all sides of $\\triangle XMP$. By Law of Cosines, $\\cos\\theta = \\frac{3\\cdot13^2+939-4\\cdot399}{2\\cdot13\\cdot3\\sqrt{313}}=\\frac{-150}{6\\cdot13\\sqrt{313}}=-\\frac{25}{13\\sqrt{313}}$. Hence, \\[\\sin\\theta = \\sqrt{1-\\cos^2\\theta}\\] \\[=\\sqrt{1-\\frac{625}{13^2\\cdot313}}\\] \\[=\\frac{\\sqrt{13^2\\cdot313-625}}{13\\sqrt{313}}\\] \\[=\\frac{\\sqrt{3\\cdot132^2}}{13\\sqrt{313}}\\] \\[=\\frac{132\\sqrt3}{13\\sqrt{313}}\\] Thus, the height of the pyramid is $PX\\sin\\theta=\\frac{\\sqrt{939}}{2}\\cdot\\frac{132\\sqrt3}{13\\sqrt{313}}=\\frac{66\\cdot3}{13}$. Since $[CPD]=3\\sqrt{3}\\cdot13\\sqrt{3}=9\\cdot13$, the volume of the pyramid is $\\frac13 \\cdot 9\\cdot13\\cdot \\frac{66\\cdot3}{13}=594.",
"Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\\overline{CD}$. We find the side lengths of $\\triangle XMP$. $MP = \\frac{13\\sqrt3}{2}$. $PX$ is half of $AC$, which is $\\frac{\\sqrt{3\\cdot13^2+3\\cdot12^2}}{2} = \\frac{\\sqrt{939}}{2}$. To find $MX$, consider right triangle $XMD$; since $XD=13\\sqrt3$ and $MD=6\\sqrt3$, we have $MX=\\sqrt{3\\cdot13^2-3\\cdot6^2} = \\sqrt{399}$. Let $\\theta=\\angle XPM$. For calculating trig, let us double all sides of $\\triangle XMP$. By Law of Cosines, $\\cos\\theta = \\frac{3\\cdot13^2+939-4\\cdot399}{2\\cdot13\\cdot3\\sqrt{313}}=\\frac{-150}{6\\cdot13\\sqrt{313}}=-\\frac{25}{13\\sqrt{313}}$. Hence, \\[\\sin\\theta = \\sqrt{1-\\cos^2\\theta}\\] \\[=\\sqrt{1-\\frac{625}{13^2\\cdot313}}\\] \\[=\\frac{\\sqrt{13^2\\cdot313-625}}{13\\sqrt{313}}\\] \\[=\\frac{\\sqrt{3\\cdot132^2}}{13\\sqrt{313}}\\] \\[=\\frac{132\\sqrt3}{13\\sqrt{313}}\\] Thus, the height of the pyramid is $PX\\sin\\theta=\\frac{\\sqrt{939}}{2}\\cdot\\frac{132\\sqrt3}{13\\sqrt{313}}=\\frac{66\\cdot3}{13}$. Since $[CPD]=3\\sqrt{3}\\cdot13\\sqrt{3}=9\\cdot13$, the volume of the pyramid is $\\frac13 \\cdot 9\\cdot13\\cdot \\frac{66\\cdot3}{13}=594.",
"Alternatively, since you can use rulers on the AIME, you can measure out a $12\\sqrt{3}$ by $13\\sqrt{3}$ rectangle and cut out a piece to form the pyramid. Then, using the ruler, you can measure the height and find the area of the base, multiply the two together, and approximate the answer $594. ~yeetdayeet",
"Alternatively, since you can use rulers on the AIME, you can measure out a $12\\sqrt{3}$ by $13\\sqrt{3}$ rectangle and cut out a piece to form the pyramid. Then, using the ruler, you can measure the height and find the area of the base, multiply the two together, and approximate the answer $594. ~yeetdayeet"
] |
1990-I-15
| 1,990 | 15 |
Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*}
| 20 | null |
[
"Set $S = (x + y)$ and $P = xy$. Then the relationship \\[(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})\\] can be exploited: \\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\\\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\\end{eqnarray*} Therefore: \\begin{eqnarray*}7S & = & 16 + 3P \\\\ 16S & = & 42 + 7P\\end{eqnarray*} Consequently, $S = - 14$ and $P = - 38$. Finally: \\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\\\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\\\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\\\ ax^5 + by^5 & = & 020\\end{eqnarray*}",
"A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$, where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms. Suppose we have such a recurrence with $T_1=3$ and $T_2=7$. Then $T_3=ax^3+by^3=16=7A+3B$, and $T_4=ax^4+by^4=42=16A+7B$. Solving these simultaneous equations for $A$ and $B$, we see that $A=-14$ and $B=38$. So, $ax^5+by^5=T_5=-14(42)+38(16)= 020.",
"Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting: $a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16$ $ax\\left( \\frac{x^3-1}{x-1} \\right) + by\\left( \\frac{y^3-1}{y-1} \\right) = 16$. Similarly take the first two terms, yielding: $ax \\left( \\frac{x^2-1}{x-1} \\right) + by \\left( \\frac{y^2-1}{y-1} \\right) = 10$. Lastly take an alternating three-term sum, $a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12$ $ax \\left( \\frac{x^3+1}{x+1} \\right) + by \\left( \\frac{y^3+1}{y+1} \\right) = 12$. Now to get the solution, let the answer be $k$, so $ax \\left( \\frac{x^4-1}{x-1} \\right) + by \\left(\\frac{y^4-1}{y-1} \\right) = 68$. Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated as done in the first solution. ~lpieleanu (reformatting and minor edits)",
"We first let the answer to this problem be $k.$ Multiplying the first equation by $x$ gives $ax^2 + bxy=3x$. Subtracting this equation from the second equation gives $by^2-bxy=7-3x$. Similarly, doing the same for the other equations, we obtain: $by^2-bxy=7-3x$, $by^3-bxy^2=16-7x$, $by^4-bxy^3=42-16x$, and $by^5-bxy^4=k-42x$ Now lets take the first equation. Multiplying this by $y$ and subtracting this from the second gives us $by^3-bxy^2=(7-3x)y$. We can also obtain $by^4-bxy^3=(16-7x)y$. Now we can solve for $x$ and $y$! $(7-3x)y = 16-7x$ and $(16-7x)y=42-16x$. Solving for $x$ and $y$ gives us $(-7+\\sqrt{87},-7-\\sqrt{87})$ (It can be switched, but since the given equations are symmetric, it doesn't matter). $k-42x= (42-16x)y$, and solving for $k$ gives us $k= 020. ~pi_is_3.141"
] |
1991-I-1
| 1,991 | 1 |
Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}
| 146 | null |
[
"Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$, which factors to $(a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\\le 16$, no two factors of $16$ can sum greater than $32$, and so there are no integral solutions for $(x,y)$. The solution is $5^2 + 11^2 = 146.",
"Since $xy + x + y + 1 = 72$, this can be factored to $(x + 1)(y + 1) = 72$. As $x$ and $y$ are integers, the possible sets for $(x,y)$ (ignoring cases where $x > y$ since it is symmetrical) are $(1, 35),\\ (2, 23),\\ (3, 17),\\ (5, 11),\\ (7,8)$. The second equation factors to $(x + y)xy = 880 = 2^4 \\cdot 5 \\cdot 11$. The only set with a factor of $11$ is $(5,11)$, and checking shows that it is correct.",
"Let $a=x+y$, $b=xy$ then we get the equations \\begin{align*} a+b&=71\\\\ ab&=880 \\end{align*} After finding the prime factorization of $880=2^4\\cdot5\\cdot11$, it's easy to obtain the solution $(a,b)=(16,55)$. Thus \\[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\\cdot55=146 which is invalid for an AIME answer. ~ Nafer",
"From the first equation, we know $x+y=71-xy$. We factor the second equation as $xy(71-xy)=880$. Let $a=xy$ and rearranging we get $a^2-71a+880=(a-16)(a-55)=0$. We have two cases: (1) $x+y=16$ and $xy=55$ OR (2) $x+y=55$ and $xy=16$. We find the former is true for $(x,y) = (5,11)$. $x^2+y^2=121+25=146$.",
"First, notice that you can factor $x^2y + xy^2$ as $xy(x + y)$. From this, we notice that $xy$ and $x + y$ is a common occurrence, so that lends itself to a simple solution by substitution. Let $xy = b$ and $x + y = a$. From this substitution, we get the following system: \\[a + b = 71\\] \\[ab = 880\\] Solving that system gives us the following two pairs $(a, b)$: $(16, 55)$ and $(55, 16)$. The second one is obviously too big as $55^2$ is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: \\[(x + y)^2 = (x^2 + y^2) + 2xy\\] We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: \\[256 = (x^2 + y^2) + 110\\] so $x^2 + y^2 = 146 would be absurdly out of bounds) ~EricShi1685"
] |
1991-I-2
| 1,991 | 2 |
Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$ , and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$ . For $1_{}^{} \le k \le 167$ , draw the segments $\overline {P_kQ_k}$ . Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$ , and then draw the diagonal $\overline {AC}$ . Find the sum of the lengths of the 335 parallel segments drawn.
| 840 | null |
[
"[asy] real r = 0.35; size(220); pointpen=black;pathpen=black+linewidth(0.65);pen f = fontsize(8); pair A=(0,0),B=(4,0),C=(4,3),D=(0,3); D(A--B--C--D--cycle); pair P1=A+(r,0),P2=A+(2r,0),P3=B-(r,0),P4=B-(2r,0); pair Q1=C-(0,r),Q2=C-(0,2r),Q3=B+(0,r),Q4=B+(0,2r); D(A--C);D(P1--Q1);D(P2--Q2);D(P3--Q3);D(P4--Q4); MP(\"A\",A,f);MP(\"B\",B,SE,f);MP(\"C\",C,NE,f);MP(\"D\",D,W,f); MP(\"P_1\",P1,f);MP(\"P_2\",P2,f);MP(\"P_{167}\",P3,f);MP(\"P_{166}\",P4,f);MP(\"Q_1\",Q1,E,f);MP(\"Q_2\",Q2,E,f);MP(\"Q_{167}\",Q3,E,f);MP(\"Q_{166}\",Q4,E,f); MP(\"4\",(A+B)/2,N,f);MP(\"\\cdots\",(A+B)/2,f); MP(\"3\",(B+C)/2,W,f);MP(\"\\vdots\",(C+B)/2,E,f); [/asy] The length of the diagonal is $\\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\\overline{P_kQ_k}$ is the hypotenuse of a $3-4-5$ right triangle with sides of $3 \\cdot \\frac{168-k}{168}, 4 \\cdot \\frac{168-k}{168}$. Thus, its length is $5 \\cdot \\frac{168-k}{168}$. Let $a_k=\\frac{5(168-k)}{168}$. We want to find $2\\sum\\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\\sum\\limits_{k=1}^{168} \\frac{5(168-k)}{168}-5 =2\\frac{(0+5)\\cdot169}{2}-5 =168\\cdot5 =840",
"Using the above diagram, we have that $\\Delta ABC \\sim \\Delta P_k B Q_k$ and each one of these is a dilated 3-4-5 right triangle (This is true since $\\Delta ABC$ is a 3-4-5 right triangle). Now, for all $k$, we have that $\\overline{P_k Q_k}$ is the hypotenuse for the triangle $P_k B Q_k$. Therefore we want to know the sum of the lengths of all $\\overline{P_k Q_k}$.This is given by the following: \\[2 \\cdot(\\sum_{k=1}^{168} P_kQ_k) + 5\\] \\[= 2 \\cdot \\frac{ 0+5+10+...+835}{168} +5\\] Then by the summation formula for the sum of the terms of an arithmetic series, \\[= \\frac{835 \\cdot 168}{168} +5 = 835+5 = 840\\] ~qwertysri987",
"First, count the diagonal which has length $5$. For the rest of the segments, think about pairing them up so that each pair makes $5$. For example, the parallel lines closest to the diagonal would have length $\\frac{167}{168}\\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\\frac{1}{168}\\cdot{5}$ by similar triangles. If you add the two lengths together, it is $\\frac{167}{168}\\cdot{5} + \\frac{1}{168}\\cdot{5} = 5.$ There are $\\frac{335-1}{2}$ pairs of these segments, for a total of $5+(167)(5)=168(5)=840 ~justlearningmathog"
] |
1991-I-4
| 1,991 | 4 |
How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$ ?
| 159 | null |
[
"The range of the sine function is $-1 \\le y \\le 1$. It is periodic (in this problem) with a period of $\\frac{2}{5}$. Thus, $-1 \\le \\frac{1}{5} \\log_2 x \\le 1$, and $-5 \\le \\log_2 x \\le 5$. The solutions for $x$ occur in the domain of $\\frac{1}{32} \\le x \\le 32$. When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$) of the sine curve and another curve that is $< 1$, so there are $\\frac{32}{2} \\cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$, there is exactly $1$ touching point between the two functions: $\\left(\\frac{1}{5},0\\right)$. When $y < 0$ or $x < 1$, we can count $4$ more solutions. The solution is $154 + 1 + 4 = 159.",
"Notice that the equation is satisfied twice for every sine period (which is $\\frac{2}{5}$), except in the sole case when the two equations equate to $0$. In that case, the equation is satisfied twice but only at the one instance when $y=0$. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: $32 \\cdot \\frac{5}{2} \\cdot 2 - 1 = 159"
] |
1991-I-5
| 1,991 | 5 |
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $20_{}^{}!$ be the resulting product?
| 128 | null |
[
"If the fraction is in the form $\\frac{a}{b}$, then $a < b$ and $\\gcd(a,b) = 1$. There are 8 prime numbers less than 20 ($2, 3, 5, 7, 11, 13, 17, 19$), and each can only be a factor of one of $a$ or $b$. There are $2^8$ ways of selecting some combination of numbers for $a$; however, since $a<b$, only half of them will be between $0 < \\frac{a}{b} < 1$. Therefore, the solution is $\\frac{2^8}{2} = 128."
] |
1991-I-8
| 1,991 | 8 |
For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$ ?
| 10 | null |
[
"Let $x^2 + ax + 6a = (x - s)(x - r)$. Vieta's yields $s + r = - a, sr = 6a$. \\begin{eqnarray*}sr + 6s + 6r &=& 0\\\\ sr + 6s + 6r + 36 &=& 36\\\\ (s + 6)(r + 6) &=& 36 \\end{eqnarray*} Without loss of generality let $r \\le s$. The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\\Rightarrow 10.",
"By Vieta's formulas, $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula, \\[x = \\frac{-a \\pm \\sqrt{a^2 - 24a}}{2}\\] Since $-a$ is an integer, we need $\\sqrt{a^2-24a}$ to be an integer (let this be $b$): $b^2 = a^2 - 24a$. Completing the square, we get \\[(a - 12)^2 = b^2 + 144\\] Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$). Then \\[c^2 - b^2 = 144 \\Longrightarrow (c+b)(c-b) = 144\\] The pairs of factors of $144$ are $(\\pm1,\\pm144),( \\pm 2, \\pm 72),( \\pm 3, \\pm 48),( \\pm 4, \\pm 36),( \\pm 6, \\pm 24),( \\pm 8, \\pm 18),( \\pm 9, \\pm 16),( \\pm 12, \\pm 12)$; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get $10 pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work."
] |
1991-I-9
| 1,991 | 9 |
Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$
| 44 | null |
[
"Use the two trigonometric Pythagorean identities $1 + \\tan^2 x = \\sec^2 x$ and $1 + \\cot^2 x = \\csc^2 x$. If we square the given $\\sec x = \\frac{22}{7} - \\tan x$, we find that \\begin{align*} \\sec^2 x &= \\left(\\frac{22}7\\right)^2 - 2\\left(\\frac{22}7\\right)\\tan x + \\tan^2 x \\\\ 1 &= \\left(\\frac{22}7\\right)^2 - \\frac{44}7 \\tan x \\end{align*} This yields $\\tan x = \\frac{435}{308}$. Let $y = \\frac mn$. Then squaring, \\[\\csc^2 x = (y - \\cot x)^2 \\Longrightarrow 1 = y^2 - 2y\\cot x.\\] Substituting $\\cot x = \\frac{1}{\\tan x} = \\frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$. It turns out that only the positive root will work, so the value of $y = \\frac{29}{15}$ and $m + n = 044.",
"Recall that $\\sec^2 x - \\tan^2 x = 1$, from which we find that $\\sec x - \\tan x = 7/22$. Adding the equations \\begin{eqnarray*} \\sec x + \\tan x & = & 22/7 \\\\ \\sec x - \\tan x & = & 7/22\\end{eqnarray*} together and dividing by 2 gives $\\sec x = 533/308$, and subtracting the equations and dividing by 2 gives $\\tan x = 435/308$. Hence, $\\cos x = 308/533$ and $\\sin x = \\tan x \\cos x = (435/308)(308/533) = 435/533$. Thus, $\\csc x = 533/435$ and $\\cot x = 308/435$. Finally, \\[\\csc x + \\cot x = \\frac {841}{435} = \\frac {29}{15},\\] so $m + n = 044$.",
"By the given, $\\frac {1}{\\cos x} + \\frac {\\sin x}{\\cos x} = \\frac {22}{7}$ and $\\frac {1}{\\sin x} + \\frac {\\cos x}{\\sin x} = k$. Multiplying the two, we have \\[\\frac {1}{\\sin x \\cos x} + \\frac {1}{\\sin x} + \\frac {1}{\\cos x} + 1 = \\frac {22}{7}k\\] Subtracting both of the two given equations from this, and simpliyfing with the identity $\\frac {\\sin x}{\\cos x} + \\frac {\\cos x}{\\sin x} = \\frac{\\sin ^2 x + \\cos ^2 x}{\\sin x \\cos x} = \\frac {1}{\\sin x \\cos x}$, we get \\[1 = \\frac {22}{7}k - \\frac {22}{7} - k.\\] Solving yields $k = \\frac {29}{15}$, and $m+n = 044$",
"By the given, $\\frac {1}{\\cos x} + \\frac {\\sin x}{\\cos x} = \\frac {22}{7}$ and $\\frac {1}{\\sin x} + \\frac {\\cos x}{\\sin x} = k$. Multiplying the two, we have \\[\\frac {1}{\\sin x \\cos x} + \\frac {1}{\\sin x} + \\frac {1}{\\cos x} + 1 = \\frac {22}{7}k\\] Subtracting both of the two given equations from this, and simpliyfing with the identity $\\frac {\\sin x}{\\cos x} + \\frac {\\cos x}{\\sin x} = \\frac{\\sin ^2 x + \\cos ^2 x}{\\sin x \\cos x} = \\frac {1}{\\sin x \\cos x}$, we get \\[1 = \\frac {22}{7}k - \\frac {22}{7} - k.\\] Solving yields $k = \\frac {29}{15}$, and $m+n = 044$",
"Make the substitution $u = \\tan \\frac x2$ (a substitution commonly used in calculus). By the half-angle identity for tangent, $\\tan \\frac x2 = \\frac{\\sin x}{1+\\cos x}$, so $\\csc x + \\cot x = \\frac{1+\\cos x}{\\sin x} = \\frac1u = \\frac mn$. Also, we have $\\sec x + \\tan x = \\frac{1 + \\sin x}{\\cos x}.$ Now note the following: \\begin{align*}\\sin x &= \\frac{2u}{1+u^2}\\\\ \\cos x &= \\frac{1-u^2}{1+u^2}\\end{align*} Plugging these into our equality gives: \\[\\frac{1+\\frac{2u}{1+u^2}}{\\frac{1-u^2}{1+u^2}} = \\frac{22}7\\] This simplifies to $\\frac{1+u}{1-u} = \\frac{22}7$, and solving for $u$ gives $u = \\frac{15}{29}$, and $\\frac mn = \\frac{29}{15}$. Finally, $m+n = 044$.",
"We are given that $\\frac{1+\\sin x}{\\cos x}=\\frac{22}7\\implies\\frac{1+\\sin x}{\\cos x}\\cdot\\frac{1-\\sin x}{1-\\sin x}=\\frac{1-\\sin^2x}{\\cos x(1-\\sin x)}=\\frac{\\cos^2x}{\\cos x(1-\\sin x)}$ $=\\frac{\\cos x}{1-\\sin x}$, or equivalently, $\\cos x=\\frac{7+7\\sin x}{22}=\\frac{22-22\\sin x}7\\implies\\sin x=\\frac{22^2-7^2}{22^2+7^2}$ $\\implies\\cos x=\\frac{2\\cdot22\\cdot7}{22^2+7^2}$. Note that what we want is just $\\frac{1+\\cos x}{\\sin x}=\\frac{1+\\frac{2\\cdot22\\cdot7}{22^2+7^2}}{\\frac{22^2-7^2}{22^2+7^2}}=\\frac{22^2+7^2+2\\cdot22\\cdot7}{22^2-7^2}=\\frac{(22+7)^2}{(22-7)(22+7)}=\\frac{22+7}{22-7}$ $=\\frac{29}{15}\\implies m+n=29+15=044.",
"Assign a right triangle with angle $x$, hypotenuse $c$, adjacent side $a$, and opposite side $b$. Then, through the given information above, we have that.. $\\frac{c}{a}+\\frac{b}{a}=\\frac{22}{7}\\implies \\frac{c+b}{a}=\\frac{22}{7}$ $\\frac{c}{b}+\\frac{a}{b}=\\frac{m}{n}\\implies \\frac{a+c}{b}=\\frac{m}{n}$ Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms. Hence, $a=7$, $b+c=22$ Furthermore, by the Pythagorean Theorem, we have that $a^2+b^2=c^2\\implies 49+b^2=c^2$ Solving for $c$ in the first equation and plugging in into the second equation... $49+b^2=(22-b)^2\\implies 49+b^2=484-44b+b^2\\implies 44b=435\\implies b=\\frac{435}{44}$ Hence, $c=22-\\frac{435}{44}=\\frac{533}{44}$ Now, we want $\\frac{a+c}{b}$ Plugging in, we find the answer is $\\frac{\\frac{7\\cdot{44}}{44}+\\frac{533}{44}}{\\frac{435}{44}}=\\frac{841}{435}=\\frac{29}{15}$ Hence, the answer is $29+15=044",
"We know that $\\sec(x) = \\frac{h}{a}$ and that $\\tan(x) = \\frac{o}{a}$ where $h$, $a$, $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\\sec(x) + \\tan(x) = \\frac{h+o}{a}$. We also have that $\\csc(x) + \\cot(x) = \\frac{h}{o} + \\frac{a}{o} = \\frac{h+a}{o}$. Set $\\sec(x) + \\tan(x) = \\alpha$ and csc(x)+cot(x) = $\\beta$. Then, notice that $\\alpha + \\beta = \\frac{h+o}{a} + \\frac{h+a}{o} = \\frac{oh+ah+o^2 + a^2}{oa} = \\frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$). But then notice that $\\alpha \\cdot \\beta = \\frac{(o+h)(a+h)}{oa} = \\frac{oa +oh +ha +h^2}{oa} = 1+ \\frac{h(o+a+h)}{oa} = 1+ \\alpha + \\beta$. From the information provided in the question, we can substitute $\\alpha$ for $\\frac{22}{7}$. Thus, $\\frac{22 \\beta}{7}= \\beta + \\frac{29}{7} \\Longrightarrow 22 \\beta = 7 \\beta + 29 \\Longrightarrow 15 \\beta = 29 \\Longrightarrow \\beta = \\frac{29}{15}$. Since, essentially we are asked to find the sum of the numerator and denominator of $\\beta$, we have $29 + 15 = 044. ~qwertysri987",
"Firstly, we write $\\sec x+\\tan x=a/b$ where $a=22$ and $b=7$. This will allow us to spot factorable expressions later. Now, since $\\sec^2x-\\tan^2x=1$, this gives us \\[\\sec x-\\tan x=\\frac{b}{a}\\] Adding this to our original expressions gives us \\[2\\sec x=\\frac{a^2+b^2}{ab}\\] or \\[\\cos x=\\frac{2ab}{a^2+b^2}\\] Now since $\\sin^2x+\\cos^2x=1$, $\\sin x=\\sqrt{1-\\cos^2x}$ So we can write \\[\\sin x=\\sqrt{1-\\frac{4a^2b^2}{(a^2+b^2)^2}}\\] Upon simplification, we get \\[\\sin x=\\frac{a^2-b^2}{a^2+b^2}\\] We are asked to find $1/\\sin x+\\cos x/\\sin x$ so we can write that as \\[\\csc x+\\cot x=\\frac{1}{\\sin x}+\\frac{\\cos x}{\\sin x}\\] \\[\\csc x+\\cot x=\\frac{a^2+b^2}{a^2-b^2}+\\frac{2ab}{a^2+b^2}\\frac{a^2+b^2}{a^2-b^2}\\] \\[\\csc x+\\cot x=\\frac{a^2+b^2+2ab}{a^2-b^2}\\] \\[\\csc x+\\cot x=\\frac{(a+b)^2}{(a-b)(a+b)}\\] \\[\\csc x+\\cot x=\\frac{a+b}{a-b}\\] Now using the fact that $a=22$ and $b=7$ yields, \\[\\csc x+\\cot x=\\frac{29}{15}=\\frac{p}{q}\\] so $p+q=15+29=44 ~Chessmaster20000",
"Rewriting $\\sec{x}$ and $\\tan{x}$ in terms of $\\sin{x}$ and $\\cos{x}$, we know that $\\frac{1+\\sin{x}}{\\cos{x}}=\\frac{22}{7}.$ Clearing fractions, \\[22\\cos{x}=7+7\\sin{x}.\\] Squaring to get an expression in terms of $\\sin^2{x}$ and $\\cos^2{x}$, \\[484\\cos^2{x}=49+49\\sin^2{x}+98\\sin{x}.\\] Substituting $\\cos^2{x}=1-\\sin^2{x},$ \\[484(1-\\sin^2{x})=49+49\\sin^2{x}+98\\sin{x}.\\] Expanding then collecting terms yields a quadratic in $\\sin{x}:$ \\[533\\sin^2{x}+98\\sin{x}-435=0.\\] To make calculations easier, let $y=\\sin{x}.$ \\[533y^2+98y-435=0.\\] Upon inspection, $y=-1$ is a root. Dividing by $y+1$, \\[533y^2+98y-435=(533y-435)(y+1).\\] Substituting $y=\\sin{x},$ we see that $\\sin{x}=-1$ doesn't work, as $\\cos{x}=0$, leaving $\\tan{x}$ undefined. We conclude that $\\sin{x}=\\frac{435}{533}.$ Since $\\sin^2{x}+\\cos^2{x}=1,$ \\[\\cos{x}=\\pm \\sqrt{\\frac{533^2-435^2}{533^2}}.\\] \\[=\\pm \\frac{308}{533}.\\] After checking via the given equation, we know that only the positive solution works. Therefore, \\[\\csc{x}+\\cot{x}=\\frac{1}{\\sin{x}}+\\frac{\\cos{x}}{\\sin{x}}\\] \\[=\\frac{533}{435}+\\frac{308}{435}\\] \\[=\\frac{29}{15}=\\frac{m}{n}.\\] Adding $m$ and $n$, our answer is $\\textbf{044} -Benedict T (countmath1)"
] |
1991-I-10
| 1,991 | 10 |
Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$ , are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$ , or as a $b^{}_{}$ when it should be an $a^{}_{}$ . However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter. Let $S_a^{}$ be the three-letter string received when $aaa^{}_{}$ is transmitted and let $S_b^{}$ be the three-letter string received when $bbb^{}_{}$ is transmitted. Let $p$ be the probability that $S_a^{}$ comes before $S_b^{}$ in alphabetical order. When $p$ is written as a fraction in lowest terms, what is its numerator?
| 532 | null |
[
"Solution 1 Let us make a chart of values in alphabetical order, where $P_a,\\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$, and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \\sum_{n=1}^{b} P_b$): \\[\\begin{array}{|r||r|r|r|} \\hline \\text{String}&P_a&P_b&S_b\\\\ \\hline aaa & 8 & 1 & 1 \\\\ aab & 4 & 2 & 3 \\\\ aba & 4 & 2 & 5 \\\\ abb & 2 & 4 & 9 \\\\ baa & 4 & 2 & 11 \\\\ bab & 2 & 4 & 15 \\\\ bba & 2 & 4 & 19 \\\\ bbb & 1 & 8 & 27 \\\\ \\hline \\end{array}\\] The probability is $p=\\sum P_a \\cdot (27 - S_b)$, so the answer turns out to be $\\frac{8\\cdot 26 + 4 \\cdot 24 \\ldots 2 \\cdot 8 + 1 \\cdot 0}{27^2} = \\frac{532}{729}$, and the solution is $532. ~qwertysri987",
"Let us make a chart of values in alphabetical order, where $P_a,\\ P_b$ are the probabilities that each string comes from $aaa$ and $bbb$ multiplied by $27$, and $S_b$ denotes the partial sums of $P_b$ (in other words, $S_b = \\sum_{n=1}^{b} P_b$): \\[\\begin{array}{|r||r|r|r|} \\hline \\text{String}&P_a&P_b&S_b\\\\ \\hline aaa & 8 & 1 & 1 \\\\ aab & 4 & 2 & 3 \\\\ aba & 4 & 2 & 5 \\\\ abb & 2 & 4 & 9 \\\\ baa & 4 & 2 & 11 \\\\ bab & 2 & 4 & 15 \\\\ bba & 2 & 4 & 19 \\\\ bbb & 1 & 8 & 27 \\\\ \\hline \\end{array}\\] The probability is $p=\\sum P_a \\cdot (27 - S_b)$, so the answer turns out to be $\\frac{8\\cdot 26 + 4 \\cdot 24 \\ldots 2 \\cdot 8 + 1 \\cdot 0}{27^2} = \\frac{532}{729}$, and the solution is $532. ~qwertysri987",
"Let $S(a,n)$ be the $n$th letter of string $S(a)$. Compare the first letter of the string $S(a)$ to the first letter of the string $S(b)$. There is a $(2/3)^2=4/9$ chance that $S(a,1)$ comes before $S(b,1)$. There is a $2(1/3)(2/3)=4/9$ that $S(a,1)$ is the same as $S(b,1)$. If $S(a,1)=S(b,1)$, then you do the same for the second letters of the strings. But you have to multiply the $4/9$ chance that $S(a,2)$ comes before $S(b,2)$ as there is a $4/9$ chance we will get to this step. Similarly, if $S(a,2)=S(b,2)$, then there is a $(4/9)^3$ chance that we will get to comparing the third letters and that $S(a)$ comes before $S(b)$. So we have $p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729$. Therefore, the answer is $532. ~qwertysri987",
"Consider $n$ letter strings instead. If the first letters all get transmitted correctly, then the $a$ string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next $n-1$ letter string following the first letter. This easily leads to a recursion: $p_n=\\frac23\\cdot\\frac23+2\\cdot\\frac23\\cdot\\frac13p_{n-1}=\\frac49+\\frac49p_{n-1}$. Clearly, $p_0=0\\implies p_1=\\frac49\\implies p_2=\\frac{52}{81}\\implies p_3=\\frac{532}{729}$. Therefore, the answer is $532. ~qwertysri987",
"The probability that $S_a$ will take the form $a$ _ _ and that $S_b$ will take the form $b$ _ _ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$. Then, the probability that both $S_a$ and $S_b$ will share the same first digit is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$. Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of $S_a$ is $a$ and that the second letter of $S_b$ is $b$ is $\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}$. Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$. Now, if the last case is true then the probability that $S_a$ precedes $S_b$ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$. Therefore the total probability would be: $\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9}\\right)\\right) = \\frac{4}{9}+\\frac{4}{9}\\left(\\frac{52}{81}\\right) = \\frac{4}{9} + \\frac{208}{729} = \\frac{532}{729}$. Therefore the answer is $532. ~qwertysri987",
"The probability that $S_a$ will take the form $a$ _ _ and that $S_b$ will take the form $b$ _ _ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$. Then, the probability that both $S_a$ and $S_b$ will share the same first digit is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$. Now if the first digits of either sequence are the same, then we must now consider these same probabilities for the second letter of each sequence. The probability that when the first two letters of both sequences are the same, that the second letter of $S_a$ is $a$ and that the second letter of $S_b$ is $b$ is $\\frac{2}{3}\\cdot\\frac{2}{3}=\\frac{4}{9}$. Similarly, the probability that when the first two letters of both sequences are the same, that the second set of letters in both sets of sequences are the same is $2\\cdot\\frac{2}{3}\\cdot\\frac{1}{3} = \\frac{4}{9}$. Now, if the last case is true then the probability that $S_a$ precedes $S_b$ is $\\frac{2}{3}\\cdot\\frac{2}{3} = \\frac{4}{9}$. Therefore the total probability would be: $\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9} + \\frac{4}{9}\\left(\\frac{4}{9}\\right)\\right) = \\frac{4}{9}+\\frac{4}{9}\\left(\\frac{52}{81}\\right) = \\frac{4}{9} + \\frac{208}{729} = \\frac{532}{729}$. Therefore the answer is $532. ~qwertysri987"
] |
1991-I-11
| 1,991 | 11 |
Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$ , no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the form $\pi(a-b\sqrt{c})$ , where $a,b,c^{}_{}$ are positive integers and $c^{}_{}$ is not divisible by the square of any prime. Find $a+b+c^{}_{}$ . [asy] unitsize(100); draw(Circle((0,0),1)); dot((0,0)); draw((0,0)--(1,0)); label("$1$", (0.5,0), S); for (int i=0; i<12; ++i) { dot((cos(i*pi/6), sin(i*pi/6))); } for (int a=1; a<24; a+=2) { dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))); draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12))); draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); } [/asy]
| 135 | null |
[
"We wish to find the radius of one circle, so that we can find the total area. Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius $1$. We thus know that the apothem of the dodecagon is equal to $1$. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote $A, M,$ and $O$ respectively. Notice that $OM=1$, and that $\\triangle OMA$ is a right triangle with hypotenuse $OA$ and $m \\angle MOA = 15^\\circ$. Thus $AM = (1) \\tan{15^\\circ} = 2 - \\sqrt {3}$, which is the radius of one of the circles. The area of one circle is thus $\\pi(2 - \\sqrt {3})^{2} = \\pi (7 - 4 \\sqrt {3})$, so the area of all $12$ circles is $\\pi (84 - 48 \\sqrt {3})$, giving an answer of $84 + 48 + 3 = 135. Second note: If you didn't know the side lengths of a 15-75-90 triangle, just use the Law of Cosines with two sides of length 1 and a 30 degree angle."
] |
1991-I-12
| 1,991 | 12 |
Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$ , $Q^{}_{}$ , $R^{}_{}$ , and $S^{}_{}$ are interior points on sides $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , and $\overline{DA}$ , respectively. It is given that $PB^{}_{}=15$ , $BQ^{}_{}=20$ , $PR^{}_{}=30$ , and $QS^{}_{}=40$ . Let $m/n^{}_{}$ , in lowest terms, denote the perimeter of $ABCD^{}_{}$ . Find $m+n^{}_{}$ .
| 677 | null |
[
"[asy]defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q=(38.4,8.8), R=(15,0), S=(0,20); O=intersectionpoint(A--C,B--D); draw(A--B--C--D--cycle);draw(P--R..Q--S); draw(P--Q--R--S--cycle); label(\"\\(A\\)\",A,NW);label(\"\\(B\\)\",B,NE);label(\"\\(C\\)\",C,SE);label(\"\\(D\\)\",D,SW); label(\"\\(P\\)\",P,N);label(\"\\(Q\\)\",Q,E);label(\"\\(R\\)\",R,SW);label(\"\\(S\\)\",S,W); label(\"\\(15\\)\",B/2+P/2,N);label(\"\\(20\\)\",B/2+Q/2,E);label(\"\\(O\\)\",O,SW); [/asy] Solution 1 Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ($\\triangle BPQ \\cong \\triangle DRS$, $\\triangle APS \\cong \\triangle CRQ$). Quickly we realize that $O$ is also the center of the rectangle. By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = \\sqrt{15^2 + 20^2} = 25$. Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$. Also, $\\angle POQ = 90^{\\circ}$, so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 \\cdot OB = 20 \\cdot 15 + 15 \\cdot 20 = 600$. By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$, $AS = y$. The Pythagorean Theorem gives us $x^2 + y^2 = 625\\quad \\mathrm{(1)}$. Ptolemy’s Theorem gives us $25 \\cdot OA = 20x + 15y$. Since the diagonals of a rectangle are equal, $OA = \\frac{1}{2}d = OB$, and $20x + 15y = 600\\quad \\mathrm{(2)}$. Solving for $y$, we get $y = 40 - \\frac 43x$. Substituting into $\\mathrm{(1)}$, \\begin{eqnarray*}x^2 + \\left(40-\\frac 43x\\right)^2 &=& 625\\\\ 5x^2 - 192x + 1755 &=& 0\\\\ x = \\frac{192 \\pm \\sqrt{192^2 - 4 \\cdot 5 \\cdot 1755}}{10} &=& 15, \\frac{117}{5}\\end{eqnarray*} We reject $15$ because then everything degenerates into squares, but the condition that $PR \\neq QS$ gives us a contradiction. Thus $x = \\frac{117}{5}$, and backwards solving gives $y = \\frac{44}5$. The perimeter of $ABCD$ is $2\\left(20 + 15 + \\frac{117}{5} + \\frac{44}5\\right) = \\frac{672}{5}$, and $m + n = 677",
"Let $O$ be the center of the rhombus. Via parallel sides and alternate interior angles, we see that the opposite triangles are congruent ($\\triangle BPQ \\cong \\triangle DRS$, $\\triangle APS \\cong \\triangle CRQ$). Quickly we realize that $O$ is also the center of the rectangle. By the Pythagorean Theorem, we can solve for a side of the rhombus; $PQ = \\sqrt{15^2 + 20^2} = 25$. Since the diagonals of a rhombus are perpendicular bisectors, we have that $OP = 15, OQ = 20$. Also, $\\angle POQ = 90^{\\circ}$, so quadrilateral $BPOQ$ is cyclic. By Ptolemy's Theorem, $25 \\cdot OB = 20 \\cdot 15 + 15 \\cdot 20 = 600$. By similar logic, we have $APOS$ is a cyclic quadrilateral. Let $AP = x$, $AS = y$. The Pythagorean Theorem gives us $x^2 + y^2 = 625\\quad \\mathrm{(1)}$. Ptolemy’s Theorem gives us $25 \\cdot OA = 20x + 15y$. Since the diagonals of a rectangle are equal, $OA = \\frac{1}{2}d = OB$, and $20x + 15y = 600\\quad \\mathrm{(2)}$. Solving for $y$, we get $y = 40 - \\frac 43x$. Substituting into $\\mathrm{(1)}$, \\begin{eqnarray*}x^2 + \\left(40-\\frac 43x\\right)^2 &=& 625\\\\ 5x^2 - 192x + 1755 &=& 0\\\\ x = \\frac{192 \\pm \\sqrt{192^2 - 4 \\cdot 5 \\cdot 1755}}{10} &=& 15, \\frac{117}{5}\\end{eqnarray*} We reject $15$ because then everything degenerates into squares, but the condition that $PR \\neq QS$ gives us a contradiction. Thus $x = \\frac{117}{5}$, and backwards solving gives $y = \\frac{44}5$. The perimeter of $ABCD$ is $2\\left(20 + 15 + \\frac{117}{5} + \\frac{44}5\\right) = \\frac{672}{5}$, and $m + n = 677",
"From above, we have $OB = 24$ and $BD = 48$. Returning to $BPQO,$ note that $\\angle PQO\\cong \\angle PBO \\cong ABD.$ Hence, $\\triangle ABD \\sim \\triangle OQP$ by $AA$ similarity. From here, it's clear that \\[\\frac {AD}{BD} = \\frac {OP}{PQ}\\implies \\frac {AD}{48} = \\frac {15}{25}\\implies AD = \\frac {144}{5}.\\] Similarly, \\[\\frac {AB}{BD} = \\frac {IQ}{PQ}\\implies \\frac {AB}{48} = \\frac {20}{25}\\implies AB = \\frac {192}{5}.\\] Therefore, the perimeter of rectangle $ABCD$ is $2(AB + AD) = 2\\left(\\frac {192}{5} + \\frac {144}{5}\\right) = \\frac {672}{5}.$ Solution 3 The triangles $QOB,OBC$ are isosceles, and similar (because they have $\\angle QOB = \\angle OBC$). Hence $\\frac {BQ}{OB} = \\frac {OB}{BC} \\Rightarrow OB^2 = BC \\cdot BQ$. The length of $OB$ could be found easily from the area of $BPQ$: \\[BP \\cdot BQ = \\frac {OB}{2} \\cdot PQ \\Rightarrow OB = \\frac {2BP\\cdot BQ}{PQ} \\Rightarrow OB = 24\\] \\[OB^2 = BC \\cdot BQ \\Rightarrow 24^2 = (20 + CQ) \\cdot 20 \\Rightarrow CQ = \\frac {44}{5}\\] From the right triangle $CRQ$ we have $RC^2 = 25^2 - \\left(\\frac {44}{5}\\right)^2\\Rightarrow RC = \\frac {117}{5}$. We could have also defined a similar formula: $OB^2 = BP \\cdot BA$, and then we found $AP$, the segment $OB$ is tangent to the circles with diameters $AO,CO$. The perimeter is $2(PB + BQ + QC + CR) = 2\\left(15 + 20 + \\frac {44 + 117}{5}\\right) = \\frac {672}{5}\\Rightarrow m+n=677$. Solution 4 For convenience, let $\\angle PQS = \\theta$. Since the opposite triangles are congruent we have that $\\angle BQR = 3\\theta$, and therefore $\\angle QRC = 3\\theta - 90$. Let $QC = a$, then we have $\\sin{(3\\theta - 90)} = \\frac {a}{25}$, or $- \\cos{3\\theta} = \\frac {a}{25}$. Expanding with the formula $\\cos{3\\theta} = 4\\cos^3{\\theta} - 3\\cos{\\theta}$, and since we have $\\cos{\\theta} = \\frac {4}{5}$, we can solve for $a$. The rest then follows similarily from above. Solution 5 We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of $15,\\ 20,\\ 25$, namely triangles $DSR, OSR, OQP,$ and $BQP$. Let the points of triangle $DSR$ be $(0,0)\\ (0,20)\\ (15,0)$. Let point $E$ be on $\\overline{SR}$, such that $SE = 16$ and $ER = 9$. Triangle $DSR$ can be split into two similar 3-4-5 right triangles, $ESD$ and $EDR$. By the Pythagorean Theorem, point $D$ is $12$ away from point $E$. Repeating the process, if we break down triangle $DER$ into two more similar triangles, we find that point $E$ is at $(9.6, 7.2)$. By reflecting point $D = (0,0)$ over point $E = (9.6, 7.2)$, we get point $O = (19.2, 14.4)$. By reflecting point $D$ over point $O$, we get point $B = (38.4, 28.8)$. Thus, the perimeter is equal to $(38.4 + 28.8)\\times 2 = \\frac {672}{5}$, making the final answer $672+5 = 677$. Solution 6 We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\\frac{44}{5} = a$, $\\frac{117}{5} = b$, from which it is straightforward to find the answer, $2(a+b+35) \\Rightarrow \\frac{672}{5}$. Thus, $m+n = \\frac{672}{5}\\implies677",
"The triangles $QOB,OBC$ are isosceles, and similar (because they have $\\angle QOB = \\angle OBC$). Hence $\\frac {BQ}{OB} = \\frac {OB}{BC} \\Rightarrow OB^2 = BC \\cdot BQ$. The length of $OB$ could be found easily from the area of $BPQ$: \\[BP \\cdot BQ = \\frac {OB}{2} \\cdot PQ \\Rightarrow OB = \\frac {2BP\\cdot BQ}{PQ} \\Rightarrow OB = 24\\] \\[OB^2 = BC \\cdot BQ \\Rightarrow 24^2 = (20 + CQ) \\cdot 20 \\Rightarrow CQ = \\frac {44}{5}\\] From the right triangle $CRQ$ we have $RC^2 = 25^2 - \\left(\\frac {44}{5}\\right)^2\\Rightarrow RC = \\frac {117}{5}$. We could have also defined a similar formula: $OB^2 = BP \\cdot BA$, and then we found $AP$, the segment $OB$ is tangent to the circles with diameters $AO,CO$. The perimeter is $2(PB + BQ + QC + CR) = 2\\left(15 + 20 + \\frac {44 + 117}{5}\\right) = \\frac {672}{5}\\Rightarrow m+n=677$. Solution 4 For convenience, let $\\angle PQS = \\theta$. Since the opposite triangles are congruent we have that $\\angle BQR = 3\\theta$, and therefore $\\angle QRC = 3\\theta - 90$. Let $QC = a$, then we have $\\sin{(3\\theta - 90)} = \\frac {a}{25}$, or $- \\cos{3\\theta} = \\frac {a}{25}$. Expanding with the formula $\\cos{3\\theta} = 4\\cos^3{\\theta} - 3\\cos{\\theta}$, and since we have $\\cos{\\theta} = \\frac {4}{5}$, we can solve for $a$. The rest then follows similarily from above. Solution 5 We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of $15,\\ 20,\\ 25$, namely triangles $DSR, OSR, OQP,$ and $BQP$. Let the points of triangle $DSR$ be $(0,0)\\ (0,20)\\ (15,0)$. Let point $E$ be on $\\overline{SR}$, such that $SE = 16$ and $ER = 9$. Triangle $DSR$ can be split into two similar 3-4-5 right triangles, $ESD$ and $EDR$. By the Pythagorean Theorem, point $D$ is $12$ away from point $E$. Repeating the process, if we break down triangle $DER$ into two more similar triangles, we find that point $E$ is at $(9.6, 7.2)$. By reflecting point $D = (0,0)$ over point $E = (9.6, 7.2)$, we get point $O = (19.2, 14.4)$. By reflecting point $D$ over point $O$, we get point $B = (38.4, 28.8)$. Thus, the perimeter is equal to $(38.4 + 28.8)\\times 2 = \\frac {672}{5}$, making the final answer $672+5 = 677$. Solution 6 We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\\frac{44}{5} = a$, $\\frac{117}{5} = b$, from which it is straightforward to find the answer, $2(a+b+35) \\Rightarrow \\frac{672}{5}$. Thus, $m+n = \\frac{672}{5}\\implies677",
"For convenience, let $\\angle PQS = \\theta$. Since the opposite triangles are congruent we have that $\\angle BQR = 3\\theta$, and therefore $\\angle QRC = 3\\theta - 90$. Let $QC = a$, then we have $\\sin{(3\\theta - 90)} = \\frac {a}{25}$, or $- \\cos{3\\theta} = \\frac {a}{25}$. Expanding with the formula $\\cos{3\\theta} = 4\\cos^3{\\theta} - 3\\cos{\\theta}$, and since we have $\\cos{\\theta} = \\frac {4}{5}$, we can solve for $a$. The rest then follows similarily from above. Solution 5 We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of $15,\\ 20,\\ 25$, namely triangles $DSR, OSR, OQP,$ and $BQP$. Let the points of triangle $DSR$ be $(0,0)\\ (0,20)\\ (15,0)$. Let point $E$ be on $\\overline{SR}$, such that $SE = 16$ and $ER = 9$. Triangle $DSR$ can be split into two similar 3-4-5 right triangles, $ESD$ and $EDR$. By the Pythagorean Theorem, point $D$ is $12$ away from point $E$. Repeating the process, if we break down triangle $DER$ into two more similar triangles, we find that point $E$ is at $(9.6, 7.2)$. By reflecting point $D = (0,0)$ over point $E = (9.6, 7.2)$, we get point $O = (19.2, 14.4)$. By reflecting point $D$ over point $O$, we get point $B = (38.4, 28.8)$. Thus, the perimeter is equal to $(38.4 + 28.8)\\times 2 = \\frac {672}{5}$, making the final answer $672+5 = 677$. Solution 6 We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\\frac{44}{5} = a$, $\\frac{117}{5} = b$, from which it is straightforward to find the answer, $2(a+b+35) \\Rightarrow \\frac{672}{5}$. Thus, $m+n = \\frac{672}{5}\\implies677",
"We can just find coordinates of the points. After drawing a picture, we can see 4 congruent right triangles with sides of $15,\\ 20,\\ 25$, namely triangles $DSR, OSR, OQP,$ and $BQP$. Let the points of triangle $DSR$ be $(0,0)\\ (0,20)\\ (15,0)$. Let point $E$ be on $\\overline{SR}$, such that $SE = 16$ and $ER = 9$. Triangle $DSR$ can be split into two similar 3-4-5 right triangles, $ESD$ and $EDR$. By the Pythagorean Theorem, point $D$ is $12$ away from point $E$. Repeating the process, if we break down triangle $DER$ into two more similar triangles, we find that point $E$ is at $(9.6, 7.2)$. By reflecting point $D = (0,0)$ over point $E = (9.6, 7.2)$, we get point $O = (19.2, 14.4)$. By reflecting point $D$ over point $O$, we get point $B = (38.4, 28.8)$. Thus, the perimeter is equal to $(38.4 + 28.8)\\times 2 = \\frac {672}{5}$, making the final answer $672+5 = 677$. Solution 6 We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\\frac{44}{5} = a$, $\\frac{117}{5} = b$, from which it is straightforward to find the answer, $2(a+b+35) \\Rightarrow \\frac{672}{5}$. Thus, $m+n = \\frac{672}{5}\\implies677",
"We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\\frac{44}{5} = a$, $\\frac{117}{5} = b$, from which it is straightforward to find the answer, $2(a+b+35) \\Rightarrow \\frac{672}{5}$. Thus, $m+n = \\frac{672}{5}\\implies677",
"We will bash with trigonometry. Firstly, by Pythagoras Theorem, $PQ=QR=RS=SP=25$. We observe that $[PQRS]=\\frac{1}{2}\\cdot30\\cdot40=600$. Thus, if we drop an altitude from $P$ to $\\overline{SR}$ to point $E$, it will have length $\\frac{600}{25}=24$. In particular, $SE=7$ since we form a 7-24-25 triangle. Now, $\\sin\\angle APS=\\sin\\angle SPB=\\sin(\\angle SPQ+\\angle QPB)=\\sin\\angle SPQ\\cos\\angle QPB+\\sin\\angle QPB\\cos\\angle SPQ=\\sin\\angle PSR\\cos\\angle QPB-\\sin\\angle QPB\\cos\\angle PSR=\\frac{24}{25}\\cdot\\frac{15}{25}-\\frac{20}{25}\\cdot\\frac{7}{25}=\\frac{44}{125}$. Thus, since $PS=25$, we get that $AS=\\frac{44}{5}$. Now, by the Pythagorean Theorem, $AP=\\frac{117}{5}$. Using the same idea, $\\cos\\angle RSD=-\\cos\\angle RSA=-\\cos(\\angle RSP+\\angle PSA)=\\sin\\angle RSP\\sin\\angle PSA-\\cos\\angle RSP\\cos\\angle PSA=\\frac{24}{25}\\cdot\\frac{117}{125}-\\frac{7}{25}\\cdot\\frac{44}{125}=\\frac{4}{5}$. Thus, since $SR=20$. Now, we can finish. We know $AB=\\frac{117}{5}+15=\\frac{192}{5}$. We also know $AD=\\frac{44}{5}+20=\\frac{144}{5}$. Thus, our perimeter is $\\frac{672}{5}\\implies677"
] |
1991-I-13
| 1,991 | 13 |
A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?
| 990 | null |
[
"Solution 1 Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by \\[\\frac{r(r-1)}{(r+b)(r+b-1)}+\\frac{b(b-1)}{(r+b)(r+b-1)}=\\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\\frac{1}{2}.\\] Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$, for $r$ in terms of $t$, one obtains that \\[r=\\frac{t\\pm\\sqrt{t}}{2}\\, .\\] Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$, with $n\\in\\mathbb{N}$. Hence, $r=n(n\\pm 1)/2$ would correspond to the general solution. For the present case $t\\leq 1991$, and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions. In summary, the solution is that the maximum number of red socks is $r=990. - AlexLikeMath",
"Let $r$ and $b$ denote the number of red and blue socks, respectively. Also, let $t=r+b$. The probability $P$ that when two socks are drawn randomly, without replacement, both are red or both are blue is given by \\[\\frac{r(r-1)}{(r+b)(r+b-1)}+\\frac{b(b-1)}{(r+b)(r+b-1)}=\\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\\frac{1}{2}.\\] Solving the resulting quadratic equation $r^{2}-rt+t(t-1)/4=0$, for $r$ in terms of $t$, one obtains that \\[r=\\frac{t\\pm\\sqrt{t}}{2}\\, .\\] Now, since $r$ and $t$ are positive integers, it must be the case that $t=n^{2}$, with $n\\in\\mathbb{N}$. Hence, $r=n(n\\pm 1)/2$ would correspond to the general solution. For the present case $t\\leq 1991$, and so one easily finds that $n=44$ is the largest possible integer satisfying the problem conditions. In summary, the solution is that the maximum number of red socks is $r=990. - AlexLikeMath",
"Let $r$ and $b$ denote the number of red and blue socks such that $r+b\\le1991$. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\\frac12=\\frac12=\\frac{2rb}{(r+b)(r+b-1)}\\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\\implies r^2+2rb+b^2-r-b=4rb\\implies r^2-2rb+b^2$ $=(r-b)^2=r+b$, so $r+b$ must be a perfect square $k^2$. Clearly, $r=\\frac{k^2+k}2$, so the larger $k$, the larger $r$: $k^2=44^2$ is the largest perfect square below $1991$, and our answer is $\\frac{44^2+44}2=\\frac12\\cdot44(44+1)=22\\cdot45=11\\cdot90=990. - AlexLikeMath",
"Let $r$ and $b$ denote the number of red and blue socks, respectively. In addition, let $t = r + b$, the total number of socks in the drawer. From the problem, it is clear that $\\frac{r(r-1)}{t(t-1)} + \\frac{b(b-1)}{t(t-1)} = \\frac{1}{2}$ Expanding, we get $\\frac{r^2 + b^2 - r - b}{t^2 - t} = \\frac{1}{2}$ Substituting $t$ for $r + b$ and cross multiplying, we get $2r^2 + 2b^2 - 2r - 2b = r^2 + 2br + b^2 - r - b$ Combining terms, we get $b^2 - 2br + r^2 - b - r = 0$ To make this expression factorable, we add $2r$ to both sides, resulting in $(b - r)^2 - 1(b-r) = (b - r - 1)(b - r) = 2r$ From this equation, we can test values for the expression $(b - r - 1)(b - r)$, which is the multiplication of two consecutive integers, until we find the highest value of $b$ or $r$ such that $b + r \\leq 1991$. By testing $(b - r - 1) = 43$ and $(b - r) = 44$, we get that $r = 43(22) = 946$ and $b = 990$. Testing values one integer higher, we get that $r = 990$ and $b = 1035$. Since $990 + 1035 = 2025$ is greater than $1991$, we conclude that $(946, 990)$ is our answer. Since it doesn't matter whether the number of blue or red socks is $990$, we take the lower value for $r$, thus the maximum number of red socks is $r=990. - AlexLikeMath",
"As above, let $r$, $b$, and $t$ denote the number of red socks, the number of blue socks, and the total number of socks, respectively. We see that $\\frac{r(r-1)}{t(t-1)}+\\frac{b(b-1)}{t(t-1)}=\\frac{1}{2}$, so $r^2+b^2-r-b=\\frac{t(t-1)}{2}=r^2+b^2-t=\\frac{t^2}{2}-\\frac{t}{2}$. Seeing that we can rewrite $r^2+b^2$ as $(r+b)^2-2rb$, and remembering that $r+b=t$, we have $\\frac{t^2}{2}-\\frac{t}{2}=t^2-2rb-t$, so $2rb=\\frac{t^2}{2}-\\frac{t}{2}$, which equals $r^2+b^2-t$. We now have $r^2+b^2-t=2rb$, so $r^2-2rb+b^2=t$ and $r-b=\\pm\\sqrt{t}$. Adding this to $r+b=t$, we have $2r=t\\pm\\sqrt{t}$. To maximize $r$, we must use the positive square root and maximize $t$. The largest possible value of $t$ is the largest perfect square less than 1991, which is $1936=44^2$. Therefore, $r=\\frac{t+\\sqrt{t}}{2}=\\frac{1936+44}{2}=990. - AlexLikeMath",
"Let $r$ be the number of socks that are red, and $t$ be the total number of socks. We get: $2(r(r-1)+(t-r)(t-r-1))=t(t-1)$ Expanding the left hand side and the right hand side, we get: $4r^2-4rt+2t^2-2t = t^2-t$ And, moving terms, we will get that: $4r^2-4rt+t^2 = t$ We notice that the left side is a perfect square. $(2r-t)^2 = t$ Thus $t$ is a perfect square. And, the higher $t$ is, the higher $r$ will be. So, we should set $t = 44^2 = 1936$ And, we see, $2r-1936 = \\pm44$ We will use the positive root, to get that $2r-1936 = 44$, $2r = 1980$, and $r = 990. - AlexLikeMath",
"Let $r$ and $b$ denote the red socks and blue socks, respectively. Thus the equation in question is: $\\frac{r(r-1)+b(b-1)}{(r+b)(r+b-1)}=\\frac{1}{2}$ $\\Rightarrow 2r^2-2r+2b^2-2b=r^2+2rb+b^2-r-b$ $\\Rightarrow r^2+b^2-r-b-2rb=0$ $\\Rightarrow (r-b)^2=r+b\\le 1991$ Because we wish to maximize $r$, we have $r\\ge b$ and thus $r-b\\le 44$ as both $r$ and $b$ must be integers and ${44}^2=1936$ is the largest square less than or equal to $1991$. We now know that the maximum difference between the number of socks is $44$. Now we return to an earlier equation: $r^2+b^2-r-b-2rb=0$ $\\Rightarrow r^2-(2b+1)r+(b^2-b)=0$ Solving by the Quadratic formula, we have: $r=\\frac{2b+1\\pm\\sqrt{4b^2+4b+1-4b^2+4b}}{2}$ $\\Rightarrow r=\\frac{2b+1\\pm\\sqrt{8b+1}}{2}$ $\\Rightarrow r-b=\\frac{1\\pm\\sqrt{8b+1}}{2}$ $\\Rightarrow 44\\ge r-b=\\frac{1\\pm\\sqrt{8b+1}}{2}$ $\\Rightarrow 44\\ge\\frac{1\\pm\\sqrt{8b+1}}{2}$ $\\Rightarrow \\pm\\sqrt{8b+1}\\le 87$ $\\Rightarrow b\\le 946$ (Here we use the positive sign to maximize $r$.) Thus the optimal case would be $b=946$ as $r$ increases only when $b$ increases. In addition, $\\sqrt{8b+1}=87$ as we are using the equality case. We then plug back in: $r=\\frac{2b+1\\pm\\sqrt{8b+1}}{2}$ $\\Rightarrow r=\\frac{1893+87}{2}$ (using the established $\\sqrt{8b+1}=87$) $\\Rightarrow r=\\frac{1980}{2}$ $\\Rightarrow r=990 ~eevee9406"
] |
1991-I-14
| 1,991 | 14 |
A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by $\overline{AB}$ , has length 31. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ .
| 384 | null |
[
"[asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--E); dot(A^^B^^C^^D^^E^^F); label(\"\\(A\\)\",A,(-1,-1));label(\"\\(B\\)\",B,(1,-1));label(\"\\(C\\)\",C,(1,0)); label(\"\\(D\\)\",D,(1,1));label(\"\\(E\\)\",E,(-1,1));label(\"\\(F\\)\",F,(-1,0)); label(\"31\",A/2+B/2,(0.7,1));label(\"81\",B/2+C/2,(0.45,-0.2)); label(\"81\",C/2+D/2,(-1,-1));label(\"81\",D/2+E/2,(0,-1)); label(\"81\",E/2+F/2,(1,-1));label(\"81\",F/2+A/2,(1,1)); label(\"\\(x\\)\",A/2+C/2,(-1,1));label(\"\\(y\\)\",A/2+D/2,(1,-1.5)); label(\"\\(z\\)\",A/2+E/2,(1,0)); [/asy] Let $x=AC=BF$, $y=AD=BE$, and $z=AE=BD$. Ptolemy's Theorem on $ABCD$ gives $81y+31\\cdot 81=xz$, and Ptolemy on $ACDF$ gives $x\\cdot z+81^2=y^2$. Subtracting these equations give $y^2-81y-112\\cdot 81=0$, and from this $y=144$. Ptolemy on $ADEF$ gives $81y+81^2=z^2$, and from this $z=135$. Finally, plugging back into the first equation gives $x=105$, so $x+y+z=105+144+135=384.",
"Let $\\theta$ be the inscribed angle in each of the 5 sides of length 81, so $d \\sin \\theta = 81$. Since the inscribed angles sum to $\\pi$, $d \\sin 5\\theta = d \\sin (\\pi - 5\\theta) = 31$. Now consider the Chebyshev polynomials that put $\\dfrac{\\sin n\\theta}{\\sin \\theta}$ in terms of $\\cos \\theta$: $\\dfrac{\\sin 2\\theta}{\\sin \\theta} = 2 \\cos \\theta, \\dfrac{\\sin 3\\theta}{\\sin \\theta} = 4 \\cos^2 \\theta - 1$ $\\dfrac{\\sin 4\\theta}{\\sin \\theta} = 8 \\cos^3 \\theta - 4 \\cos \\theta, \\dfrac{\\sin 5\\theta}{\\sin \\theta} = 16 \\cos^4 \\theta - 12 \\cos^2 \\theta + 1$ The sum of the diagonals is $d\\sin 2\\theta + d\\sin 3\\theta + d\\sin 4\\theta$, which becomes $(d \\sin \\theta)(8\\cos^3 \\theta + 4\\cos^2 \\theta - 2\\cos \\theta - 1)$, and we're given $16 \\cos^4 \\theta - 12 \\cos^2 \\theta + 1 = \\dfrac{31}{81}$ Solve for $\\cos \\theta$: $16 \\cos^4 \\theta - 12 \\cos^2 \\theta + \\dfrac{9}{4} = \\dfrac{9}{4} - \\dfrac{50}{81}$ $\\left(4 \\cos^2 \\theta - \\dfrac{3}{2}\\right)^2 = \\dfrac{729}{324} - \\dfrac{200}{324} = \\left(\\dfrac{23}{18}\\right)^2$ $8 \\cos^2 \\theta - 3 = \\dfrac{23}{9}$ or $-\\dfrac{23}{9}$, so $8 \\cos^2 \\theta = \\dfrac{50}{9}$ or $\\dfrac{4}{9}$ $\\cos^2 \\theta = \\dfrac{25}{36}$ or $\\dfrac{1}{18}$, which means $\\cos \\theta$ must be $\\dfrac{5}{6}$ if $5 \\theta < \\pi$. Now $(d \\sin \\theta)\\left(8\\cos^3 \\theta + 4\\cos^2 \\theta - 2\\cos \\theta - 1\\right) = 81\\left(8 \\cdot \\dfrac{125}{216} + 4 \\cdot \\dfrac{25}{36} - 2 \\cdot \\dfrac{5}{6} - 1\\right)$ $= 3 \\left(8 \\cdot \\dfrac{125}{8} + 4 \\cdot \\dfrac{75}{4} - 2 \\cdot \\dfrac{45}{2} - 27\\right) = 3(125 + 75 - 45 - 27) = 384"
] |
1992-I-1
| 1,992 | 1 |
Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.
| 400 | null |
[
"Solution 1 There are 8 fractions which fit the conditions between 0 and 1: $\\frac{1}{30},\\frac{7}{30},\\frac{11}{30},\\frac{13}{30},\\frac{17}{30},\\frac{19}{30},\\frac{23}{30},\\frac{29}{30}$ Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\\frac{19}{30}=\\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\\cdots+9)=400 1992 AIME (Problems • Answer Key • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
"There are 8 fractions which fit the conditions between 0 and 1: $\\frac{1}{30},\\frac{7}{30},\\frac{11}{30},\\frac{13}{30},\\frac{17}{30},\\frac{19}{30},\\frac{23}{30},\\frac{29}{30}$ Their sum is 4. Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\\frac{19}{30}=\\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\\cdots+9)=400 1992 AIME (Problems • Answer Key • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
"By Euler's Totient Function, there are $8$ numbers that are relatively prime to $30$, less than $30$. Note that they come in pairs $(m,30-m)$ which result in sums of $1$; thus the sum of the smallest $8$ rational numbers satisfying this is $\\frac12\\cdot8\\cdot1=4$. Now refer to solution 1. Solution 3 Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\\phi(300) = 80.$ Therefore the answer is $\\frac{80}{2} \\cdot\\frac{300}{30} = 400 1992 AIME (Problems • Answer Key • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
"Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\\phi(300) = 80.$ Therefore the answer is $\\frac{80}{2} \\cdot\\frac{300}{30} = 400 1992 AIME (Problems • Answer Key • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
] |
1992-I-2
| 1,992 | 2 |
A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there?
| 502 | null |
[
"Note that an ascending number is exactly determined by its digits: for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits. So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Note that each digit may be present or may not be present. Hence, there are $2^9=512$ potential ascending numbers, one for each subset of $\\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$. However, we've counted one-digit numbers and the empty set, so we must subtract them off to get our answer, $512-10=502",
"We see that we can express the number of ways for a 2 digit number as $\\binom{8}{1}+\\binom{7}{1}...\\binom{1}{1}$ and a 3 digit number as $\\binom{8}{2}+\\binom{7}{2}+...\\binom{2}{2}$ this is because once we choose the first number the next two numbers can be any two of the numbers above it. Using the hockey stick identity repeatedly we can get $\\binom{9}{2}+\\binom{9}{3}+...+\\binom{9}{9}$. We know that $\\binom{n}{0}+\\binom{n}{1}+...\\binom{n}{n}=2^n$ this can be proved easily by story. If you have n people and you want to know how many ways you can make a group of any size you could do this $2$ ways. You could count everyone is binary in or out so $2^n$ or you could count each case individually: $\\binom{n}{0}+\\binom{n}{1}+...\\binom{n}{n}$ so the 2 statements are equivalent. Therfore we have $2^9-\\binom{9}{1}-\\binom{9}{0}=502 Note how you could have skipped the hockey stick step by choosing all the numbers first. 1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
] |
1992-I-3
| 1,992 | 3 |
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $0.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$ . What's the largest number of matches she could've won before the weekend began?
| 164 | null |
[
"Let $n$ be the number of matches won, so that $\\frac{n}{2n}=\\frac{1}{2}$, and $\\frac{n+3}{2n+4}>\\frac{503}{1000}$. Cross multiplying, $1000n+3000>1006n+2012$, so $n<\\frac{988}{6}=164 \\dfrac {4}{6}=164 \\dfrac{2}{3}$. Thus, the answer is $164.",
"Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly .$500 = \\tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that\\[\\frac{n+3}{2n+4} > .503 = \\frac{503}{1000}.\\]Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \\frac{988}{6} = 164.\\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $164 1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
"Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly .$500 = \\tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that\\[\\frac{n+3}{2n+4} > .503 = \\frac{503}{1000}.\\]Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \\frac{988}{6} = 164.\\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $164 1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
] |
1992-I-4
| 1,992 | 4 |
In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below. \[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$ ?
| 62 | null |
[
"Consider what the ratio means. Since we know that they are consecutive terms, we can say \\[\\frac{\\dbinom{n}{k-1}}{3} = \\frac{\\dbinom{n}{k}}{4} = \\frac{\\dbinom{n}{k+1}}{5}.\\] Taking the first part, and using our expression for $n$ choose $k$, \\[\\frac{n!}{3(k-1)!(n-k+1)!} = \\frac{n!}{4k!(n-k)!}\\] \\[\\frac{1}{3(k-1)!(n-k+1)!} = \\frac{1}{4k!(n-k)!}\\] \\[\\frac{1}{3(n-k+1)} = \\frac{1}{4k}\\] \\[n-k+1 = \\frac{4k}{3}\\] \\[n = \\frac{7k}{3} - 1\\] \\[\\frac{3(n+1)}{7} = k\\] Then, we can use the second part of the equation. \\[\\frac{n!}{4k!(n-k)!} = \\frac{n!}{5(k+1)!(n-k-1)!}\\] \\[\\frac{1}{4k!(n-k)!} = \\frac{1}{5(k+1)!(n-k-1)!}\\] \\[\\frac{1}{4(n-k)} = \\frac{1}{5(k+1)}\\] \\[\\frac{4(n-k)}{5} = k+1\\] \\[\\frac{4n}{5}-\\frac{4k}{5} = k+1\\] \\[\\frac{4n}{5} = \\frac{9k}{5} +1.\\] Since we know $k = \\frac{3(n+1)}{7}$ we can plug this in, giving us \\[\\frac{4n}{5} = \\frac{9\\left(\\frac{3(n+1)}{7}\\right)}{5} +1\\] \\[4n = 9\\left(\\frac{3(n+1)}{7}\\right)+5\\] \\[7(4n - 5) = 27n+27\\] \\[28n - 35 = 27n+27\\] \\[n = 62\\] We can also evaluate for $k$, and find that $k = \\frac{3(62+1)}{7} = 27.$ Since we want $n$, however, our final answer is $062. by ciceronii",
"Call the row $x=t+k$, and the position of the terms $t-1, t, t+1$. Call the middle term in the ratio $N = \\dbinom{t+k}{t} = \\frac{(t+k)!}{k!t!}$. The first term is $N \\frac{t}{k+1}$, and the final term is $N \\frac{k}{t+1}$. Because we have the ratio $3:4:5$, $\\frac{t}{k+1} = \\frac{3}{4}$ and $\\frac{k}{t+1} = \\frac{5}{4}$. $4t = 3k+3$ and $4k= 5t+5$ $4t-3k=3$ $5t-4k=-5$ Solve the equations to get $t= 27, k=35$ and $x = t+k = 062. -Solution and LaTeX by jackshi2006, variables and algebra simplified by oinava"
] |
1992-I-5
| 1,992 | 5 |
Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$ , $0^{}_{}<r<1$ , that have a repeating decimal expansion in the form $0.abcabcabc\ldots=0.\overline{abc}$ , where the digits $a^{}_{}$ , $b^{}_{}$ , and $c^{}_{}$ are not necessarily distinct. To write the elements of $S^{}_{}$ as fractions in lowest terms, how many different numerators are required?
| 660 | null |
[
"We consider the method in which repeating decimals are normally converted to fractions with an example: $x=0.\\overline{176}$ $\\Rightarrow 1000x=176.\\overline{176}$ $\\Rightarrow 999x=1000x-x=176$ $\\Rightarrow x=\\frac{176}{999}$ Thus, let $x=0.\\overline{abc}$ $\\Rightarrow 1000x=abc.\\overline{abc}$ $\\Rightarrow 999x=1000x-x=abc$ $\\Rightarrow x=\\frac{abc}{999}$ If $abc$ is not divisible by $3$ or $37$, then this is in lowest terms. Let us consider the other multiples: $333$ multiples of $3$, $27$ of $37$, and $9$ of both $3$ and $37$, so $999-333-27+9 = 648$, which is the amount that are neither. The $12$ numbers that are multiples of $81$ reduce to multiples of $3$. We have to count these since it will reduce to a multiple of $3$ which we have removed from $999$, but, this cannot be removed since the numerator cannot cancel the $3$.There aren't any numbers which are multiples of $37^2$, so we can't get numerators which are multiples of $37$. Therefore $648 + 12 = 660. Note: You can use Euler's totient function to find the numbers relatively prime to 999 which gives 648. 1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
] |
1992-I-6
| 1,992 | 6 |
For how many pairs of consecutive integers in $\{1000,1001,1002^{}_{},\ldots,2000\}$ is no carrying required when the two integers are added?
| 156 | null |
[
"For one such pair of consecutive integers, let the smaller integer be $\\underline{1ABC},$ where $A,B,$ and $C$ are digits from $0$ through $9.$ We wish to count the ordered triples $(A,B,C).$ By casework, we consider all possible forms of the larger integer, as shown below. \\[\\begin{array}{c|c|c|c|c|c|c} & & & & & & \\\\ [-2.5ex] \\textbf{Case} & \\textbf{Thousands} & \\textbf{Hundreds} & \\textbf{Tens} & \\textbf{Ones} & \\textbf{Conditions for No Carrying} & \\boldsymbol{\\#}\\textbf{ of Ordered Triples} \\\\ [0.5ex] \\hline & & & & & & \\\\ [-2ex] 0\\leq C\\leq 8 & 1 & A & B & C+1 & 0\\leq A,B,C\\leq 4 & 5^3 \\\\ 0\\leq B\\leq 8; \\ C=9 & 1 & A & B+1 & 0 & 0\\leq A,B\\leq 4; \\ C=9 & 5^2 \\\\ 0\\leq A\\leq 8; \\ B=C=9 & 1 & A+1 & 0 & 0 & 0\\leq A\\leq 4; \\ B=C=9 & 5 \\\\ A=B=C=9 & 2 & 0 & 0 & 0 & A=B=C=9 & 1 \\end{array}\\] Together, the answer is $5^3+5^2+5+1=156 ~MRENTHUSIASM",
"Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$, then $h+d\\ge 10$ or $c+g\\ge 10$ or $b+f\\ge 10$. 6. Consider $c \\in \\{0, 1, 2, 3, 4\\}$. $1abc + 1ab(c+1)$ has no carry if $a, b \\in \\{0, 1, 2, 3, 4\\}$. This gives $5^3=125$ possible solutions. With $c \\in \\{5, 6, 7, 8\\}$, there obviously must be a carry. Consider $c = 9$. $a, b \\in \\{0, 1, 2, 3, 4\\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$, $a \\in \\{0, 1, 2, 3, 4, 9\\}$ have no carry. Thus, the solution is $125 + 25 + 6=156.",
"Consider the ordered pair $(1abc , 1abc - 1)$ where $a,b$ and $c$ are digits. We are trying to find all ordered pairs where $(1abc) + (1abc - 1)$ does not require carrying. For the addition to require no carrying, $2a,2b < 10$, so $a,b < 5$ unless $1abc$ ends in $00$, which we will address later. Clearly, if $c \\in \\{0, 1, 2, 3, 4 ,5\\}$, then adding $(1abc) + (1abc - 1)$ will require no carrying. We have $5$ possibilities for the value of $a$, $5$ for $b$, and $6$ for $c$, giving a total of $(5)(5)(6) = 150$, but we are not done yet. We now have to consider the cases where $b,c = 0$, specifically when $1abc \\in \\{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\\}$. We can see that $1100, 1200, 1300, 1400, 1500$, and $2000$ all work, giving a grand total of $150 + 6 = 156 ordered pairs.",
"Since there are $2000-1000+1 = 1001$ numbers in the set, this means that there are $1000$ consecutive pairs of integers $(a, b)$ in the set. For a pair to carry over, either the hundreds digits will carry over, the tens digits will carry over, or the ones digits will carry over. For the hundreds digits, every consecutive pair in $\\{1500, 1501, ..., 2000\\}$ will have to carry over, except for the pair $\\{1999, 2000\\}$ when no carrying occurs, removing $500-1 = 499$ pairs. Now, consider the case when the tens digit is more than 5. Every pair of consecutive integers from $\\{1050,1051, ..., 1100\\}$ will have to carry over except for the pair $\\{1099, 1100\\}$, and since there are 5 of these cases less than $1500$ (namely the integers from $1050-1100$, $1150-1200$, $1250-1300$, $1350-1400$, and $1450-1500$) this removes $49 \\cdot 5 = 245$ pairs from the list. Now, consider the case where the ones digits carry over. In this case, every pair of consecutive integers from $\\{1005, 1006, 1007, 1008, 1009\\}$ carries over, removing 4 pairs. Since there are 5 possible cases for the tens digit ($0-4$) and 5 possible cases for the hundreds digit, there are $4 \\cdot 5 \\cdot 5 = 100$ more cases to be excluded. In total, there are $1000 - 499 - 245 - 100 = 156 consecutive pairs. ~Soupboy0",
"Since there are $2000-1000+1 = 1001$ numbers in the set, this means that there are $1000$ consecutive pairs of integers $(a, b)$ in the set. For a pair to carry over, either the hundreds digits will carry over, the tens digits will carry over, or the ones digits will carry over. For the hundreds digits, every consecutive pair in $\\{1500, 1501, ..., 2000\\}$ will have to carry over, except for the pair $\\{1999, 2000\\}$ when no carrying occurs, removing $500-1 = 499$ pairs. Now, consider the case when the tens digit is more than 5. Every pair of consecutive integers from $\\{1050,1051, ..., 1100\\}$ will have to carry over except for the pair $\\{1099, 1100\\}$, and since there are 5 of these cases less than $1500$ (namely the integers from $1050-1100$, $1150-1200$, $1250-1300$, $1350-1400$, and $1450-1500$) this removes $49 \\cdot 5 = 245$ pairs from the list. Now, consider the case where the ones digits carry over. In this case, every pair of consecutive integers from $\\{1005, 1006, 1007, 1008, 1009\\}$ carries over, removing 4 pairs. Since there are 5 possible cases for the tens digit ($0-4$) and 5 possible cases for the hundreds digit, there are $4 \\cdot 5 \\cdot 5 = 100$ more cases to be excluded. In total, there are $1000 - 499 - 245 - 100 = 156 consecutive pairs. ~Soupboy0"
] |
1992-I-7
| 1,992 | 7 |
Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$ . The area of face $ABC^{}_{}$ is $120^{}_{}$ , the area of face $BCD^{}_{}$ is $80^{}_{}$ , and $BC=10^{}_{}$ . Find the volume of the tetrahedron.
| 320 | null |
[
"Since the area $BCD=80=\\frac{1}{2}\\cdot10\\cdot16$, the perpendicular from $D$ to $BC$ has length $16$. The perpendicular from $D$ to $ABC$ is $16 \\cdot \\sin 30^\\circ=8$. Therefore, the volume is $\\frac{8\\cdot120}{3}=320.",
"The area of $ABC$ is 120 and $BC$=10, the slant height is 24. Height from $A$ to $BCD$ is $24 \\cdot \\sin 30^\\circ=12$. Since area of $BCD$ is 80, the volume of tetrahedron $ABCD$= $\\frac{80\\cdot12}{3}=320."
] |
1992-I-8
| 1,992 | 8 |
For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$ , define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$ , whose $n^{th}$ term is $a_{n+1}-a_n^{}$ . Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$ , and that $a_{19}=a_{92}^{}=0$ . Find $a_1^{}$ .
| 819 | null |
[
"Note that the $\\Delta$s are reminiscent of differentiation; from the condition $\\Delta(\\Delta{A}) = 1$, we are led to consider the differential equation \\[\\frac{d^2 A}{dn^2} = 1\\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \\[a_{n} = \\frac{1}{2}(n-19)(n-92)\\] as we must have roots at $n = 19$ and $n = 92$. Thus, $a_1=\\frac{1}{2}(1-19)(1-92)=819.",
"Note that the $\\Delta$s are reminiscent of differentiation; from the condition $\\Delta(\\Delta{A}) = 1$, we are led to consider the differential equation \\[\\frac{d^2 A}{dn^2} = 1\\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \\[a_{n} = \\frac{1}{2}(n-19)(n-92)\\] as we must have roots at $n = 19$ and $n = 92$. Thus, $a_1=\\frac{1}{2}(1-19)(1-92)=819.",
"Let $\\Delta^1 A=\\Delta A$, and $\\Delta^n A=\\Delta(\\Delta^{(n-1)}A)$. Note that in every sequence of $a_i$, $a_n=\\binom{n-1}{1}\\Delta a_n + \\binom{n-1}{2}\\Delta^2 a_n +\\binom{n-1}{3}\\Delta^3 a_n + ...$ Then $a_n=a_1 +\\binom{n-1}{1}\\Delta a_1 +\\binom{n-1}{2}\\Delta^2 a_1 +\\binom{n-1}{3}\\Delta^3 a_1 + ...$ Since $\\Delta a_1 =a_2 -a_1$, $a_n=a_1 +\\binom{n-1}{1}(a_2-a_1) +\\binom{n-1}{2}\\cdot 1=a_1 +n(a_2-a_1) +\\binom{n-1}{2}$ $a_{19}=0=a_1+18(a_2-a_1)+\\binom{18}{2}=18a_2-17a_1+153$ $a_{92}=0=a_1+91(a_2-a_1)+\\binom{91}{2}=91a_2-90a_1+4095$ Solving, $a_1=819.",
"The sequence $\\Delta(\\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below. $\\begin{array}{rcl} a_3+a_1-2a_2&=&1\\\\ a_4+a_2-2a_3&=&1\\\\ &\\vdots\\\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\\\ \\end{array}$ Adding the above $k-1$ equations we find that \\[(a_{k+1} - a_k) = k-1 + (a_2-a_1).\\tag{1}\\] We can sum equation $(1)$ from $k=1$ to $18$, finding \\[18(a_1-a_2) - a_1 = 153.\\tag{2}\\] We can also sum equation $(1)$ from $k=1$ to $91$, finding \\[91(a_1-a_2) - a_1 = 4095.\\tag{3}\\] Finally, $18\\cdot (3) - 91\\cdot(2)$ gives $a_1=819. Kris17",
"Since all terms of $\\Delta(\\Delta A)$ are 1, we know that $\\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$. This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$. More specifically, $A_n=a_1+k(n-1)+\\frac{(n-1)(n-2)}{2}$. Plugging in $a_{19}=a_{92}=0$, we have the following linear system: \\[a_1+91k=-4095\\] \\[a_1+18k=-153\\] From this, we can easily find that $k=-54$ and $a_1=819. Solution by Zeroman",
"Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\\frac{1}{2}.$ Therefore $f(1) = 9 \\cdot 91 = 819",
"Let $a_1=a,a_2=b.$ From the conditions, we have \\[a_{n-1}+a_{n+1}=2a_n+1,\\] for all $n>1.$ From this, we find that \\begin{align*} a_3&=2b+1-a \\\\ a_4&=3b+3-2a\\\\ a_5&=4b+6-3a, \\end{align*} or, in general, \\[a_n=(n-1)b+\\frac{(n-2)(n-1)}{2}-(n-2)a.\\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \\begin{align*} 0=&18b+\\frac{17\\cdot18}{2}-17a\\\\ 0=&91b+\\frac{90\\cdot91}{2}-90a\\\\ b=&\\frac{17a-\\frac{17\\cdot18}{2}}{18}=\\frac{90a-\\frac{90\\cdot91}{2}}{91}. \\end{align*} Now, cross multiplying, we find that \\begin{align*} 91\\left(17a-\\frac{17\\cdot18}{2}\\right)&=18\\left(90a-\\frac{90\\cdot91}{2}\\right)\\\\ 1547a-\\frac{(18\\cdot91)\\cdot17}{2}&=1620-\\frac{(18\\cdot91)\\cdot90}{2}\\\\ 73a&=\\frac{18\\cdot91}{2}\\cdot(90-17=73)\\\\ a=\\frac{18\\cdot91}{2}=9\\cdot91=819. \\end{align*}"
] |
1992-I-9
| 1,992 | 9 |
Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$ , $BC=50^{}_{}$ , $CD=19^{}_{}$ , and $AD=70^{}_{}$ , with $AB^{}_{}$ parallel to $CD^{}_{}$ . A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$ . Given that $AP^{}_{}=\frac mn$ , where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive integers, find $m+n^{}_{}$ .
| 164 | null |
[
"Let $AP=x$ so that $PB=92-x.$ Extend $AD, BC$ to meet at $X,$ and note that $XP$ bisects $\\angle AXB;$ let it meet $CD$ at $E.$ Using the angle bisector theorem, we let $XB=y(92-x), XA=xy$ for some $y.$ Then $XD=xy-70, XC=y(92-x)-50,$ thus \\[\\frac{xy-70}{y(92-x)-50} = \\frac{XD}{XC} = \\frac{ED}{EC}=\\frac{AP}{PB} = \\frac{x}{92-x},\\] which we can rearrange, expand and cancel to get $120x=70\\cdot 92,$ hence $AP=x=\\frac{161}{3}$. This gives us a final answer of $161+3=164",
"Let $AB$ be the base of the trapezoid and consider angles $A$ and $B$. Let $x=AP$ and let $h$ equal the height of the trapezoid. Let $r$ equal the radius of the circle. Then \\[\\sin{A}= \\frac{r}{x} = \\frac{h}{70}\\qquad\\text{ and }\\qquad\\sin{B}= \\frac{r}{92-x} = \\frac{h}{50}.\\tag{1}\\] Let $z$ be the distance along $AB$ from $A$ to where the perp from $D$ meets $AB$. Then $h^2 +z^2 =70^2$ and $(73-z)^2 + h^2 =50^2$ so $h =\\frac{\\sqrt{44710959}}{146}$. We can substitute this into $(1)$ to find that $x= \\frac{11753}{219} = \\frac{161}{3}$ and $m+n = 164$. Remark: One can come up with the equations in $(1)$ without directly resorting to trig. From similar triangles, $h/r = 70/x$ and $h/r = 50/ (92-x)$. This implies that $70/x =50/(92-x)$, so $x = 161/3$.",
"From $(1)$ above, $x = \\frac{70r}{h}$ and $92-x = \\frac{50r}{h}$. Adding these equations yields $92 = \\frac{120r}{h}$. Thus, $x = \\frac{70r}{h} = \\frac{7}{12}\\cdot\\frac{120r}{h} = \\frac{7}{12}\\cdot92 = \\frac{161}{3}$, and $m+n = 164",
"Extend $AD$ and $BC$ to meet at a point $X$. Since $AB$ and $CD$ are parallel, $\\triangle XCD \\sim \\triangle XAB$. If $XA$ is further extended to a point $A'$ and $XB$ is extended to a point $B'$ such that $A'B'$ is tangent to circle $P$, we discover that circle $P$ is the incircle of triangle $XA'B'$. Then line $XP$ is the angle bisector of $\\angle AXB$. By homothety, $P$ is the intersection of the angle bisector of $\\triangle XAB$ with $AB$. By the angle bisector theorem, \\begin{align*} \\frac{AX}{AP} &= \\frac{XB}{BP}\\\\ \\frac{AX}{AP} - \\frac{XD}{AP} &= \\frac{XB}{BP} - \\frac{XC}{BP}\\\\ \\frac{AD}{AP} &= \\frac{BC}{BP}\\\\ &=\\frac{7}{5} \\end{align*} Let $7a = AP$, then $AB = 7a + 5a = 12a$. $AP = \\frac{7}{12}(AB) = \\frac{92\\times 7}{12} = \\frac{161}{3}$. Thus, $m + n = 164$. Note: this solution shows that the length of $CD$ is irrelevant as long as there still exists a circle as described in the problem.",
"The area of the trapezoid is $\\frac{(19+92)h}{2}$, where $h$ is the height of the trapezoid. Draw lines $CP$ and $BP$. We can now find the area of the trapezoid as the sum of the areas of the three triangles $BPC$, $CPD$, and $PBA$. $[BPC] = \\frac{1}{2} \\cdot 50 \\cdot r$ (where $r$ is the radius of the tangent circle.) $[CPD] = \\frac{1}{2} \\cdot 19 \\cdot h$ $[PBA] = \\frac{1}{2} \\cdot 70 \\cdot r$ $[BPC] + [CPD] + [PBA] = 60r + \\frac{19h}{2} = [ABCD] = \\frac{(19+92)h}{2}$ $60r = 46h$ $r = \\frac{23h}{30}$ From Solution 1 above, $\\frac{h}{70} = \\frac{r}{x}$ Substituting $r = \\frac{23h}{30}$, we find $x = \\frac{161}{3}$, hence the answer is $164.",
"As the problem tells, the circle is tangent to both sides $AD,BC$, we can make it up to a triangle $QAB$ and $P$ must lie on its angular bisector. Then we know that $AP:BP=7:5$, which makes $AP = \\frac{7}{12}(AB) = \\frac{92\\times 7}{12} = \\frac{161}{3}$. Thus, $m + n = 164$."
] |
1992-I-10
| 1,992 | 10 |
Consider the region $A^{}_{}$ in the complex plane that consists of all points $z^{}_{}$ such that both $\frac{z^{}_{}}{40}$ and $\frac{40^{}_{}}{\overline{z}}$ have real and imaginary parts between $0^{}_{}$ and $1^{}_{}$ , inclusive. What is the integer that is nearest the area of $A^{}_{}$ ?
| 572 | null |
[
"Let $z=a+bi \\implies \\frac{z}{40}=\\frac{a}{40}+\\frac{b}{40}i$. Since $0\\leq \\frac{a}{40},\\frac{b}{40}\\leq 1$ we have the inequality \\[0\\leq a,b \\leq 40\\]which is a square of side length $40$. Also, $\\frac{40}{\\overline{z}}=\\frac{40}{a-bi}=\\frac{40a}{a^2+b^2}+\\frac{40b}{a^2+b^2}i$ so we have $0\\leq a,b \\leq \\frac{a^2+b^2}{40}$, which leads to:\\[(a-20)^2+b^2\\geq 20^2\\] \\[a^2+(b-20)^2\\geq 20^2\\] We graph them: To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap: 1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$, and then doubled (to consider the entire overlapped area) 2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\\theta$ and $r = 40cos\\theta$. Using calculus with the appropriate bounds, we can compute the overlapped area. Using either method, we compute the overlapped area to be $200\\pi + 400$, and so the area of the intersection of those three graphs is $40^2-(200\\pi + 400) \\Rightarrow 1200 - 200\\pi \\approx 571.68$ $572"
] |
1992-I-11
| 1,992 | 11 |
Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$ , the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$ , and the resulting line is reflected in $l_2^{}$ . Let $R^{(1)}(l)=R(l)^{}_{}$ and $R^{(n)}(l)^{}_{}=R\left(R^{(n-1)}(l)\right)$ . Given that $l^{}_{}$ is the line $y=\frac{19}{92}x^{}_{}$ , find the smallest positive integer $m^{}_{}$ for which $R^{(m)}(l)=l^{}_{}$ .
| 945 | null |
[
"Let $l$ be a line that makes an angle of $\\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$. The angle between $l$ and $l_1$ is $\\theta - \\frac{\\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\\theta - \\frac{\\pi}{70}$. Thus, $l'$ makes an angle of $\\frac{\\pi}{70}-\\left(\\theta-\\frac{\\pi}{70}\\right) = \\frac{\\pi}{35}-\\theta$ with the positive $x$-axis. Similarly, since the angle between $l'$ and $l_2$ is $\\left(\\frac{\\pi}{35}-\\theta\\right)-\\frac{\\pi}{54}$, the angle between $l''$ and the positive $x$-axis is $\\frac{\\pi}{54}-\\left(\\left(\\frac{\\pi}{35}-\\theta\\right)-\\frac{\\pi}{54}\\right) = \\frac{\\pi}{27}-\\frac{\\pi}{35}+\\theta = \\frac{8\\pi}{945} + \\theta$. Thus, $R(l)$ makes an $\\frac{8\\pi}{945} + \\theta$ angle with the positive $x$-axis. So $R^{(n)}(l)$ makes an $\\frac{8n\\pi}{945} + \\theta$ angle with the positive $x$-axis. Therefore, $R^{(m)}(l)=l$ iff $\\frac{8m\\pi}{945}$ is an integral multiple of $\\pi$. Thus, $8m \\equiv 0\\pmod{945}$. Since $\\gcd(8,945)=1$, $m \\equiv 0 \\pmod{945}$, so the smallest positive integer $m$ is $945."
] |
1992-I-12
| 1,992 | 12 |
In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one of the remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed from the original board in previous moves.) AIME 1992 Problem 12.png The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.
| 792 | null |
[
"By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts. One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right \"carved\" out by the path a possible subset. Therefore, the total number of such paths is $\\binom{12}{5}=792",
"If any square is eaten, the squares to the right of it must also be eaten. Thus, if $a_i$ is the number of squares remaining on row $i$, there is exactly one way the row can be configured (the leftmost $a_i$ squares are uneaten and the ones to the right are eaten.) Additionally, the squares above an eaten square must be eaten, so $\\{a_i\\}$ is nondecreasing. We can thus write $0 \\leq a_1 \\leq a_2 \\leq \\dots \\leq a_5 \\leq 7$; the number of sequences $\\{a_i\\}$ satisfying this inequality may be found by Stars and Bars to be $\\binom{7+6-1}{6-1} = 792. Note This game is similar to an AoPS book."
] |
1992-I-13
| 1,992 | 13 |
Triangle $ABC^{}_{}$ has $AB=9^{}_{}$ and $BC: AC=40: 41^{}_{}$ . What's the largest area that this triangle can have?
| 820 | null |
[
"Solution 1 First, consider the triangle in a coordinate system with vertices at $(0,0)$, $(9,0)$, and $(a,b)$. Applying the distance formula, we see that $\\frac{ \\sqrt{a^2 + b^2} }{ \\sqrt{ (a-9)^2 + b^2 } } = \\frac{40}{41}$. We want to maximize $b$, the height, with $9$ being the base. Simplifying gives $-a^2 -\\frac{3200}{9}a +1600 = b^2$. To maximize $b$, we want to maximize $b^2$. So if we can write: $b^2=-(a+n)^2+m$, then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality, because if ${x^2 \\ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \\ge 0}$). $b^2=-a^2 -\\frac{3200}{9}a +1600=-\\left(a +\\frac{1600}{9}\\right)^2 +1600+\\left(\\frac{1600}{9}\\right)^2$. $\\Rightarrow b\\le\\sqrt{1600+\\left(\\frac{1600}{9}\\right)^2}=40\\sqrt{1+\\frac{1600}{81}}=\\frac{40}{9}\\sqrt{1681}=\\frac{40\\cdot 41}{9}$. Then the area is $9\\cdot\\frac{1}{2} \\cdot \\frac{40\\cdot 41}{9} = 820",
"First, consider the triangle in a coordinate system with vertices at $(0,0)$, $(9,0)$, and $(a,b)$. Applying the distance formula, we see that $\\frac{ \\sqrt{a^2 + b^2} }{ \\sqrt{ (a-9)^2 + b^2 } } = \\frac{40}{41}$. We want to maximize $b$, the height, with $9$ being the base. Simplifying gives $-a^2 -\\frac{3200}{9}a +1600 = b^2$. To maximize $b$, we want to maximize $b^2$. So if we can write: $b^2=-(a+n)^2+m$, then $m$ is the maximum value of $b^2$ (this follows directly from the trivial inequality, because if ${x^2 \\ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \\ge 0}$). $b^2=-a^2 -\\frac{3200}{9}a +1600=-\\left(a +\\frac{1600}{9}\\right)^2 +1600+\\left(\\frac{1600}{9}\\right)^2$. $\\Rightarrow b\\le\\sqrt{1600+\\left(\\frac{1600}{9}\\right)^2}=40\\sqrt{1+\\frac{1600}{81}}=\\frac{40}{9}\\sqrt{1681}=\\frac{40\\cdot 41}{9}$. Then the area is $9\\cdot\\frac{1}{2} \\cdot \\frac{40\\cdot 41}{9} = 820",
"Let the three sides be $9,40x,41x$, so the area is $\\frac14\\sqrt {(81^2 - 81x^2)(81x^2 - 1)}$ by Heron's formula. By AM-GM, $\\sqrt {(81^2 - 81x^2)(81x^2 - 1)}\\le\\frac {81^2 - 1}2$, and the maximum possible area is $\\frac14\\cdot\\frac {81^2 - 1}2 = \\frac18(81 - 1)(81 + 1) = 10\\cdot82 = 820",
"Let $A, B$ be the endpoints of the side with length $9$. Let $\\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$; let this intersect $AB$ at $P$ and $Q$, where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\\implies \\dfrac{\\frac{41}{9}}{BQ+9}=\\dfrac{\\frac{40}{9}}{BQ}\\implies BQ=360$. Finally, this means that the radius of $\\Gamma$ is $\\dfrac{360+\\frac{40}{9}}{2}=180+\\dfrac{20}{9}$. Since the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\\Gamma$, which just makes the altitude have length $180+\\dfrac{20}{9}$. Thus, the area of the triangle is $\\dfrac{9\\cdot \\left(180+\\frac{20}{9}\\right)}{2}=820",
"We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$. Heron's gives $\\sqrt{\\left(\\frac{81x+9}{2} \\right) \\left(\\frac{81x-9}{2} \\right) \\left(\\frac{x+9}{2} \\right) \\left(\\frac{-x+9}{2} \\right)}$. This can be simplified to $\\frac{9}{4} \\cdot \\sqrt{(81x^2-1)(81-x^2)}$. We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0. We have that $-324x^3+13124x=0$, so $x=\\frac{\\sqrt{3281}}{9}$. Plugging this into the expression, we have that the area is $820",
"We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$. Heron's gives $\\sqrt{\\left(\\frac{81x+9}{2} \\right) \\left(\\frac{81x-9}{2} \\right) \\left(\\frac{x+9}{2} \\right) \\left(\\frac{-x+9}{2} \\right)}$. This can be simplified to $\\frac{9}{4} \\cdot \\sqrt{(81x^2-1)(81-x^2)}$. We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0. We have that $-324x^3+13124x=0$, so $x=\\frac{\\sqrt{3281}}{9}$. Plugging this into the expression, we have that the area is $820",
"We can start how we did above in solution 4 to get $\\frac{9}{4} * \\sqrt{(81x^2-1)(81-x^2)}$. Then, we can notice the inside is a quadratic in terms of $x^2$, which is $-81(x^2)^2+6562x^2-81$. This is maximized when $x^2 = \\frac{3281}{81}$.If we plug it into the equation, we get $\\frac{9}{4} \\cdot \\frac{3280}{9} = 820"
] |
1992-I-14
| 1,992 | 14 |
In triangle $ABC^{}_{}$ , $A'$ , $B'$ , and $C'$ are on the sides $BC$ , $AC^{}_{}$ , and $AB^{}_{}$ , respectively. Given that $AA'$ , $BB'$ , and $CC'$ are concurrent at the point $O^{}_{}$ , and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$ , find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$ .
| 94 | null |
[
"Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base, \\[\\frac{AO}{OA'}+1=\\frac{AA'}{OA'}=\\frac{[ABC]}{[BOC]}=\\frac{K_A+K_B+K_C}{K_A}.\\] Therefore, we have \\[\\frac{AO}{OA'}=\\frac{K_B+K_C}{K_A}\\] \\[\\frac{BO}{OB'}=\\frac{K_A+K_C}{K_B}\\] \\[\\frac{CO}{OC'}=\\frac{K_A+K_B}{K_C}.\\] Thus, we are given \\[\\frac{K_B+K_C}{K_A}+\\frac{K_A+K_C}{K_B}+\\frac{K_A+K_B}{K_C}=92.\\] Combining and expanding gives \\[\\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}=92.\\] We desire $\\frac{(K_B+K_C)(K_C+K_A)(K_A+K_B)}{K_AK_BK_C}.$ Expanding this gives \\[\\frac{K_A^2K_B+K_AK_B^2+K_A^2K_C+K_AK_C^2+K_B^2K_C+K_BK_C^2}{K_AK_BK_C}+2=094.\\]",
"Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively. Thus, $\\frac{AO^{}_{}}{OA'} = \\frac{b+c}{a}$, $\\frac{BO^{}_{}}{OB'} = \\frac{c+a}{b}$, and $\\frac{CO^{}_{}}{OC'} = \\frac{a+b}{c}$. Therefore: $\\frac{AO}{OA'}\\cdot \\frac{BO}{OB'}\\cdot \\frac{CO}{OC'} = \\frac{b+c}{a} \\cdot \\frac{c+a}{b} \\cdot \\frac{a+b}{c}$ $= \\frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$ $2+\\frac{bc(b+c)}{abc}+\\frac{ca(c+a)}{abc}+\\frac{ab(a+b)}{abc} = 2 + \\frac{b+c}{a} + \\frac{c+a}{b} + \\frac{a+b}{c}$ $= 2 + \\frac{AO^{}_{}}{OA'}+\\frac{BO}{OB'}+\\frac{CO}{OC'} = 2+92 = 094.",
"Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively. Thus, $\\frac{AO^{}_{}}{OA'} = \\frac{b+c}{a}$, $\\frac{BO^{}_{}}{OB'} = \\frac{c+a}{b}$, and $\\frac{CO^{}_{}}{OC'} = \\frac{a+b}{c}$. Therefore: $\\frac{AO}{OA'}\\cdot \\frac{BO}{OB'}\\cdot \\frac{CO}{OC'} = \\frac{b+c}{a} \\cdot \\frac{c+a}{b} \\cdot \\frac{a+b}{c}$ $= \\frac{2abc+b^2c+bc^2+c^2a+ca^2+a^2b+ab^2}{abc} =$ $2+\\frac{bc(b+c)}{abc}+\\frac{ca(c+a)}{abc}+\\frac{ab(a+b)}{abc} = 2 + \\frac{b+c}{a} + \\frac{c+a}{b} + \\frac{a+b}{c}$ $= 2 + \\frac{AO^{}_{}}{OA'}+\\frac{BO}{OB'}+\\frac{CO}{OC'} = 2+92 = 094.",
"As in above solutions, find $\\sum_{cyc} \\frac{y+z}{x}=92$ (where $O=(x:y:z)$ in barycentric coordinates). Now letting $y=z=1$ we get $\\frac{2}{x}+2(x+1)=92 \\implies x+\\frac{1}{x}=45$, and so $\\frac{2}{x}(x+1)^2=2(x+\\frac{1}{x}+2)=2 \\cdot 47 = 94$. ~Lcz",
"A consequence of Ceva's theorem sometimes attributed to Gergonne is that $\\frac{AO}{OA'}=\\frac{AC'}{C'B}+\\frac{AB'}{B'C}$, and similarly for cevians $BB'$ and $CC'$. Now we apply Gergonne several times and do algebra: \\begin{align*} \\frac{AO}{OA'}\\frac{BO}{OB'}\\frac{CO}{OC'} &= \\left(\\frac{AB'}{B'C}+\\frac{AC'}{C'B}\\right) \\left(\\frac{BC'}{C'A}+\\frac{BA'}{A'C}\\right) \\left(\\frac{CB'}{B'A}+\\frac{CA'}{A'B}\\right)\\\\ &=\\underbrace{\\frac{AB'\\cdot CA'\\cdot BC'}{B'C\\cdot A'B\\cdot C'A}}_{\\text{Ceva}} + \\underbrace{\\frac{AC'\\cdot BA'\\cdot CB'}{C'B\\cdot A'C\\cdot B'A}}_{\\text{Ceva}} + \\underbrace{\\frac{AB'}{B'C} + \\frac{AC'}{C'B}}_{\\text{Gergonne}} + \\underbrace{\\frac{BA'}{A'C} + \\frac{BC'}{C'A}}_{\\text{Gergonne}} + \\underbrace{\\frac{CA'}{A'B} + \\frac{CB'}{B'A}}_{\\text{Gergonne}}\\\\ &= 1 + 1 + \\underbrace{\\frac{AO}{OA'} + \\frac{BO}{OB'} + \\frac{CO}{OC'}}_{92} = 94 \\end{align*} ~ proloto",
"A consequence of Ceva's theorem sometimes attributed to Gergonne is that $\\frac{AO}{OA'}=\\frac{AC'}{C'B}+\\frac{AB'}{B'C}$, and similarly for cevians $BB'$ and $CC'$. Now we apply Gergonne several times and do algebra: \\begin{align*} \\frac{AO}{OA'}\\frac{BO}{OB'}\\frac{CO}{OC'} &= \\left(\\frac{AB'}{B'C}+\\frac{AC'}{C'B}\\right) \\left(\\frac{BC'}{C'A}+\\frac{BA'}{A'C}\\right) \\left(\\frac{CB'}{B'A}+\\frac{CA'}{A'B}\\right)\\\\ &=\\underbrace{\\frac{AB'\\cdot CA'\\cdot BC'}{B'C\\cdot A'B\\cdot C'A}}_{\\text{Ceva}} + \\underbrace{\\frac{AC'\\cdot BA'\\cdot CB'}{C'B\\cdot A'C\\cdot B'A}}_{\\text{Ceva}} + \\underbrace{\\frac{AB'}{B'C} + \\frac{AC'}{C'B}}_{\\text{Gergonne}} + \\underbrace{\\frac{BA'}{A'C} + \\frac{BC'}{C'A}}_{\\text{Gergonne}} + \\underbrace{\\frac{CA'}{A'B} + \\frac{CB'}{B'A}}_{\\text{Gergonne}}\\\\ &= 1 + 1 + \\underbrace{\\frac{AO}{OA'} + \\frac{BO}{OB'} + \\frac{CO}{OC'}}_{92} = 94 \\end{align*} ~ proloto"
] |
1992-I-15
| 1,992 | 15 |
Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?
| 396 | null |
[
"Let the number of zeros at the end of $m!$ be $f(m)$. We have $f(m) = \\left\\lfloor \\frac{m}{5} \\right\\rfloor + \\left\\lfloor \\frac{m}{25} \\right\\rfloor + \\left\\lfloor \\frac{m}{125} \\right\\rfloor + \\left\\lfloor \\frac{m}{625} \\right\\rfloor + \\left\\lfloor \\frac{m}{3125} \\right\\rfloor + \\cdots$. Note that if $m$ is a multiple of $5$, $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$. Since $f(m) \\le \\frac{m}{5} + \\frac{m}{25} + \\frac{m}{125} + \\cdots = \\frac{m}{4}$, a value of $m$ such that $f(m) = 1991$ is greater than $7964$. Testing values greater than this yields $f(7975)=1991$. There are $\\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$, less than $1992$. Thus, there are $1991-1595 = 396 that are not factorial tails.",
"After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$, we find that $m \\in \\{7980, 7981, 7982, 7983, 7984\\}$. WLOG, we select $7980$. Furthermore, note that every time $k$ reaches a multiple of $25$, $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers. With this logic, we realize that the desired quantity is simply $\\left \\lfloor \\frac{7980}{25} \\right \\rfloor + \\left \\lfloor \\frac{7980}{125} \\right \\rfloor \\cdots$, where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = 396. - Spacesam(edited by srisainandan6)"
] |
1993-I-1
| 1,993 | 1 |
How many even integers between 4000 and 7000 have four different digits?
| 728 | null |
[
"The thousands digit is $\\in \\{4,5,6\\}$. Case $1$: Thousands digit is even $4, 6$, two possibilities, then there are only $\\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \\cdot 8 \\cdot 7 \\cdot 4 = 448$. Case $2$: Thousands digit is odd $5$, one possibility, then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \\cdot 8 \\cdot 7 \\cdot 5= 280$ possibilities. Together, the solution is $448 + 280 = 728.",
"Firstly, we notice that the thousands digit could be $4$, $5$ or $6$. Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework. Case $1$: Here we let thousands digit be $4$. 4 _ _ _ We take care of restrictions first, and realize that there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$. Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 4 = 224$ numbers that satisfy the conditions posed by the problem. Case $2$ Here we let thousands digit be $5$. 5 _ _ _ Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd. Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 5 = 280$ numbers that satisfy the conditions posed by the problem. Case $3$ Here we let thousands digit be $6$. 6 _ _ _ Once again, we take care of restrictions first. Since 6 is even, there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$ Now we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 4 = 224$ numbers that satisfy the conditions posed by the problem. Now that we have our answers for each possible thousands place digit, we add up our answers and get $224+280+224$= $728 ~PEKKA",
"Firstly, we notice that the thousands digit could be $4$, $5$ or $6$. Since the parity of the possibilities are different, we cannot cover all cases in one operation. Therefore, we must do casework. Case $1$: Here we let thousands digit be $4$. 4 _ _ _ We take care of restrictions first, and realize that there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$. Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 4 = 224$ numbers that satisfy the conditions posed by the problem. Case $2$ Here we let thousands digit be $5$. 5 _ _ _ Again, we take care of restrictions first. This time there are 5 choices for the last digit, which are all the even numbers because 5 is odd. Now that there are no restrictions we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 5 = 280$ numbers that satisfy the conditions posed by the problem. Case $3$ Here we let thousands digit be $6$. 6 _ _ _ Once again, we take care of restrictions first. Since 6 is even, there are 4 choices for the last digit, namely $2$,$6$,$8$ and $0$ Now we proceed to find that there are $8$ choices for the hundreds digit and $7$ choices for the tens digit. Therefore we have $8 \\cdot 7 \\cdot 4 = 224$ numbers that satisfy the conditions posed by the problem. Now that we have our answers for each possible thousands place digit, we add up our answers and get $224+280+224$= $728 ~PEKKA"
] |
1993-I-2
| 1,993 | 2 |
During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $\frac{n^{2}}{2}$ miles on the $n^{\mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\mbox{th}}_{}$ day?
| 580 | null |
[
"On the first day, the candidate moves $[4(0) + 1]^2/2\\ \\text{east},\\, [4(0) + 2]^2/2\\ \\text{north},\\, [4(0) + 3]^2/2\\ \\text{west},\\, [4(0) + 4]^2/2\\ \\text{south}$, and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \\ldots +37^2 - 39^2 = \\left|\\sum_{i=0}^9 \\frac{(4i+1)^2}{2} - \\sum_{i=0}^9 \\frac{(4i+3)^2}{2}\\right|$. Applying difference of squares, we see that \\begin{align*} \\left|\\sum_{i=0}^9 \\frac{(4i+1)^2 - (4i+3)^2}{2}\\right| &= \\left|\\sum_{i=0}^9 \\frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\\right|\\\\ &= \\left|\\sum_{i=0}^9 -(8i+4) \\right|. \\end{align*} The N/S displacement is \\[\\left|\\sum_{i=0}^9 \\frac{(4i+2)^2}{2} - \\sum_{i=0}^9 \\frac{(4i+4)^2}{2}\\right| = \\left|\\sum_{i=0}^9 -(8i+6) \\right|.\\] Since $\\sum_{i=0}^{9} i = \\frac{9(10)}{2} = 45$, the two distances evaluate to $8(45) + 10\\cdot 4 = 400$ and $8(45) + 10\\cdot 6 = 420$. By the Pythagorean Theorem, the answer is $\\sqrt{400^2 + 420^2} = 29 \\cdot 20 = 580."
] |
1993-I-4
| 1,993 | 4 |
How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$ ?
| 870 | null |
[
"Solution 1 Let $k = a + d = b + c$ so $d = k-a, b=k-c$. It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$. Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$. Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$. The last two solutions don't follow $a < b < c < d$, so we only need to consider the first two solutions. The first solution gives us $c - 93\\geq 1$ and $c + 1\\leq 499$ $\\implies 94\\leq c\\leq 498$, and the second one gives us $32\\leq c\\leq 496$. So the total number of such quadruples is $405 + 465 = 870 four-tuples.",
"Let $k = a + d = b + c$ so $d = k-a, b=k-c$. It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$. Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$. Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$. The last two solutions don't follow $a < b < c < d$, so we only need to consider the first two solutions. The first solution gives us $c - 93\\geq 1$ and $c + 1\\leq 499$ $\\implies 94\\leq c\\leq 498$, and the second one gives us $32\\leq c\\leq 496$. So the total number of such quadruples is $405 + 465 = 870 four-tuples.",
"Let $b = a + m$ and $c = a + m + n$. From $a + d = b + c$, $d = b + c - a = a + 2m + n$. Substituting $b = a + m$, $c = a + m + n$, and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$, \\[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\\] Hence, $(m,n) = (1,92)$ or $(3,28)$. For $(m,n) = (1,92)$, we know that $0 < a < a + 1 < a + 93 < a + 94 < 500$, so there are $405$ four-tuples. For $(m,n) = (3,28)$, $0 < a < a + 3 < a + 31 < a + 34 < 500$, and there are $465$ four-tuples. In total, we have $405 + 465 = 870 four-tuples.",
"Square both sides of the first equation in order to get $bc$ and $ad$ terms, which we can plug $93$ in for. \\begin{align*} (a+d)^2 = (b+c)^2 &\\implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \\\\ &\\implies 2bc-2ad = a^2-b^2 + d^2-c^2 \\\\ &\\implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c) \\end{align*} We can plug $93$ in for $bc - ad$ to get $186$ on the left side, and also observe that $a-b = c-d$ after rearranging the first equation. Plug in $c-d$ for $a-b$. $186 = (c-d)(a+b) + (d-c)(d+c) \\implies 186 = -(d-c)(a+b) + (d-c)(d+c) \\implies 186 = (d-c)(d+c-a-b)$ Now observe the possible factors of $186$, which are $1 \\cdot 186, 2\\cdot 93, 3 \\cdot 62, 6\\cdot 31$. $(d-c)$ and $(d+c-a-b)$ must be factors of $186$, and $(d+c-a-b)$ must be greater than $(d-c)$. $1 \\cdot 186$ work, and yields $405$ possible solutions. $2 \\cdot 93$ does not work, because if $c-d = 2$, then $a+b$ must differ by 2 as well, but an odd number $93$ can only result from two numbers of different parity. $c-d$ will be even, and $a+b$ will be even, so $c+d - (a+b)$ must be even. $3 \\cdot 62$ works, and yields $465$ possible solutions, while $6 \\cdot 31$ fails for the same reasoning above. Thus, the answer is $405 + 465 = 870 four-tuples.",
"Add the two conditions together to get $a+d+ad+93=b+c+bc$. Rearranging and factorising with SFFT, $(a+1)(d+1)+93=(b+1)(c+1)$. This implies that for every quadruple $(a,b,c,d)$, we can replace $a\\longrightarrow a+1$, $b\\longrightarrow b+1$, etc. and this will still produce a valid quadruple. This means, that we can fix $a=1$, and then just repeatedly add $1$ to get the other quadruples. Now, our conditions are $b+c=d+1$ and $bc=d+93$. Replacing $d$ in the first equation, we get $bc-b-c=92$. Factorising again with SFFT gives $(b-1)(c-1)=93$. Since $b<c$, we have two possible cases to consider. Case 1: $b=2$, $c=94$. This produces the quadruple $(1,2,94,95)$, which indeed works. Case 2: $b=4$, $c=32$. This produces the quadruple $(1,4,32,35)$, which indeed works. Now, for case 1, we can add $1$ to each term exactly $404$ times (until we get the quadruple $(405,406,498,499)$), until we violate $d<500$. This gives $405$ quadruples for case 1. For case 2, we can add $1$ to each term exactly $464$ times (until we get the quadruple $(465,468,496,499)$). this gives $465$ quadruples for case 2. In conclusion, having exhausted all cases, we can finish. There are hence $405+465=870 four-tuples.",
"Let $r = d-c$. From the equation $a+d = b+c$, we have \\[r = d-c = b-a ,\\] so $b = a+r$ and $c = d-r$. We then have \\[93 = (a+r)(d-r) - ad = rd - ra - r^2 = r(d-a-r) .\\] Since $c > b$, $d-r > a+r$, or $d-a-r > r$. Since the prime factorization of 93 is $3 \\cdot 31$, we must either have $r=1$ and $d-a-r = 93$, or $r=3$ and $d-a-r = 31$. We consider these cases separately. If $r=1$, then $d-a = 94$, $b= a+1$, and $c = d-1$. Thus $d$ can be any integer between 95 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$. We therefore have $499-95+1 = 405$ possibilities in this case. If $r=3$, then $d-a = 34$, $b = a+3$, and $c=d-3$. Thus $d$ can be any integer between 35 and 499, inclusive, and our choice of $d$ determines the four-tuple $(a,b,c,d)$, as before. We therefore have $499-35+1 = 465$ possibilities in this case. Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are $405 + 465 = 870 four-tuples.",
"Assume $d = x+m, a = x-m, c = x+n$, and $b = x-n$. This clearly satisfies the condition that $a+d = b+c$ since ($2x = 2x$) . Now plug this into $bc-ad = 93$. You get $(x+n)(x-n) - (x+m)(x-m) = 93 \\Rightarrow m^2 - n^2 = 93 \\Rightarrow (m-n)(m+n) = 93$ Since $m>n$ (as given by the condition that $a<b<c<d$), $m+n>m-n$ and $m$ and $n$ are integers, there are two cases we have to consider since $93 = 3\\cdot 31$. We first have to consider $m-n = 1, m+n = 93$, and then consider $m-n=3, m+n = 31$. In the first case, we get $m=47, n=46$ and in the second case we get $m=17, n=14$. Now plug these values (in separate cases) back into $a,b,c,d$. Since the only restriction is that all numbers have to be greater than $0$ or less than $500$, we can write two inequalities. Either $x+47 < 500, x-47 > 0$, or $x+17 < 500, x-17 > 0$ (using the inequalities given by $d$ and $a$, and since $b$ and $c$ are squeezed in between $d$ and $a$, we only have to consider these two inequalities). This gives us either $47 < x < 453$ or $17 < x < 483$, and using simple counting, there are $405$ values for $x$ in the first case and $465$ values for $x$ in the second case, and hence our answer is $405+465 = 870"
] |
1993-I-5
| 1,993 | 5 |
Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$ . For integers $n \ge 1\,$ , define $P_n(x) = P_{n - 1}(x - n)\,$ . What is the coefficient of $x\,$ in $P_{20}(x)\,$ ?
| 763 | null |
[
"Notice that \\begin{align*}P_{20}(x) &= P_{19}(x - 20)\\\\ &= P_{18}((x - 20) - 19)\\\\ &= P_{17}(((x - 20) - 19) - 18)\\\\ &= \\cdots\\\\ &= P_0(x - (20 + 19 + 18 + \\ldots + 2 + 1)).\\end{align*} Using the formula for the sum of the first $n$ numbers, $1 + 2 + \\cdots + 20 = \\frac{20(20+1)}{2} = 210$. Therefore, \\[P_{20}(x) = P_0(x - 210).\\] Substituting $x - 210$ into the function definition, we get $P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8$. We only need the coefficients of the linear terms, which we can find by the binomial theorem. $(x-210)^3$ will have a linear term of ${3\\choose1}210^2x = 630 \\cdot 210x$. $313(x-210)^2$ will have a linear term of $-313 \\cdot {2\\choose1}210x = -626 \\cdot 210x$. $-77(x-210)$ will have a linear term of $-77x$. Adding up the coefficients, we get $630 \\cdot 210 - 626 \\cdot 210 - 77 = 763.",
"Notice the transformation of $P_{n-1}(x)\\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\\cdots+20=210$ to them. (Indeed, this is very easy to check in general.) Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/common sense, you see the coefficient of $x$ is $(r_1+210)(r_2+210)+(r_2+210)(r_3+210)+(r_3+210)(r_1+210).$ Expanding yields $r_1r_2+r_2r_3+r_3r_1+210\\cdot 2(r_1+r_2+r_3)+3\\cdot 210^2.$ Using Vieta's (again) and plugging stuff in yields $-77+210\\cdot 2\\cdot -313+3\\cdot 210^2=763",
"Denote the coefficient of term $x^m$ of polynomial $P_n(x)$ as $c_{m, n}$. We can see that $c_{2, 0} = 313$, $c_{1, 0} = -77$, and $c_{0, 0} = -8$. Note that $(x-n)^3 = x^3-3nx^2+3n^2x-n^3$ and $(x-n)^2 = x^2-2nx+n^2$ When substituting $x-1$ for $x$ in $P_0(x)$, we get that $P_1(x) = P_0(x-1) = (x^3-3x^2+3x-1) + 313(x^2-2x+1) -77(x-1) -8$. Observe that \\[c_{2,1} = c_{2,0} - 3(1) = 313 - 3 = 310\\] It is evident that \\[c_{2,n} = c_{2,n-1} - 3n\\] Define the generating function $C_2(X)$ as \\[C_2(X) := \\sum_{n \\geq 0} c_{2,n}X^n\\] By multiplying both sides of the previous recurrence relation and taking the sum of the terms $\\forall n\\geq 1$, we get that \\[\\implies \\sum_{n\\geq1} c_{2,n}X^n = \\sum_{n\\geq1}c_{2,n-1}X^n + 3\\sum_{n\\geq1}nX^n\\] \\[\\implies \\sum_{n\\geq0} c_{2,n}X^n - c_{2,0} = X\\sum_{n\\geq0}c_{2,n}X^n + 3\\sum_{n\\geq1}nX^n\\] \\[\\implies C_2(X) - 313 = XC_2(X) + 3\\cdot \\frac{X}{(1-X)^2}^*\\] \\[\\implies C_2(X) = \\frac{313}{1-X} - \\frac{3X}{(1-X)^3}\\] \\[= \\frac{313X^2 - 629X + 313}{(1-X)^3} = (313X^2 - 629X + 313) \\cdot \\sum_{n\\geq0}{n+2 \\choose 2}X^n\\ ^\\dagger\\] \\[\\implies c_{2,n} = 313{n+2 \\choose 2} - 629{n+1 \\choose 2} + 313{n \\choose 2}^\\ddagger = \\frac{-3n^2 - 3n + 626}{2}\\] We can perform a similar analysis on $c_{1,n}$ to get the recurrence relation \\[c_{1,n} = c_{1, n-1} + c_{2,n-1}\\cdot(-2n) + 3n^2\\] \\[= c_{1, n-1} +3n^3 - 626n\\] Define the generating function $C_1(X)$ as \\[C_1(X) := \\sum_{n\\geq0}c_{1,n}X^n\\] We can then perform this process again on our new recurrence relation: \\[\\implies \\sum_{n\\geq1} c_{1,n}X^n = \\sum_{n\\geq1}c_{1,n-1}X^n + 3\\sum_{n\\geq1}n^3X^n - 626\\sum_{n\\geq1}nX^n\\] \\[\\implies \\sum_{n\\geq0} c_{1,n}X^n - c_{1,0} = X\\sum_{n\\geq0}c_{1,n}X^n + 3\\sum_{n\\geq1}n^3X^n - 626\\sum_{n\\geq1}nX^n\\] \\[\\implies C_1(X) + 77 = XC_1(X) + 3\\cdot \\frac{X(X^2+4X+1)}{(1-X)^4} - 626\\cdot \\frac{X}{(1-X)^2}\\] \\[\\implies C_1(X) = \\frac{-77(1-X)^4 + 3X(X^2+4X+1) - 626X(1-X)^2}{(1-X)^5}\\] \\[= \\frac{-77X^4-315X^3+802X^2-315X-77}{(1-X)^5}\\] \\[=(-77X^4-315X^3+802X^2-315X-77) \\cdot \\sum_{n\\geq0}{n+5 \\choose 5}X^n\\] \\[\\implies c_{1,n} = -77{n+4 \\choose 4} - 315{n+3 \\choose 4} + 802{n+2 \\choose 4} - 315{n+1 \\choose 4} -77{n \\choose 4}\\] \\[= \\frac{3n^4+6n^3-1249n^2-1252n-308}{4}\\] Finally, we can plug $n=20$ into our new explicit formula to get \\[c_{1,20} = 763This formula can be found by convolving the polynomial with the series. - MathCactus0_0 (don't try this on a test unless you can't think of anything else!!!)",
"Denote the coefficient of term $x^m$ of polynomial $P_n(x)$ as $c_{m, n}$. We can see that $c_{2, 0} = 313$, $c_{1, 0} = -77$, and $c_{0, 0} = -8$. Note that $(x-n)^3 = x^3-3nx^2+3n^2x-n^3$ and $(x-n)^2 = x^2-2nx+n^2$ When substituting $x-1$ for $x$ in $P_0(x)$, we get that $P_1(x) = P_0(x-1) = (x^3-3x^2+3x-1) + 313(x^2-2x+1) -77(x-1) -8$. Observe that \\[c_{2,1} = c_{2,0} - 3(1) = 313 - 3 = 310\\] It is evident that \\[c_{2,n} = c_{2,n-1} - 3n\\] Define the generating function $C_2(X)$ as \\[C_2(X) := \\sum_{n \\geq 0} c_{2,n}X^n\\] By multiplying both sides of the previous recurrence relation and taking the sum of the terms $\\forall n\\geq 1$, we get that \\[\\implies \\sum_{n\\geq1} c_{2,n}X^n = \\sum_{n\\geq1}c_{2,n-1}X^n + 3\\sum_{n\\geq1}nX^n\\] \\[\\implies \\sum_{n\\geq0} c_{2,n}X^n - c_{2,0} = X\\sum_{n\\geq0}c_{2,n}X^n + 3\\sum_{n\\geq1}nX^n\\] \\[\\implies C_2(X) - 313 = XC_2(X) + 3\\cdot \\frac{X}{(1-X)^2}^*\\] \\[\\implies C_2(X) = \\frac{313}{1-X} - \\frac{3X}{(1-X)^3}\\] \\[= \\frac{313X^2 - 629X + 313}{(1-X)^3} = (313X^2 - 629X + 313) \\cdot \\sum_{n\\geq0}{n+2 \\choose 2}X^n\\ ^\\dagger\\] \\[\\implies c_{2,n} = 313{n+2 \\choose 2} - 629{n+1 \\choose 2} + 313{n \\choose 2}^\\ddagger = \\frac{-3n^2 - 3n + 626}{2}\\] We can perform a similar analysis on $c_{1,n}$ to get the recurrence relation \\[c_{1,n} = c_{1, n-1} + c_{2,n-1}\\cdot(-2n) + 3n^2\\] \\[= c_{1, n-1} +3n^3 - 626n\\] Define the generating function $C_1(X)$ as \\[C_1(X) := \\sum_{n\\geq0}c_{1,n}X^n\\] We can then perform this process again on our new recurrence relation: \\[\\implies \\sum_{n\\geq1} c_{1,n}X^n = \\sum_{n\\geq1}c_{1,n-1}X^n + 3\\sum_{n\\geq1}n^3X^n - 626\\sum_{n\\geq1}nX^n\\] \\[\\implies \\sum_{n\\geq0} c_{1,n}X^n - c_{1,0} = X\\sum_{n\\geq0}c_{1,n}X^n + 3\\sum_{n\\geq1}n^3X^n - 626\\sum_{n\\geq1}nX^n\\] \\[\\implies C_1(X) + 77 = XC_1(X) + 3\\cdot \\frac{X(X^2+4X+1)}{(1-X)^4} - 626\\cdot \\frac{X}{(1-X)^2}\\] \\[\\implies C_1(X) = \\frac{-77(1-X)^4 + 3X(X^2+4X+1) - 626X(1-X)^2}{(1-X)^5}\\] \\[= \\frac{-77X^4-315X^3+802X^2-315X-77}{(1-X)^5}\\] \\[=(-77X^4-315X^3+802X^2-315X-77) \\cdot \\sum_{n\\geq0}{n+5 \\choose 5}X^n\\] \\[\\implies c_{1,n} = -77{n+4 \\choose 4} - 315{n+3 \\choose 4} + 802{n+2 \\choose 4} - 315{n+1 \\choose 4} -77{n \\choose 4}\\] \\[= \\frac{3n^4+6n^3-1249n^2-1252n-308}{4}\\] Finally, we can plug $n=20$ into our new explicit formula to get \\[c_{1,20} = 763This formula can be found by convolving the polynomial with the series. - MathCactus0_0 (don't try this on a test unless you can't think of anything else!!!)"
] |
1993-I-6
| 1,993 | 6 |
What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?
| 495 | null |
[
"Solution 1 Denote the first of each of the series of consecutive integers as $a,\\ b,\\ c$. Therefore, $n = a + (a + 1) \\ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$. Simplifying, $9a = 10b + 9 = 11c + 19$. The relationship between $a,\\ b$ suggests that $b$ is divisible by $9$. Also, $10b -10 = 10(b-1) = 11c$, so $b-1$ is divisible by $11$. We find that the least possible value of $b = 45$, so the answer is $10(45) + 45 = 495$. Solution 2 Let the desired integer be $n$. From the information given, it can be determined that, for positive integers $a, \\ b, \\ c$: $n = 9a + 36 = 10b + 45 = 11c + 55$ This can be rewritten as the following congruences: $n \\equiv 0 \\pmod{9}$ $n \\equiv 5 \\pmod{10}$ $n \\equiv 0 \\pmod{11}$ Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $495",
"Denote the first of each of the series of consecutive integers as $a,\\ b,\\ c$. Therefore, $n = a + (a + 1) \\ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$. Simplifying, $9a = 10b + 9 = 11c + 19$. The relationship between $a,\\ b$ suggests that $b$ is divisible by $9$. Also, $10b -10 = 10(b-1) = 11c$, so $b-1$ is divisible by $11$. We find that the least possible value of $b = 45$, so the answer is $10(45) + 45 = 495$. Solution 2 Let the desired integer be $n$. From the information given, it can be determined that, for positive integers $a, \\ b, \\ c$: $n = 9a + 36 = 10b + 45 = 11c + 55$ This can be rewritten as the following congruences: $n \\equiv 0 \\pmod{9}$ $n \\equiv 5 \\pmod{10}$ $n \\equiv 0 \\pmod{11}$ Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $495",
"Let the desired integer be $n$. From the information given, it can be determined that, for positive integers $a, \\ b, \\ c$: $n = 9a + 36 = 10b + 45 = 11c + 55$ This can be rewritten as the following congruences: $n \\equiv 0 \\pmod{9}$ $n \\equiv 5 \\pmod{10}$ $n \\equiv 0 \\pmod{11}$ Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $495",
"Let $n$ be the desired integer. From the given information, we have \\begin{align*}9x &= a \\\\ 11y &= a \\\\ 10z + 5 &= a, \\end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = 495",
"By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$. From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that $n$ cannot be divisible by 10, so $n$ must equal $495",
"First note that the integer clearly must be divisible by $9$ and $11$ since we can use the \"let the middle number be x\" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \\cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \\equiv 0 \\pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \\cdot 5 = 495",
"From the first equation solution 1, we get the three part equation 9x=10y+9=11z+19 after subtracting 36 from each of the equations 9x+36=10y+45=11z+55 knowing that the integer we are looking for has a last digit of 9 from the second part of our first equation, and also knowing that our mystery number must be 19 more the a multiple of 11 that ends in 0, the only way this could happen is if the number is able to be deduced in the form 110a+19, where a is between the range 1-9, inclusive. so, also knowing our mystery number is divisible by 9, our only answer being 129,239,349,459 ... etc. until 899. Brushing through our act of casework for this problem shows that the answer is 459... but we are not done yet! We have to add the 36 we subtracted at the beggining of our solution to get the equation 459+36, which is equal to our final answer... 495!!! $michaellin16$"
] |
1993-I-7
| 1,993 | 7 |
Three numbers, $a_1, a_2, a_3$ , are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$ . Three other numbers, $b_1, b_2, b_3$ , are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3$ can be enclosed in a box of dimension $b_1 \times b_2 \times b_3$ , with the sides of the brick parallel to the sides of the box. If $p$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
| 5 | null |
[
"There is a total of $P(1000,6)$ possible ordered $6$-tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$ There are $C(1000,6)$ possible sets $\\{a_1,a_2,a_3,b_1,b_2,b_3\\}.$ We have five valid cases for the increasing order of these six elements: $aaabbb$ $aababb$ $aabbab$ $abaabb$ $ababab$ Note that the $a$'s are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$'s are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$ So, there is a total of $C(1000,6)\\cdot5\\cdot6^2$ valid ordered $6$-tuples. The requested probability is \\[p=\\frac{C(1000,6)\\cdot5\\cdot6^2}{P(1000,6)}=\\frac{C(1000,6)\\cdot5\\cdot6^2}{C(1000,6)\\cdot6!}=\\frac14,\\] from which the answer is $1+4=005 ~MRENTHUSIASM",
"There is a total of $P(1000,6)$ possible ordered $6$-tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$ There are $C(1000,6)$ possible sets $\\{a_1,a_2,a_3,b_1,b_2,b_3\\}.$ We have five valid cases for the increasing order of these six elements: $aaabbb$ $aababb$ $aabbab$ $abaabb$ $ababab$ Note that the $a$'s are different from each other, as there are $3!=6$ ways to permute them as $a_1,a_2,$ and $a_3.$ Similarly, the $b$'s are different from each other, as there are $3!=6$ ways to permute them as $b_1,b_2,$ and $b_3.$ So, there is a total of $C(1000,6)\\cdot5\\cdot6^2$ valid ordered $6$-tuples. The requested probability is \\[p=\\frac{C(1000,6)\\cdot5\\cdot6^2}{P(1000,6)}=\\frac{C(1000,6)\\cdot5\\cdot6^2}{C(1000,6)\\cdot6!}=\\frac14,\\] from which the answer is $1+4=005 ~MRENTHUSIASM",
"Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick. If $x_2$ is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us $3$ possibilities. If $x_2$ is not a dimension of the box but $x_3$ is, then both remaining dimensions will work as a dimension of the box. That gives us $2$ possibilities. If $x_4$ is a dimension of the box but $x_2,x_3$ aren’t, there are no possibilities (same for $x_5$). The total number of arrangements is ${6\\choose3} = 20$; therefore, $p = \\frac{3 + 2}{20} = \\frac{1}{4}$, and the answer is $1 + 4 = 005 in some order.",
"Call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Clearly, $x_1$ must be a dimension of the box, and $x_6$ must be a dimension of the brick. If $x_2$ is a dimension of the box, then any of the other three remaining dimensions will work as a dimension of the box. That gives us $3$ possibilities. If $x_2$ is not a dimension of the box but $x_3$ is, then both remaining dimensions will work as a dimension of the box. That gives us $2$ possibilities. If $x_4$ is a dimension of the box but $x_2,x_3$ aren’t, there are no possibilities (same for $x_5$). The total number of arrangements is ${6\\choose3} = 20$; therefore, $p = \\frac{3 + 2}{20} = \\frac{1}{4}$, and the answer is $1 + 4 = 005 in some order.",
"Like Solution 2, call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Using the Hook Length Formula, the number of valid configuration is $\\frac{6!}{4\\cdot3\\cdot2\\cdot3\\cdot2}=5$. We proceed as Solution 2 does.",
"Like Solution 2, call the six numbers selected $x_1 > x_2 > x_3 > x_4 > x_5 > x_6$. Using the Hook Length Formula, the number of valid configuration is $\\frac{6!}{4\\cdot3\\cdot2\\cdot3\\cdot2}=5$. We proceed as Solution 2 does.",
"As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the $n$th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \\[\\frac{\\binom{2n}{n}}{n+1}\\] Thus, there are $\\frac{\\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are $\\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\\frac{\\binom{6}{3}/4}{\\binom{6}{3}}=\\frac14\\Longrightarrow 005.",
"As in Solutions 2 and 3, we let $x_1>x_2>x_3>x_4>x_5>x_6$ where each $x_i$ is a number selected. It is clear that when choosing whether each number must be in the set with larger dimensions (the box) or the set with smaller dimensions (the brick) there must always be at least as many numbers in the former set as the latter. We realize that this resembles Catalan numbers, where the indices of the numbers in the first set can be replaced with rising sections of a mountain, and the other indices representing falling sections of a mountain. The formula for the $n$th Catalan number (where $n$ is the number of pairs of rising and falling sections) is \\[\\frac{\\binom{2n}{n}}{n+1}\\] Thus, there are $\\frac{\\binom{6}{3}}{4}$ ways to pick which of $x_1,x_2,x_3,x_4,x_5,$ and $x_6$ are the dimensions of the box, and which are the dimensions of the brick, such that the condition is fulfilled. There are $\\binom{6}{3}$ total ways to choose which numbers make up the brick and box, so the probability of the condition being fulfilled is $\\frac{\\binom{6}{3}/4}{\\binom{6}{3}}=\\frac14\\Longrightarrow 005."
] |
1993-I-8
| 1,993 | 8 |
Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$ ? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{b, c, d, e, f\},\{a, c\}.$
| 365 | null |
[
"Call the two subsets $m$ and $n.$ For each of the elements in $S,$ we can assign it to either $m,n,$ or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $S.$ So our final answer is then $\\frac {3^6 - 1}{2} + 1 = 365",
"Call the two subsets $m$ and $n.$ For each of the elements in $S,$ we can assign it to either $m,n,$ or both. This gives us $3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $m$ and $n$ contain all $6$ elements of $S.$ So our final answer is then $\\frac {3^6 - 1}{2} + 1 = 365",
"Given one of ${6 \\choose k}$ subsets with $k$ elements, the other also has $2^k$ possibilities; this is because it must contain all of the \"missing\" $n - k$ elements and thus has a choice over the remaining $k.$ We want $\\sum_{k = 0}^6 {6 \\choose k}2^k = (2 + 1)^6 = 729$ by the Binomial Theorem. But the order of the sets doesn't matter, so we get $\\dfrac{729 - 1}{2} + 1 = 365",
"Given one of ${6 \\choose k}$ subsets with $k$ elements, the other also has $2^k$ possibilities; this is because it must contain all of the \"missing\" $n - k$ elements and thus has a choice over the remaining $k.$ We want $\\sum_{k = 0}^6 {6 \\choose k}2^k = (2 + 1)^6 = 729$ by the Binomial Theorem. But the order of the sets doesn't matter, so we get $\\dfrac{729 - 1}{2} + 1 = 365",
"For all nonnegative integers $n,$ let $n$ be the number of elements in $S,$ and $f(n)$ be the number of unordered pairs $\\{A,B\\}$ of subsets of $S$ for which $A\\cup B=S.$ We wish to find $f(6).$ Without the loss of generality, let the elements of $S$ be $1,2,\\ldots,n.$ Based on the value of $n,$ we construct the following table: \\[\\begin{array}{c|c|c} & & \\\\ [-2.5ex] \\boldsymbol{n} & \\textbf{Unordered Pairs }\\boldsymbol{\\{A,B\\}}\\textbf{ of subsets of }\\boldsymbol{S}\\textbf{ for which }\\boldsymbol{A\\cup B=S} & \\boldsymbol{f(n)} \\\\ [0.5ex] \\hline & & \\\\ [-2ex] 0 & \\{\\varnothing,\\varnothing\\} & 1 \\\\ 1 & \\{\\{1\\},\\{1\\}\\}, \\ \\{\\{1\\},\\varnothing\\} & 2 \\\\ 2 & \\color{red}\\{\\{1,2\\},\\{1,2\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2\\},\\{1\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2\\},\\{2\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2\\},\\varnothing\\}\\color{black}, \\ \\color{magenta}\\{\\{1\\},\\{2\\}\\} & 5 \\\\ 3 & \\color{red}\\{\\{1,2,3\\},\\{1,2,3\\}\\}\\color{black}, \\ \\color{red}\\{\\{1,2,3\\},\\{1,2\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2,3\\},\\{1,3\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2,3\\},\\{1\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2\\},\\{1,3\\}\\}\\color{black}, & 14 \\\\ & \\color{blue}\\{\\{1,2,3\\},\\{2,3\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2,3\\},\\{2\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2\\},\\{2,3\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2,3\\},\\{3\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2,3\\},\\varnothing\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2\\},\\{3\\}\\}\\color{black}, & \\\\ & \\color{magenta}\\{\\{1,3\\},\\{2,3\\}\\}\\color{black}, \\ \\color{magenta}\\{\\{1,3\\},\\{2\\}\\}\\color{black}, \\ \\color{magenta}\\{\\{1\\},\\{2,3\\}\\} & \\\\ \\vdots & \\vdots & \\vdots \\end{array}\\] Note that for all $n\\geq1,$ each unordered pair of the $(n-1)$-element set contributes three unordered pairs to the $n$-element set, as the element $n$ can be added to either or both of the subsets. The only exception is that the unordered pair of identical subsets of the $(n-1)$-element set, namely $\\{\\{1,2,\\ldots,n-1\\},\\{1,2,\\ldots,n-1\\}\\},$ only contributes two unordered pairs to the $n$-element set. The table above shows this relationship between the $2$-element set and the $3$-element set. Together, the recursive formula for $f(n)$ is \\[f(n) = \\begin{cases} 1 & \\mathrm{if} \\ n=0 \\\\ 3f(n-1)-1 & \\mathrm{if} \\ n\\geq1 \\end{cases}.\\] Two solutions follow from here: Solution 3.1 (Recursive Formula) We evaluate $f(6)$ recursively: \\begin{alignat*}{6} f(0)&=1, \\\\ f(1)&=3f(0)-1&&=2, \\\\ f(2)&=3f(1)-1&&=5, \\\\ f(3)&=3f(2)-1&&=14, \\\\ f(4)&=3f(3)-1&&=41, \\\\ f(5)&=3f(4)-1&&=122, \\\\ f(6)&=3f(5)-1&&=365 ~MRENTHUSIASM",
"For all nonnegative integers $n,$ let $n$ be the number of elements in $S,$ and $f(n)$ be the number of unordered pairs $\\{A,B\\}$ of subsets of $S$ for which $A\\cup B=S.$ We wish to find $f(6).$ Without the loss of generality, let the elements of $S$ be $1,2,\\ldots,n.$ Based on the value of $n,$ we construct the following table: \\[\\begin{array}{c|c|c} & & \\\\ [-2.5ex] \\boldsymbol{n} & \\textbf{Unordered Pairs }\\boldsymbol{\\{A,B\\}}\\textbf{ of subsets of }\\boldsymbol{S}\\textbf{ for which }\\boldsymbol{A\\cup B=S} & \\boldsymbol{f(n)} \\\\ [0.5ex] \\hline & & \\\\ [-2ex] 0 & \\{\\varnothing,\\varnothing\\} & 1 \\\\ 1 & \\{\\{1\\},\\{1\\}\\}, \\ \\{\\{1\\},\\varnothing\\} & 2 \\\\ 2 & \\color{red}\\{\\{1,2\\},\\{1,2\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2\\},\\{1\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2\\},\\{2\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2\\},\\varnothing\\}\\color{black}, \\ \\color{magenta}\\{\\{1\\},\\{2\\}\\} & 5 \\\\ 3 & \\color{red}\\{\\{1,2,3\\},\\{1,2,3\\}\\}\\color{black}, \\ \\color{red}\\{\\{1,2,3\\},\\{1,2\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2,3\\},\\{1,3\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2,3\\},\\{1\\}\\}\\color{black}, \\ \\color{green}\\{\\{1,2\\},\\{1,3\\}\\}\\color{black}, & 14 \\\\ & \\color{blue}\\{\\{1,2,3\\},\\{2,3\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2,3\\},\\{2\\}\\}\\color{black}, \\ \\color{blue}\\{\\{1,2\\},\\{2,3\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2,3\\},\\{3\\}\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2,3\\},\\varnothing\\}\\color{black}, \\ \\color{cyan}\\{\\{1,2\\},\\{3\\}\\}\\color{black}, & \\\\ & \\color{magenta}\\{\\{1,3\\},\\{2,3\\}\\}\\color{black}, \\ \\color{magenta}\\{\\{1,3\\},\\{2\\}\\}\\color{black}, \\ \\color{magenta}\\{\\{1\\},\\{2,3\\}\\} & \\\\ \\vdots & \\vdots & \\vdots \\end{array}\\] Note that for all $n\\geq1,$ each unordered pair of the $(n-1)$-element set contributes three unordered pairs to the $n$-element set, as the element $n$ can be added to either or both of the subsets. The only exception is that the unordered pair of identical subsets of the $(n-1)$-element set, namely $\\{\\{1,2,\\ldots,n-1\\},\\{1,2,\\ldots,n-1\\}\\},$ only contributes two unordered pairs to the $n$-element set. The table above shows this relationship between the $2$-element set and the $3$-element set. Together, the recursive formula for $f(n)$ is \\[f(n) = \\begin{cases} 1 & \\mathrm{if} \\ n=0 \\\\ 3f(n-1)-1 & \\mathrm{if} \\ n\\geq1 \\end{cases}.\\] Two solutions follow from here: Solution 3.1 (Recursive Formula) We evaluate $f(6)$ recursively: \\begin{alignat*}{6} f(0)&=1, \\\\ f(1)&=3f(0)-1&&=2, \\\\ f(2)&=3f(1)-1&&=5, \\\\ f(3)&=3f(2)-1&&=14, \\\\ f(4)&=3f(3)-1&&=41, \\\\ f(5)&=3f(4)-1&&=122, \\\\ f(6)&=3f(5)-1&&=365 ~MRENTHUSIASM",
"We evaluate $f(6)$ recursively: \\begin{alignat*}{6} f(0)&=1, \\\\ f(1)&=3f(0)-1&&=2, \\\\ f(2)&=3f(1)-1&&=5, \\\\ f(3)&=3f(2)-1&&=14, \\\\ f(4)&=3f(3)-1&&=41, \\\\ f(5)&=3f(4)-1&&=122, \\\\ f(6)&=3f(5)-1&&=365 ~MRENTHUSIASM",
"We evaluate $f(6)$ recursively: \\begin{alignat*}{6} f(0)&=1, \\\\ f(1)&=3f(0)-1&&=2, \\\\ f(2)&=3f(1)-1&&=5, \\\\ f(3)&=3f(2)-1&&=14, \\\\ f(4)&=3f(3)-1&&=41, \\\\ f(5)&=3f(4)-1&&=122, \\\\ f(6)&=3f(5)-1&&=365 ~MRENTHUSIASM",
"For all $n\\geq1,$ we have \\begin{align*} f(n) &= 3f(n-1)-1 \\\\ &= 3\\left(3f(n-2)-1\\right)-1 \\\\ &= 3^2f(n-2)-3-1 \\\\ &= 3^2\\left(3f(n-3)-1\\right)-3-1 \\\\ &= 3^3f(n-3)-3^2-3-1 \\\\ & \\ \\vdots \\\\ &= 3^nf(0)-3^{n-1}-3^{n-2}-3^{n-3}-\\cdots-1 \\\\ &= 3^n-\\left(3^{n-1}+3^{n-2}+3^{n-3}+\\cdots+1\\right) \\\\ &= 3^n-\\frac{3^n-1}{2} \\\\ &= \\frac{3^n+1}{2} \\\\ &= \\frac{3^n-1}{2}+1, \\end{align*} which resembles the result in Solutions 1 and 2. The answer is $f(6)=365 ~MRENTHUSIASM",
"For all $n\\geq1,$ we have \\begin{align*} f(n) &= 3f(n-1)-1 \\\\ &= 3\\left(3f(n-2)-1\\right)-1 \\\\ &= 3^2f(n-2)-3-1 \\\\ &= 3^2\\left(3f(n-3)-1\\right)-3-1 \\\\ &= 3^3f(n-3)-3^2-3-1 \\\\ & \\ \\vdots \\\\ &= 3^nf(0)-3^{n-1}-3^{n-2}-3^{n-3}-\\cdots-1 \\\\ &= 3^n-\\left(3^{n-1}+3^{n-2}+3^{n-3}+\\cdots+1\\right) \\\\ &= 3^n-\\frac{3^n-1}{2} \\\\ &= \\frac{3^n+1}{2} \\\\ &= \\frac{3^n-1}{2}+1, \\end{align*} which resembles the result in Solutions 1 and 2. The answer is $f(6)=365 ~MRENTHUSIASM",
"We can perform casework based on the number of overlapping elements. If no elements overlap, there is $\\binom60=1$ way to choose the overlapping elements, and $2^{6-0}$ ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by $2$ because the order of the subsets does not matter. Proceeding similarly for the other cases, our sum is \\[\\dbinom{6}0 \\cdot 2^5+\\dbinom{6}1 \\cdot 2^4+\\cdots +\\dbinom{6}4 \\cdot 2+\\dbinom{6}5 \\cdot 1+\\dbinom{6}6 \\cdot 1.\\] (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of $365 ~vaporwave ~MrThinker (LaTeX improvements)",
"We can perform casework based on the number of overlapping elements. If no elements overlap, there is $\\binom60=1$ way to choose the overlapping elements, and $2^{6-0}$ ways to distribute the remaining elements--each element can go in one subset or the other. We must also divide by $2$ because the order of the subsets does not matter. Proceeding similarly for the other cases, our sum is \\[\\dbinom{6}0 \\cdot 2^5+\\dbinom{6}1 \\cdot 2^4+\\cdots +\\dbinom{6}4 \\cdot 2+\\dbinom{6}5 \\cdot 1+\\dbinom{6}6 \\cdot 1.\\] (Note that we have to be especially careful with the last case, as it does not follow the pattern of the other cases.) Adding these up gives a total of $365 ~vaporwave ~MrThinker (LaTeX improvements)",
"If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is $x\\rightarrow y,$ which implies that we pick $x$ numbers for the first set which then the second set can have $y$ numbers. Clearly: \\begin{align*} 0&\\rightarrow6 \\\\ 1&\\rightarrow5\\mid6 \\\\ 2&\\rightarrow4\\mid5\\mid6 \\\\ 3&\\rightarrow3\\mid4\\mid5\\mid6 \\\\ 4&\\rightarrow2\\mid3\\mid4\\mid5\\mid6 \\\\ 5&\\rightarrow1\\mid2\\mid3\\mid4\\mid5\\mid6 \\\\ 6&\\rightarrow0\\mid1\\mid2\\mid3\\mid4\\mid5\\mid6 \\end{align*} However notice that many of the cases are double counted as direction does not matter, e.g. $2\\rightarrow4$ is the same as $4\\rightarrow2.$ Get rid of those cases so we are just left with: \\begin{align*} 0&\\rightarrow6 \\\\ 1&\\rightarrow5\\mid6 \\\\ 2&\\rightarrow4\\mid5\\mid6 \\\\ 3&\\rightarrow3\\mid4\\mid5\\mid6 \\\\ 4&\\rightarrow4\\mid5\\mid6 \\\\ 5&\\rightarrow5\\mid6 \\\\ 6&\\rightarrow6 \\end{align*} Now we start computing, $0\\rightarrow6$ is just $1$ case, $1\\rightarrow5\\mid6$ is just $\\binom{6}{1}\\cdot2$ cases, $2\\rightarrow4\\mid5\\mid6$ is just $\\binom{6}{2}\\cdot2^2$ cases, and $3\\rightarrow3\\mid4\\mid5\\mid6$ is just $\\binom{6}{3}\\cdot2^3$ cases (If you have trouble understanding this, write down the six letters and then try to understand what $x\\rightarrow y$ really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that. One quick thing though. We have a double counted case with the $3\\rightarrow3,$ as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are $\\frac{\\binom{6}{3}}{2} = 10$ cases of this. For computing $4\\rightarrow4\\mid5\\mid6,$ we use the same process as before. We have $\\binom{6}{4}\\cdot(3+4+1)$ (Note, the $3$ comes from $\\frac{\\binom{4}{2}}{2}$), and for $5\\rightarrow5\\mid6$ we have $\\binom{6}{5}\\cdot\\left(\\frac{5}{2}+1\\right),$ and then for $6\\rightarrow6$ we just have $\\binom{6}{6}$ (there is no double counted case since ABCDEF, ABCDEF is only counted once). Summing case by case, we have $1+12+60+150+120+21+1 = 365 ~Tost (Solution) ~MRENTHUSIASM (Reformatting)",
"If we wanted to, we could use casework, being very careful not to double count cases. Let's first figure out what cases we need to look at. The notation I will be using is $x\\rightarrow y,$ which implies that we pick $x$ numbers for the first set which then the second set can have $y$ numbers. Clearly: \\begin{align*} 0&\\rightarrow6 \\\\ 1&\\rightarrow5\\mid6 \\\\ 2&\\rightarrow4\\mid5\\mid6 \\\\ 3&\\rightarrow3\\mid4\\mid5\\mid6 \\\\ 4&\\rightarrow2\\mid3\\mid4\\mid5\\mid6 \\\\ 5&\\rightarrow1\\mid2\\mid3\\mid4\\mid5\\mid6 \\\\ 6&\\rightarrow0\\mid1\\mid2\\mid3\\mid4\\mid5\\mid6 \\end{align*} However notice that many of the cases are double counted as direction does not matter, e.g. $2\\rightarrow4$ is the same as $4\\rightarrow2.$ Get rid of those cases so we are just left with: \\begin{align*} 0&\\rightarrow6 \\\\ 1&\\rightarrow5\\mid6 \\\\ 2&\\rightarrow4\\mid5\\mid6 \\\\ 3&\\rightarrow3\\mid4\\mid5\\mid6 \\\\ 4&\\rightarrow4\\mid5\\mid6 \\\\ 5&\\rightarrow5\\mid6 \\\\ 6&\\rightarrow6 \\end{align*} Now we start computing, $0\\rightarrow6$ is just $1$ case, $1\\rightarrow5\\mid6$ is just $\\binom{6}{1}\\cdot2$ cases, $2\\rightarrow4\\mid5\\mid6$ is just $\\binom{6}{2}\\cdot2^2$ cases, and $3\\rightarrow3\\mid4\\mid5\\mid6$ is just $\\binom{6}{3}\\cdot2^3$ cases (If you have trouble understanding this, write down the six letters and then try to understand what $x\\rightarrow y$ really means.). Now what you can do is continue on this same pattern like Solution 2 does, and then use simple symmetry to figure out the double counted cases. However, the purpose of this solution is to bash out the double counted cases, so we will do exactly that. One quick thing though. We have a double counted case with the $3\\rightarrow3,$ as choosing ABC and DEF is the same thing as choosing DEF and then ABC. There are $\\frac{\\binom{6}{3}}{2} = 10$ cases of this. For computing $4\\rightarrow4\\mid5\\mid6,$ we use the same process as before. We have $\\binom{6}{4}\\cdot(3+4+1)$ (Note, the $3$ comes from $\\frac{\\binom{4}{2}}{2}$), and for $5\\rightarrow5\\mid6$ we have $\\binom{6}{5}\\cdot\\left(\\frac{5}{2}+1\\right),$ and then for $6\\rightarrow6$ we just have $\\binom{6}{6}$ (there is no double counted case since ABCDEF, ABCDEF is only counted once). Summing case by case, we have $1+12+60+150+120+21+1 = 365 ~Tost (Solution) ~MRENTHUSIASM (Reformatting)"
] |
1993-I-9
| 1,993 | 9 |
Two thousand points are given on a circle. Label one of the points 1. From this point, count 2 points in the clockwise direction and label this point 2. From the point labeled 2, count 3 points in the clockwise direction and label this point 3. (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as 1993? AIME 1993 Problem 9.png
| 118 | null |
[
"The label $1993$ will occur on the $\\frac12(1993)(1994) \\pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$. Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\\equiv 0\\pmod{2000}$. Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$. For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$, $n \\equiv 118 \\pmod {125}$ and $n\\equiv 6 \\pmod {16}$. The smallest $n$ for this case is $118$. In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$, $n\\equiv 9\\pmod{16}$ and $n\\equiv 6\\pmod{125}$. The smallest $n$ for this case is larger than $118$, so $118 to simplify calculations.",
"Two labels $a$ and $b$ occur on the same point if $\\ a(a+1)/2\\equiv \\ b(b+1)/2\\pmod{2000}$. If we assume the final answer be $n$, then we have $\\frac12(n)(n + 1)\\equiv \\frac12(1993)(1994) \\pmod{2000}$. Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\\equiv 0\\pmod{4000}$. As they have different parities, the even one must be divisible by $32$. As $(1993 - n)+(1994 + n)\\equiv 2\\pmod{5}$, one of them is divisible by $5$, which indicates it's divisible by $125$. Which leads to four different cases: $1993-n\\equiv 0\\pmod{4000}$ ; $1994+n\\equiv 0\\pmod{4000}$ ; $1993-n\\equiv 0\\pmod{32}$ and $1994+n\\equiv 0\\pmod{125}$ ; $1993-n\\equiv 0\\pmod{125}$ and $1994+n\\equiv 0\\pmod{32}$. Which leads to $n\\equiv 1993,2006,3881$ and $118\\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $118.(by ZJY)"
] |
1993-I-10
| 1,993 | 10 |
Euler's formula states that for a convex polyhedron with $V\,$ vertices, $E\,$ edges, and $F\,$ faces, $V-E+F=2\,$ . A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its $V\,$ vertices, $T\,$ triangular faces and $P^{}_{}$ pentagonal faces meet. What is the value of $100P+10T+V\,$ ?
| 250 | null |
[
"Solution 1 The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\\frac{3t+5p}{2}$, (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$. Finally, using Euler's formula we have $V=E-30=30$. In summary, the solution to the problem is $100P+10T+V=250. ~Williamgolly",
"The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\\frac{3t+5p}{2}$, (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$. Finally, using Euler's formula we have $V=E-30=30$. In summary, the solution to the problem is $100P+10T+V=250. ~Williamgolly",
"As seen above, $E=V+30$. Every vertex $V$, there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\\frac{VT}3+\\frac{VP}5=32\\implies V(5T+3P)=480$. Also, at every vertex $V$, there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\\frac{V(T+P)}2=E\\implies V(T+P)=2E\\implies V(5T+5P)=10E$, and subtracting the other equation involving the vertices from this gives $2VP=10E-480\\implies VP=5E-240=5(V+30)-240=5V-90$ $\\implies V(5-P)=90$. Since $V|480$ from the first vertex-related observation and $P>0\\implies5-P<5$, and it quickly follows that $V=30\\implies E=60\\implies P=2\\implies T=2\\implies100P+10T+V=250. ~Williamgolly",
"Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron. Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \\frac{3t+5p}{2}$ and $V = \\frac{3t}{T}=\\frac{5p}{P}.$ Now, we do casework on what $P$ is. Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral. Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions. Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = 250. ~Williamgolly",
"We know that $V-E = -30 \\implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \\frac{4V}{2} = 2V$. Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = 250."
] |
1993-I-11
| 1,993 | 11 |
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is $m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. What are the last three digits of $m+n\,$ ?
| 93 | null |
[
"The probability that the $n$th flip in each game occurs and is a head is $\\frac{1}{2^n}$. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $\\frac{1}{2}+\\frac{1}{8}+\\frac{1}{32}+\\cdots = \\frac{\\frac{1}{2}}{1-\\frac{1}{4}}=\\frac{2}{3}$, and the probability of the second person winning is $\\frac{1}{3}$. Let $a_n$ be the probability that Alfred wins the $n$th game, and let $b_n$ be the probability that Bonnie wins the $n$th game. If Alfred wins the $n$th game, then the probability that Alfred wins the $n+1$th game is $\\frac{1}{3}$. If Bonnie wins the $n$th game, then the probability that Alfred wins the $n+1$th game is $\\frac{2}{3}$. Thus, $a_{n+1}=\\frac{1}{3}a_n+\\frac{2}{3}b_n$. Similarly, $b_{n+1}=\\frac{2}{3}a_n+\\frac{1}{3}b_n$. Since Alfred goes first in the $1$st game, $(a_1,b_1)=\\left(\\frac{2}{3}, \\frac{1}{3}\\right)$. Using these recursive equations: $(a_2,b_2)=\\left(\\frac{4}{9}, \\frac{5}{9}\\right)$ $(a_3,b_3)=\\left(\\frac{14}{27}, \\frac{13}{27}\\right)$ $(a_4,b_4)=\\left(\\frac{40}{81}, \\frac{41}{81}\\right)$ $(a_5,b_5)=\\left(\\frac{122}{243}, \\frac{121}{243}\\right)$ $(a_6,b_6)=\\left(\\frac{364}{729}, \\frac{365}{729}\\right)$ Since $a_6=\\frac{364}{729}$, $m+n = 1093 \\equiv 093.",
"In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a $\\frac{1}{2}$ chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a $\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}$, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is $\\frac{1}{4} \\times \\frac{1}{2} = \\frac{1}{8}$ From this, we can see that Alfred’s (who goes first) chance of winning the first round is: $\\frac{1}{2} + \\frac{1}{8} + \\frac{1}{32} + \\cdots = \\frac{2}{3}$ Bonnie’s (who goes second) chance of winning the first round is then $1 - \\frac{2}{3} = \\frac{1}{3}$. This means that the person who goes first has a $\\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\\frac{1}{3}$ chance of winning. Now, through casework, we can calculate Alfred’s chance of winning the second round. Case 1: Alfred wins twice; $\\frac{2}{3} \\times \\frac{1}{3}$ (Bonnie goes first this round) $=\\frac{2}{9}$. Case 2: Alfred loses the first round, but wins the second; $\\frac{1}{3} \\times \\frac{2}{3} = \\frac{2}{9}$. Adding up the cases, we get $\\frac{2}{9} + \\frac{2}{9} = \\frac{4}{9}$. Alfred, therefore, has a $\\frac{4}{9}$ of winnning the second round, and Bonnie has a $1-\\frac{4}{9} = \\frac{5}{9}$ chance of winning this round. From here, it is not difficult to see that the probabilities alternate in a pattern. Make $A$ the probability that Alfred wins a round. The chances of Alfred and Bonnie, respectively, winning the first round are $A$ and $A - \\frac{1}{3}$, which can be written as $2A - \\frac{1}{3} = 1$ The second round’s for chances are $A$ and $A + \\frac{1}{9}$, which can also be written as $2A + \\frac{1}{9}$ From this, we can conclude that for the $n$th even round, the probability that Alfred ($A$) wins can be calculated through the equation $2A + \\frac{1}{3^n} = 1$. Solving the equation for $n = 6$, we get $A = \\frac{364}{729}$. $364 + 729 = 1093$. So our answer is $093.",
"In order to begin this problem, we need to calculate the probability that Alfred will win on the first round. Because he goes first, Alfred has a $\\frac{1}{2}$ chance of winning (getting heads) on his first flip. Then, Bonnie, who goes second, has a $\\frac{1}{2} \\times \\frac{1}{2} = \\frac{1}{4}$, chance of winning on her first coin toss. Therefore Alfred’s chance of winning on his second flip is $\\frac{1}{4} \\times \\frac{1}{2} = \\frac{1}{8}$ From this, we can see that Alfred’s (who goes first) chance of winning the first round is: $\\frac{1}{2} + \\frac{1}{8} + \\frac{1}{32} + \\cdots = \\frac{2}{3}$ Bonnie’s (who goes second) chance of winning the first round is then $1 - \\frac{2}{3} = \\frac{1}{3}$. This means that the person who goes first has a $\\frac{2}{3}$ chance of winning the round, while the person who goes second has a $\\frac{1}{3}$ chance of winning. Now, through casework, we can calculate Alfred’s chance of winning the second round. Case 1: Alfred wins twice; $\\frac{2}{3} \\times \\frac{1}{3}$ (Bonnie goes first this round) $=\\frac{2}{9}$. Case 2: Alfred loses the first round, but wins the second; $\\frac{1}{3} \\times \\frac{2}{3} = \\frac{2}{9}$. Adding up the cases, we get $\\frac{2}{9} + \\frac{2}{9} = \\frac{4}{9}$. Alfred, therefore, has a $\\frac{4}{9}$ of winnning the second round, and Bonnie has a $1-\\frac{4}{9} = \\frac{5}{9}$ chance of winning this round. From here, it is not difficult to see that the probabilities alternate in a pattern. Make $A$ the probability that Alfred wins a round. The chances of Alfred and Bonnie, respectively, winning the first round are $A$ and $A - \\frac{1}{3}$, which can be written as $2A - \\frac{1}{3} = 1$ The second round’s for chances are $A$ and $A + \\frac{1}{9}$, which can also be written as $2A + \\frac{1}{9}$ From this, we can conclude that for the $n$th even round, the probability that Alfred ($A$) wins can be calculated through the equation $2A + \\frac{1}{3^n} = 1$. Solving the equation for $n = 6$, we get $A = \\frac{364}{729}$. $364 + 729 = 1093$. So our answer is $093.",
"Rather than categorizing games as wins or losses, we can categorize them as starters (S), where Alfred starts, and non-starters (NS), where Bonnie starts. Game 1 is a starter, and since Alfred must win Game 6, Game 7 is a non-starter. As shown in Solution 1, if a player starts a certain game, the probability $P(NS)$ that they will not start the next game (i.e. win the current game) is $\\frac{2}{3}$, and the probability $P(S)$ that they will start the next game (i.e. lose the current game) is $\\frac{1}{3}$. Similarly, if a player does not start a certain game, $P(NS) = \\frac{1}{3}$ and $P(S) = \\frac{2}{3}$. We conclude that the probability of switching from S to NS or vice versa is always $\\frac{2}{3}$, and the probability of staying the same is always $\\frac{1}{3}$. Listing out all the games from Game 1 to Game 7, we get: S, ?, ?, ?, ?, ?, NS. Between the 7 games, there are 6 opportunities to either switch or stay the same. We need to eventually switch from S to NS, so there must be an odd number of switches. Furthermore, this number is less than 6, so it must be 1, 3, or 5. 1 switch: There are ${6 \\choose 1} = 6$ ways to place the switch, so $P = 6\\left(\\frac{1}{3}\\right)^5\\left(\\frac{2}{3}\\right) = \\frac{12}{729}$. 3 switches: There are ${6 \\choose 3} = 20$ ways to place the switches, so $P = 20\\left(\\frac{1}{3}\\right)^3\\left(\\frac{2}{3}\\right)^3 = \\frac{160}{729}$. 5 switches: There are ${6 \\choose 5} = 6$ ways to place the switch, so $P = 6\\left(\\frac{1}{3}\\right)\\left(\\frac{2}{3}\\right)^5 = \\frac{192}{729}$. Add up all these probabilities to get $\\frac{12+160+192}{729} = \\frac{364}{729}$. $364+729=1093$, so the answer is $093."
] |
1993-I-12
| 1,993 | 12 |
The vertices of $\triangle ABC$ are $A = (0,0)\,$ , $B = (0,420)\,$ , and $C = (560,0)\,$ . The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ , $P_3\,$ , $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ . Given that $P_7 = (14,92)\,$ , what is $k + m\,$ ?
| 344 | null |
[
"Solution 1 If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$, then $u=2r-p$ and $v=2s-q$. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \\[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\\bmod{560},\\ 2y_n\\bmod{420})\\] Then $P_7=(14,92)$, so $x_7=14$ and $y_7=92$, and we get \\[\\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\\\ \\hline\\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\\\ \\hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \\end{array}\\] So the answer is $344.",
"If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$, then $u=2r-p$ and $v=2s-q$. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \\[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\\bmod{560},\\ 2y_n\\bmod{420})\\] Then $P_7=(14,92)$, so $x_7=14$ and $y_7=92$, and we get \\[\\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\\\ \\hline\\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\\\ \\hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \\end{array}\\] So the answer is $344.",
"Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$. Note that $P_7 = \\cfrac{\\cfrac{\\cfrac{P_1 + L_1}2 + L_2}2 + \\cdots}{2\\ldots} = \\frac {(k,m)}{64} + \\frac {L_1}{64} + \\frac {L_2}{32} + \\frac {L_3}{16} + \\frac {L_4}8 + \\frac {L_5}4 + \\frac {L_6}2 = (14,92)$. Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$, we can just ignore it!): for $\\frac {L_6}2,\\frac {L_5}4$, since all addends are nonnegative, a non-$(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$, respectively, and we can ignore them, for $\\frac {L_4}8,\\frac {L_3}{16},\\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities, and for $\\frac {L_1}{64}$, all three work. Also, to be in the triangle, $0\\le k\\le560$ and $0\\le m\\le420$. Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$, we look at that first. If $L_1 = (0,0)$, then $k = 2^6\\cdot14 = 64\\cdot14 > 40\\cdot14 = 560$, so it can only be that $L_1 = (560,0)$, and $k + 560 = 2^6\\cdot14$ $\\implies k = 64\\cdot14 - 40\\cdot14 = 24\\cdot14 = 6\\cdot56 = 336$. Now, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ ($L_2$ would influence the least, so we test that), then $\\frac {L_2}{32} + \\frac {L_3}{16} + \\frac {L_4}8 < \\frac {420}{16} + \\frac {420}8 = 79\\pm\\epsilon < 80$, which would mean that $P_1 > 2^6\\cdot(92 - 80) = 64\\cdot12 > 42\\cdot10 = 420\\ge m$, so $L_2,L_3,L_4 = (0,420)$, and now $\\frac {P_1}{64} + \\frac {420}{2^5} + \\frac {420}{2^4} + \\frac {420}{2^3} = 92$ $\\implies P_1$ $= 64\\cdot92 - 420(2 + 4 + 8)$ $= 64\\cdot92 - 420\\cdot14= 64(100 - 8) - 14^2\\cdot30$ $= 6400 - 512 - (200 - 4)\\cdot30$ $= 6400 - 512 - 6000 + 120$ $= - 112 + 120$ $= 8$, and finally, $k + m = 336 + 8 = 344."
] |
1993-I-13
| 1,993 | 13 |
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
| 163 | null |
[
"Solution 1 Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \\begin{align}y&=-\\frac{100}{t}x+200-\\frac{5000}{t}\\\\50^2&=x^2+y^2\\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \\[-\\frac{x}{y}=-\\frac{100}{t}\\] or $xt=100y.$ Now substitute \\[y= \\frac{xt}{100}\\] into $(2)$ and get \\[x=\\frac{5000}{\\sqrt{100^2+t^2}}.\\] Now substitute this and \\[y=\\frac{xt}{100}\\] into $(1)$ and solve for $t$ to get \\[t=\\frac{160}{3}.\\] Finally, the sum of the numerator and denominator is $160+3=163.",
"Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \\begin{align}y&=-\\frac{100}{t}x+200-\\frac{5000}{t}\\\\50^2&=x^2+y^2\\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \\[-\\frac{x}{y}=-\\frac{100}{t}\\] or $xt=100y.$ Now substitute \\[y= \\frac{xt}{100}\\] into $(2)$ and get \\[x=\\frac{5000}{\\sqrt{100^2+t^2}}.\\] Now substitute this and \\[y=\\frac{xt}{100}\\] into $(1)$ and solve for $t$ to get \\[t=\\frac{160}{3}.\\] Finally, the sum of the numerator and denominator is $160+3=163.",
"Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$. Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively. [asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,50)); label(\"$A$\",A,SW); label(\"$B$\",B,NNW); label(\"$C$\",C,S); label(\"$D$\",D,NE); label(\"$P$\",P,S); label(\"$Q$\",Q,NE); label(\"$O$\",O,W); label(\"$X$\",X,ESE); [/asy] From the problem statement, $AB=3t$, and $CD=t$. Since $\\Delta ABX \\sim \\Delta CDX$, $CX=AC\\cdot\\left(\\frac{CD}{AB-CD}\\right)=200\\cdot\\left(\\frac{t}{3t-t}\\right)=100$. Since $PC=100$, $PX=200$. So, $\\tan(\\angle OXP)=\\frac{OP}{PX}=\\frac{50}{200}=\\frac{1}{4}$. Since circle $O$ is tangent to $BX$ and $AX$, $OX$ is the angle bisector of $\\angle BXA$. Thus, $\\tan(\\angle BXA)=\\tan(2\\angle OXP)=\\frac{2\\tan(\\angle OXP)}{1- \\tan^2(\\angle OXP)} = \\frac{2\\cdot \\left(\\frac{1}{4}\\right)}{1-\\left(\\frac{1}{4}\\right)^2}=\\frac{8}{15}$. Therefore, $t = CD = CX\\cdot\\tan(\\angle BXA) = 100 \\cdot \\frac{8}{15} = \\frac{160}{3}$, and the answer is $163.",
"[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O--Q--R--cycle); draw(Circle(O,50)); label(\"$A$\",A,SW); label(\"$B$\",B,NNW); label(\"$C$\",(200,-205),S); label(\"$D$\",D,NE); label(\"$P$\",(100,-205),S); label(\"$Q$\",Q,NE); label(\"$O$\",O,SW); label(\"$R$\",R,NE); label(\"$S$\",S,W); [/asy] Let $t$ be the time they walk. Then $CD=t$ and $AB=3t$. Draw a line from point $O$ to $Q$ such that $OQ$ is perpendicular to $BD$. Further, draw a line passing through points $O$ and $P$, so $OP$ is parallel to $AB$ and $CD$ and is midway between those two lines. Then $PR=\\dfrac{AB+CD}{2}=\\dfrac{3t+t}{2}=2t$. Draw another line passing through point $D$ and parallel to $AC$, and call the point of intersection of this line with $AB$ as $S$. Then $SB=AB-AS=3t-t=2t$. We see that $m\\angle SBD=m\\angle ORQ$ since they are corresponding angles, and thus by angle-angle similarity, $\\triangle QOR\\sim\\triangle SDB$. Then \\begin{align*} \\dfrac{OQ}{DS}=\\dfrac{RO}{BD}&\\implies\\dfrac{50}{200}=\\dfrac{RO}{\\sqrt{200^2+4t^2}}\\\\ &\\implies RO=\\dfrac{1}{4}\\left(\\sqrt{200^2+4t^2}\\right)\\\\ &\\implies RO=\\dfrac{1}{2}\\left(\\sqrt{100^2+t^2}\\right) \\end{align*} And we obtain \\begin{align*} PR-OP&=RO\\\\ 2t-50&=\\dfrac{1}{2}\\left(\\sqrt{100^2+t^2}\\right)\\\\ 4t-100&=\\sqrt{100^2+t^2}\\\\ (4t-100)^2&=\\left(\\sqrt{100^2+t^2}\\right)^2\\\\ 16t^2-800t+100^2&=t^2+100^2\\\\ 15t^2&=800t\\\\ t&=\\dfrac{800}{15} \\end{align*} so we have $t=\\frac{160}{3}$, and our answer is thus $160+3=163.",
"We can use areas to find the answer. Since Jenny and Kenny are 200 feet apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building. We know that the radius of the circle is 50, and that $\\overline{AJ} = x$, $\\overline{BK} = 3x$. Illustration: [asy] size(8cm); pair A = (0,0), B = (200,0), K = (200,160), J = (0,53.3333), O = (100,50); pair D = tangent(J,O,50,2), F = tangent(A,O,50); pair[] p = {A,B,K,J,O,D,F}; for (pair point : p) { dot(point); } label(\"A\",A,SW); label(\"B\",B,SE); label(\"K\",K,NE); label(\"J\",J,NW); label(\"O\",O,SW); label(\"D\",D,N); label(\"F\",F,S); draw(Circle(O,50)); draw(A--B--K--J--cycle); draw(F--O--J); draw(D--O); draw(O--K); label(\"$50$\",(100,26.666),0.5E); label(\"$50$\",midpoint(D--O),0.5NE); label(\"$x$\",midpoint(A--J),0.5W); label(\"$3x$\",midpoint(B--K),0.5E); label(\"$200$\",midpoint(A--B),4S); draw(rightanglemark(J,A,F,150)); draw(rightanglemark(J,D,O,150)); draw(rightanglemark(A,F,O,150)); draw(rightanglemark(F,B,K,150)); [/asy] By areas, $[OJK] + [AJOF] + [OFBK] = [ABKJ]$. Having right trapezoids, $[AFOJ] = \\frac{x+50}{2} \\cdot 100$. The other areas of right trapezoids can be calculated in the same way. We just need to find $[OJK]$ in terms of $x$. If we bring $\\overline{AB}$ up to where the point J is, we have by the Pythagorean Theorem, $\\overline{JK} = 2\\sqrt{x^2+10000} \\Rightarrow [OJK] = \\frac{1}{2} \\overline{JK} \\cdot \\overline{OD}$. Now we have everything to solve for $x$. \\[[OJK] + [AJOF] + [OFBK] = [ABKJ]\\] \\[50 \\sqrt{x^2 + 10000} + \\frac{x+50}{2} \\cdot 100 + \\frac{3x+50}{2} \\cdot 100 = \\frac{x+3x}{2} \\cdot 200\\] After isolating the radical, dividing by 50, and squaring, we obtain: $15x^2 - 800x = 0 \\Rightarrow x = \\frac{160}{3}$. Since Jenny walks $\\frac{160}{3}$ feet at 1 foot/s, our answer is $160+3 = 163.",
"[asy] import math; import geometry; import olympiad; point A,B,C,D,F,G,H,I,M,O,P; A=(0,0); B=(0,200); C=(160,200); D=(160/3,0); F=(0,150); G=(400/3,150); H=(80,50); I=(0,50); M=(0,100); O=(320/3,100); P=(80,150); draw(I--H--D--A--B--C--G--F); draw(M--O--H); draw(circle((50,100),50)); draw(G--H--P); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",F,W); label(\"F\",G,E); label(\"G\",H,E); label(\"H\",I,W); label(\"M\",M,W); label(\"N\",O,E); label(\"P\",P,N); [/asy] Consider our diagram here, where Jenny goes from $A$ to $D$ and Kenny goes from $B$ to $C$. Really what we are asking is the length of $AD$, knowing that $AB$ and $CD$ are tangents and $AB$ is perpendicular to the parallel lines. Draw lines $EF$ and $GH$ as shown such that they are parallel to $BC$ and $AD$ and are tangent to the circle. Additionally, let $MN$ be the median of trapezoid $ABCD$. Notice how $AH=MH=50$ and $BE=ME=50$ as well, so $EF$ and $GH$ are medians of trapezoids $MBCN$ and $AMND$, respectively. If we set $AD=x$ and $BC=3x$, then $MN=2x$ so $HG=\\frac{3}{2}x$ and $EF=\\frac{5}{2}x$. Now we will find $FG$ in two different ways. For the first way, notice that $EFGH$ has its inscribed circle, so $EF+GH=EH+FG$. This means that $FG=\\frac{3}{2}x+\\frac{5}{2}x-100=4x-100$. However, if we let $P$ be the altitude from $G$ to $EF$, then notice how $PF=x$. Since $PG=100$, then by the Pythagorean Theorem, we also have that $FG=\\sqrt{x^2+10000}$. After that, then we have an equation in $x$, which is \\[\\sqrt{x^2+10000}=4x-100.\\] Squaring both sides, it expands to \\[x^2+10000=16x^2-800x+10000,\\] and since $x$ is nonzero, then once we cancel $10000$ from both sides, we can divide $x$ and get the new equation \\[x=16x-800\\implies x=\\frac{160}{3}.\\] Because Jenny travels this at $1$ foot per second, then it will take $\\frac{160}{3}$ seconds, so our answer is $163. ~~ethanzhang1001",
"Let the point where Jenny started be $A = (0,0)$. Then let $D = (t,0)$ the point Jenny is when she can see Kenny. Similarly, let $B = (0,200)$ be where Kenny started, and let $C = (3t,200)$ the point Kenny is when she can see Jenny. Then the equation of the line passing through $C, D$ is $100x-ty-100t=0$. Now, note that the center of the building is at $O = (50,100)$, and the distance from $O$ to $CD$ is $50$, so, using the distance from a point to a line formula, we get \\[\\frac{|100\\cdot50-100a-100a|}{\\sqrt{10000+a^2}} = 50,\\] which simplifies to $15a^2-800a = 0$ from clearing denominators, squaring to remove absolute value and radical, then cancelling and rearranging. But this is equivalent to $a = \\frac{160}{3}$, so our answer is $163. ~Yiyj1"
] |
1993-I-14
| 1,993 | 14 |
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$ , for a positive integer $N\,$ . Find $N\,$ .
| 448 | null |
[
"Put the rectangle on the coordinate plane so its vertices are at $(\\pm4,\\pm3)$, for all four combinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, $O$. Note that such a rectangle is unstuck if its four vertices are in or on the edge of all four quadrants, and it is not the same rectangle as the big one. Let the four vertices of this rectangle be $A(4,y)$, $B(-x,3)$, $C(-4,-y)$ and $D(x,-3)$ for nonnegative $x,y$. Then this is a rectangle, so $OA=OB$, or $16+y^2=9+x^2$, so $x^2=y^2+7$. Reflect $D$ across the side of the rectangle containing $C$ to $D'(-8-x,-3)$. Then $BD'=\\sqrt{(-8-x-(-x))^2+(3-(-3))^2}=10$ is constant, and the perimeter of the rectangle is equal to $2(BC+CD')$. The midpoint of $\\overline{BD'}$ is $(-4-x,0)$, and since $-4>-4-x$ and $-y\\le0$, $C$ always lies below $\\overline{BD'}$. If $y$ is positive, it can be decreased to $y'<y$. This causes $x$ to decrease as well, to $x'$, where $x'^2=y'^2+7$ and $x'$ is still positive. If $B$ and $D'$ are held in place as everything else moves, then $C$ moves $(y-y')$ units up and $(x-x')$ units left to $C'$, which must lie within $\\triangle BCD'$. Then we must have $BC'+C'D'<BC+CD'$, and the perimeter of the rectangle is decreased. Therefore, the minimum perimeter must occur with $y=0$, so $x=\\sqrt7$. By the distance formula, this minimum perimeter is \\[2\\left(\\sqrt{(4-\\sqrt7)^2+3^2}+\\sqrt{(4+\\sqrt7)^2+3^2}\\right)=4\\left(\\sqrt{8-2\\sqrt7}+\\sqrt{8+2\\sqrt7}\\right)\\] \\[=4(\\sqrt7-1+\\sqrt7+1)=8\\sqrt7=\\sqrt{448}.\\] Therefore $N$ would equal $448 ~minor edit by Yiyj1 and Root three over two",
"Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$. So $(a+b)^2 = 64+2(24) = 112$, so $2(a+b) = \\sqrt{448."
] |
1993-I-15
| 1,993 | 15 |
Let $\overline{CH}$ be an altitude of $\triangle ABC$ . Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$ . If $AB = 1995\,$ , $AC = 1994\,$ , and $BC = 1993\,$ , then $RS\,$ can be expressed as $m/n\,$ , where $m\,$ and $n\,$ are relatively prime integers. Find $m + n\,$ .
| 997 | null |
[
"[asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label(\"$A$\",A,SE); label(\"$B$\",B,SW); label(\"$C$\",C,N); label(\"$H$\",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label(\"$S$\",x[0],SW); draw(circle((4.29843788128,1.29843788128),1.29843788128)); pair [] y=intersectionpoints(C--H,circle((4.29843788128,1.29843788128),1.29843788128)); dot(y[0]); label(\"$R$\",y[0],NE); label(\"$1993$\",(1.5,2),NW); label(\"$1994$\",(5.5,2),NE); label(\"$1995$\",(4,0),S); [/asy] From the Pythagorean Theorem, $AH^2+CH^2=1994^2$, and $(1995-AH)^2+CH^2=1993^2$. Subtracting those two equations yields $AH^2-(1995-AH)^2=3987$. After simplification, we see that $2*1995AH-1995^2=3987$, or $AH=\\frac{1995}{2}+\\frac{3987}{2*1995}$. Note that $AH+BH=1995$. Therefore we have that $BH=\\frac{1995}{2}-\\frac{3987}{2*1995}$. Therefore $AH-BH=\\frac{3987}{1995}$. $\\textbf{Note: }$An easier method is to use both facts together: First, by the Pythagorean Theorem, we find that $1993^2-BH^2=1994^2-AH^2$, so $AH^2-BH^2=1994^2-1993^2=3987$. We know that $AH+BH=1995$, so by difference of squares and dividing we find that $AH-BH=\\frac{3987}{1995}=\\frac{1329}{665}$. ~eevee9406 Now note that $RS=|HR-HS|$, $RH=\\frac{AH+CH-AC}{2}$, and $HS=\\frac{CH+BH-BC}{2}$. Therefore we have $RS=\\left| \\frac{AH+CH-AC-CH-BH+BC}{2} \\right|=\\frac{|AH-BH-1994+1993|}{2}$. Plugging in $AH-BH$ and simplifying, we have $RS=\\frac{1992}{1995*2}=\\frac{332}{665} \\rightarrow 332+665=997 by the triangle inequality. ~Yiyj1"
] |
1994-I-1
| 1,994 | 1 |
The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000?
| 63 | null |
[
"One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$. Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \\equiv -1,\\ 1 \\equiv 2,\\ 1 \\pmod{3}$. Since 1994 is even, $n$ must be congruent to $1 \\pmod{3}$. It will be the $\\frac{1994}{2} = 997$th such term, so $n = 4 + (997-1) \\cdot 3 = 2992$. The value of $n^2 - 1 = 2992^2 - 1 \\pmod{1000}$ is $063. ~minor edit by Yiyj1"
] |
1994-I-2
| 1,994 | 2 |
A circle with diameter $\overline{PQ}\,$ of length 10 is internally tangent at $P^{}_{}$ to a circle of radius 20. Square $ABCD\,$ is constructed with $A\,$ and $B\,$ on the larger circle, $\overline{CD}\,$ tangent at $Q\,$ to the smaller circle, and the smaller circle outside $ABCD\,$ . The length of $\overline{AB}\,$ can be written in the form $m + \sqrt{n}\,$ , where $m\,$ and $n\,$ are integers. Find $m + n\,$ .
| 312 | null |
[
"Call the center of the larger circle $O$. Extend the diameter $\\overline{PQ}$ to the other side of the square (at point $E$), and draw $\\overline{AO}$. We now have a right triangle, with hypotenuse of length $20$. Since $OQ = OP - PQ = 20 - 10 = 10$, we know that $OE = AB - OQ = AB - 10$. The other leg, $AE$, is just $\\frac 12 AB$. Apply the Pythagorean Theorem: $(AB - 10)^2 + \\left(\\frac 12 AB\\right)^2 = 20^2$ $AB^2 - 20 AB + 100 + \\frac 14 AB^2 - 400 = 0$ $AB^2 - 16 AB - 240 = 0$ The quadratic formula shows that the answer is $\\frac{16 \\pm \\sqrt{16^2 + 4 \\cdot 240}}{2} = 8 \\pm \\sqrt{304}$. Discard the negative root, so our answer is $8 + 304 = 312."
] |
1994-I-8
| 1,994 | 8 |
The points $(0,0)\,$ , $(a,11)\,$ , and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$ .
| 315 | null |
[
"Solution 1 Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so: \\[(a+11i)\\left(\\mathrm{cis}\\,60^{\\circ}\\right) = (a+11i)\\left(\\frac 12+\\frac{\\sqrt{3}i}2\\right)=b+37i.\\] Equating the real and imaginary parts, we have: \\begin{align*}b&=\\frac{a}{2}-\\frac{11\\sqrt{3}}{2}\\\\37&=\\frac{11}{2}+\\frac{a\\sqrt{3}}{2} \\end{align*} Solving this system, we find that $a=21\\sqrt{3}, b=5\\sqrt{3}$. Thus, the answer is $315. ~Arcticturn",
"Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so: \\[(a+11i)\\left(\\mathrm{cis}\\,60^{\\circ}\\right) = (a+11i)\\left(\\frac 12+\\frac{\\sqrt{3}i}2\\right)=b+37i.\\] Equating the real and imaginary parts, we have: \\begin{align*}b&=\\frac{a}{2}-\\frac{11\\sqrt{3}}{2}\\\\37&=\\frac{11}{2}+\\frac{a\\sqrt{3}}{2} \\end{align*} Solving this system, we find that $a=21\\sqrt{3}, b=5\\sqrt{3}$. Thus, the answer is $315. ~Arcticturn",
"Using the Pythagorean theorem with these beastly numbers doesn't seem promising. How about properties of equilateral triangles? $\\sqrt{3}$ and perpendiculars inspires this solution: First, drop a perpendicular from $O$ to $AB$. Call this midpoint of $AB M$. Thus, $M=\\left(\\frac{a+b}{2}, 24\\right)$. The vector from $O$ to $M$ is $\\left[\\frac{a+b}{2}, 24\\right]$. Meanwhile from point $M$ we can use a vector with $\\frac{\\sqrt{3}}{3}$ the distance; we have to switch the $x$ and $y$ and our displacement is $\\left[8\\sqrt{3}, \\frac{(a+b)\\sqrt{3}}{6}\\right]$. (Do you see why we switched $x$ and $y$ due to the rotation of 90 degrees?) We see this displacement from $M$ to $A$ is $\\left[\\frac{a-b}{2}, 13\\right]$ as well. Equating the two vectors, we get $a+b=26\\sqrt{3}$ and $a-b=16\\sqrt{3}$. Therefore, $a=21\\sqrt{3}$ and $b=5\\sqrt{3}$. And the answer is $315. ~Arcticturn",
"Plot this equilateral triangle on the complex plane. Translate the equilateral triangle so that its centroid is located at the origin. (The centroid can be found by taking the average of the three vertices of the triangle, which gives $\\left(\\frac{a+b}{3}, 16i\\right)$. The new coordinates of the equilateral triangle are $\\left(-\\frac{a+b}{3}-16i\\right), \\left(a-\\frac{a+b}{3}-5i\\right), \\left(b-\\frac{a+b}{3}+21i\\right)$. These three vertices are solutions of a cubic polynomial of form $x^3 + C$. By Vieta's Formulas, the sum of the paired roots of the cubic polynomial are zero. (Or for the three roots $r_1, r_2,$ and $r_3,$$\\, r_1r_2 + r_2r_3 + r_3r_1 = 0$.) The vertices of the equilateral triangle represent the roots of a polynomial, so the vertices can be plugged into the above equation. Because both the real and complex components of the equation have to sum to zero, you really have two equations. Multiply out the equation given by Vieta's Formulas and isolate the ones with imaginary components. Simplify that equation, and that gives the equation $5a = 21b.$ Now use the equation with only real parts. This should give you a quadratic $a^2 - ab + b^2 = 1083$. Use your previously obtained equation to plug in for $a$ and solve for $b$, which should yield $5\\sqrt{3}$. $a$ is then $\\frac{21}{5}\\sqrt{3}$. Multiplying $a$ and $b$ yields $315. ~Arcticturn",
"Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 + 37^2$. $a^2 + 121 = b^2 + 1369 \\longrightarrow a^2 = b^2 + 1248$. Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$. $b^2 = 2ab - 555 \\longrightarrow a^2 = 2ab + 693$. We have $b^2 + 1248 = 2b\\sqrt{b^2+1248} + 693$. Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$. $3b^4 + 3882b^2 - 308025 = 0$. This factors to $(3b^2-225)(b^2+1369)$, so $3b^2 = 225, b = 5\\sqrt 3$. Plugging it back in, we find that a = $\\sqrt{1323}$ which is $21\\sqrt3$, so the product $ab$ is $315. ~Arcticturn",
"Just using the Pythagorean Theorem, we get that $a^2 + 11^2 = (b-a)^2 + 26^2 = b^2 + 37^2$. $a^2 + 121 = b^2 + 1369 \\longrightarrow a^2 = b^2 + 1248$. Expanding the second and subtracting the first equation from it we get $b^2 = 2ab - 555$. $b^2 = 2ab - 555 \\longrightarrow a^2 = 2ab + 693$. We have $b^2 + 1248 = 2b\\sqrt{b^2+1248} + 693$. Moving the square root to one side and non square roots to the other we eventually get $b^4 + 1110b^2 + 308025 = 4b^4 + 4992b^2$. $3b^4 + 3882b^2 - 308025 = 0$. This factors to $(3b^2-225)(b^2+1369)$, so $3b^2 = 225, b = 5\\sqrt 3$. Plugging it back in, we find that a = $\\sqrt{1323}$ which is $21\\sqrt3$, so the product $ab$ is $315. ~Arcticturn"
] |
1994-I-9
| 1,994 | 9 |
A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers. Find $p+q.\,$
| 394 | null |
[
"Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards: Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability $\\frac {1}{2k - 1}$, then we have $k - 1$ pairs left. So this contributes probability $\\frac {P_{k - 1}}{2k - 1}$. Case 2. We draw a pair on the first and third cards. The second card is different from the first with probability $\\frac {2k - 2}{2k - 1}$ and the third is the same as the first with probability $\\frac {1}{2k - 2}$. We are left with $k - 1$ pairs but one card already drawn. However, having drawn one card doesn't affect the game, so this also contributes probability $\\frac {P_{k - 1}}{2k - 1}$. Case 3. We draw a pair on the second and third cards. This is pretty much the same as case 2, so we get $\\frac {P_{k - 1}}{2k - 1}$. Therefore, we obtain the recursion $P_k = \\frac {3}{2k - 1}P_{k - 1}$. Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$), we get $\\frac {3^5}{11*9*7*5*3} = \\frac {9}{385}$, and $p+q=394.",
"Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here. 1. We pick the other A. There's a $\\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this is the case. 2. We pick any other letter that is not an A. There's a $\\frac{10}{11}$ chance for this to happen. WLOG, assume we pick the letter B. Now in order for us to continue the game, we must choose either the other A or B. There's a $\\frac{2}{10}$ chance for this to happen. WLOG, assume we choose A. Now we have BCDEFCDEF left. Notice however that in the first case, the probability of emptying the bag with [BCDEF] is the same thing as with BCDEFCDEF, as the only difference is you've removed one of the letters (and it doesn't matter which you chose). Hence for this case, there is a $\\frac{1}{11} + \\frac{10}{11}*\\frac{2}{10} = \\frac{3}{11}$ * [BCDEF] chance to empty the bag. Continuing this process, we get that: [BCDEF] = $\\frac{3}{9}$ * [CDEF] [CDEF] = $\\frac{3}{7}$ * [DEF] [DEF] = $\\frac{3}{5}$ * [EF] [EF] = 1 (clearly, since if we are only left with EFEF then we are going to empty the bag). And hence [ABCDEF] = $1*\\frac{3}{5}*\\frac{3}{7}*\\frac{3}{9}*\\frac{3}{11} = \\frac{9}{385}$ so our answer is $9+385=394",
"Let $a_{m,n}$ represent the situation in which $n$ tiles are in your hand and $m$ pairs have been removed; i.e., there are $2m$ tiles remaining in play. We know that $a_{m,0}=a_{m,1}$ to begin with, so we find some other recursions: When the player has $1$ tile in their hand with $m$ pairs remaining, there is a $\\frac{1}{2m-1}$ chance that the player finds the one matching tile and the game state changes to $a_{m-1,0}$. There is a $\\frac{2m-2}{2m-1}$ chance that the player does not find the matching tile and the game state changes to $a_{m,2}$. As a result, $a_{m,1}=\\frac{1}{2m-1}a_{m-1,0}+\\frac{2m-2}{2m-1}a_{m,2}$. When the player has $2$ tiles in their hand with $m$ pairs remaining, there is a $\\frac{1}{m-1}$ chance that the player finds a matching tile and the game moves on to game state $a_{m-1,1}$. As a result, $a_{m,2}=\\frac{1}{m-1}a_{m-1,1}$. We also know that $a_0=a_1=a_2=1$ (i.e., you are guaranteed to win with two tiles remaining), so we are able to find an end to the recursion. Now we are left with four rules: 1) $a_{m,0}=a_{m,1}$ 2) $a_{m,1}=\\frac{1}{2m-1}a_{m-1,0}+\\frac{2m-2}{2m-1}a_{m,2}$ 3) $a_{m,2}=\\frac{1}{m-1}a_{m-1,1}$ 4) $a_0=a_1=a_2=1$ We then calculate the terms of the multi-variable recursion and arrive at $a_{6,0}=\\frac{9}{385}$, so the answer is $385+9=394. ~eevee9406"
] |
1994-I-10
| 1,994 | 10 |
In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$
| 450 | null |
[
"Since $\\triangle ABC \\sim \\triangle CBD$, we have $\\frac{BC}{AB} = \\frac{29^3}{BC} \\Longrightarrow BC^2 = 29^3 AB$. It follows that $29^2 | BC$ and $29 | AB$, so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$, respectively, where $x$ is an integer. By the Pythagorean Theorem, we find that $AC^2 + BC^2 = AB^2 \\Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2$, so $29x | AC$. Letting $y = AC / 29x$, we obtain after dividing through by $(29x)^2$, $29^2 = x^2 - y^2 = (x-y)(x+y)$. As $x,y \\in \\mathbb{Z}$, the pairs of factors of $29^2$ are $(1,29^2)(29,29)$; clearly $y = \\frac{AC}{29x} \\neq 0$, so $x-y = 1, x+y= 29^2$. Then, $x = \\frac{1+29^2}{2} = 421$. Thus, $\\cos B = \\frac{BC}{AB} = \\frac{29^2 x}{29x^2} = \\frac{29}{421}$, and $m+n = 450.",
"We will solve for $\\cos B$ using $\\triangle CBD$, which gives us $\\cos B = \\frac{29^3}{BC}$. By the Pythagorean Theorem on $\\triangle CBD$, we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$. Trying out factors of $29^6$, we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The other pairs give answers over 999). Adding these, we have $2BC = 29^4 + 29^2$ and $\\frac{29^3}{BC} = \\frac{2*29^3}{29^2 (29^2 +1)} = \\frac{58}{842} = \\frac{29}{421}$, and our answer is $450.",
"Using similar right triangles, we identify that $CD = \\sqrt{AD \\cdot BD}$. Let $AD$ be $29 \\cdot k^2$, to avoid too many radicals, getting $CD = k \\cdot 29^2$. Next we know that $AC = \\sqrt{AB \\cdot AD}$ and that $BC = \\sqrt{AB \\cdot BD}$. Applying the logic with the established values of k, we get $AC = 29k \\cdot \\sqrt{29^2 + k^2}$ and $BC = 29^2 \\cdot \\sqrt{29^2 + k^2}$. Next we look to the integer requirement. Since $k$ is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let $y$ be $\\sqrt{29^2 + k^2}$, thus $29^2 = y^2 - k^2$, and $29^2 = (y + k)(y - k)$. Since $29$ is prime, and $k$ cannot be zero, we find $k = 420$ and $y = 421$ as the smallest integers satisfying this quadratic Diophantine equation. Then, since $cos B$ = $\\frac{29}{\\sqrt{29^2 + k^2}}$. Plugging in we get $cos B = \\frac{29}{421}$, thus our answer is $450."
] |
1994-I-11
| 1,994 | 11 |
Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all 94 of the bricks?
| 465 | null |
[
"Solution 1 We have the smallest stack, which has a height of $94 \\times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$'s, $6$'s, and $15$'s. Because $0$, $6$, and $15$ are all multiples of $3$, the change will always be a multiple of $3$, so we just need to find the number of changes we can get from $0$'s, $2$'s, and $5$'s. From here, we count what we can get: \\[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, \\ldots\\] It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 \\times 2 - 5 - 2=3$. But we also have a maximum change ($94 \\times 5$), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$'s, $3$'s, or $5$'s. The maximum we can't get is $5 \\times 3-5-3=7$, so the numbers $94 \\times 5-8$ and below, except $3$ and $1$, work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 \\times 5-8$ and $94 \\times 5$. $94 \\times 5-7$ obviously doesn't work, $94 \\times 5-6$ does since 6 is a multiple of 3, $94 \\times 5-5$ does because it is a multiple of $5$ (and $3$), $94 \\times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$, $94 \\times 5-3$ does since $3=3$, and $94 \\times 5-2$ and $94 \\times 5-1$ don't, and $94 \\times 5$ does. Thus the numbers $0$, $2$, $4$ all the way to $94 \\times 5-8$, $94 \\times 5-6$, $94 \\times 5-5$, $94 \\times 5-3$, and $94\\times 5$ work. That's $2+(94 \\times 5 - 8 - 4 +1)+4=465 tower heights are different.",
"We have the smallest stack, which has a height of $94 \\times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$'s, $6$'s, and $15$'s. Because $0$, $6$, and $15$ are all multiples of $3$, the change will always be a multiple of $3$, so we just need to find the number of changes we can get from $0$'s, $2$'s, and $5$'s. From here, we count what we can get: \\[0, 2 = 2, 4 = 2+2, 5 = 5, 6 = 2+2+2, 7 = 5+2, 8 = 2+2+2+2, 9 = 5+2+2, \\ldots\\] It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of $2m + 5n$ for $m,n$ being positive integers is $5 \\times 2 - 5 - 2=3$. But we also have a maximum change ($94 \\times 5$), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$'s, $3$'s, or $5$'s. The maximum we can't get is $5 \\times 3-5-3=7$, so the numbers $94 \\times 5-8$ and below, except $3$ and $1$, work. Now there might be ones that we haven't counted yet, so we check all numbers between $94 \\times 5-8$ and $94 \\times 5$. $94 \\times 5-7$ obviously doesn't work, $94 \\times 5-6$ does since 6 is a multiple of 3, $94 \\times 5-5$ does because it is a multiple of $5$ (and $3$), $94 \\times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$, $94 \\times 5-3$ does since $3=3$, and $94 \\times 5-2$ and $94 \\times 5-1$ don't, and $94 \\times 5$ does. Thus the numbers $0$, $2$, $4$ all the way to $94 \\times 5-8$, $94 \\times 5-6$, $94 \\times 5-5$, $94 \\times 5-3$, and $94\\times 5$ work. That's $2+(94 \\times 5 - 8 - 4 +1)+4=465 tower heights are different.",
"Using bricks of dimensions $4''\\times10''\\times19''$ is comparable to using bricks of dimensions $0''\\times6''\\times15''$ which is comparable to using bricks of dimensions $0''\\times2''\\times5''$. Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$. It follows that all tower heights can be made by using 4 or fewer bricks of height $2''$. There are $95+94+93+92+91=465$ ways to build a tower using 4 or fewer bricks of height $2''$. Taking the heights $\\mod 5$, we see that towers using a different number of bricks of height $2''$ have unequal heights. Thus, all of the $465 tower heights are different."
] |
1994-I-12
| 1,994 | 12 |
A fenced, rectangular field measures 24 meters by 52 meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?
| 702 | null |
[
"Suppose there are $n$ squares in every column of the grid, so there are $\\frac{52}{24}n = \\frac {13}6n$ squares in every row. Then $6|n$, and our goal is to maximize the value of $n$. Each vertical fence has length $24$, and there are $\\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$, and there are $n-1$ such fences. Then the total length of the internal fencing is $24\\left(\\frac{13n}{6}-1\\right) + 52(n-1) = 104n - 76 \\le 1994 \\Longrightarrow n \\le \\frac{1035}{52} \\approx 19.9$, so $n \\le 19$. The largest multiple of $6$ that is $\\le 19$ is $n = 18$, which we can easily verify works, and the answer is $\\frac{13}{6}n^2 = 702.",
"Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$. (If you are confused, just draw the square out). This is clearly greater than $1994$, so the actual side length that we are looking for is greater than $1$. Now we can convert this into an equation. The equation is simply $(\\frac{52}{x})(\\frac{24}{x}+1) + (\\frac{24}{x})(\\frac{52}{x}+1)$ (The intutition comes from considering partioning the field into side lengths of $1$ and then partitioning those squares). This is equivalent to $\\frac{2496}{x^2} + \\frac{76}{x}$, which should be less than or equal to $1994$. Now we can just find possible lengths of the square that are greater than $1$ and test them out. A viable side length would mean that $\\frac{24}{x}, \\frac{52}{x} \\in$ N. Since $\\gcd(24,52) = 4$, then the smallest value greater than $1$ that we satisfies the conditions has $4$ in the numerator, and hence $3$ in the denominator. Test out $x=\\frac{4}{3}$. This will equate to something less than $1994$, and hence the smallest square length that is plausible is $\\frac{4}{3}$. Now the rest is elementary, we do $\\frac{52}{\\frac{4}{3}} * \\frac{24}{\\frac{4}{3}} \\Rightarrow 39*18 = 702"
] |
1994-I-14
| 1,994 | 14 |
A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection. Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments. Include the first reflection at $C\,$ in your count. AIME 1994 Problem 14.png
| 71 | null |
[
"At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP(\"B\",(0,0),NW), C = MP(\"C\",D((1,0))), A = MP(\"A\",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); D(A--B--(1.5,0));D(r);D(anglemark(C,B,D(A)));D(anglemark((1.5,0),C,C+.4*expi(beta*pi/180)));MP(\"\\beta\",B,(5,1.2),fontsize(9));MP(\"\\alpha\",C,(4,1.2),fontsize(9)); for(int i = 0; i < 180/alpha; ++i){ path l = B -- (1+i/2)*expi(-i * alpha * pi / 180); D(l, linetype(\"4 4\")); D(IP(l,r)); } D(B); [/asy] Note that after $k$ reflections (excluding the first one at $C$) the extended line will form an angle $k \\beta$ at point $B$. For the $k$th reflection to be just inside or at point $B$, we must have $k\\beta \\le 180 - 2\\alpha \\Longrightarrow k \\le \\frac{180 - 2\\alpha}{\\beta} = 70.27$. Thus, our answer is, including the first intersection, $\\left\\lfloor \\frac{180 - 2\\alpha}{\\beta} \\right\\rfloor + 1 = 071."
] |
1994-I-15
| 1,994 | 15 |
Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppose that $AB=36, AC=72,\,$ and $\angle B=90^\circ.\,$ Then the area of the set of all fold points of $\triangle ABC\,$ can be written in the form $q\pi-r\sqrt{s},\,$ where $q, r,\,$ and $s\,$ are positive integers and $s\,$ is not divisible by the square of any prime. What is $q+r+s\,$ ?
| 597 | null |
[
"Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\\overline{PA}$ and $\\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\\triangle PAB, PBC, PCA$. According to the problem statement, the circumcenters of the triangles cannot lie within the interior of the respective triangles, since they are not on the paper. It follows that $\\angle APB, \\angle BPC, \\angle CPA > 90^{\\circ}$; the locus of each of the respective conditions for $P$ is the region inside the (semi)circles with diameters $\\overline{AB}, \\overline{BC}, \\overline{CA}$. We note that the circle with diameter $AC$ covers the entire triangle because it is the circumcircle of $\\triangle ABC$, so it suffices to take the intersection of the circles about $AB, BC$. We note that their intersection lies entirely within $\\triangle ABC$ (the chord connecting the endpoints of the region is in fact the altitude of $\\triangle ABC$ from $B$). Thus, the area of the locus of $P$ (shaded region below) is simply the sum of two segments of the circles. If we construct the midpoints of $M_1, M_2 = \\overline{AB}, \\overline{BC}$ and note that $\\triangle M_1BM_2 \\sim \\triangle ABC$, we see that thse segments respectively cut a $120^{\\circ}$ arc in the circle with radius $18$ and $60^{\\circ}$ arc in the circle with radius $18\\sqrt{3}$. [asy] pair project(pair X, pair Y, real r){return X+r*(Y-X);} path endptproject(pair X, pair Y, real a, real b){return project(X,Y,a)--project(X,Y,b);} pathpen = linewidth(1); size(250); pen dots = linetype(\"2 3\") + linewidth(0.7), dashes = linetype(\"8 6\")+linewidth(0.7)+blue, bluedots = linetype(\"1 4\") + linewidth(0.7) + blue; pair B = (0,0), A=(36,0), C=(0,36*3^.5), P=D(MP(\"P\",(6,25), NE)), F = D(foot(B,A,C)); D(D(MP(\"A\",A)) -- D(MP(\"B\",B)) -- D(MP(\"C\",C,N)) -- cycle); fill(arc((A+B)/2,18,60,180) -- arc((B+C)/2,18*3^.5,-90,-30) -- cycle, rgb(0.8,0.8,0.8)); D(arc((A+B)/2,18,0,180),dots); D(arc((B+C)/2,18*3^.5,-90,90),dots); D(arc((A+C)/2,36,120,300),dots); D(B--F,dots); D(D((B+C)/2)--F--D((A+B)/2),dots); D(C--P--B,dashes);D(P--A,dashes); pair Fa = bisectorpoint(P,A), Fb = bisectorpoint(P,B), Fc = bisectorpoint(P,C); path La = endptproject((A+P)/2,Fa,20,-30), Lb = endptproject((B+P)/2,Fb,12,-35); D(La,bluedots);D(Lb,bluedots);D(endptproject((C+P)/2,Fc,18,-15),bluedots);D(IP(La,Lb),blue); [/asy] The diagram shows $P$ outside of the grayed locus; notice that the creases [the dotted blue] intersect within the triangle, which is against the problem conditions. The area of the locus is the sum of two segments of two circles; these segments cut out $120^{\\circ}, 60^{\\circ}$ angles by simple similarity relations and angle-chasing. Hence, the answer is, using the $\\frac 12 ab\\sin C$ definition of triangle area, $\\left[\\frac{\\pi}{3} \\cdot 18^2 - \\frac{1}{2} \\cdot 18^2 \\sin \\frac{2\\pi}{3} \\right] + \\left[\\frac{\\pi}{6} \\cdot \\left(18\\sqrt{3}\\right)^2 - \\frac{1}{2} \\cdot (18\\sqrt{3})^2 \\sin \\frac{\\pi}{3}\\right] = 270\\pi - 324\\sqrt{3}$, and $q+r+s = 597."
] |
1995-I-1
| 1,995 | 1 |
Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$ The total area enclosed by at least one of $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m-n.$ AIME 1995 Problem 1.png
| 255 | null |
[
"The sum of the areas of the squares if they were not interconnected is a geometric sequence: $1^2 + \\left(\\frac{1}{2}\\right)^2 + \\left(\\frac{1}{4}\\right)^2 + \\left(\\frac{1}{8}\\right)^2 + \\left(\\frac{1}{16}\\right)^2$ Then subtract the areas of the intersections, which is $\\left(\\frac{1}{4}\\right)^2 + \\ldots + \\left(\\frac{1}{32}\\right)^2$: $1^2 + \\left(\\frac{1}{2}\\right)^2 + \\left(\\frac{1}{4}\\right)^2 + \\left(\\frac{1}{8}\\right)^2 + \\left(\\frac{1}{16}\\right)^2 - \\left[\\left(\\frac{1}{4}\\right)^2 + \\left(\\frac{1}{8}\\right)^2 + \\left(\\frac{1}{16}\\right)^2 + \\left(\\frac{1}{32}\\right)^2\\right]$ $= 1 + \\frac{1}{2}^2 - \\frac{1}{32}^2$ The majority of the terms cancel, leaving $1 + \\frac{1}{4} - \\frac{1}{1024}$, which simplifies down to $\\frac{1024 + \\left(256 - 1\\right)}{1024}$. Thus, $m-n = 255, which when simplified will produce the same answer."
] |
1995-I-2
| 1,995 | 2 |
Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2.$
| 25 | null |
[
"Taking the $\\log_{1995}$ (logarithm) of both sides and then moving to one side yields the quadratic equation $2(\\log_{1995}x)^2 - 4(\\log_{1995}x) + 1 = 0$. Applying the quadratic formula yields that $\\log_{1995}x = 1 \\pm \\frac{\\sqrt{2}}{2}$. Thus, the product of the two roots (both of which are positive) is $1995^{1+\\sqrt{2}/2} \\cdot 1995^{1 - \\sqrt{2}/2} = 1995^2$, making the solution $(2000-5)^2 \\equiv 025.",
"Instead of taking $\\log_{1995}$, we take $\\log_x$ of both sides and simplify: $\\log_x(\\sqrt{1995}x^{\\log_{1995}x})=\\log_x(x^{2})$ $\\log_x\\sqrt{1995}+\\log_x x^{\\log_{1995}x}=2$ $\\dfrac{1}{2} \\log_x 1995 + \\log_{1995} x = 2$ We know that $\\log_x 1995$ and $\\log_{1995} x$ are reciprocals, so let $a=\\log_{1995} x$. Then we have $\\dfrac{1}{2}\\left(\\dfrac{1}{a}\\right) + a = 2$. Multiplying by $2a$ and simplifying gives us $2a^2-4a+1=0$, as shown above. Because $a=\\log_{1995} x$, $x=1995^a$. By the quadratic formula, the two roots of our equation are $a=\\frac{2\\pm\\sqrt2}{2}$. This means our two roots in terms of $x$ are $1995^\\frac{2+\\sqrt2}{2}$ and $1995^\\frac{2-\\sqrt2}{2}.$ Multiplying these gives $1995^2$ $1995^2\\pmod{1000}\\equiv 995^2\\pmod{1000}\\equiv (-5)^2\\pmod{1000}\\equiv 25\\pmod{1000}$, so our answer is $025.",
"Let $y=\\log_{1995}x$. Rewriting the equation in terms of $y$, we have \\[\\sqrt{1995}\\left(1995^y\\right)^y=1995^{2y}\\] \\[1995^{y^2+\\frac{1}{2}}=1995^{2y}\\] \\[y^2+\\frac{1}{2}=2y\\] \\[2y^2-4y+1=0\\] \\[y=\\frac{4\\pm\\sqrt{16-\\left(4\\right)\\left(2\\right)\\left(1\\right)}}{4}=\\frac{4\\pm\\sqrt{8}}{4}=\\frac{2\\pm\\sqrt{8}}{2}={1\\pm\\sqrt{2}}\\] Thus, the product of the positive roots is $\\left(1995^{\\frac{2+\\sqrt{8}}{2}}\\right)\\left(1995^{\\frac{2-\\sqrt{8}}{2}}\\right)=1995^2=\\left(2000-5\\right)^2$, so the last three digits are $025."
] |
1995-I-3
| 1,995 | 3 |
Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$
| 67 | null |
[
"It takes an even number of steps for the object to reach $(2,2)$, so the number of steps the object may have taken is either $4$ or $6$. If the object took $4$ steps, then it must have gone two steps N and two steps E, in some permutation. There are $\\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and the probability is $\\frac{6}{4^{4}}$. If the object took $6$ steps, then it must have gone two steps N and two steps E, and an additional pair of moves that would cancel out, either N/S or W/E. The sequences N,N,N,E,E,S can be permuted in $\\frac{6!}{3!2!1!} = 60$ ways. However, if the first four steps of the sequence are N,N,E,E in some permutation, it would have already reached the point $(2,2)$ in four moves. There are $\\frac{4!}{2!2!}$ ways to order those four steps and $2!$ ways to determine the order of the remaining two steps, for a total of $12$ sequences that we have to exclude. This gives $60-12=48$ sequences of steps. There are the same number of sequences for the steps N,N,E,E,E,W, so the probability here is $\\frac{2 \\times 48}{4^6}$. The total probability is $\\frac{6}{4^4} + \\frac{96}{4^6} = \\frac{3}{64}$, and $m+n= 067.",
"Let's let the object wander for 6 steps so we get a constant denominator of $4^{6}$ In the first case, we count how many ways the object can end at (2,2), at the end of 6 steps. We will also count it even if we go to (2,2), and go back to (2,2). So, there are 2 different paths for the object to end at (2,2): 1.To go a permutation of R,R,R,U,U,L or 2.To go a permutation of R,R,U,U,U,D. There are 60 ways to permute for each case, giving a total of 120 ways for the object to succeed and end at (2,2). In these 120 ways the object could reach (2,2) first and then come back to (2,2). This will be a factor in our second case. In the second case, the object can get to (2,2) first in 4 moves, then move away from (2,2) with the remaing 2 moves. So, there are 6 ways to get to (2,2) in 4 moves, then there are 16 ways the object can \"move around\", but 4 of the ways will return the object back to (2,2). Those 4 ways were already counted in the first case, so we should only count 12 of the 16 ways to prevent over-counting. Thus, there are $12*6 =$ 72 ways in the second case. So, in all, there are 120+72 ways for the object to achieve it's goal of moving to (2,2). Put that over our denominator, we get $\\frac{192}{4^{6}} = \\frac{3}{4^{3}} = \\frac{3}{64}$, in which adding the numerator and denominator get us an answer of $067 - AlexLikeMath"
] |
1995-I-4
| 1,995 | 4 |
Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$ . The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
| 224 | null |
[
"We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\\overline{O_3A_3} \\parallel \\overline{O_6A_6} \\parallel \\overline{O_9A_9}$, and $O_6O_9 : O_9O_3 = 3:6 = 1:2$. Thus, $O_9A_9 = \\frac{2 \\cdot O_6A_6 + 1 \\cdot O_3A_3}{3} = 5$ (consider similar triangles). Applying the Pythagorean Theorem to $\\triangle O_9A_9P$, we find that \\[PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = 224\\] [asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP(\"O_9\",A)),9)); D(CR(D(MP(\"O_3\",B)),3)); D(CR(D(MP(\"O_6\",C)),6)); D(MP(\"P\",P,NW)--MP(\"Q\",Q,NE)); D((-9,0)--(9,0)); D(A--MP(\"A_9\",G,N)); D(B--MP(\"A_3\",F,N)); D(C--MP(\"A_6\",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); [/asy]",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP(\"E\",A)),9)); D(CR(D(MP(\"F\",B)),3)); D(CR(D(MP(\"D\",C)),6)); D((-9,0)--(9,0)); D(MP(\"\",P,NW)--MP(\"\",Q,NE)); D(A--MP(\"B\",G,N)); D(B--MP(\"C\",F,N)); D(C--MP(\"A\",D,N)); D(rightanglemark(A,G,P,12)); D(rightanglemark(C,D,P,12)); D(rightanglemark(B,F,P,12)); [/asy] Let $A$ be defined as the origin of a coordinate plane with the $y$-axis running across the chord and $C(6\\sqrt{2},0)$ by the Pythagorean Theorem. Then we have $D(0,-6)$ and $F(6\\sqrt{2},-3)$, and since $\\frac{DE}{DF}=\\frac{1}{3}$, the point $E$ is one-third of the way from $D$ to $F$, so point $E$ has coordinates $(2\\sqrt{2},-5)$. $E$ is the center of the circle with radius $9$, so the equation of this circle is $(x-2\\sqrt{2})^2+(y+5)^2=81$. Since the chord's equation is $y=0$, we must find all values of $x$ satisfying the equation of the circle such that $y=0$. We find that $x-2\\sqrt{2}=\\pm\\sqrt{56}$, so the chord has length $|\\sqrt{56}+2\\sqrt{2}-(-\\sqrt{56}+2\\sqrt{2})|=2\\sqrt{56}$ and the answer is $(2\\sqrt{56})^2=224. ~eevee9406",
"[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP(\"E\",A)),9)); D(CR(D(MP(\"F\",B)),3)); D(CR(D(MP(\"D\",C)),6)); D((-9,0)--(9,0)); D(MP(\"\",P,NW)--MP(\"\",Q,NE)); D(A--MP(\"B\",G,N)); D(B--MP(\"C\",F,N)); D(C--MP(\"A\",D,N)); D(rightanglemark(A,G,P,12)); D(rightanglemark(C,D,P,12)); D(rightanglemark(B,F,P,12)); [/asy] Let $A$ be defined as the origin of a coordinate plane with the $y$-axis running across the chord and $C(6\\sqrt{2},0)$ by the Pythagorean Theorem. Then we have $D(0,-6)$ and $F(6\\sqrt{2},-3)$, and since $\\frac{DE}{DF}=\\frac{1}{3}$, the point $E$ is one-third of the way from $D$ to $F$, so point $E$ has coordinates $(2\\sqrt{2},-5)$. $E$ is the center of the circle with radius $9$, so the equation of this circle is $(x-2\\sqrt{2})^2+(y+5)^2=81$. Since the chord's equation is $y=0$, we must find all values of $x$ satisfying the equation of the circle such that $y=0$. We find that $x-2\\sqrt{2}=\\pm\\sqrt{56}$, so the chord has length $|\\sqrt{56}+2\\sqrt{2}-(-\\sqrt{56}+2\\sqrt{2})|=2\\sqrt{56}$ and the answer is $(2\\sqrt{56})^2=224. ~eevee9406"
] |
1995-I-5
| 1,995 | 5 |
For certain real values of $a, b, c,$ and $d_{},$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots. The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$ Find $b.$
| 51 | null |
[
"Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be $m,n$. Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$. Let $m'$ be the conjugate of $m$, and $n'$ be the conjugate of $n$. Then, \\[m\\cdot n = 13 + i,m' + n' = 3 + 4i\\Longrightarrow m'\\cdot n' = 13 - i,m + n = 3 - 4i.\\] By Vieta's formulas, we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = 051.",
"Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get \\[b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\\] \\[(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\\] \\[(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2\\] We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of $13+i$, and the two roots that are added give $3+4i$. This gets three equations necessary for solving the problem. \\[p+r = 3\\] \\[pr-qs = 13\\] \\[-q-s = 4\\] So, alright. Let's use the first equation to get that $(p+r)^2 = 9$, and substitute that in. Now, the equation becomes: \\[b = 9 + 2pr + q^2 + s^2\\] We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be $pr = 13+qs$, and substitute that in. \\[b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2\\] We can square both sides of the third equation, and get $(q+s)^2 = 16$ We substitute that in and we get \\[b = 35+16 = 051\\] - AlexLikeMath - Corrections by VSPuzzler - small correction be Marshall_Huang"
] |
1995-I-6
| 1,995 | 6 |
Let $n=2^{31}3^{19}.$ How many positive integer divisors of $n^2$ are less than $n_{}$ but do not divide $n_{}$ ?
| 589 | null |
[
"We know that $n^2 = 2^{62}3^{38}$ must have $(62+1)\\times (38+1)$ factors by its prime factorization. If we group all of these factors (excluding $n$) into pairs that multiply to $n^2$, then one factor per pair is less than $n$, and so there are $\\frac{63\\times 39-1}{2} = 1228$ factors of $n^2$ that are less than $n$. There are $32\\times20-1 = 639$ factors of $n$, which clearly are less than $n$, but are still factors of $n$. Therefore, using complementary counting, there are $1228-639=589.",
"Let $n=p_1^{k_1}p_2^{k_2}$ for some prime $p_1,p_2$. Then $n^2$ has $\\frac{(2k_1+1)(2k_2+1)-1}{2}$ factors less than $n$. This simplifies to $\\frac{4k_1k_2+2k_1+2k_2}{2}=2k_1k_2+k_1+k_2$. The number of factors of $n$ less than $n$ is equal to $(k_1+1)(k_2+1)-1=k_1k_2+k_1+k_2$. Thus, our general formula for $n=p_1^{k_1}p_2^{k_2}$ is Number of factors that satisfy the above $=(2k_1k_2+k_1+k_2)-(k_1k_2+k_1+k_2)=k_1k_2$ Incorporating this into our problem gives $19\\times31=589.",
"Consider divisors of $n^2: a,b$ such that $ab=n^2$. WLOG, let $b\\ge{a}$ and $b=\\frac{n}{a}$ Then, it is easy to see that $a$ will always be less than $b$ as we go down the divisor list of $n^2$ until we hit $n$. Therefore, the median divisor of $n^2$ is $n$. Then, there are $(63)(39)=2457$ divisors of $n^2$. Exactly $\\frac{2457-1}{2}=1228$ of these divisors are $<n$ There are $(32)(20)-1=639$ divisors of $n$ that are $<n$. Therefore, the answer is $1228-639=589."
] |
1995-I-8
| 1,995 | 8 |
For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?
| 85 | null |
[
"Since $y|x$, $y+1|x+1$, then $\\text{gcd}\\,(y,x)=y$ (the bars indicate divisibility) and $\\text{gcd}\\,(y+1,x+1)=y+1$. By the Euclidean algorithm, these can be rewritten respectively as $\\text{gcd}\\,(y,x-y)=y$ and $\\text{gcd}\\, (y+1,x-y)=y+1$, which implies that both $y,y+1 | x-y$. Also, as $\\text{gcd}\\,(y,y+1) = 1$, it follows that $y(y+1)|x-y$. [1] Thus, for a given value of $y$, we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \\le 100$). It follows that there are $\\left\\lfloor\\frac{100-y}{y(y+1)} \\right\\rfloor$ satisfactory positive integers for all integers $y \\le 100$. The answer is \\[\\sum_{y=1}^{99} \\left\\lfloor\\frac{100-y}{y(y+1)} \\right\\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = 085 must also be an integer.",
"We know that $x \\equiv 0 \\mod y$ and $x+1 \\equiv 0 \\mod y+1$. Write $x$ as $ky$ for some integer $k$. Then, $ky+1 \\equiv 0\\mod y+1$. We can add $k$ to each side in order to factor out a $y+1$. So, $ky+k+1 \\equiv k \\mod y+1$ or $k(y+1)+1 \\equiv k \\mod y+1$. We know that $k(y+1) \\equiv 0 \\mod y+1$. We finally achieve the congruence $1-k \\equiv 0 \\mod y+1$. We can now write $k$ as $(y+1)a+1$. Plugging this back in, if we have a value for $y$, then $x = ky = ((y+1)a+1)y = y(y+1)a+y$. We only have to check values of $y$ when $y(y+1)<100$. This yields the equations $x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9$. Finding all possible values of $a$ such that $y<x<100$, we get $49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = 085",
"We use casework: For $y=1$, we have $3,5,\\cdots ,99$, or $49$ cases. For $y=2$, we have $8,14,\\cdots ,98$, or $16$ cases. For $y=3$, we have $15,27,\\cdots ,99$, or $8$ cases. For $y=4$, we have $24,44\\cdots ,84$, or $4$ cases. For $y=5$, we have $35,65,95$, or $3$ cases. For $y=6$, we have $48,90$, or $2$ cases. For $y=7$, we have $63$, or $1$ case. For $y=8$, we have $80$, or $1$ case. For $y=9$, we have $99$, or $1$ case. Adding, we get our final result of $085. ~Yiyj1",
"We use casework: For $y=1$, we have $3,5,\\cdots ,99$, or $49$ cases. For $y=2$, we have $8,14,\\cdots ,98$, or $16$ cases. For $y=3$, we have $15,27,\\cdots ,99$, or $8$ cases. For $y=4$, we have $24,44\\cdots ,84$, or $4$ cases. For $y=5$, we have $35,65,95$, or $3$ cases. For $y=6$, we have $48,90$, or $2$ cases. For $y=7$, we have $63$, or $1$ case. For $y=8$, we have $80$, or $1$ case. For $y=9$, we have $99$, or $1$ case. Adding, we get our final result of $085. ~Yiyj1"
] |
1995-I-9
| 1,995 | 9 |
Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$ AIME 1995 Problem 9.png
| 616 | null |
[
"Let $x=\\angle CAM$, so $3x=\\angle CDM$. Then, $\\frac{\\tan 3x}{\\tan x}=\\frac{CM/1}{CM/11}=11$. Expanding $\\tan 3x$ using the angle sum identity gives \\[\\tan 3x=\\tan(2x+x)=\\frac{3\\tan x-\\tan^3x}{1-3\\tan^2x}.\\] Thus, $\\frac{3-\\tan^2x}{1-3\\tan^2x}=11$. Solving, we get $\\tan x= \\frac 12$. Hence, $CM=\\frac{11}2$ and $AC= \\frac{11\\sqrt{5}}2$ by the Pythagorean Theorem. The total perimeter is $2(AC + CM) = \\sqrt{605}+11$. The answer is thus $a+b=616.",
"In a similar fashion, we encode the angles as complex numbers, so if $BM=x$, then $\\angle BAD=\\text{Arg}(11+xi)$ and $\\angle BDM=\\text{Arg}(1+xi)$. So we need only find $x$ such that $\\text{Arg}((11+xi)^3)=\\text{Arg}(1331-33x^2+(363x-x^3)i)=\\text{Arg}(1+xi)$. This will happen when $\\frac{363x-x^3}{1331-33x^2}=x$, which simplifies to $121x-4x^3=0$. Therefore, $x=\\frac{11}{2}$. By the Pythagorean Theorem, $AB=\\frac{11\\sqrt{5}}{2}$, so the perimeter is $11+11\\sqrt{5}=11+\\sqrt{605}$, giving us our answer, $616.",
"Let $\\angle BAD=\\alpha$, so $\\angle BDM=3\\alpha$, $\\angle BDA=180-3\\alpha$, and thus $\\angle ABD=2\\alpha.$ We can then draw the angle bisector of $\\angle ABD$, and let it intersect $\\overline{AM}$ at $E.$ Since $\\angle BAE=\\angle ABE$, $AE=BE.$ Let $AE=x$. Then we see by the Pythagorean Theorem, $BM=\\sqrt{BE^2-ME^2}=\\sqrt{x^2-(11-x)^2}=\\sqrt{22x-121}$, $BD=\\sqrt{BM^2+1}=\\sqrt{22x-120}$, $BA=\\sqrt{BM^2+121}=\\sqrt{22x}$, and $DE=10-x.$ By the angle bisector theorem, $BA/BD=EA/ED.$ Substituting in what we know for the lengths of those segments, we see that \\[\\frac{\\sqrt{22x}}{\\sqrt{22x-120}}=\\frac{x}{10-x}.\\] multiplying by both denominators and squaring both sides yields \\[22x(10-x)^2=x^2(22x-120)\\] which simplifies to $x=\\frac{55}{8}.$ Substituting this in for x in the equations for $BA$ and $BM$ yields $BA=\\frac{\\sqrt{605}}{2}$ and $BM=\\frac{11}{2}.$ Thus the perimeter is $11+\\sqrt{605}$, and the answer is $616.",
"The triangle is symmetrical so we can split it in half ($\\triangle ABM$ and $\\triangle ACM$). Let $\\angle BAM = y$ and $\\angle BDM = 3y$. By the Law of Sines on triangle $BAD$, $\\frac{10}{\\sin 2y} = \\frac{BD}{\\sin y}$. Using $\\sin 2y = 2\\sin y\\cos y$ we can get $BD = \\frac{5}{\\cos y}$. We can use this information to relate $BD$ to $DM$ by using the Law of Sines on triangle $BMD$. \\[\\frac{\\frac{5}{\\cos y}}{\\sin BMD} = \\frac{1}{\\sin 90^\\circ - 3y}\\] $\\sin BMD = 1$ (as $\\angle BMD$ is a right angle), so $\\frac{1}{\\sin 90^\\circ - 3y} = \\frac{5}{\\cos y}$. Using the identity $\\sin 90^\\circ - x = \\cos x$, we can turn the equation into:: \\[\\frac{1}{\\cos 3y} = \\frac{5}{\\cos y}\\] \\[5\\cos 3y = \\cos y\\] \\[5(4\\cos ^3 y - 3\\cos y) = \\cos y\\] \\[20\\cos ^3 y = 16 \\cos y\\] \\[5\\cos ^3 y = 4\\cos y\\] \\[5\\cos ^2 y = 4\\] \\[\\cos ^2 y = \\frac{4}{5}\\] Now that we've found $\\cos y$, we can look at the side lengths of $BM$ and $AB$ (since they are symmetrical, the perimeter of $\\triangle ABC$ is $2(BM + AB)$. We note that $BM = 11\\tan y$ and $AB = 11\\sec y$. \\[\\sin ^2 y = 1 - \\cos ^2 y\\] \\[\\sin ^2 y = \\frac{1}{5}\\] \\[\\tan ^2 y = \\frac{1}{4}\\] \\[\\tan y = \\frac{1}{2}\\] (Note it is positive since $BM > 0$). \\[\\sec ^2 y = \\frac{5}{4}\\] \\[\\sec y = \\frac{\\sqrt{5}}{2}\\] \\[BM + AB = 11\\frac{\\sqrt{5}+1}{2}\\] \\[2(BM + AB) = 11(\\sqrt{5} + 1)\\] \\[2(BM + AB) = 11\\sqrt{5} + 11\\] \\[2(BM + AB) = \\sqrt{605} + 11\\] The answer is $616.",
"Suppose $\\angle BAM=\\angle CAM =x$, since $\\angle BDC=3\\angle BAC$, we have $\\angle BDM=\\angle MDC = 3x$. Therefore, $\\angle DBC=\\angle DCB = 90^\\circ -3x$ and $\\angle ABD=\\angle DCA=2x$. As a result, $\\triangle KAC$ is isosceles, $KC=KA$. Let $H$ be a point on the extension of $CD$ through $D$ such that $\\overline{HB}\\perp\\overline{BC}$ and denote the intersection of $\\overline{HC}$ and $\\overline{AB}$ as $K$. Then, $BH=2DM=2, \\overline{HB}\\parallel\\overline{DM}$, and $HD=DC$ by the Midpoint Theorem. So, $\\angle HBA=x$ and $\\angle CDM=\\angle CHB=\\angle HDA= 3x$. Consequently, $\\triangle HBK\\sim \\triangle DAK$, \\[\\frac{BK}{KA}=\\frac{HK}{KD}=\\frac{1}{5}\\] Assume $BK=a$ and $HK=b$, then $KA=5a$ and $KD = 5b$. Since $KC=KA, KC=5a$, and since $HD=DC$, $KC=11b$. Therefore, $a=\\frac{11}{5}b$. In $\\triangle BDM$, by the Pythagorean Theorem, $BM=\\sqrt{36b^2-1}$. Similarly in $\\triangle BAM$, $BM=\\sqrt{36a^2-121}$. So \\[\\sqrt{36a^2-121}=\\sqrt{36b^2-1}\\] Since $a=\\frac{11}{5}b$, we have $b=\\frac{5\\sqrt{5}}{12}$ and $a=\\frac{11\\sqrt{5}}{12}$. Consequently, $BM=\\frac{11}{2}$ and $AB=\\frac{11\\sqrt{5}}{2}$. Thus, the perimeter of $\\triangle ABC$ is $11+\\sqrt{605}$, and the answer is $616."
] |
1995-I-10
| 1,995 | 10 |
What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?
| 215 | null |
[
"The requested number $\\mod {42}$ must be a prime number. Also, every number that is a multiple of $42$ greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to $42$ and the numbers that are multiples of $42$ greater than them, until they reach a composite number. \\[\\begin{tabular}{|r||r|r|r|r|r|} \\hline 1 & 43 & 85&&& \\\\ 2&44&&&& \\\\ 3&45&&&& \\\\ 5&47&89&131&173&215 \\\\ 7&49&&&& \\\\ 11&53&95&&& \\\\ 13&55&&&& \\\\ 17&59&101&143&& \\\\ 19&61&103&145&& \\\\ 23&65&&&& \\\\ 29&71&113&155&& \\\\ 31&73&115&&& \\\\ 37&79&121&&& \\\\ 41&83&125&&& \\\\ \\hline \\end{tabular}\\] Since $215. -jackshi2006",
"Let our answer be $n$. Write $n = 42a + b$, where $a, b$ are positive integers and $0 \\leq b < 42$. Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes. If $b$ is $0\\mod{5}$, then $b = 5$ because $5$ is the only prime divisible by $5$. We get $n = 215$ as our largest possibility in this case. If $b$ is $1\\mod{5}$, then $b + 2 \\times 42$ is divisible by $5$ and thus $a \\leq 2$. Thus, $n \\leq 3 \\times 42 = 126 < 215$. If $b$ is $2\\mod{5}$, then $b + 4 \\times 42$ is divisible by $5$ and thus $a \\leq 4$. Thus, $n \\leq 5 \\times 42 = 210 < 215$. If $b$ is $3\\mod{5}$, then $b + 1 \\times 42$ is divisible by $5$ and thus $a = 1$. Thus, $n \\leq 2 \\times 42 = 84 < 215$. If $b$ is $4\\mod{5}$, then $b + 3 \\times 42$ is divisible by $5$ and thus $a \\leq 3$. Thus, $n \\leq 4 \\times 42 = 168 < 215$. Our answer is $215."
] |
1995-I-11
| 1,995 | 11 |
A right rectangular prism $P_{}$ (i.e., a rectangular parallelepiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P_{}$ cuts $P_{}$ into two prisms, one of which is similar to $P_{},$ and both of which have nonzero volume. Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist?
| 40 | null |
[
"Let $P'$ be the prism similar to $P$, and let the sides of $P'$ be of length $x,y,z$, such that $x \\le y \\le z$. Then \\[\\frac{x}{a} = \\frac{y}{b} = \\frac zc < 1.\\] Note that if the ratio of similarity was equal to $1$, we would have a prism with zero volume. As one face of $P'$ is a face of $P$, it follows that $P$ and $P'$ share at least two side lengths in common. Since $x < a, y < b, z < c$, it follows that the only possibility is $y=a,z=b=1995$. Then, \\[\\frac{x}{a} = \\frac{a}{1995} = \\frac{1995}{c} \\Longrightarrow ac = 1995^2 = 3^25^27^219^2.\\] The number of factors of $3^25^27^219^2$ is $(2+1)(2+1)(2+1)(2+1) = 81$. Only in $\\left\\lfloor \\frac {81}2 \\right\\rfloor = 40$ of these cases is $a < c$ (for $a=c$, we end with a prism of zero volume). We can easily verify that these will yield nondegenerate prisms, so the answer is $040."
] |
1995-I-12
| 1,995 | 12 |
Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m_{}$ and $n_{}$ are integers, find $m+n.$
| 5 | null |
[
"Solution 1 (trigonometry) [asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, NW); label(\"$D$\", D, W); label(\"$O$\", O, N); dot(\"$P$\", P, NE); [/asy] The angle $\\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\\overline{OB}$ meet $\\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\\triangle OPA$ is a $45-45-90$ right triangle, so $OP = AP = 1,$ $OB = OA = \\sqrt {2},$ and $AB = \\sqrt {4 - 2\\sqrt {2}}.$ Therefore, $AC = \\sqrt {8 - 4\\sqrt {2}}.$ From the Law of Cosines, $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\\cos \\theta,$ so \\[8 - 4\\sqrt {2} = 1 + 1 - 2\\cos \\theta \\Longrightarrow \\cos \\theta = - 3 + 2\\sqrt {2} = - 3 + \\sqrt{8}.\\] Thus $m + n = 005 as well. ~RubixMaster21",
"[asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, NW); label(\"$D$\", D, W); label(\"$O$\", O, N); dot(\"$P$\", P, NE); [/asy] The angle $\\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\\overline{OB}$ meet $\\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\\triangle OPA$ is a $45-45-90$ right triangle, so $OP = AP = 1,$ $OB = OA = \\sqrt {2},$ and $AB = \\sqrt {4 - 2\\sqrt {2}}.$ Therefore, $AC = \\sqrt {8 - 4\\sqrt {2}}.$ From the Law of Cosines, $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\\cos \\theta,$ so \\[8 - 4\\sqrt {2} = 1 + 1 - 2\\cos \\theta \\Longrightarrow \\cos \\theta = - 3 + 2\\sqrt {2} = - 3 + \\sqrt{8}.\\] Thus $m + n = 005 as well. ~RubixMaster21",
"[asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, NW); label(\"$D$\", D, W); label(\"$O$\", O, N); dot(\"$P$\", P, NE); [/asy] The angle $\\theta$ is the angle formed by two perpendiculars drawn to $BO$, one on the plane determined by $OAB$ and the other by $OBC$. Let the perpendiculars from $A$ and $C$ to $\\overline{OB}$ meet $\\overline{OB}$ at $P.$ Without loss of generality, let $AP = 1.$ It follows that $\\triangle OPA$ is a $45-45-90$ right triangle, so $OP = AP = 1,$ $OB = OA = \\sqrt {2},$ and $AB = \\sqrt {4 - 2\\sqrt {2}}.$ Therefore, $AC = \\sqrt {8 - 4\\sqrt {2}}.$ From the Law of Cosines, $AC^{2} = AP^{2} + PC^{2} - 2(AP)(PC)\\cos \\theta,$ so \\[8 - 4\\sqrt {2} = 1 + 1 - 2\\cos \\theta \\Longrightarrow \\cos \\theta = - 3 + 2\\sqrt {2} = - 3 + \\sqrt{8}.\\] Thus $m + n = 005 as well. ~RubixMaster21",
"Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown. We first find $z.$ Note that \\[\\overrightarrow{OA}\\cdot \\overrightarrow{OB} = \\parallel \\overrightarrow{OA}\\parallel \\parallel \\overrightarrow{OB}\\parallel \\cos 45^\\circ.\\] Since $\\overrightarrow{OA} =\\, <1,0, - z>$ and $\\overrightarrow{OB} =\\, <0,1, - z> ,$ this simplifies to \\[z^{2}\\sqrt {2} = 1 + z^{2}\\implies z^{2} = 1 + \\sqrt {2}.\\] Now let's find $\\cos \\theta.$ Let $\\vec{u}$ and $\\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\\vec{u}\\cdot \\vec{v} = \\parallel \\vec{u}\\parallel \\parallel \\vec{v}\\parallel \\cos \\theta$, we will be able to solve for $\\cos \\theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$, and then to find their normal vectors) \\[\\vec{u} = \\overrightarrow{OA}\\times \\overrightarrow{OB} = \\left| \\begin{array}{ccc}\\hat{i} & \\hat{j} & \\hat{k} \\\\ 1 & 0 & - z \\\\ 0 & 1 & - z \\end{array}\\right| =\\, < z,z,1 > .\\] Similarly, \\[\\vec{v} = \\overrightarrow{OB}\\times \\overrightarrow{OC} - \\left|\\begin{array}{ccc}\\hat{i} & \\hat{j} & \\hat{k} \\\\ 0 & 1 & - z \\\\ - 1 & 0 & - z \\end{array}\\right| =\\, < - z,z,1 > .\\] Hence, taking the dot product of $\\vec{u}$ and $\\vec{v}$ yields \\[\\cos \\theta = \\frac{ \\vec{u} \\cdot \\vec{v} }{ \\parallel \\vec{u} \\parallel \\parallel \\vec{v} \\parallel } = \\frac{- z^{2} + z^{2} + 1}{(\\sqrt {1 + 2z^{2}})^{2}} = \\frac {1}{3 + 2\\sqrt {2}} = 3 - 2\\sqrt {2} = 3 - \\sqrt {8}.\\] Flipping the signs (we found the cosine of the supplement angle) yields $\\cos \\theta = - 3 + \\sqrt {8},$ so the answer is $005 as well. ~RubixMaster21",
"Without loss of generality, place the pyramid in a 3-dimensional coordinate system such that $A = (1,0,0),$ $B = (0,1,0),$ $C = ( - 1,0,0),$ $D = (0, - 1,0),$ and $O = (0,0,z),$ where $z$ is unknown. We first find $z.$ Note that \\[\\overrightarrow{OA}\\cdot \\overrightarrow{OB} = \\parallel \\overrightarrow{OA}\\parallel \\parallel \\overrightarrow{OB}\\parallel \\cos 45^\\circ.\\] Since $\\overrightarrow{OA} =\\, <1,0, - z>$ and $\\overrightarrow{OB} =\\, <0,1, - z> ,$ this simplifies to \\[z^{2}\\sqrt {2} = 1 + z^{2}\\implies z^{2} = 1 + \\sqrt {2}.\\] Now let's find $\\cos \\theta.$ Let $\\vec{u}$ and $\\vec{v}$ be normal vectors to the planes containing faces $OAB$ and $OBC,$ respectively. From the definition of the dot product as $\\vec{u}\\cdot \\vec{v} = \\parallel \\vec{u}\\parallel \\parallel \\vec{v}\\parallel \\cos \\theta$, we will be able to solve for $\\cos \\theta.$ A cross product yields (alternatively, it is simple to find the equation of the planes $OAB$ and $OAC$, and then to find their normal vectors) \\[\\vec{u} = \\overrightarrow{OA}\\times \\overrightarrow{OB} = \\left| \\begin{array}{ccc}\\hat{i} & \\hat{j} & \\hat{k} \\\\ 1 & 0 & - z \\\\ 0 & 1 & - z \\end{array}\\right| =\\, < z,z,1 > .\\] Similarly, \\[\\vec{v} = \\overrightarrow{OB}\\times \\overrightarrow{OC} - \\left|\\begin{array}{ccc}\\hat{i} & \\hat{j} & \\hat{k} \\\\ 0 & 1 & - z \\\\ - 1 & 0 & - z \\end{array}\\right| =\\, < - z,z,1 > .\\] Hence, taking the dot product of $\\vec{u}$ and $\\vec{v}$ yields \\[\\cos \\theta = \\frac{ \\vec{u} \\cdot \\vec{v} }{ \\parallel \\vec{u} \\parallel \\parallel \\vec{v} \\parallel } = \\frac{- z^{2} + z^{2} + 1}{(\\sqrt {1 + 2z^{2}})^{2}} = \\frac {1}{3 + 2\\sqrt {2}} = 3 - 2\\sqrt {2} = 3 - \\sqrt {8}.\\] Flipping the signs (we found the cosine of the supplement angle) yields $\\cos \\theta = - 3 + \\sqrt {8},$ so the answer is $005 as well. ~RubixMaster21",
"[asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, NW); label(\"$D$\", D, W); label(\"$O$\", O, N); dot(\"$P$\", P, NE); [/asy] Similar to Solution 1, $\\angle APC$ is the dihedral angle we want. WLOG, we will let $AB=1,$ meaning $AC=\\sqrt{2}$. Because $\\triangle OAB,\\triangle OBC$ are isosceles, $\\angle ABP = 67.5^{\\circ}$ $PC=PA=\\cos(\\angle PAB)=\\cos(22.5^{\\circ})$. Thus by the half-angle identity, \\[PA=\\cos\\left(\\frac{45}{2}\\right) = \\sqrt{\\frac{1+\\cos(45^{\\circ})}{2}}\\] \\[= \\sqrt{\\frac{2+\\sqrt{2}}{4}}.\\] Now looking at triangle $\\triangle PAC,$ we drop the perpendicular from $P$ to $AC$, and call the foot $H$. Then $\\angle CPH = \\theta / 2.$ By Pythagoreas, \\[PH=\\sqrt{\\frac{2+\\sqrt{2}}{4}-\\frac{1}{2}}=\\frac{\\sqrt[4]{2}}{2}.\\] [asy] // if you see this // hello // gap for label on P--H: https://tex.stackexchange.com/questions/475945/asymptote-how-do-i-make-a-gap-in-a-segment-to-include-a-label pair P,C,A,H; H = (0, 0); C = (-0.71, 0); A = (0.71, 0); P = (0,0.59); draw(P--C--A--cycle); draw(P--H); label(\"$A$\", A, SE); label(\"$C$\", C, SW); label(\"$P$\", P, N); label(\"$H$\", H, S); label(\"$\\sqrt{\\frac{2+\\sqrt{2}}{4}}$\",align=NE,point(P--A,0.5)); label(\"$\\sqrt{\\frac{2+\\sqrt{2}}{4}}$\",align=NW,point(P--C,0.5)); label(\"$\\frac{\\sqrt{2}}{2}$\",align=S,point(C--H,0.5)); label(\"$\\frac{\\sqrt{2}}{2}$\",align=S,point(A--H,0.5)); pen fillpen = white; Label mylabel = Label(\"$\\frac{\\sqrt[4]{2}}{2}$\", align=(0,0), position=MidPoint, filltype=Fill(fillpen)); draw(P--H, L=mylabel); [/asy] We have that \\[\\cos\\left(\\frac{\\theta}{2}\\right)=\\frac{\\sqrt[4]{2}}{\\sqrt{2+\\sqrt{2}}},\\text{ so}\\] \\[\\cos(\\theta)=2\\cos^{2}\\left(\\frac{\\theta}{2}\\right)-1\\] \\[=2\\left(\\frac{\\sqrt{2}}{2+\\sqrt{2}}\\right)-1\\] \\[=2(\\frac{2\\sqrt{2}-2}{2})-1\\] \\[=-3+\\sqrt{8}.\\] Because $m$ and $n$ can be negative integers, our answer is $(-3)+8=005. as well. ~RubixMaster21",
"[asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); } currentprojection=perspective(2,1,1); triple A, B, C, D, O, P; A = (sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); B = (-sqrt(2 - sqrt(2)), sqrt(2 - sqrt(2)), 0); C = (-sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); D = (sqrt(2 - sqrt(2)), -sqrt(2 - sqrt(2)), 0); O = (0,0,sqrt(2*sqrt(2))); P = projectionofpointontoline(A,O,B); draw(D--A--B); draw(B--C--D,dashed); draw(A--O); draw(B--O); draw(C--O,dashed); draw(D--O); draw(A--P); draw(P--C,dashed); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, NW); label(\"$D$\", D, W); label(\"$O$\", O, N); dot(\"$P$\", P, NE); [/asy] Similar to Solution 1, $\\angle APC$ is the dihedral angle we want. WLOG, we will let $AB=1,$ meaning $AC=\\sqrt{2}$. Because $\\triangle OAB,\\triangle OBC$ are isosceles, $\\angle ABP = 67.5^{\\circ}$ $PC=PA=\\cos(\\angle PAB)=\\cos(22.5^{\\circ})$. Thus by the half-angle identity, \\[PA=\\cos\\left(\\frac{45}{2}\\right) = \\sqrt{\\frac{1+\\cos(45^{\\circ})}{2}}\\] \\[= \\sqrt{\\frac{2+\\sqrt{2}}{4}}.\\] Now looking at triangle $\\triangle PAC,$ we drop the perpendicular from $P$ to $AC$, and call the foot $H$. Then $\\angle CPH = \\theta / 2.$ By Pythagoreas, \\[PH=\\sqrt{\\frac{2+\\sqrt{2}}{4}-\\frac{1}{2}}=\\frac{\\sqrt[4]{2}}{2}.\\] [asy] // if you see this // hello // gap for label on P--H: https://tex.stackexchange.com/questions/475945/asymptote-how-do-i-make-a-gap-in-a-segment-to-include-a-label pair P,C,A,H; H = (0, 0); C = (-0.71, 0); A = (0.71, 0); P = (0,0.59); draw(P--C--A--cycle); draw(P--H); label(\"$A$\", A, SE); label(\"$C$\", C, SW); label(\"$P$\", P, N); label(\"$H$\", H, S); label(\"$\\sqrt{\\frac{2+\\sqrt{2}}{4}}$\",align=NE,point(P--A,0.5)); label(\"$\\sqrt{\\frac{2+\\sqrt{2}}{4}}$\",align=NW,point(P--C,0.5)); label(\"$\\frac{\\sqrt{2}}{2}$\",align=S,point(C--H,0.5)); label(\"$\\frac{\\sqrt{2}}{2}$\",align=S,point(A--H,0.5)); pen fillpen = white; Label mylabel = Label(\"$\\frac{\\sqrt[4]{2}}{2}$\", align=(0,0), position=MidPoint, filltype=Fill(fillpen)); draw(P--H, L=mylabel); [/asy] We have that \\[\\cos\\left(\\frac{\\theta}{2}\\right)=\\frac{\\sqrt[4]{2}}{\\sqrt{2+\\sqrt{2}}},\\text{ so}\\] \\[\\cos(\\theta)=2\\cos^{2}\\left(\\frac{\\theta}{2}\\right)-1\\] \\[=2\\left(\\frac{\\sqrt{2}}{2+\\sqrt{2}}\\right)-1\\] \\[=2(\\frac{2\\sqrt{2}-2}{2})-1\\] \\[=-3+\\sqrt{8}.\\] Because $m$ and $n$ can be negative integers, our answer is $(-3)+8=005. as well. ~RubixMaster21"
] |
1995-I-13
| 1,995 | 13 |
Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$
| 400 | null |
[
"When $\\left(k - \\frac {1}{2}\\right)^4 \\leq n < \\left(k + \\frac {1}{2}\\right)^4$, $f(n) = k$. Thus there are $\\left \\lfloor \\left(k + \\frac {1}{2}\\right)^4 - \\left(k - \\frac {1}{2}\\right)^4 \\right\\rfloor$ values of $n$ for which $f(n) = k$. Expanding using the binomial theorem, \\begin{align*} \\left(k + \\frac {1}{2}\\right)^4 - \\left(k - \\frac {1}{2}\\right)^4 &= \\left(k^4 + 2k^3 + \\frac 32k^2 + \\frac 12k + \\frac 1{16}\\right) - \\left(k^4 - 2k^3 + \\frac 32k^2 - \\frac 12k + \\frac 1{16}\\right)\\\\ &= 4k^3 + k. \\end{align*} Thus, $\\frac{1}{k}$ appears in the summation $4k^3 + k$ times, and the sum for each $k$ is then $(4k^3 + k) \\cdot \\frac{1}{k} = 4k^2 + 1$. From $k = 1$ to $k = 6$, we get $\\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370$ (either adding or using the sum of consecutive squares formula). But this only accounts for $\\sum_{k = 1}^{6} (4k^3 + k) = 4\\left(\\frac{6(6+1)}{2}\\right)^2 + \\frac{6(6+1)}{2} = 1764 + 21 = 1785$ terms, so we still have $1995 - 1785 = 210$ terms with $f(n) = 7$. This adds $210 \\cdot \\frac {1}{7} = 30$ to our summation, giving ${400}$.",
"This is a pretty easy problem just to bash. Since the max number we can get is $7$, we just need to test $n$ values for $1.5,2.5,3.5,4.5,5.5$ and $6.5$. Then just do how many numbers there are times $\\frac{1}{\\lfloor n \\rfloor}$, which should be $5+17+37+65+101+145+30 = 400"
] |
1995-I-14
| 1,995 | 14 |
In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18. The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d_{}$ are positive integers and $d_{}$ is not divisible by the square of any prime number. Find $m+n+d.$
| 378 | null |
[
"Let the center of the circle be $O$, and the two chords be $\\overline{AB}, \\overline{CD}$ and intersecting at $E$, such that $AE = CE < BE = DE$. Let $F$ be the midpoint of $\\overline{AB}$. Then $\\overline{OF} \\perp \\overline{AB}$. [asy] size(200); pathpen = black + linewidth(0.7); pen d = dashed+linewidth(0.7); pair O = (0,0), E=(0,18), B=E+48*expi(11*pi/6), D=E+48*expi(7*pi/6), A=E+30*expi(5*pi/6), C=E+30*expi(pi/6), F=foot(O,B,A); D(CR(D(MP(\"O\",O)),42)); D(MP(\"A\",A,NW)--MP(\"B\",B,SE)); D(MP(\"C\",C,NE)--MP(\"D\",D,SW)); D(MP(\"E\",E,N)); D(C--B--O--E,d);D(O--D(MP(\"F\",F,NE)),d); MP(\"39\",(B+F)/2,NE);MP(\"30\",(C+E)/2,NW);MP(\"42\",(B+O)/2); [/asy] By the Pythagorean Theorem, $OF = \\sqrt{OB^2 - BF^2} = \\sqrt{42^2 - 39^2} = 9\\sqrt{3}$, and $EF = \\sqrt{OE^2 - OF^2} = 9$. Then $OEF$ is a $30-60-90$ right triangle, so $\\angle OEB = \\angle OED = 60^{\\circ}$. Thus $\\angle BEC = 60^{\\circ}$, and by the Law of Cosines, $BC^2 = BE^2 + CE^2 - 2 \\cdot BE \\cdot CE \\cos 60^{\\circ} = 42^2.$ It follows that $\\triangle BCO$ is an equilateral triangle, so $\\angle BOC = 60^{\\circ}$. The desired area can be broken up into two regions, $\\triangle BCE$ and the region bounded by $\\overline{BC}$ and minor arc $\\stackrel{\\frown}{BC}$. The former can be found by Heron's formula to be $[BCE] = \\sqrt{60(60-48)(60-42)(60-30)} = 360\\sqrt{3}$. The latter is the difference between the area of sector $BOC$ and the equilateral $\\triangle BOC$, or $\\frac{1}{6}\\pi (42)^2 - \\frac{42^2 \\sqrt{3}}{4} = 294\\pi - 441\\sqrt{3}$. Thus, the desired area is $360\\sqrt{3} + 294\\pi - 441\\sqrt{3} = 294\\pi - 81\\sqrt{3}$, and $m+n+d = 378 -NL008"
] |
1995-I-15
| 1,995 | 15 |
Let $p_{}$ be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before one encounters a run of 2 tails. Given that $p_{}$ can be written in the form $m/n$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n$ .
| 37 | null |
[
"Solution 1 Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the successful sequences are composed of blocks of $1$ to $4$ H's separated by T's and end with $5$ H's. Since the probability that the sequence starts with TH is $1/4$, the total probability is that $3/2$ of the probability given that the sequence starts with an H. The answer to the problem is then the sum of all numbers of the form $\\frac 32 \\left( \\frac 1{2^a} \\cdot \\frac 12 \\cdot \\frac 1{2^b} \\cdot \\frac 12 \\cdot \\frac 1{2^c} \\cdots \\right) \\cdot \\left(\\frac 12\\right)^5$, where $a,b,c \\ldots$ are all numbers $1-4$, since the blocks of H's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$, so if there are n blocks of H's before the final five H's, the answer can be rewritten as the sum of all numbers of the form $\\frac 32\\left( \\left(\\frac {15}{16}\\right)^n \\cdot \\left(\\frac 12\\right)^n \\right) \\cdot \\left(\\frac 1{32}\\right)=\\frac 3{64}\\left(\\frac{15}{32}\\right)^n$, where $n$ ranges from $0$ to $\\infty$, since that's how many blocks of H's there can be before the final five. This is an infinite geometric series whose sum is $\\frac{3/64}{1-(15/32)}=\\frac{3}{34}$, so the answer is $037 instead. The repeated back-substitution is cleaner because we used the complement.",
"Think of the problem as a sequence of H's and T's. No two T's can occur in a row, so the successful sequences are composed of blocks of $1$ to $4$ H's separated by T's and end with $5$ H's. Since the probability that the sequence starts with TH is $1/4$, the total probability is that $3/2$ of the probability given that the sequence starts with an H. The answer to the problem is then the sum of all numbers of the form $\\frac 32 \\left( \\frac 1{2^a} \\cdot \\frac 12 \\cdot \\frac 1{2^b} \\cdot \\frac 12 \\cdot \\frac 1{2^c} \\cdots \\right) \\cdot \\left(\\frac 12\\right)^5$, where $a,b,c \\ldots$ are all numbers $1-4$, since the blocks of H's can range from $1-4$ in length. The sum of all numbers of the form $(1/2)^a$ is $1/2+1/4+1/8+1/16=15/16$, so if there are n blocks of H's before the final five H's, the answer can be rewritten as the sum of all numbers of the form $\\frac 32\\left( \\left(\\frac {15}{16}\\right)^n \\cdot \\left(\\frac 12\\right)^n \\right) \\cdot \\left(\\frac 1{32}\\right)=\\frac 3{64}\\left(\\frac{15}{32}\\right)^n$, where $n$ ranges from $0$ to $\\infty$, since that's how many blocks of H's there can be before the final five. This is an infinite geometric series whose sum is $\\frac{3/64}{1-(15/32)}=\\frac{3}{34}$, so the answer is $037 instead. The repeated back-substitution is cleaner because we used the complement.",
"Let $p_H, p_T$ respectively denote the probabilities that a string beginning with H's and T's are successful. Thus, $p_T = \\frac 12p_H.$ A successful string can either start with 1 to 4 H's, start with a T and then continue with a string starting with H (as there cannot be $2$ T's in a row, or be the string HHHHH. There is a $\\frac{1}{16} \\cdot \\frac{1}{2} = \\frac{1}{32}$ probability we roll HHHH and then T. On the other hand, there is a $\\frac{15}{16}$ probability we roll a H, HH, HHH, or HHHH, and then a T. Thus, $p_H = \\left(\\frac{15}{16}\\right) \\cdot \\left(\\frac 12\\right) p_H + \\frac{1}{32} \\Longrightarrow p_H = \\frac{1}{17}.$ The answer is $p_H + p_T = \\frac{3}{2}p_H = \\frac{3}{34}$, and $m+n = 037 instead. The repeated back-substitution is cleaner because we used the complement.",
"For simplicity, let's compute the complement, namely the probability of getting to $2$ tails before $5$ heads. Let $h_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive heads. Similarly, let $t_{i}$ denote the probability that we get $2$ tails before $5$ heads, given that we have $i$ consecutive tails. Specifically, $h_{5} = 0$ and $t_{2} = 1$. If we can solve for $h_{1}$ and $t_{1}$, we are done; the answer is simply $1/2 * (h_{1} + t_{1})$, since on our first roll, we have equal chances of getting a string with \"1 consecutive head\" or \"1 consecutive tail.\" Consider solving for $t_{1}$. If we have 1 consecutive tail, then (a) rolling a head gets us to 1 consecutive head and (b) rolling a tail gets us to 2 consecutive tails. So, we must have: $t_{1} = \\frac{1}{2} t_{2} + \\frac{1}{2} h_{1}$ Applying similar logic, we get the equations: \\begin{align*} h_{1} &= \\frac{1}{2} h_{2} + \\frac{1}{2} t_{1}\\\\ h_{2} &= \\frac{1}{2} h_{3} + \\frac{1}{2} t_{1}\\\\ h_{3} &= \\frac{1}{2} h_{4} + \\frac{1}{2} t_{1}\\\\ h_{4} &= \\frac{1}{2} h_{5} + \\frac{1}{2} t_{1} \\end{align*} Since $h_{5} = 0$, we get $h_{4} = \\frac{1}{2} t_{1}$ $\\Rightarrow h_{3} = \\frac{3}{4} t_{1}$ $\\Rightarrow h_{2} = \\frac{7}{8} t_{1}$ $\\Rightarrow h_{1} = \\frac{15}{16} t_{1}$ $\\Rightarrow t_{1} = \\frac{1}{2} t_{2} + \\frac{1}{2} \\cdot \\frac{15}{16} t_{1} = \\frac{1}{2} + \\frac{15}{32} t_{1} \\Rightarrow t_{1} = \\frac{16}{17}$ $\\Rightarrow h_{1} = \\frac{15}{16} \\cdot \\frac{16}{17} = \\frac{15}{17}$. So, the probability of reaching $2$ tails before $5$ heads is $\\frac{1}{2} (h_{1} + t_{1}) = \\frac{31}{34}$; we want the complement, $\\frac{3}{34}$, yielding an answer of $3 + 34 = 037 instead. The repeated back-substitution is cleaner because we used the complement.",
"Consider what happens in the \"endgame\" or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are $\\frac{1}{32},\\frac{1}{64},\\frac{1}{4},\\frac{1}{8},\\frac{1}{16},\\frac{1}{32},\\frac{1}{64}$ respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $\\frac{3}{64}$. The sum of all these endgame outcomes are $\\frac{34}{64}$, hence the desired probability is $\\frac{3}{34}$, and in this case $m=3,n=34$ so we have $m+n=037 -vsamc",
"Consider what happens in the \"endgame\" or what ultimately leads to the end. Let A denote that a head has been flipped, and let B denote that a tail has been flipped. The endgame outcomes are AAAAA, BAAAAA, BB, ABB, AABB, AAABB, AAAABB. The probabilities of each of these are $\\frac{1}{32},\\frac{1}{64},\\frac{1}{4},\\frac{1}{8},\\frac{1}{16},\\frac{1}{32},\\frac{1}{64}$ respectively. The ones where there is five heads are AAAAA and BAAAAA. The sum of these probabilities are $\\frac{3}{64}$. The sum of all these endgame outcomes are $\\frac{34}{64}$, hence the desired probability is $\\frac{3}{34}$, and in this case $m=3,n=34$ so we have $m+n=037 -vsamc",
"There can be 1-4 heads and 1 tails in any grouping, so we write $h_5,th_5,h_nth_5,th_nth_5,…$ etc., where $1\\leq n\\leq 4$ and subscript $n$ means there are $n$ of that state in a row. We see that this gives us 2 geometric sequences: one with first term $\\frac{1}{2^5}$ and common ratio $\\frac{1}{2}*\\left(1/2+1/4+1/8+1/16\\right)$ and one with first tem $\\frac{1}{2^6}$ and the same common ratio. Therefore we get $\\frac{3}{34}$ or $037 ~joeythetoey"
] |
1996-I-2
| 1,996 | 2 |
For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed $x$ . For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer?
| 340 | null |
[
"For integers $k$, we want $\\lfloor \\log_2 n\\rfloor = 2k$, or $2k \\le \\log_2 n < 2k+1 \\Longrightarrow 2^{2k} \\le n < 2^{2k+1}$. Thus, $n$ must satisfy these inequalities (since $n < 1000$): $4\\leq n <8$ $16\\leq n<32$ $64\\leq n<128$ $256\\leq n<512$ There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+16+64+256=340."
] |
1996-I-3
| 1,996 | 3 |
Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms.
| 44 | null |
[
"Using Simon's Favorite Factoring Trick, we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$. Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \\ge 1996$, the smallest square after $1996$ is $2025 = 45^2$, so our answer is $45 - 1 = 044.",
"Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of $(xy+x+y+1)^n$ (we do this to simplify the problem). If we expand the exponent the expression becomes $\\underbrace{(xy+x+y+1)\\cdot(xy+x+y+1)\\cdots(xy+x+y+1)}_{n\\;times}$. This is equivalent to starting off with a $x^0y^0$ terms and choosing between $4$ options $n$ different times: $\\bullet$ Adding nothing to either exponent (choosing $1$). $\\bullet$ Adding $1$ to the $x$ exponent (choosing $x$). $\\bullet$ Adding $1$ to the $y$ exponent (choosing $y$). $\\bullet$ Adding $1$ to the $x$ exponent and adding $1$ to the $y$ exponent (choosing $xy$). Doing this $n$ times, you can see that you end up with a term in the form $k\\cdot x^i\\cdot y^j$ where $k$ is some coefficient (which we don't care about) and $0\\le i,j\\le n$ and $i,j \\in \\mathbb{N}$. Repeating this for all possible combinations of choices yields $n+1$ options for each of $i$ and $j$ which means there are a total of $(n+1)^2$ possible terms in the form $x^i\\cdot y^j$. Therefore $(xy+x+y+1)^n$ has $(n+1)^2$ terms. $(n+1)^2 \\ge 1996$ which yields $n=44$. $044 ~coolishu",
"Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of $(xy+x+y+1)^n$ (we do this to simplify the problem). If we expand the exponent the expression becomes $\\underbrace{(xy+x+y+1)\\cdot(xy+x+y+1)\\cdots(xy+x+y+1)}_{n\\;times}$. This is equivalent to starting off with a $x^0y^0$ terms and choosing between $4$ options $n$ different times: $\\bullet$ Adding nothing to either exponent (choosing $1$). $\\bullet$ Adding $1$ to the $x$ exponent (choosing $x$). $\\bullet$ Adding $1$ to the $y$ exponent (choosing $y$). $\\bullet$ Adding $1$ to the $x$ exponent and adding $1$ to the $y$ exponent (choosing $xy$). Doing this $n$ times, you can see that you end up with a term in the form $k\\cdot x^i\\cdot y^j$ where $k$ is some coefficient (which we don't care about) and $0\\le i,j\\le n$ and $i,j \\in \\mathbb{N}$. Repeating this for all possible combinations of choices yields $n+1$ options for each of $i$ and $j$ which means there are a total of $(n+1)^2$ possible terms in the form $x^i\\cdot y^j$. Therefore $(xy+x+y+1)^n$ has $(n+1)^2$ terms. $(n+1)^2 \\ge 1996$ which yields $n=44$. $044 ~coolishu"
] |
1996-I-4
| 1,996 | 4 |
A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of a shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$ .
| 166 | null |
[
"[asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5; real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit)); draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); label(\"$x$\",(0,0,unit+unit/(r-1)/2),WSW); label(\"$1$\",(unit/2,0,unit),N); label(\"$1$\",(unit,0,unit/2),W); label(\"$1$\",(unit/2,0,0),N); label(\"$6$\",(unit*(r+1)/2,0,0),N); label(\"$7$\",(unit*r,unit*r/2,0),SW); [/asy] (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$, and so the sides of the shadow are $7$. Using the similar triangles in blue, $\\frac {x}{1} = \\frac {1}{6}$, and $\\left\\lfloor 1000x \\right\\rfloor = 166."
] |
1996-I-5
| 1,996 | 5 |
Suppose that the roots of $x^3+3x^2+4x-11=0$ are $a$ , $b$ , and $c$ , and that the roots of $x^3+rx^2+sx+t=0$ are $a+b$ , $b+c$ , and $c+a$ . Find $t$ .
| 23 | null |
[
"By Vieta's formulas on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$, we have $a + b + c = s = -3$, $ab + bc + ca = 4$, and $abc = 11$. Then $t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$ This is just the definition for $-P(-3) = 23. Alternatively, we can expand the expression to get \\begin{align*} t &= -(-3-a)(-3-b)(-3-c)\\\\ &= (a+3)(b+3)(c+3)\\\\ &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\\\ t &= 11 + 3(4) + 9(-3) + 27 = 23\\end{align*}",
"Each term in the expansion of $(a+b)(b+c)(c+a)$ has a total degree of 3. Another way to get terms with degree 3 is to multiply out $(a+b+c)(ab+bc+ca)$. Expanding both of these expressions and comparing them shows that: $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$ $t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23$ A way to realize $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$: $(a + b)(b+c)(c+a) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + abc$ (Add an extra $abc$) $a^2b + a^2c + abc + b^2a + b^2c + abc + c^2a + c^2b + abc - abc =$ $a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc =$ $(a + b + c)(ab + bc + ac) - abc = (-3)(4) - 11 = -23$. The value of $t$ is the negation of this, which is $-(-23) = 23 ~Extremelysupercooldude",
"We have that $x^3+3x^2+4x-11=0$ for roots $a,b,c.$ In the second cubic function $x^3+rx^2+sx+t=0,$ the roots are $a+b,b+c,c+a.$ By Vieta's formulae, we see that $t= -(a+b)(b+c)(a+c).$ As we know that the sum of the roots of the first polynomial, $a+b+c$ is $-3$ by applying Vieta's again. Using this fact, we can rewrite $t$ as $-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).$ Seeing this, we can find the value of the product of the roots by applying this to the first equation. This can be done by setting $x'=a+3,$ so, from this, we see that we should plug in $x'-3$ for $x$ in $x^3+3x^2+4x-11=0$. After simplifying, we get that the polynomial is $x'^3 - 6x'^2 + 13x' - 23 = 0.$ Given that the product of the roots of this equation is equivalent to our desired value for $t$, we can apply Vieta's formulae for a third time to find that $t = -(\\frac{-23}{1}) = 23 ~MathWhiz35"
] |
1996-I-6
| 1,996 | 6 |
In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$ .
| 49 | null |
[
"Solution 1 We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games. No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games. Now we use PIE: The probability that one team wins all games is $5\\cdot \\left(\\frac{1}{2}\\right)^4=\\frac{5}{16}$. Similarity, the probability that one team loses all games is $\\frac{5}{16}$. The probability that one team wins all games and another team loses all games is $\\left(5\\cdot \\left(\\frac{1}{2}\\right)^4\\right)\\left(4\\cdot \\left(\\frac{1}{2}\\right)^3\\right)=\\frac{5}{32}$. $\\frac{5}{16}+\\frac{5}{16}-\\frac{5}{32}=\\frac{15}{32}$ Since this is the opposite of the probability we want, we subtract that from 1 to get $\\frac{17}{32}$. $17+32=049 Solution by Ilikeapos",
"We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games. No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games. Now we use PIE: The probability that one team wins all games is $5\\cdot \\left(\\frac{1}{2}\\right)^4=\\frac{5}{16}$. Similarity, the probability that one team loses all games is $\\frac{5}{16}$. The probability that one team wins all games and another team loses all games is $\\left(5\\cdot \\left(\\frac{1}{2}\\right)^4\\right)\\left(4\\cdot \\left(\\frac{1}{2}\\right)^3\\right)=\\frac{5}{32}$. $\\frac{5}{16}+\\frac{5}{16}-\\frac{5}{32}=\\frac{15}{32}$ Since this is the opposite of the probability we want, we subtract that from 1 to get $\\frac{17}{32}$. $17+32=049 Solution by Ilikeapos",
"There are $\\dbinom{5}{2} = 10$ games in total, and every game can either end in a win or a loss. Therefore, there are $2^{10} = 1024$ possible outcomes. Now, computing this probability directly seems a little hard, so let's compute the complement -- the probability that there is an undefeated team, a winless team, or both. The number of ways that we can have an undefeated team is $5 \\cdot 2^6,$ as there are five ways to choose the team itself and $2^6$ outcomes for the games that are not concerning the undefeated team. Reversing all the wins to losses yields a winless team, so this situation is symmetrical to the first one. Therefore, there are $5 \\cdot 2^6 + 5 \\cdot 2^6$ ways for an undefeated or winless team. In the process, however, we have accidentally overcounted the scenario of both an undefeated and winless team. There are $5 \\cdot 4 \\cdot 2^3$ ways for this scenario to happen, because there are five ways to choose the undefeated team, four ways for the winless, and three games that don't concern either of the teams. Therefore, there are $5 \\cdot 2^6 + 5 \\cdot 2^6 - 5 \\cdot 4 \\cdot 2^3 = 480$ ways to have an undefeated and/or winless team, so there are $1024 - 480 = 544$ to not have any. Our probability is thus $\\dfrac{544}{1024},$ which simplifies to $\\dfrac{17}{32},$ for our answer of $17 + 32 = 049 Solution by Ilikeapos"
] |
1996-I-7
| 1,996 | 7 |
Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?
| 300 | null |
[
"There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. [asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy] [asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]For most pairs, there will be three other equivalent boards.For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \\choose 2}-24$ pairs, there are three other pairs that yield an equivalent board. Thus, the number of inequivalent boards is $\\frac{{49 \\choose 2} - 24}{4} + \\frac{24}{2} = 300 inequivalent configurations.",
"There are ${49 \\choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards. [asy] pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(2,3)*unitsquare,rgb(.8,.8,.5));fill(shift(2,1)*unitsquare,rgb(.8,.8,.5)); fill(shift(3,2)*unitsquare,rgb(.8,.8,.5));fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,1,280,350),EndArrow(4)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,1,10,80),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,1,190,260),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,1,100,170),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy] [asy]pathpen = black; pair O = (3.5,3.5); D(O); for(int i=0;i<7;++i) for(int j=0;j<7;++j) D(shift(i,j)*unitsquare); fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(2,1)*unitsquare,rgb(1,1,.4)); fill(shift(1,4)*unitsquare,rgb(.8,.8,.5)); fill(shift(5,2)*unitsquare,rgb(.8,.8,.5)); D(arc(O,5^.5,-20,50),EndArrow(4)); D(arc(O,5^.5,70,140),EndArrow(4)); D(arc(O,5^.5,250,320),EndArrow(4)); D(arc(O,5^.5,160,230),EndArrow(4)); [/asy]For most pairs, there will be three other equivalent boards.For those symmetric about the center, there is only one other. Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\\frac{49-1}{2}=24$ such pairs. There are then ${49 \\choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \\choose 2}-24$ pairs, there are three other pairs that yield an equivalent board. Thus, the number of inequivalent boards is $\\frac{{49 \\choose 2} - 24}{4} + \\frac{24}{2} = 300 inequivalent configurations.",
"There are 4 cases: 1. The center square is occupied, in which there are $12$ cases. 2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases. 3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases. 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\\dbinom{12}{2} \\cdot \\frac{16}{4} = 264$ cases. Add up all the values for each case to get $300 as your answer. ~First",
"There are 4 cases: 1. The center square is occupied, in which there are $12$ cases. 2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases. 3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases. 4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\\dbinom{12}{2} \\cdot \\frac{16}{4} = 264$ cases. Add up all the values for each case to get $300 as your answer. ~First",
"Consider the group $G=\\mathbb{Z}/4\\mathbb{Z}$ with addition acting on set $X$, where $X$ is the set of color schemes on the checkerboard. We now apply Burnside’s Lemma. Since $\\mathbb{Z}/4\\mathbb{Z}=\\langle r \\mid r^4=1\\rangle$, we only need to consider 4 cases for $g$ an element of $\\mathbb{Z}/4\\mathbb{Z}$. 1. $g=1$. There are clearly ${49 \\choose 2}$ cases. 2. $g=r$. Since this is a $90$ degree rotation, there are no fixed points under this transformation. 3. $g=r^2$. There are $24$ cases (which can be manually checked). 4. $g=r^3$. There are no cases since this is just Case $2$ but with a $-90$ degree rotation. Therefore, by Burnside’s Lemma there are $\\frac{1176+24+0+0}{|G|}=\\frac{1200}{4}=300 inequivalent configurations (also known as orbits in group-theoretic terms).",
"Consider the group $G=\\mathbb{Z}/4\\mathbb{Z}$ with addition acting on set $X$, where $X$ is the set of color schemes on the checkerboard. We now apply Burnside’s Lemma. Since $\\mathbb{Z}/4\\mathbb{Z}=\\langle r \\mid r^4=1\\rangle$, we only need to consider 4 cases for $g$ an element of $\\mathbb{Z}/4\\mathbb{Z}$. 1. $g=1$. There are clearly ${49 \\choose 2}$ cases. 2. $g=r$. Since this is a $90$ degree rotation, there are no fixed points under this transformation. 3. $g=r^2$. There are $24$ cases (which can be manually checked). 4. $g=r^3$. There are no cases since this is just Case $2$ but with a $-90$ degree rotation. Therefore, by Burnside’s Lemma there are $\\frac{1176+24+0+0}{|G|}=\\frac{1200}{4}=300 inequivalent configurations (also known as orbits in group-theoretic terms)."
] |
1996-I-8
| 1,996 | 8 |
The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$ ?
| 799 | null |
[
"The harmonic mean of $x$ and $y$ is equal to $\\frac{1}{\\frac{\\frac{1}{x}+\\frac{1}{y}}2} = \\frac{2xy}{x+y}$, so we have $xy=(x+y)(3^{20}\\cdot2^{19})$, and by SFFT, $(x-3^{20}\\cdot2^{19})(y-3^{20}\\cdot2^{19})=3^{40}\\cdot2^{38}$. Now, $3^{40}\\cdot2^{38}$ has $41\\cdot39=1599$ factors, one of which is the square root ($3^{20}2^{19}$). Since $x<y$, the answer is half of the remaining number of factors, which is $\\frac{1599-1}{2}= 799."
] |
1996-I-9
| 1,996 | 9 |
A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?
| 342 | null |
[
"Solution 1 On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \\pmod{8}$ and $6 \\pmod{8}$. Then he goes ahead and opens all lockers $2 \\pmod {8}$, leaving lockers either $6 \\pmod {16}$ or $14 \\pmod {16}$. He then goes ahead and opens all lockers $14 \\pmod {16}$, leaving the lockers either $6 \\pmod {32}$ or $22 \\pmod {32}$. He then goes ahead and opens all lockers $6 \\pmod {32}$, leaving $22 \\pmod {64}$ or $54 \\pmod {64}$. He then opens $54 \\pmod {64}$, leaving $22 \\pmod {128}$ or $86 \\pmod {128}$. He then opens $22 \\pmod {128}$ and leaves $86 \\pmod {256}$ and $214 \\pmod {256}$. He then opens all $214 \\pmod {256}$, so we have $86 \\pmod {512}$ and $342 \\pmod {512}$, leaving lockers $86, 342, 598$, and $854$, and he is at where he started again. He then opens $86$ and $598$, and then goes back and opens locker number $854$, leaving locker number $342. (Note: If you try to do this, first look through all the problems! -Guy) Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496",
"On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \\pmod{8}$ and $6 \\pmod{8}$. Then he goes ahead and opens all lockers $2 \\pmod {8}$, leaving lockers either $6 \\pmod {16}$ or $14 \\pmod {16}$. He then goes ahead and opens all lockers $14 \\pmod {16}$, leaving the lockers either $6 \\pmod {32}$ or $22 \\pmod {32}$. He then goes ahead and opens all lockers $6 \\pmod {32}$, leaving $22 \\pmod {64}$ or $54 \\pmod {64}$. He then opens $54 \\pmod {64}$, leaving $22 \\pmod {128}$ or $86 \\pmod {128}$. He then opens $22 \\pmod {128}$ and leaves $86 \\pmod {256}$ and $214 \\pmod {256}$. He then opens all $214 \\pmod {256}$, so we have $86 \\pmod {512}$ and $342 \\pmod {512}$, leaving lockers $86, 342, 598$, and $854$, and he is at where he started again. He then opens $86$ and $598$, and then goes back and opens locker number $854$, leaving locker number $342. (Note: If you try to do this, first look through all the problems! -Guy) Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496",
"We can also solve this with recursion. Let $L_n$ be the last locker he opens given that he started with $2^n$ lockers. Let there be $2^n$ lockers. After he first reaches the end of the hallway, there are $2^{n-1}$ lockers remaining. There is a correspondence between these unopened lockers and if he began with $2^{n-1}$ lockers. The locker $y$ (if he started with $2^{n-1}$ lockers) corresponds to the locker $2^n+2-2y$ (if he started with $2^n$ lockers). It follows that $L_{n} = 2^{n} +2 -2L_{n-1}$ as they are corresponding lockers. We can compute $L_1=2$ and use the recursion to find $L_{10}=342. (Note: If you try to do this, first look through all the problems! -Guy) Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496",
"List all the numbers from $1$ through $1024$, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $342. (Note: If you try to do this, first look through all the problems! -Guy) Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496",
"List all the numbers from $1$ through $1024$, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $342. (Note: If you try to do this, first look through all the problems! -Guy) Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496"
] |
1996-I-10
| 1,996 | 10 |
Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$ .
| 159 | null |
[
"$\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{186^{\\circ}}+\\sin{96^{\\circ}}}{\\sin{186^{\\circ}}-\\sin{96^{\\circ}}} =$ $\\dfrac{\\sin{(141^{\\circ}+45^{\\circ})}+\\sin{(141^{\\circ}-45^{\\circ})}}{\\sin{(141^{\\circ}+45^{\\circ})}-\\sin{(141^{\\circ}-45^{\\circ})}} =$ $\\dfrac{2\\sin{141^{\\circ}}\\cos{45^{\\circ}}}{2\\cos{141^{\\circ}}\\sin{45^{\\circ}}} = \\tan{141^{\\circ}}$. The period of the tangent function is $180^\\circ$, and the tangent function is one-to-one over each period of its domain. Thus, $19x \\equiv 141 \\pmod{180}$. Since $19^2 \\equiv 361 \\equiv 1 \\pmod{180}$, multiplying both sides by $19$ yields $x \\equiv 141 \\cdot 19 \\equiv (140+1)(18+1) \\equiv 0+140+18+1 \\equiv 159 \\pmod{180}$. Therefore, the smallest positive solution is $x = 159.",
"$\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\dfrac{1 + \\tan{96^{\\circ}}}{1-\\tan{96^{\\circ}}}$ which is the same as $\\dfrac{\\tan{45^{\\circ}} + \\tan{96^{\\circ}}}{1-\\tan{45^{\\circ}}\\tan{96^{\\circ}}} = \\tan{141{^\\circ}}$. So $19x = 141 +180n$, for some integer $n$. Multiplying by $19$ gives $x \\equiv 141 \\cdot 19 \\equiv 2679 \\equiv 159 \\pmod{180}$. The smallest positive solution of this is $x = 159",
"It seems reasonable to assume that $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\tan{\\theta}$ for some angle $\\theta$. This means \\[\\dfrac{\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}})}{\\alpha (\\cos{96^{\\circ}}-\\sin{96^{\\circ}})} = \\frac{\\sin{\\theta}}{\\cos{\\theta}}\\] for some constant $\\alpha$. We can set $\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}}) = \\sin{\\theta}$.Note that if we have $\\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\\sin{45^{\\circ}} = \\cos{45^{\\circ}} = \\tfrac{\\sqrt{2}}{2}$, if $\\alpha = \\tfrac{\\sqrt{2}}{2}$ we have \\[\\alpha (\\cos{96^{\\circ}} + \\sin{96^{\\circ}}) = \\cos{96^{\\circ}} \\frac{\\sqrt{2}}{2} + \\sin{96^{\\circ}} \\frac{\\sqrt{2}}{2} = \\cos{96^{\\circ}} \\sin{45^{\\circ}} + \\sin{96^{\\circ}} \\cos{45^{\\circ}} = \\sin({45^{\\circ} + 96^{\\circ}}) = \\sin{141^{\\circ}}\\] from the sine sum formula. For the denominator, from the cosine sum formula, we have \\[\\alpha (\\cos{96^{\\circ}} - \\sin{96^{\\circ}}) = \\cos{96^{\\circ}} \\frac{\\sqrt{2}}{2} + \\sin{96^{\\circ}} \\frac{\\sqrt{2}}{2} = \\cos{96^{\\circ}} \\cos{45^{\\circ}} + \\sin{96^{\\circ}} \\sin{45^{\\circ}} = \\cos({96^{\\circ} + 45^{\\circ}}) = \\cos{141^{\\circ}}.\\] This means $\\theta = 141^{\\circ},$ so $19x = 141 + 180k$ for some positive integer $k$ (since the period of tangent is $180^{\\circ}$), or $19 x \\equiv 141 \\pmod{180}$. Note that the inverse of $19$ modulo $180$ is itself as $19^2 \\equiv 361 \\equiv 1 \\pmod {180}$, so multiplying this congruence by $19$ on both sides gives $x \\equiv 2679 \\equiv 159 \\pmod{180}.$ For the smallest possible $x$, we take $x = 159",
"It seems reasonable to assume that $\\dfrac{\\cos{96^{\\circ}}+\\sin{96^{\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}} = \\tan{\\theta}$ for some angle $\\theta$. This means \\[\\dfrac{\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}})}{\\alpha (\\cos{96^{\\circ}}-\\sin{96^{\\circ}})} = \\frac{\\sin{\\theta}}{\\cos{\\theta}}\\] for some constant $\\alpha$. We can set $\\alpha (\\cos{96^{\\circ}}+\\sin{96^{\\circ}}) = \\sin{\\theta}$.Note that if we have $\\alpha$ equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since $\\sin{45^{\\circ}} = \\cos{45^{\\circ}} = \\tfrac{\\sqrt{2}}{2}$, if $\\alpha = \\tfrac{\\sqrt{2}}{2}$ we have \\[\\alpha (\\cos{96^{\\circ}} + \\sin{96^{\\circ}}) = \\cos{96^{\\circ}} \\frac{\\sqrt{2}}{2} + \\sin{96^{\\circ}} \\frac{\\sqrt{2}}{2} = \\cos{96^{\\circ}} \\sin{45^{\\circ}} + \\sin{96^{\\circ}} \\cos{45^{\\circ}} = \\sin({45^{\\circ} + 96^{\\circ}}) = \\sin{141^{\\circ}}\\] from the sine sum formula. For the denominator, from the cosine sum formula, we have \\[\\alpha (\\cos{96^{\\circ}} - \\sin{96^{\\circ}}) = \\cos{96^{\\circ}} \\frac{\\sqrt{2}}{2} + \\sin{96^{\\circ}} \\frac{\\sqrt{2}}{2} = \\cos{96^{\\circ}} \\cos{45^{\\circ}} + \\sin{96^{\\circ}} \\sin{45^{\\circ}} = \\cos({96^{\\circ} + 45^{\\circ}}) = \\cos{141^{\\circ}}.\\] This means $\\theta = 141^{\\circ},$ so $19x = 141 + 180k$ for some positive integer $k$ (since the period of tangent is $180^{\\circ}$), or $19 x \\equiv 141 \\pmod{180}$. Note that the inverse of $19$ modulo $180$ is itself as $19^2 \\equiv 361 \\equiv 1 \\pmod {180}$, so multiplying this congruence by $19$ on both sides gives $x \\equiv 2679 \\equiv 159 \\pmod{180}.$ For the smallest possible $x$, we take $x = 159",
"Multiplying the numerator and denominator of the right-hand side by $\\cos(96^{\\circ})+\\sin(96^{\\circ})$, we get ${\\tan(19x^{\\circ})} ={\\frac{\\cos(96^{\\circ}) +\\sin(96^{\\circ})}{\\cos(96^{\\circ})-\\sin(96^{\\circ})}}\\times{\\frac{\\cos(96^{\\circ})+\\sin(96^{\\circ})}{\\cos(96^{\\circ})+\\sin(96^{\\circ})}} \\\\ ={\\frac{(\\cos(96^{\\circ})+\\sin(96^{\\circ}))^2}{\\cos^2(96^{\\circ})-\\sin^2(96^{\\circ})}} \\\\ ={\\frac{\\cos^2(96^{\\circ}) + 2\\cos(96^{\\circ})\\sin(96^{\\circ}) + \\sin^2(96^{\\circ})}{\\cos(192^{\\circ})}} \\\\ ={\\frac{1+\\sin(192^{\\circ})}{\\cos(192^{\\circ})}}$ Using the fact that $\\tan(\\theta)=\\frac{\\sin(\\theta)}{\\cos(\\theta)}$, we get $\\tan(19x^{\\circ})=\\frac{\\sin(19x^{\\circ})}{\\cos(19x^{\\circ})}=\\frac{1+\\sin(192^{\\circ})}{\\cos(192^{\\circ})}$. Cross-multiplying, we find that $\\sin(19x^{\\circ})\\cos(192^{\\circ})=\\cos(19x^{\\circ})+\\cos(19x^{\\circ})\\sin(192^{\\circ})$. Rearranging the equation gives us $\\cos(19x^{\\circ})=\\sin(19x^{\\circ})\\cos(192^{\\circ})-\\cos(19x^{\\circ})\\sin(192^{\\circ})$ which leads us to $\\cos(19x^{\\circ})=\\sin(19x-192^{\\circ})$ by the sine difference formula. Using the identity that $\\cos(\\theta)=\\sin(90^{\\circ}-\\theta)$, we find that $\\sin(90-19x^{\\circ})=\\sin(19x-192^{\\circ})$. Therefore, $90-19x \\equiv 19x-192 \\pmod{360}$, or $38x \\equiv 282 \\pmod{360}$. We know that $38 \\times 9=342$ and $38 \\times 10 \\equiv 20 \\pmod{360}$ (by simple arithmetic). To \"make\" $282$ we subtract $10$ three times from $9$, giving us $-21$. Finally, because $360|38 \\times 180$, we can add $180$ to get that $x=180-21=159 which is the final answer. ~primenumbersfun"
] |
1996-I-11
| 1,996 | 11 |
Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$ , where $0<r$ and $0\leq \theta <360$ . Find $\theta$ .
| 276 | null |
[
"\\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \\frac{z^5-1}{z-1}\\\\ 0 &=& \\frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \\frac{(z^2-z+1)(z^5-1)}{z-1} \\end{eqnarray*} Thus $z^5 = 1, z \\neq 1 \\Longrightarrow z = \\mathrm{cis}\\ 72, 144, 216, 288$, or $z^2 - z + 1 = 0 \\Longrightarrow z = \\frac{1 \\pm \\sqrt{-3}}{2} = \\mathrm{cis}\\ 60, 300$ (see cis). Discarding the roots with negative imaginary parts (leaving us with $\\mathrm{cis} \\theta,\\ 0 < \\theta < 180$), we are left with $\\mathrm{cis}\\ 60, 72, 144$; their product is $P = \\mathrm{cis} (60 + 72 + 144) = \\mathrm{cis} 276.",
"Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1 | z^6 + z^4 + z^3 + z^2 + 1$. Long division quickly gives the other factor to be $z^2 - z + 1$. The solution follows as above.",
"Divide through by $z^3$. We get the equation $z^3 + \\frac {1}{z^3} + z + \\frac {1}{z} + 1 = 0$. Let $x = z + \\frac {1}{z}$. Then $z^3 + \\frac {1}{z^3} = x^3 - 3x$. Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$, with solutions $x = 1, \\frac { - 1\\pm\\sqrt {5}}{2}$. For $x = 1$, we get $z = \\text{cis}60,\\text{cis}300$. For $x = \\frac { - 1 + \\sqrt {5}}{2}$, we get $z = \\text{cis}{72},\\text{cis}{292}$ (using exponential form of $\\cos$). For $x = \\frac { - 1 - \\sqrt {5}}{2}$, we get $z = \\text{cis}144,\\text{cis}216$. The ones with positive imaginary parts are ones where $0\\le\\theta\\le180$, so we have $60 + 72 + 144 = 276.",
"This is just a slight variation of Solution 1. We start off by adding $z^5$ to both sides, to get a neat geometric sequence with $a = 1$ and $r = z$, which gives us $\\frac{z^7 - 1}{z - 1} = z^5$. From here, multiply by $z - 1$ to both sides, noting that then $z \\neq \\cos 0 + i\\sin 0$ since, then we are multiplying by $0$ which makes it undefined. We now note that $z^7 - 1 = z^6 - z^5 \\implies z^7 - z^6 + z^5 = 1$. (This is the part that it becomes almost identical to Solution 1). Factor $z^5$ from the LHS, to get $z^5( z^2 - z + 1) = 1$. Call the set of roots from $z^5$ as $A$, and set of roots from $z^2 - z + 1$ as $B$. We are to find $|A \\cup B| = A + B - |A \\cap B|$. (Essentially, we are to find the roots that are not common to both equations/sets, or else we are overcounting a root two times, rather than once. Try out some equation to see where this might apply) Thankfully in this case, none of the roots are overcounted. From here, proceed with Solution 1.",
"We recognize that $z^6+z^5+z^4+z^3+z^2+z+1=\\frac{z^7-1}{z-1}$ and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting $z^5$ and $z$ as shown below. $z^6+z^5+z^4+z^3+z^2+z+1-(z^5+z)=0\\implies \\frac{z^7-1}{z-1}-(z^5+z)=0$ Now, knowing that $z=1$ is not a root, we multiply by $z-1$ to obtain $z^7-1-(z-1)(z^5+z)=0\\implies z^7-1-(z^6+z^2-z^5-z)=0\\implies z^7-z^6+z^5-z^2+z-1=0$ Now, we see the $z^2+z-1$ and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. $z^5(z^2-z)+z^5-(z^2-z+1)=0\\implies z^5(z^2-z+1)-(z^2-z+1)=0\\implies (z^5-1)(z^2-z+1)=0$. Now, it is clear that we have two cases to consider. Case $1$: $z^5-1=0$ We obtain that $z^5=1$ or $z^5=e^{2\\pi{n}{i}}$ Obviously, the answers to this case are $e^{ia}, a\\in{\\frac{2\\pi}{5}, \\frac{4\\pi}{5}}$ Case $2$: $z^2-z+1=0$ Completing the square and then algebra allows us to find that $z=\\frac{1}{2}\\pm\\frac{i\\sqrt{3}}{2}$ which has $\\arg$ $\\frac{\\pi}{3}$ Hence, the answer is $\\frac{18\\pi}{15}+\\frac{5\\pi}{15}=\\frac{23\\pi}{15}\\cdot\\frac{180}{\\pi}=276",
"Add 1 to both sides of the equation to get $x^6+x^4+x^3+x^2+1+1=1$. We can rearrange to find that $(x^6+x^3+1)+(x^4+x^2+1)=1$. Then, using sum of a geometric series, $\\frac{x^9-1}{x^3-1}+\\frac{x^6-1}{x^2-1}=1$. Combining the two terms of the LHS, we get that $\\frac{x^{11}-x^9-x^2+1+x^9-x^6-x^3+1}{x^5-x^3-x^2+1}=1$, so $x^{11}-x^6-x^3-x^2+2=x^5-x^3-x^2+1$, and simplifying, we see that $x^{11}-x^6-x^5+1=0$, so by SFFT, $(x^6-1)(x^5-1)=0$. Then, the roots of our polynomial are the fifth and sixth roots of unity. However, looking back at our expression when it had fractions, we realize that if $x$ is a second or third root of unity, it would cause a denominator to be zero, so the roots of our polynomial are the fifth and sixth roots of unity that are not the second or third roots of unity. The only roots in this category are $\\mathrm{cis} (60, 72, 144)$, so our desired sum is $276, and we are done. -coolak",
"Take accout of the $Z^3$ and use the formula $(x^n-1)(x^m+x^{m-n}+...+1) = x^{m+n}-1$ Then factorise the function to solve $Z^5=1$, $Z^3 \\neq 1$, $Z^2 \\neq 1$ ~JiYang"
] |
1996-I-12
| 1,996 | 12 |
For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum $|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|$ . The average value of all such sums can be written in the form $\dfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 58 | null |
[
"Solution 1 Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \\cdot 8!$, because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n + 1}$ can be. To find all possible values for $|a_n - a_{n - 1}|$ we have to compute \\begin{eqnarray*} |1 - 10| + |1 - 9| + \\ldots + |1 - 2|\\\\ + |2 - 10| + \\ldots + |2 - 3| + |2 - 1|\\\\ + \\ldots\\\\ + |10 - 9| \\end{eqnarray*} This is equivalent to \\[2\\sum\\limits_{k = 1}^{9}\\sum\\limits_{j = 1}^{k}j = 330\\] The total number of permutations is $10!$, so the average value is $\\frac {330 \\cdot 8! \\cdot 5}{10!} = \\frac {55}{3}$, and $m+n = 058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).",
"Because of symmetry, we may find all the possible values for $|a_n - a_{n - 1}|$ and multiply by the number of times this value appears. Each occurs $5 \\cdot 8!$, because if you fix $a_n$ and $a_{n + 1}$ there are still $8!$ spots for the others and you can do this $5$ times because there are $5$ places $a_n$ and $a_{n + 1}$ can be. To find all possible values for $|a_n - a_{n - 1}|$ we have to compute \\begin{eqnarray*} |1 - 10| + |1 - 9| + \\ldots + |1 - 2|\\\\ + |2 - 10| + \\ldots + |2 - 3| + |2 - 1|\\\\ + \\ldots\\\\ + |10 - 9| \\end{eqnarray*} This is equivalent to \\[2\\sum\\limits_{k = 1}^{9}\\sum\\limits_{j = 1}^{k}j = 330\\] The total number of permutations is $10!$, so the average value is $\\frac {330 \\cdot 8! \\cdot 5}{10!} = \\frac {55}{3}$, and $m+n = 058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).",
"Without loss of generality, let $a_1 > a_2;\\, a_3 > a_4;\\, \\ldots ;\\, a_9 > a_{10}$. We may do this because all sums obtained from these paired sequences are also obtained in another $2^5-1$ ways by permuting the adjacent terms $\\{a_1,a_2\\},\\{a_3,a_4\\}, \\cdots , \\{a_9, a_{10}\\}$, and thus are canceled when the average is taken. So now we only have to form the sum $S= (a_1 + a_3 + a_5 + a_7 + a_9) - (a_2 + a_4 + a_6 + a_8 + a_{10})$. Due to the symmetry of this situation, we only need to compute the expected value of the result. $10$ must always be the greatest number in its pair; $9$ will be the greater number in its pair $\\frac{8}{9}$ of the time and the lesser number $\\frac 19$ of the time; $8$ will be the greater number in its pair $\\frac 79$ of the time and the lesser $\\frac 29$ of the time; and so forth. Each number either adds or subtracts from the sum depending upon whether it is one of the five greater or five lesser numbers in the pairs, respectively. Thus \\begin{align*} \\overline{S} &= 10 + \\left(\\frac{8}{9} \\cdot 9\\right) - \\left(\\frac{1}{9} \\cdot 9\\right) + \\left(\\frac{7}{9} \\cdot 8\\right) - \\left(\\frac 29 \\cdot 8\\right) + \\cdots + \\left(\\frac{1}{9} \\cdot 2\\right) - \\left(\\frac{8}{9} \\cdot 2\\right) - 1 \\\\ &= \\frac{9 \\cdot 10 + 7 \\cdot 9 + 5 \\cdot 8 + 3 \\cdot 7 + 1 \\cdot 6 - 1 \\cdot 5 - 3 \\cdot 4 - 5 \\cdot 3 - 7 \\cdot 2 - 9 \\cdot 1}{9} \\\\ &= \\frac{55}{3} \\end{align*} And the answer is $m+n = 058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).",
"Similar to Solution 1, we can find the average value of $|a_2 - a_1|$, and multiply this by 5 due to symmetry. And again due to symmetry, we can arbitrarily choose $a_2 > a_1$. Thus there are $\\binom{10}{2} = 45$ ways to pick the two values of $a_2$ and $a_1$ from the set $\\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\\}$ such that $a_2 > a_1$. First fix $a_2 = 10$, and vary $a_1$ from $1$ to $9$. Then fix $a_2 = 9$, and vary $a_1$ from $1$ to $8$. Continue, and you find that the sum of these $45$ ways to pick $|a_2 - a_1|$ is: $\\sum\\limits_{k = 1}^{9}\\sum\\limits_{j = 1}^{k}j = 45+36+28+21+15+10+6+3+1 = 165$. Thus, each term contributes on average $\\frac{165}{45}$, and the sum will be five times this, or $\\frac{165}{9} = \\frac{55}{3}$. The final answer is $p+q = 058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).",
"We use expected value of one of the sums, say $|a_2 - a_1|$, since all five are similar, so the average or expected value of the sum is just 5 times the expected value of one of them. To pick two random expected numbers from 1 to 10, we use a well known expected value trick by tying the two ends 1 and 10 with an extra number 0.* This gives 11 spaces, and we distribute 3 gaps evenly to choose our two numbers. We get $\\frac{11}{3}$ and $\\frac{22}{3}$ giving an absolute difference of $\\frac{11}{3}$. Therefore, the average/expected value of the sum of the five would be $\\frac{55}{3}$. This gives $55+3=058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together).",
"We use expected value of one of the sums, say $|a_2 - a_1|$, since all five are similar, so the average or expected value of the sum is just 5 times the expected value of one of them. To pick two random expected numbers from 1 to 10, we use a well known expected value trick by tying the two ends 1 and 10 with an extra number 0.* This gives 11 spaces, and we distribute 3 gaps evenly to choose our two numbers. We get $\\frac{11}{3}$ and $\\frac{22}{3}$ giving an absolute difference of $\\frac{11}{3}$. Therefore, the average/expected value of the sum of the five would be $\\frac{55}{3}$. This gives $55+3=058. This is like a bendable stick that has equally spaced marks \"end-1-2-3-4-5-6-7-8-9-10-end\", and the ends are bent together giving a circle with 11 equally spaced marks(the ends produce 1 mark together)."
] |
1996-I-13
| 1,996 | 13 |
In triangle $ABC$ , $AB=\sqrt{30}$ , $AC=\sqrt{6}$ , and $BC=\sqrt{15}$ . There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ , and $\angle ADB$ is a right angle. The ratio \[\dfrac{\text{Area}(\triangle ADB)}{\text{Area}(\triangle ABC)}\] can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 65 | null |
[
"[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP(\"A\",A)--MP(\"B\",B,SW)--MP(\"C\",C)--A--MP(\"D\",D)--B); D(MP(\"E\",E)); MP(\"\\sqrt{30}\",(A+B)/2,NW); MP(\"\\sqrt{6}\",(A+C)/2,SE); MP(\"\\frac{\\sqrt{15}}2\",(E+C)/2); D(rightanglemark(B,D,A)); [/asy] Let $E$ be the midpoint of $\\overline{BC}$. Since $BE = EC$, then $\\triangle ABE$ and $\\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$, so $\\frac{[BDE]}{[BAE]} = \\frac{DE}{AE}$ (see area ratios). Now, $\\dfrac{[ADB]}{[ABC]} = \\frac{[ABE] + [BDE]}{2[ABE]} = \\frac{1}{2} + \\frac{DE}{2AE}.$ By Stewart's Theorem, $AE = \\frac{\\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \\frac{\\sqrt {57}}{2}$, and by the Pythagorean Theorem on $\\triangle ABD, \\triangle EBD$, \\begin{align*} BD^2 + \\left(DE + \\frac {\\sqrt{57}}2\\right)^2 &= 30 \\\\ BD^2 + DE^2 &= \\frac{15}{4} \\\\ \\end{align*} Subtracting the two equations yields $DE\\sqrt{57} + \\frac{57}{4} = \\frac{105}{4} \\Longrightarrow DE = \\frac{12}{\\sqrt{57}}$. Then $\\frac mn = \\frac{1}{2} + \\frac{DE}{2AE} = \\frac{1}{2} + \\frac{\\frac{12}{\\sqrt{57}}}{2 \\cdot \\frac{\\sqrt{57}}{2}} = \\frac{27}{38}$, and $m+n = 065.",
"Because the problem asks for a ratio, we can divide each side length by $\\sqrt{3}$ to make things simpler. We now have a triangle with sides $\\sqrt{10}$, $\\sqrt{5}$, and $\\sqrt{2}$. We use the same graph as above. Draw perpendicular from $C$ to $AE$. Denote this point as $F$. We know that $DE = EF = x$ and $BD = CF = z$ and also let $AE = y$. Using Pythagorean theorem, we get three equations, $(y+x)^2 + z^2 = 10$ $(y-x)^2 + z^2 = 2$ $x^2 + z^2 = \\frac{5}{4}$ Adding the first and second, we obtain $x^2 + y^2 + z^2 = 6$, and then subtracting the third from this we find that $y = \\frac{\\sqrt{19}}{2}$. (Note, we could have used Stewart's Theorem to achieve this result). Subtracting the first and second, we see that $xy = 2$, and then we find that $x = \\frac{4}{\\sqrt{19}}$ Using base ratios, we then quickly find that the desired ratio is $\\frac{27}{38}$ so our answer is $065"
] |
1996-I-14
| 1,996 | 14 |
A $150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes?
| 768 | null |
[
"Solution 1 Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \\in \\mathbb{Z_{+}}$. We consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$, where $m \\in \\mathbb{R}, 0 < m \\le 1$. Consider a point on the diagonal with coordinates $(ma, mb, mc)$. We have 3 key observations as this point moves from $(0,0,0)$ towards $(a,b,c)$: When exactly one of $ma$, $mb$, $mc$ becomes a positive integer, the point crosses one of the faces of a $1 \\times 1 \\times 1$ cube from the outside to the inside of the cube. When exactly two of $ma$, $mb$, $mc$ become positive integers, the point enters a new cube through one of the edges of the cube from the outside to the inside of the cube. When all three $ma$, $mb$, $mc$ become positive integers, the point enters a cube through one of its vertices from the outside to the inside of the cube. The number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates. If we slice the solid up by the $x$-planes defined by $x=1,2,3,4, \\ldots, a$, the diagonal will cut these $x$-planes exactly $a$ times (plane of $x=0$ is not considered since $m \\ne 0$). Similar arguments for slices along $y$-planes and $z$-planes give diagonal cuts of $b$, and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$, if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception. But for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point. There are $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$. And for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$; ie $a+b+c$ over-counts each of such points by $2$. But since $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we have the final count for each such point as $1 = 3-3+k \\Rightarrow k=1$, where $k$ is our correction term. That is, we need to add $k=1$ time $\\gcd(a,b,c)$ back to account for the case of 3 simultaneous integers. Therefore, the total diagonal cube passes is: $D = a+b+c-\\left[ \\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a) \\right]+\\gcd(a,b,c)$. For $(a,b,c) = (150, 324, 375)$, we have: $\\gcd(150,324)=6$, $\\gcd(324,375)=3$, $\\gcd(150,375)=75$, $\\gcd(150,324,375)=3$. Therefore $D = 150+324+375-(6+3+75)+3 = 768",
"Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \\in \\mathbb{Z_{+}}$. We consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$, where $m \\in \\mathbb{R}, 0 < m \\le 1$. Consider a point on the diagonal with coordinates $(ma, mb, mc)$. We have 3 key observations as this point moves from $(0,0,0)$ towards $(a,b,c)$: When exactly one of $ma$, $mb$, $mc$ becomes a positive integer, the point crosses one of the faces of a $1 \\times 1 \\times 1$ cube from the outside to the inside of the cube. When exactly two of $ma$, $mb$, $mc$ become positive integers, the point enters a new cube through one of the edges of the cube from the outside to the inside of the cube. When all three $ma$, $mb$, $mc$ become positive integers, the point enters a cube through one of its vertices from the outside to the inside of the cube. The number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates. If we slice the solid up by the $x$-planes defined by $x=1,2,3,4, \\ldots, a$, the diagonal will cut these $x$-planes exactly $a$ times (plane of $x=0$ is not considered since $m \\ne 0$). Similar arguments for slices along $y$-planes and $z$-planes give diagonal cuts of $b$, and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$, if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception. But for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point. There are $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$. And for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$; ie $a+b+c$ over-counts each of such points by $2$. But since $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we have the final count for each such point as $1 = 3-3+k \\Rightarrow k=1$, where $k$ is our correction term. That is, we need to add $k=1$ time $\\gcd(a,b,c)$ back to account for the case of 3 simultaneous integers. Therefore, the total diagonal cube passes is: $D = a+b+c-\\left[ \\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a) \\right]+\\gcd(a,b,c)$. For $(a,b,c) = (150, 324, 375)$, we have: $\\gcd(150,324)=6$, $\\gcd(324,375)=3$, $\\gcd(150,375)=75$, $\\gcd(150,324,375)=3$. Therefore $D = 150+324+375-(6+3+75)+3 = 768",
"Consider a point travelling across the internal diagonal, and let the internal diagonal have a length of $d$. The point enters a new unit cube in the $x,y,z$ dimensions at multiples of $\\frac{d}{150}, \\frac{d}{324}, \\frac{d}{375}$ respectively. We proceed by using PIE. The point enters a new cube in the $x$ dimension $150$ times, in the $y$ dimension $324$ times and in the $z$ dimension, $375$ times. The point enters a new cube in the $x$ and $y$ dimensions whenever a multiple of $\\frac{d}{150}$ equals a multiple of $\\frac{d}{324}$. This occurs $\\gcd(150, 324)$ times. Similarly, a point enters a new cube in the $y,z$ dimensions $\\gcd(324, 375)$ times and a point enters a new cube in the $z,x$ dimensions $\\gcd(375, 150)$ times. The point enters a new cube in the $x,y$ and $z$ dimensions whenever some multiples of $\\frac{d}{150}, \\frac{d}{324}, \\frac{d}{375}$ are equal. This occurs $\\gcd(150, 324, 375)$ times. The total number of unit cubes entered is then $150+324+375-[\\gcd(150, 324)+\\gcd(324, 375)+\\gcd(375, 150)] + \\gcd(150, 324, 375) = 768"
] |
1996-I-15
| 1,996 | 15 |
In parallelogram $ABCD,$ let $O$ be the intersection of diagonals $\overline{AC}$ and $\overline{BD}$ . Angles $CAB$ and $DBC$ are each twice as large as angle $DBA,$ and angle $ACB$ is $r$ times as large as angle $AOB$ . Find the greatest integer that does not exceed $1000r$ .
| 777 | null |
[
"Solution 1 (trigonometry) [asy]size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--MP(\"D\",D,N)--cycle); D(B--D); D(A--C); D(MP(\"O\",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(anglemark(C,B,D,3.5));D(anglemark(C,B,D,4.5)); [/asy] Let $\\theta = \\angle DBA$. Then $\\angle CAB = \\angle DBC = 2\\theta$, $\\angle AOB = 180 - 3\\theta$, and $\\angle ACB = 180 - 5\\theta$. Since $ABCD$ is a parallelogram, it follows that $OA = OC$. By the Law of Sines on $\\triangle ABO,\\, \\triangle BCO$, $\\frac{\\sin \\angle CBO}{OC} = \\frac{\\sin \\angle ACB}{OB} \\quad \\text{and} \\quad \\frac{\\sin \\angle DBA}{OC} = \\frac{\\sin \\angle BAC}{OB}.$ Dividing the two equalities yields \\[\\frac{\\sin 2\\theta}{\\sin \\theta} = \\frac{\\sin (180 - 5\\theta)}{\\sin 2\\theta} \\Longrightarrow \\sin^2 2\\theta = \\sin 5\\theta \\sin \\theta.\\] Pythagorean and product-to-sum identities yield \\[1 - \\cos^2 2 \\theta = \\frac{\\cos 4\\theta - \\cos 6 \\theta}{2},\\] and the double and triple angle ($\\cos 3x = 4\\cos^3 x - 3\\cos x$) formulas further simplify this to \\[4\\cos^3 2\\theta - 4\\cos^2 2\\theta - 3\\cos 2\\theta + 3 = (4\\cos^2 2\\theta - 3)(\\cos 2\\theta - 1) = 0\\] The only value of $\\theta$ that fits in this context comes from $4 \\cos^2 2\\theta - 3 = 0 \\Longrightarrow \\cos 2\\theta = \\frac{\\sqrt{3}}{2} \\Longrightarrow \\theta = 15^{\\circ}$. The answer is $\\lfloor 1000r \\rfloor = \\left\\lfloor 1000 \\cdot \\frac{180 - 5\\theta}{180 - 3\\theta} \\right\\rfloor = \\left \\lfloor \\frac{7000}{9} \\right \\rfloor = 777. The answer follows as above.",
"[asy]size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--MP(\"D\",D,N)--cycle); D(B--D); D(A--C); D(MP(\"O\",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(anglemark(C,B,D,3.5));D(anglemark(C,B,D,4.5)); [/asy] Let $\\theta = \\angle DBA$. Then $\\angle CAB = \\angle DBC = 2\\theta$, $\\angle AOB = 180 - 3\\theta$, and $\\angle ACB = 180 - 5\\theta$. Since $ABCD$ is a parallelogram, it follows that $OA = OC$. By the Law of Sines on $\\triangle ABO,\\, \\triangle BCO$, $\\frac{\\sin \\angle CBO}{OC} = \\frac{\\sin \\angle ACB}{OB} \\quad \\text{and} \\quad \\frac{\\sin \\angle DBA}{OC} = \\frac{\\sin \\angle BAC}{OB}.$ Dividing the two equalities yields \\[\\frac{\\sin 2\\theta}{\\sin \\theta} = \\frac{\\sin (180 - 5\\theta)}{\\sin 2\\theta} \\Longrightarrow \\sin^2 2\\theta = \\sin 5\\theta \\sin \\theta.\\] Pythagorean and product-to-sum identities yield \\[1 - \\cos^2 2 \\theta = \\frac{\\cos 4\\theta - \\cos 6 \\theta}{2},\\] and the double and triple angle ($\\cos 3x = 4\\cos^3 x - 3\\cos x$) formulas further simplify this to \\[4\\cos^3 2\\theta - 4\\cos^2 2\\theta - 3\\cos 2\\theta + 3 = (4\\cos^2 2\\theta - 3)(\\cos 2\\theta - 1) = 0\\] The only value of $\\theta$ that fits in this context comes from $4 \\cos^2 2\\theta - 3 = 0 \\Longrightarrow \\cos 2\\theta = \\frac{\\sqrt{3}}{2} \\Longrightarrow \\theta = 15^{\\circ}$. The answer is $\\lfloor 1000r \\rfloor = \\left\\lfloor 1000 \\cdot \\frac{180 - 5\\theta}{180 - 3\\theta} \\right\\rfloor = \\left \\lfloor \\frac{7000}{9} \\right \\rfloor = 777. The answer follows as above.",
"[asy]size(180); pathpen = black+linewidth(0.7); pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP(\"A\",A,N)--MP(\"B\",B)--MP(\"C\",C)--MP(\"D\",D,N)--cycle); D(B--D); D(A--C); D(MP(\"O\",O,SE)); D(anglemark(D,B,A,4));D(anglemark(B,A,C,3.5));D(anglemark(B,A,C,4.5));D(anglemark(C,B,D,3.5));D(anglemark(C,B,D,4.5)); [/asy] Let $\\theta = \\angle DBA$. Then $\\angle CAB = \\angle DBC = 2\\theta$, $\\angle AOB = 180 - 3\\theta$, and $\\angle ACB = 180 - 5\\theta$. Since $ABCD$ is a parallelogram, it follows that $OA = OC$. By the Law of Sines on $\\triangle ABO,\\, \\triangle BCO$, $\\frac{\\sin \\angle CBO}{OC} = \\frac{\\sin \\angle ACB}{OB} \\quad \\text{and} \\quad \\frac{\\sin \\angle DBA}{OC} = \\frac{\\sin \\angle BAC}{OB}.$ Dividing the two equalities yields \\[\\frac{\\sin 2\\theta}{\\sin \\theta} = \\frac{\\sin (180 - 5\\theta)}{\\sin 2\\theta} \\Longrightarrow \\sin^2 2\\theta = \\sin 5\\theta \\sin \\theta.\\] Pythagorean and product-to-sum identities yield \\[1 - \\cos^2 2 \\theta = \\frac{\\cos 4\\theta - \\cos 6 \\theta}{2},\\] and the double and triple angle ($\\cos 3x = 4\\cos^3 x - 3\\cos x$) formulas further simplify this to \\[4\\cos^3 2\\theta - 4\\cos^2 2\\theta - 3\\cos 2\\theta + 3 = (4\\cos^2 2\\theta - 3)(\\cos 2\\theta - 1) = 0\\] The only value of $\\theta$ that fits in this context comes from $4 \\cos^2 2\\theta - 3 = 0 \\Longrightarrow \\cos 2\\theta = \\frac{\\sqrt{3}}{2} \\Longrightarrow \\theta = 15^{\\circ}$. The answer is $\\lfloor 1000r \\rfloor = \\left\\lfloor 1000 \\cdot \\frac{180 - 5\\theta}{180 - 3\\theta} \\right\\rfloor = \\left \\lfloor \\frac{7000}{9} \\right \\rfloor = 777. The answer follows as above.",
"Define $\\theta$ as above. Since $\\angle CAB = \\angle CBO$, it follows that $\\triangle COB \\sim \\triangle CBA$, and so $\\frac{CO}{BC} = \\frac{BC}{AC} \\Longrightarrow BC^2 = AC \\cdot CO = 2CO^2 \\Longrightarrow BC = CO\\sqrt{2}$. The Law of Sines on $\\triangle BOC$ yields that \\[\\frac{BC}{CO} = \\frac{\\sin (180-3\\theta)}{\\sin 2\\theta} = \\frac{\\sin 3\\theta}{\\sin 2\\theta} = \\sqrt{2}\\] Expanding using the sine double and triple angle formulas, we have \\[2\\sqrt {2} \\sin \\theta \\cos \\theta = \\sin\\theta( - 4\\sin ^2 \\theta + 3) \\Longrightarrow \\sin \\theta\\left(4\\cos^2\\theta - 2\\sqrt {2} \\cos \\theta - 1\\right) = 0.\\] By the quadratic formula, we have $\\cos \\theta = \\frac {2\\sqrt {2} \\pm \\sqrt {8 + 4 \\cdot 1 \\cdot 4}}{8} = \\frac {\\sqrt {6} \\pm \\sqrt {2}}{4}$, so $\\theta = 15^{\\circ}$ (as the other roots are too large to make sense in context). The answer follows as above. Solution 3 We will focus on $\\triangle ABC$. Let $\\angle ABO = x$, so $\\angle BAO = \\angle OBC = 2x$. Draw the perpendicular from $C$ intersecting $AB$ at $H$. Without loss of generality, let $AO = CO = 1$. Then $HO = 1$, since $O$ is the circumcenter of $\\triangle AHC$. Then $\\angle OHA = 2x$. By the Exterior Angle Theorem, $\\angle COB = 3x$ and $\\angle COH = 4x$. That implies that $\\angle HOB = x$. That makes $HO = HB = 1$. Then since by AA ($\\angle HBC = \\angle HOC = 3x$ and reflexive on $\\angle OCB$), $\\triangle OCB \\sim \\triangle BCA$. $\\frac {CO}{BC} = \\frac {BC}{AC} \\implies 2 = BC^2 = \\implies BC = \\sqrt {2}.$ Then by the Pythagorean Theorem, $1^2 + HC^2 = \\left(\\sqrt {2}\\right)^2\\implies HC = 1$. That makes $\\triangle HOC$ equilateral. Then $\\angle HOC = 4x = 60 \\implies x = 15$. The answer follows as above.",
"Define $\\theta$ as above. Since $\\angle CAB = \\angle CBO$, it follows that $\\triangle COB \\sim \\triangle CBA$, and so $\\frac{CO}{BC} = \\frac{BC}{AC} \\Longrightarrow BC^2 = AC \\cdot CO = 2CO^2 \\Longrightarrow BC = CO\\sqrt{2}$. The Law of Sines on $\\triangle BOC$ yields that \\[\\frac{BC}{CO} = \\frac{\\sin (180-3\\theta)}{\\sin 2\\theta} = \\frac{\\sin 3\\theta}{\\sin 2\\theta} = \\sqrt{2}\\] Expanding using the sine double and triple angle formulas, we have \\[2\\sqrt {2} \\sin \\theta \\cos \\theta = \\sin\\theta( - 4\\sin ^2 \\theta + 3) \\Longrightarrow \\sin \\theta\\left(4\\cos^2\\theta - 2\\sqrt {2} \\cos \\theta - 1\\right) = 0.\\] By the quadratic formula, we have $\\cos \\theta = \\frac {2\\sqrt {2} \\pm \\sqrt {8 + 4 \\cdot 1 \\cdot 4}}{8} = \\frac {\\sqrt {6} \\pm \\sqrt {2}}{4}$, so $\\theta = 15^{\\circ}$ (as the other roots are too large to make sense in context). The answer follows as above. Solution 3 We will focus on $\\triangle ABC$. Let $\\angle ABO = x$, so $\\angle BAO = \\angle OBC = 2x$. Draw the perpendicular from $C$ intersecting $AB$ at $H$. Without loss of generality, let $AO = CO = 1$. Then $HO = 1$, since $O$ is the circumcenter of $\\triangle AHC$. Then $\\angle OHA = 2x$. By the Exterior Angle Theorem, $\\angle COB = 3x$ and $\\angle COH = 4x$. That implies that $\\angle HOB = x$. That makes $HO = HB = 1$. Then since by AA ($\\angle HBC = \\angle HOC = 3x$ and reflexive on $\\angle OCB$), $\\triangle OCB \\sim \\triangle BCA$. $\\frac {CO}{BC} = \\frac {BC}{AC} \\implies 2 = BC^2 = \\implies BC = \\sqrt {2}.$ Then by the Pythagorean Theorem, $1^2 + HC^2 = \\left(\\sqrt {2}\\right)^2\\implies HC = 1$. That makes $\\triangle HOC$ equilateral. Then $\\angle HOC = 4x = 60 \\implies x = 15$. The answer follows as above.",
"We will focus on $\\triangle ABC$. Let $\\angle ABO = x$, so $\\angle BAO = \\angle OBC = 2x$. Draw the perpendicular from $C$ intersecting $AB$ at $H$. Without loss of generality, let $AO = CO = 1$. Then $HO = 1$, since $O$ is the circumcenter of $\\triangle AHC$. Then $\\angle OHA = 2x$. By the Exterior Angle Theorem, $\\angle COB = 3x$ and $\\angle COH = 4x$. That implies that $\\angle HOB = x$. That makes $HO = HB = 1$. Then since by AA ($\\angle HBC = \\angle HOC = 3x$ and reflexive on $\\angle OCB$), $\\triangle OCB \\sim \\triangle BCA$. $\\frac {CO}{BC} = \\frac {BC}{AC} \\implies 2 = BC^2 = \\implies BC = \\sqrt {2}.$ Then by the Pythagorean Theorem, $1^2 + HC^2 = \\left(\\sqrt {2}\\right)^2\\implies HC = 1$. That makes $\\triangle HOC$ equilateral. Then $\\angle HOC = 4x = 60 \\implies x = 15$. The answer follows as above."
] |
1997-I-1
| 1,997 | 1 |
How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?
| 750 | null |
[
"Notice that all odd numbers can be obtained by using $(a+1)^2-a^2=2a+1,$ where $a$ is a nonnegative integer. All multiples of $4$ can be obtained by using $(b+1)^2-(b-1)^2 = 4b$, where $b$ is a positive integer. Numbers congruent to $2 \\pmod 4$ cannot be obtained because squares are $0, 1 \\pmod 4.$ Thus, the answer is $500+250 = 750"
] |
1997-I-2
| 1,997 | 2 |
The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares. The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 125 | null |
[
"To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\\choose 2} = 36$. Similarly, there are ${9\\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles. For $s$, there are $8^2$ unit squares, $7^2$ of the $2\\times2$ squares, and so on until $1^2$ of the $8\\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\\cdots+8^2=\\dfrac{(8)(8+1)(2\\cdot8+1)}{6}=12*17=204$. Thus $\\frac sr = \\dfrac{204}{1296}=\\dfrac{17}{108}$, and $m+n=125.",
"First, to find the number of squares, we can look case by case by the side length of the possible squares on the checkerboard. We see that there are $8^2$ ways to place a $1$ x $1$ square and $7^2$ for a $2$ x $2$ square. This pattern can be easily generalized and we see that the number of squares is just $\\sum^8_{i=1}{i^2}$. This can be simplified by using the well-known formula for the sum of consecutive squares $\\frac{n(n+1)(2n+1)}{6}$ to get $204$. Then, to find the number of rectangles, first note that a square falls under the definition of a rectangle. We can break up the rectangles into cases for the length x width. As we note down the cases for $1$x$1, 1$x$2 , 2$x$1, 2$x$2,...,$ we see they are respectively $8$x$8, 8$x$7, 7$x$8, 7$x$7, ...$. We can quickly generalize this pattern to basically just ${\\sum^8_{i=1}{i}}\\cdot{\\sum^8_{i=1}{i}}$. This gets us ${(\\frac{9\\cdot8}{2})}^2,$ which is just $1296.$ Now, to calculate the ratio of $s/r,$ we divide $204$ by $1296$ to get a simplified fraction of $\\frac{17}{108}.$ Thus, our answer is just $s + r = 17+108 = 125 ~MathWhiz35"
] |
1997-I-3
| 1,997 | 3 |
Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?
| 126 | null |
[
"Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \\Longrightarrow 9xy-1000x-y=0$. Using SFFT, this factorizes to $(9x-1)\\left(y-\\dfrac{1000}{9}\\right)=\\dfrac{1000}{9}$, and $(9x-1)(9y-1000)=1000$. Since $89 < 9x-1 < 890$, we can use trial and error on factors of 1000. If $9x - 1 = 100$, we get a non-integer. If $9x - 1 = 125$, we get $x=14$ and $y=112$, which satisifies the conditions. Hence the answer is $112 + 14 = 126.",
"As shown above, we have $1000x+y=9xy$, so $1000/y=9-1/x$. $1000/y$ must be just a little bit smaller than 9, so we find $y=112$, $x=14$, and the solution is $126.",
"To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$ as $(90a+9b-1)(100x+10y+z) = 10000a + 1000b$ and $(90a+9b-1)(100x+10y+z) = 1000(10a + b)$ This is the most important part: Notice $(90a+9b-1)$ is $-1 \\pmod{10a+b}$ and $1000(10a + b)$ is $0\\pmod{10a+b}$. That means $(100x+10y+z)$ is also $0\\pmod{10+b}$. Rewrite $(100x+10y+z)$ as $n\\times(10a+b)$. $(90a+9b-1)\\times n(10a+b)= 1000(10a + b)$ $(90a+9b-1)\\times n= 1000$ Now we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\\pmod9$. It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$. Therefore the answer is $112 + 14 = 126. -jackshi2006"
] |
1997-I-4
| 1,997 | 4 |
Circles of radii 5, 5, 8, and $m/n$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 17 | null |
[
"If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\\frac mn = r$, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$. Then we form two right triangles, of lengths $5, x, 5+r$ and $5, 8+r+x, 13$, wher $x$ is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii $5$. By the Pythagorean Theorem, we now have two equations with two unknowns: \\begin{eqnarray*} 5^2 + x^2 &=& (5+r)^2 \\\\ x &=& \\sqrt{10r + r^2} \\\\ && \\\\ (8 + r + \\sqrt{10r+r^2})^2 + 5^2 &=& 13^2\\\\ 8 + r + \\sqrt{10r+r^2} &=& 12\\\\ \\sqrt{10r+r^2}&=& 4-r\\\\ 10r+r^2 &=& 16 - 8r + r^2\\\\ r &=& \\frac{8}{9} \\end{eqnarray*} So $m+n = 17. NOTE: It can be seen that there is no apparent need to use the variable x as a 5,12,13 right triangle has been formed.",
"We may also use Descartes' theorem, $k_4=k_1+k_2+k_3\\pm 2\\sqrt{k_1k_2+k_2k_3+k_3k_1}$ where each of $k_i$ is the curvature of a circle with radius $r_i$, and the curvature is defined as $k_i=\\frac{1}{r_i}$. The larger solution for $k_4$ will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get $k_4=\\frac15+\\frac15+\\frac18+2\\sqrt{\\frac{1}{40}+\\frac{1}{40}+\\frac{1}{25}}=\\frac{21}{40}+2\\sqrt{\\frac{45}{500}}=\\frac{45}{40}$. Thus, $r_4=\\frac{1}{k_4}=\\frac{40}{45}=\\frac89$, and the answer is $017"
] |
1997-I-5
| 1,997 | 5 |
The number $r$ can be expressed as a four-place decimal $0.abcd,$ where $a, b, c,$ and $d$ represent digits, any of which could be zero. It is desired to approximate $r$ by a fraction whose numerator is 1 or 2 and whose denominator is an integer. The closest such fraction to $r$ is $\frac 27.$ What is the number of possible values for $r$ ?
| 417 | null |
[
"The nearest fractions to $\\frac 27$ with numerator $1$ are $\\frac 13, \\frac 14$; and with numerator $2$ are $\\frac 26, \\frac 28 = \\frac 13, \\frac 14$ anyway. For $\\frac 27$ to be the best approximation for $r$, the decimal must be closer to $\\frac 27 \\approx .28571$ than to $\\frac 13 \\approx .33333$ or $\\frac 14 \\approx .25$. Thus $r$ can range between $\\frac{\\frac 14 + \\frac{2}{7}}{2} \\approx .267857$ and $\\frac{\\frac 13 + \\frac{2}{7}}{2} \\approx .309523$. At $r = .2678, .3096$, it becomes closer to the other fractions, so $.2679 \\le r \\le .3095$ and the number of values of $r$ is $3095 - 2679 + 1 = 417."
] |
1997-I-6
| 1,997 | 6 |
Point $B$ is in the exterior of the regular $n$ -sided polygon $A_1A_2\cdots A_n$ , and $A_1A_2B$ is an equilateral triangle. What is the largest value of $n$ for which $A_1$ , $A_n$ , and $B$ are consecutive vertices of a regular polygon?
| 42 | null |
[
"Let the other regular polygon have $m$ sides. Using the interior angle of a regular polygon formula, we have $\\angle A_2A_1A_n = \\frac{(n-2)180}{n}$, $\\angle A_nA_1B = \\frac{(m-2)180}{m}$, and $\\angle A_2A_1B = 60^{\\circ}$. Since those three angles add up to $360^{\\circ}$, \\begin{eqnarray*} \\frac{(n-2)180}{n} + \\frac{(m-2)180}{m} &=& 300\\\\ m(n-2)180 + n(m-2)180 &=& 300mn\\\\ 360mn - 360m - 360n &=& 300mn\\\\ mn - 6m - 6n &=& 0 \\end{eqnarray*} Using SFFT, \\begin{eqnarray*} (m-6)(n-6) &=& 36 \\end{eqnarray*} Clearly $n$ is maximized when $m = 7, n = 042.",
"As above, find that $mn - 6m - 6n = 0$ using the formula for the interior angle of a polygon. Solve for $n$ to find that $n = \\frac{6m}{m-6}$. Clearly, $m>6$ for $n$ to be positive. With this restriction of $m>6$, the larger $m$ gets, the smaller the fraction $\\frac{6m}{m-6}$ becomes. This can be proven either by calculus, by noting that $n = \\frac{6m}{m-6}$ is a transformed hyperbola, or by dividing out the rational function to get $n = 6 + \\frac{36}{m - 6}.$ Either way, minimizng $m$ will maximize $n$, and the smallest integer $m$ such that $n$ is positive is $m=7$, giving $n = 042",
"From the formula for the measure for an individual angle of a regular n-gon, $180 - \\frac{360}{n}$, the measure of $\\angle A_2A_1A_n = 180 - \\frac{360}{n}$. Together with the fact that an equilateral triangle has angles measuring 60 degrees, the measure of $\\angle A_nA_1B = 120 + \\frac{360}{n}$ (Notice that this value decreases as $n$ increases; hence, we are looking for the least possible value of $\\angle A_nA_1B$). For $A_n, A_1, B$ to be vertices of a regular polygon, $\\angle A_nA_1B$ must be of the form $180 - \\frac{360}{n}$, where $n$ is a natural number greater than or equal to 3. It is obvious that $\\angle A_nA_1B > 120$. The least angle satisfying this condition is $180 - \\frac{360}{7}$. Equating this with $120 + \\frac{360}{n}$ and solving yields $n = 042"
] |
1997-I-7
| 1,997 | 7 |
A car travels due east at $\frac 23$ miles per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac 12\sqrt{2}$ miles per minute. At time $t=0$ , the center of the storm is $110$ miles due north of the car. At time $t=t_1$ minutes, the car enters the storm circle, and at time $t=t_2$ minutes, the car leaves the storm circle. Find $\frac 12(t_1+t_2)$ .
| 198 | null |
[
"We set up a coordinate system, with the starting point of the car at the origin. At time $t$, the car is at $\\left(\\frac 23t,0\\right)$ and the center of the storm is at $\\left(\\frac{t}{2}, 110 - \\frac{t}{2}\\right)$. Using the distance formula, \\begin{eqnarray*} \\sqrt{\\left(\\frac{2}{3}t - \\frac 12t\\right)^2 + \\left(110-\\frac{t}{2}\\right)^2} &\\le& 51\\\\ \\frac{t^2}{36} + \\frac{t^2}{4} - 110t + 110^2 &\\le& 51^2\\\\ \\frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\\le& 0\\\\ \\end{eqnarray*} Noting that $\\frac 12 \\left(t_1+t_2 \\right)$ is at the maximum point of the parabola, we can use $-\\frac{b}{2a} = \\frac{110}{2 \\cdot \\frac{5}{18}} = 198.",
"First do the same process for assigning coordinates to the car. The car moves $\\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $\\left(\\frac{2}{3}t, 0\\right)$. Take the storm as circle. Given southeast movement, split the vector into component, getting position $\\left(\\frac{1}{2}t, 110 - \\frac{1}{2}t\\right)$ for the storm's center. This circle with radius 51 yields $\\left(x - \\frac{1}{2}t\\right)^2 + \\left(y -110 + \\frac{1}{2}t\\right)^2 = 51^2$. Now substitute the car's coordinates into the circle's: $\\left(\\frac{2}{3}t - \\frac{1}{2}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$. Simplifying and then squaring: $\\left(\\frac{1}{6}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$. $\\frac{1}{36}t^2 + \\frac{1}{4}t^2 - 110t + 110^2$ Forming into a quadratic we get the following, then set equal to 0, since the first time the car hits the circumference of the storm is $t_{1}$ and the second is $t_{2}$. $\\frac{5}{18}t^2 - 110t + 110^2 - 51^2 = 0$ The problem asks for sum of solutions divided by 2 so sum is equal to: $-\\frac{b}{a} = -\\frac{-110}{\\frac{5}{18}} = 110\\cdot{\\frac{18}{5}} = 396\\cdot{\\frac{1}{2}} = 198",
"First do the same process for assigning coordinates to the car. The car moves $\\frac{2}{3}$ miles per minute to the right, so the position starting from $(0,0)$ is $\\left(\\frac{2}{3}t, 0\\right)$. Take the storm as circle. Given southeast movement, split the vector into component, getting position $\\left(\\frac{1}{2}t, 110 - \\frac{1}{2}t\\right)$ for the storm's center. This circle with radius 51 yields $\\left(x - \\frac{1}{2}t\\right)^2 + \\left(y -110 + \\frac{1}{2}t\\right)^2 = 51^2$. Now substitute the car's coordinates into the circle's: $\\left(\\frac{2}{3}t - \\frac{1}{2}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$. Simplifying and then squaring: $\\left(\\frac{1}{6}t\\right)^2 + \\left(-110 + \\frac{1}{2}t\\right)^2 = 51^2$. $\\frac{1}{36}t^2 + \\frac{1}{4}t^2 - 110t + 110^2$ Forming into a quadratic we get the following, then set equal to 0, since the first time the car hits the circumference of the storm is $t_{1}$ and the second is $t_{2}$. $\\frac{5}{18}t^2 - 110t + 110^2 - 51^2 = 0$ The problem asks for sum of solutions divided by 2 so sum is equal to: $-\\frac{b}{a} = -\\frac{-110}{\\frac{5}{18}} = 110\\cdot{\\frac{18}{5}} = 396\\cdot{\\frac{1}{2}} = 198",
"We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\\left(\\frac{1}{6}, \\frac{1}{2}\\right)$. Labeling the car's starting position A, the storm center B, and the right triangle formed by AB with a right angle at B and the car's path, we get the following diagram, with AD as our desired length since D is the average of the points where the car enters and exits the storm. [asy] size(200,200); draw((0,0)--(0,110)); label(\"A\",(0,0),S); dot((0,0)); dot((0,110)); label(\"B\",(0,110),NE); draw(circle((0,110),51)); draw((0,0)--(161/3,161.0),EndArrow); draw((0,110)--(110/3,110.0)); label(\"C\",(110/3,110.0),SE); dot((110/3,110.0)); label(\"D\",(33,99),SE); dot((33,99)); draw((0,110)--(33,99)); markscalefactor=1; draw(rightanglemark((0,110),(33,99),(0,0))); [/asy] $AB = 110$, so $CB = \\frac{110}{3}$. The Pythagorean Theorem then gives $AC = \\frac{110\\sqrt{10}}{3}$, and since $\\bigtriangleup ABC \\sim \\bigtriangleup ADB$, $AD = (AB)\\frac{AB}{AC} = 33\\sqrt{10}$. The Pythagorean Theorem now gives the car's speed as $\\sqrt{\\frac{5}{18}}$, and finally $\\frac{33\\sqrt{10}}{\\sqrt{\\frac{5}{18}}} = 198.",
"We only need to know how the storm and car move relative to each other, so we can find this by subtracting the storm's movement vector from the car's. This gives the car's movement vector as $\\left(\\frac{1}{6}, \\frac{1}{2}\\right)$. Labeling the car's starting position A, the storm center B, and the right triangle formed by AB with a right angle at B and the car's path, we get the following diagram, with AD as our desired length since D is the average of the points where the car enters and exits the storm. [asy] size(200,200); draw((0,0)--(0,110)); label(\"A\",(0,0),S); dot((0,0)); dot((0,110)); label(\"B\",(0,110),NE); draw(circle((0,110),51)); draw((0,0)--(161/3,161.0),EndArrow); draw((0,110)--(110/3,110.0)); label(\"C\",(110/3,110.0),SE); dot((110/3,110.0)); label(\"D\",(33,99),SE); dot((33,99)); draw((0,110)--(33,99)); markscalefactor=1; draw(rightanglemark((0,110),(33,99),(0,0))); [/asy] $AB = 110$, so $CB = \\frac{110}{3}$. The Pythagorean Theorem then gives $AC = \\frac{110\\sqrt{10}}{3}$, and since $\\bigtriangleup ABC \\sim \\bigtriangleup ADB$, $AD = (AB)\\frac{AB}{AC} = 33\\sqrt{10}$. The Pythagorean Theorem now gives the car's speed as $\\sqrt{\\frac{5}{18}}$, and finally $\\frac{33\\sqrt{10}}{\\sqrt{\\frac{5}{18}}} = 198."
] |
1997-I-8
| 1,997 | 8 |
How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
| 90 | null |
[
"For more detailed explanations, see related problem (AIME I 2007, 10). The problem is asking us for all configurations of $4\\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns: The first two columns share no two numbers in the same row. There are ${4\\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\\choose 2}$ ways. This gives $6^2 = 36$. The first two columns share one number in the same row. There are ${4\\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \\cdot 3 \\cdot 2 \\cdot 2 = 48$. The first two columns share two numbers in the same row. There are ${4\\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed. Adding these cases up, we get $36 + 48 + 6 = 090.",
"Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $090. -pi_is_3.141",
"Notice that for every arrangement $A$ of the first rows of $-1$s and $1$s, we have the inverse of that row $A^{-1}$ so that the sum of the rows and columns of $A$ and $A^{-1}$ is $0$. Therefore if we have another arrangement $B$, we have $B^{-1}$. For instance, if $A=(-1,1,1,-1)$, $A^{-1}=(1,-1,-1,1)$. We then have that if we fix the first row $A$, we have that first there are $\\binom{4}{2}$ values of the fixed $A$. We then have the following cases: Case $1$: ($AA^{-1}BB^{-1}$). $\\binom{4}{2}$. Case $2$: ($ABA^{-1}B^{-1}$). $\\binom{4}{2}$. Case $3$: ($AAA^{-1}A^{-1}$). $\\binom{4}{2}/2$, where here we divided by $2$ because then we would overcount $AA$ and $A^{-1}A^{-1}$. Therefore the answer is $\\binom{4}{2}(\\binom{4}{2}+\\binom{4}{2}+\\binom{4}{2}/2)$=$6(6+6+3)$=$090. ~th1nq3r"
] |
1997-I-9
| 1,997 | 9 |
Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ .
| 233 | null |
[
"Looking at the properties of the number, it is immediately guess-able that $a = \\phi = \\frac{1+\\sqrt{5}}2$ (the golden ratio) is the answer. The following is the way to derive that: Since $\\sqrt{2} < a < \\sqrt{3}$, $0 < \\frac{1}{\\sqrt{3}} < a^{-1} < \\frac{1}{\\sqrt{2}} < 1$. Thus $\\langle a^2 \\rangle = a^{-1}$, and it follows that $a^2 - 2 = a^{-1} \\Longrightarrow a^3 - 2a - 1 = 0$. Noting that $-1$ is a root, this factors to $(a+1)(a^2 - a - 1) = 0$, so $a = \\frac{1 + \\sqrt{5}}{2}$ (we discard the negative root). Our answer is $(a^2)^{6}-144a^{-1} = \\left(\\frac{3+\\sqrt{5}}2\\right)^6 - 144\\left(\\frac{2}{1 + \\sqrt{5}}\\right)$. Complex conjugates reduce the second term to $-72(\\sqrt{5}-1)$. The first term we can expand by the binomial theorem to get $\\frac 1{2^6}\\left(3^6 + 6\\cdot 3^5\\sqrt{5} + 15\\cdot 3^4 \\cdot 5 + 20\\cdot 3^3 \\cdot 5\\sqrt{5} + 15 \\cdot 3^2 \\cdot 25 + 6 \\cdot 3 \\cdot 25\\sqrt{5} + 5^3\\right)$ $= \\frac{1}{64}\\left(10304 + 4608\\sqrt{5}\\right) = 161 + 72\\sqrt{5}$. The answer is $161 + 72\\sqrt{5} - 72\\sqrt{5} + 72 = 233.",
"Find $a$ as shown above. Note that, since $a$ is a root of the equation $a^3 - 2a - 1 = 0$, $a^3 = 2a + 1$, and $a^{12} = (2a + 1)^4$. Also note that, since $a$ is a root of $a^2 - a - 1 = 0$, $\\frac{1}{a} = a - 1$. The expression we wish to calculate then becomes $(2a + 1)^4 - 144(a - 1)$. Plugging in $a = \\frac{1 + \\sqrt{5}}{2}$, we plug in to get an answer of $(161 + 72\\sqrt{5}) + 72 - 72\\sqrt{5} = 161 + 72 = 233.",
"Find $a$ as shown above. Note that $a$ satisfies the equation $a^2 = a+1$ (this is the equation we solved to get it). Then, we can simplify $a^{12}$ as follows using the fibonacci numbers: $a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^9+ 2a^8 = ... = 144a^1+89a^0 = 144a+89$ So we want $144(a-\\frac1a)+89 = 144(1)+89 = 233.",
"As Solution 1 stated, $a^3 - 2a - 1 = 0$. $a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)$. So, $a^2 - a - 1 = 0$, $1 = a^2 - a$, $\\frac1a = a-1$, $a^3 = 2a+1$, $a^2 = a+1$. $a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5$ $a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 144a + 89$ Therefore, $a^{12} - 144 a^{-1} = 144a + 89 - 144(a-1) = 89 + 144 = 233 ~isabelchen"
] |
1997-I-10
| 1,997 | 10 |
Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true: i. Either each of the three cards has a different shape or all three of the card have the same shape. ii. Either each of the three cards has a different color or all three of the cards have the same color. iii. Either each of the three cards has a different shade or all three of the cards have the same shade. How many different complementary three-card sets are there?
| 117 | null |
[
"Case 1: All three attributes are the same. This is impossible since sets contain distinct cards. Case 2: Two of the three attributes are the same. There are ${3\\choose 2}$ ways to pick the two attributes in question. Then there are $3$ ways to pick the value of the first attribute, $3$ ways to pick the value of the second attribute, and $1$ way to arrange the positions of the third attribute, giving us ${3\\choose 2} \\cdot 3 \\cdot 3 = 27$ ways. Case 3: One of the three attributes are the same. There are ${3\\choose 1}$ ways to pick the one attribute in question, and then $3$ ways to pick the value of that attribute. Then there are $3!$ ways to arrange the positions of the next two attributes, giving us ${3\\choose 1} \\cdot 3 \\cdot 3! = 54$ ways. Case 4: None of the three attributes are the same. We fix the order of the first attribute, and then there are $3!$ ways to pick the ordering of the second attribute and $3!$ ways to pick the ordering of the third attribute. This gives us $(3!)^2 = 36$ ways. Adding the cases up, we get $27 + 54 + 36 = 117.",
"Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of $\\binom{27}{2} = 27*13 = 351$ possibilities. Note, however, that each set is generated by ${3\\choose 2} = 3$ pairs, so we've overcounted by a multiple of 3 and the answer is $\\frac{351}{3} = 117.",
"Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out $3^3=27$ possibilities. Notice that we have counted every set $3!=6$ times by treating the set as ordered. The final solution is then $\\frac{729-27}{6}=117"
] |
1997-I-11
| 1,997 | 11 |
Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$ ?
| 241 | null |
[
"Note that $\\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=1}^{44} \\sin n} = \\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=46}^{89} \\cos n} = \\frac {\\cos 1 + \\cos 2 + \\dots + \\cos 44}{\\cos 89 + \\cos 88 + \\dots + \\cos 46}$ by the cofunction identities.(We could have also written it as $\\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=1}^{44} \\sin n} = \\frac{\\sum_{n=46}^{89} \\sin n}{\\sum_{n=1}^{44} \\sin n} = \\frac {\\sin 89 + \\sin 88 + \\dots + \\sin 46}{\\sin 1 + \\sin 2 + \\dots + \\sin 44}$.) Now use the sum-product formula $\\cos x + \\cos y = 2\\cos\\left(\\frac{x+y}{2}\\right)\\cos\\left(\\frac{x-y}{2}\\right)$ We want to pair up $[1, 44]$, $[2, 43]$, $[3, 42]$, etc. from the numerator and $[46, 89]$, $[47, 88]$, $[48, 87]$ etc. from the denominator. Then we get: \\[\\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=1}^{44} \\sin n} = \\frac{\\sum_{n=1}^{44} \\cos n}{\\sum_{n=46}^{89} \\cos n} = \\frac{2\\cos(\\frac{45}{2})[\\cos(\\frac{43}{2})+\\cos(\\frac{41}{2})+\\dots+\\cos(\\frac{1}{2})]}{2\\cos(\\frac{135}{2})[\\cos(\\frac{43}{2})+\\cos(\\frac{41}{2})+\\dots+\\cos(\\frac{1}{2})]} \\Rightarrow \\frac{\\cos(\\frac{45}{2})}{\\cos(\\frac{135}{2})}\\] To calculate this number, use the half angle formula. Since $\\cos\\left(\\frac{x}{2}\\right) = \\pm \\sqrt{\\frac{\\cos x + 1}{2}}$, then our number becomes: \\[\\frac{\\sqrt{\\frac{\\frac{\\sqrt{2}}{2} + 1}{2}}}{\\sqrt{\\frac{\\frac{-\\sqrt{2}}{2} + 1}{2}}}\\] in which we drop the negative roots (as it is clear cosine of $22.5$ and $67.5$ are positive). We can easily simplify this: \\begin{eqnarray*} \\frac{\\sqrt{\\frac{\\frac{\\sqrt{2}}{2} + 1}{2}}}{\\sqrt{\\frac{\\frac{-\\sqrt{2}}{2} + 1}{2}}} &=& \\sqrt{\\frac{\\frac{2+\\sqrt{2}}{4}}{\\frac{2-\\sqrt{2}}{4}}} \\\\ &=& \\sqrt{\\frac{2+\\sqrt{2}}{2-\\sqrt{2}}} \\cdot \\sqrt{\\frac{2+\\sqrt{2}}{2+\\sqrt{2}}} \\\\ &=& \\sqrt{\\frac{(2+\\sqrt{2})^2}{2}} \\\\ &=& \\frac{2+\\sqrt{2}}{\\sqrt{2}} \\\\ &=& \\sqrt{2}+1 \\end{eqnarray*} And hence our answer is $\\lfloor 100x \\rfloor = \\lfloor 100(1 + \\sqrt {2}) \\rfloor = 241.",
"\\begin{eqnarray*} x &=& \\frac {\\sum_{n = 1}^{44} \\cos n^\\circ}{\\sum_{n = 1}^{44} \\sin n^\\circ} = \\frac {\\cos 1 + \\cos 2 + \\dots + \\cos 44}{\\sin 1 + \\sin 2 + \\dots + \\sin 44}\\\\ &=& \\frac {\\cos (45 - 1) + \\cos(45 - 2) + \\dots + \\cos(45 - 44)}{\\sin 1 + \\sin 2 + \\dots + \\sin 44} \\end{eqnarray*} Using the identity $\\sin a + \\sin b = 2\\sin \\frac{a+b}2 \\cos \\frac{a-b}{2}$ $\\Longrightarrow \\sin x + \\cos x$ $= \\sin x + \\sin (90-x)$ $= 2 \\sin 45 \\cos (45-x)$ $= \\sqrt{2} \\cos (45-x)$, that summation reduces to \\begin{eqnarray*}x &=& \\left(\\frac {1}{\\sqrt {2}}\\right)\\left(\\frac {(\\cos 1 + \\cos2 + \\dots + \\cos44) + (\\sin1 + \\sin2 + \\dots + \\sin44)}{\\sin1 + \\sin2 + \\dots + \\sin44}\\right)\\\\ &=& \\left(\\frac {1}{\\sqrt {2}}\\right)\\left(1 + \\frac {\\cos 1 + \\cos 2 + \\dots + \\cos 44}{\\sin 1 + \\sin 2 + \\dots + \\sin 44}\\right) \\end{eqnarray*} This fraction is equivalent to $x$. Therefore, \\begin{eqnarray*} x &=& \\left(\\frac {1}{\\sqrt {2}}\\right)\\left(1 + x\\right)\\\\ \\frac {1}{\\sqrt {2}} &=& x\\left(\\frac {\\sqrt {2} - 1}{\\sqrt {2}}\\right)\\\\ x &=& \\frac {1}{\\sqrt {2} - 1} = 1 + \\sqrt {2}\\\\ \\lfloor 100x \\rfloor &=& \\lfloor 100(1 + \\sqrt {2}) \\rfloor = 241\\\\ \\end{eqnarray*}",
"A slight variant of the above solution, note that \\begin{eqnarray*} \\sum_{n=1}^{44} \\cos n + \\sum_{n=1}^{44} \\sin n &=& \\sum_{n=1}^{44} \\sin n + \\sin(90-n)\\\\ &=& \\sqrt{2}\\sum_{n=1}^{44} \\cos(45-n) = \\sqrt{2}\\sum_{n=1}^{44} \\cos n\\\\ \\sum_{n=1}^{44} \\sin n &=& (\\sqrt{2}-1)\\sum_{n=1}^{44} \\cos n \\end{eqnarray*} This is the ratio we are looking for. $x$ reduces to $\\frac{1}{\\sqrt{2} - 1} = \\sqrt{2} + 1$, and $\\lfloor 100(\\sqrt{2} + 1)\\rfloor = 241.",
"Consider the sum $\\sum_{n = 1}^{44} \\text{cis } n^\\circ$. The fraction is given by the real part divided by the imaginary part. The sum can be written $- 1 + \\sum_{n = 0}^{44} \\text{cis } n^\\circ = - 1 + \\frac {\\text{cis } 45^\\circ - 1}{\\text{cis } 1^\\circ - 1}$ (by De Moivre's Theorem with geometric series) $= - 1 + \\frac {\\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2}}{\\text{cis } 1^\\circ - 1} = - 1 + \\frac {\\left( \\frac {\\sqrt {2}}{2} - 1 + \\frac {i \\sqrt {2}}{2} \\right) (\\text{cis } ( - 1^\\circ) - 1)}{(\\cos 1^\\circ - 1)^2 + \\sin^2 1^\\circ}$ (after multiplying by complex conjugate) $= - 1 + \\frac {\\left( \\frac {\\sqrt {2}}{2} - 1 \\right) (\\cos 1^\\circ - 1) + \\frac {\\sqrt {2}}{2}\\sin 1^\\circ + i\\left( \\left(1 - \\frac {\\sqrt {2}}{2} \\right) \\sin 1^\\circ + \\frac {\\sqrt {2}}{2} (\\cos 1^\\circ - 1)\\right)}{2(1 - \\cos 1^\\circ)}$ $= - \\frac {1}{2} - \\frac {\\sqrt {2}}{4} - \\frac {i\\sqrt {2}}{4} + \\frac {\\sin 1^\\circ \\left( \\frac {\\sqrt {2}}{2} + i\\left( 1 - \\frac {\\sqrt {2}}{2} \\right) \\right)}{2(1 - \\cos 1^\\circ)}$ Using the tangent half-angle formula, this becomes $\\left( - \\frac {1}{2} + \\frac {\\sqrt {2}}{4}[\\cot (1/2^\\circ) - 1] \\right) + i\\left( \\frac {1}{2}\\cot (1/2^\\circ) - \\frac {\\sqrt {2}}{4}[\\cot (1/2^\\circ) + 1] \\right)$. Dividing the two parts and multiplying each part by 4, the fraction is $\\frac { - 2 + \\sqrt {2}[\\cot (1/2^\\circ) - 1]}{2\\cot (1/2^\\circ) - \\sqrt {2}[\\cot (1/2^\\circ) + 1]}$. Although an exact value for $\\cot (1/2^\\circ)$ in terms of radicals will be difficult, this is easily known: it is really large! So treat it as though it were $\\infty$. The fraction is approximated by $\\frac {\\sqrt {2}}{2 - \\sqrt {2}} = \\frac {\\sqrt {2}(2 + \\sqrt {2})}{2} = 1 + \\sqrt {2}\\Rightarrow \\lfloor 100(1+\\sqrt2)\\rfloor=241.",
"Consider the sum $\\sum_{n = 1}^{44} \\text{cis } n^\\circ$. The fraction is given by the real part divided by the imaginary part. The sum can be written as $\\sum_{n=1}^{22} (\\text{cis } n^\\circ + \\text{cis } 45-n^\\circ)$. Consider the rhombus $OABC$ on the complex plane such that $O$ is the origin, $A$ represents $\\text{cis } n^\\circ$, $B$ represents $\\text{cis } n^\\circ + \\text{cis } 45-n^\\circ$ and $C$ represents $\\text{cis } n^\\circ$. Simple geometry shows that $\\angle BOA = 22.5-k^\\circ$, so the angle that $\\text{cis } n^\\circ + \\text{cis } 45-n^\\circ$ makes with the real axis is simply $22.5^\\circ$. So $\\sum_{n=1}^{22} (\\text{cis } n^\\circ + \\text{cis } 45-n^\\circ)$ is the sum of collinear complex numbers, so the angle the sum makes with the real axis is $22.5^\\circ$. So our answer is $\\lfloor 100 \\cot(22.5^\\circ) \\rfloor = 241 can be shown easily through half-angle formula.",
"We write $x =\\frac{\\sum_{n=46}^{89} \\sin n^{\\circ}}{\\sum_{n=1}^{44} \\sin n^{\\circ}}$ since $\\cos x = \\sin (90^{\\circ}-x).$ Now we by the sine angle sum we know that $\\sin (x+45^{\\circ}) = \\sin 45^{\\circ}(\\sin x + \\cos x).$ So the expression simplifies to $\\sin 45^{\\circ}\\left(\\frac{\\sum_{n=1}^{44} (\\sin n^{\\circ}+\\cos n^{\\circ})}{\\sum_{n=1}^{44} \\sin n^{\\circ}}\\right) = \\sin 45^{\\circ}\\left(1+\\frac{\\sum_{n=1}^{44} \\cos n^{\\circ}}{\\sum_{n=1}^{44} \\sin n^{\\circ}}\\right)=\\sin 45^{\\circ}(1+x).$ Therefore we have the equation $x = \\sin 45^{\\circ}(1+x) \\implies x = \\sqrt{2}+1.$ Finishing, we have $\\lfloor 100x \\rfloor = 241",
"We can pair the terms of the summations as below. \\[\\dfrac{(\\cos{1} + \\cos{44}) + (\\cos{2} + \\cos{43}) + (\\cos{3} + \\cos{42}) + \\cdots + (\\cos{22} + \\cos{23})}{(\\sin{1} + \\sin{44}) + (\\sin{2} + \\sin{43}) + (\\sin{3} + \\sin{42}) + \\cdots + (\\sin{22} + \\sin{23})}.\\] From here, we use the cosine and sine subtraction formulas as shown. \\begin{align*} &\\dfrac{(\\cos(45-44) + \\cos(45-1)) + (\\cos(45-43) + \\cos(45-2))+ \\cdots (\\cos(45-23) + \\cos(45-22))}{(\\sin(45-44) + \\sin(45-1)) + (\\sin(45-43) + \\sin(45-2))+ \\cdots (\\sin(45-23) + \\sin(45-22))} \\\\ &= \\dfrac{(\\cos{45}\\cos{44} +\\sin{45}\\sin{44} + \\cos{45}\\cos{1} + \\sin{45}\\sin{1})+ \\cdots + (\\cos{45}\\cos{23} + \\sin{45}\\sin{23} + \\cos{45}\\cos{22} + \\sin{45}\\sin{22})}{(\\sin{45}\\cos{44}-\\cos{45}\\sin{44} + \\sin{45}\\cos{1} - \\cos{45}\\sin{1}) + \\cdots + (\\sin{45}\\cos{23} -\\cos{45}\\sin{23} + \\sin{45}\\cos{22} - \\cos{45}\\sin{22})} \\\\ &= \\dfrac{\\sqrt{2}/2(\\cos{44} + \\sin{44} + \\cos{1}+\\sin{1} +\\cos{43} + \\sin{43} + \\cos{2} + \\sin{2} + \\cdots + \\cos{23} + \\sin{23} + \\cos{22} + \\sin{22})}{\\sqrt{2}/2(\\cos{44} -\\sin{44} +\\cos{1} - \\sin{1} + \\cos{43}-\\sin{43} + \\cos{2} -\\sin{2} +\\cdots + \\cos{23}-\\sin{23} + \\cos{22} - \\sin{22})} \\\\ &=\\dfrac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ + \\sum\\limits_{n=1}^{44} \\sin n^\\circ}{\\sum\\limits_{n=1}^{44} \\cos n^\\circ - \\sum\\limits_{n=1}^{44} \\sin n^\\circ}. \\end{align*} Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression. \\[\\dfrac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ + \\sum\\limits_{n=1}^{44} \\sin n^\\circ}{\\sum\\limits_{n=1}^{44} \\cos n^\\circ - \\sum\\limits_{n=1}^{44} \\sin n^\\circ} = \\frac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ}.\\] For the sake of clarity, let $\\sum\\limits_{n=1}^{44} \\cos n^\\circ = C$ and $\\sum\\limits_{n=1}^{44} \\sin n^\\circ = S$. Then, we have \\[\\dfrac{C+S}{C-S} = \\dfrac{C}{S} \\implies CS+S^2 = C^2-CS.\\] Finishing, we have $S^2+2CS=C^2$. Adding $C^2$ to both sides gives $(C+S)^2 = 2C^2$, or $C+S = \\pm C\\sqrt{2}$. Taking the positive case gives $S= C(\\sqrt{2}-1)$. Finally, \\[x=\\frac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ} = \\dfrac{C}{S} = \\dfrac{1}{\\sqrt{2}-1} = \\sqrt{2} +1 \\implies \\lfloor{100x\\rfloor} = 241.\\] ~NTfish",
"Because the numerator and denominator are both sums, we can replace each with $44$ times the average (i.e. arithmetic mean) of the terms of the respective sums. These $44$s will cancel out, leaving simply the ratio of the averages of the respective sums. Given that the sums span from $1^{\\circ}\\approx 0$ radians to $44^{\\circ}\\approx \\tfrac\\pi4$ radians, we would expect the average heights of cosine and sine on $[0,\\tfrac\\pi4]$ to be very close if not equal to the given ratio. We can use definite integrals to calculate these averages. The average height of $\\cos x$ on $[0,\\tfrac\\pi4]$ is $\\frac1{\\pi/4}\\int^{\\pi/4}_0(\\cos x) dx = \\frac4\\pi(\\sin\\frac\\pi4-\\sin0) = \\frac4\\pi\\cdot\\frac{\\sqrt2}2$. Similarly, the average height of $\\sin x$ on $[0,\\tfrac\\pi4]$ is $\\frac1{\\pi/4}\\int^{\\pi/4}_0(\\sin x) dx = \\frac4\\pi(-\\cos\\frac\\pi4+\\cos0) = \\frac4\\pi\\cdot(1-\\frac{\\sqrt2}2)$. Taking the ratio of these two values yields $\\frac{\\sqrt2/2}{1-\\sqrt2/2}=\\frac1{\\sqrt2-1}\\cdot\\frac{\\sqrt2+1}{\\sqrt2+1}=\\frac{\\sqrt2+1}{2-1}=\\sqrt2+1 \\approx 1.41+1 = 2.41$. Thus, our answer is $241.",
"Because the numerator and denominator are both sums, we can replace each with $44$ times the average (i.e. arithmetic mean) of the terms of the respective sums. These $44$s will cancel out, leaving simply the ratio of the averages of the respective sums. Given that the sums span from $1^{\\circ}\\approx 0$ radians to $44^{\\circ}\\approx \\tfrac\\pi4$ radians, we would expect the average heights of cosine and sine on $[0,\\tfrac\\pi4]$ to be very close if not equal to the given ratio. We can use definite integrals to calculate these averages. The average height of $\\cos x$ on $[0,\\tfrac\\pi4]$ is $\\frac1{\\pi/4}\\int^{\\pi/4}_0(\\cos x) dx = \\frac4\\pi(\\sin\\frac\\pi4-\\sin0) = \\frac4\\pi\\cdot\\frac{\\sqrt2}2$. Similarly, the average height of $\\sin x$ on $[0,\\tfrac\\pi4]$ is $\\frac1{\\pi/4}\\int^{\\pi/4}_0(\\sin x) dx = \\frac4\\pi(-\\cos\\frac\\pi4+\\cos0) = \\frac4\\pi\\cdot(1-\\frac{\\sqrt2}2)$. Taking the ratio of these two values yields $\\frac{\\sqrt2/2}{1-\\sqrt2/2}=\\frac1{\\sqrt2-1}\\cdot\\frac{\\sqrt2+1}{\\sqrt2+1}=\\frac{\\sqrt2+1}{2-1}=\\sqrt2+1 \\approx 1.41+1 = 2.41$. Thus, our answer is $241.",
"$x=\\frac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ}=\\frac{\\sum\\limits_{n=46}^{89} \\sin n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ}$ $=\\frac{\\sum\\limits_{n=46}^{89} \\sin n^\\circ\\sin0.5^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ\\sin0.5^\\circ}$ $=\\frac{\\sum\\limits_{n=46}^{89}\\frac{1}{2}\\left(\\cos(n-0.5)^\\circ-\\cos(n+0.5)^\\circ\\right)}{\\sum\\limits_{n=1}^{44}\\frac{1}{2}\\left(\\cos(n-0.5)^\\circ-\\cos(n+0.5)^\\circ\\right)}$ $=\\frac{\\frac{1}{2}(\\cos45.5^\\circ-\\cos89.5^\\circ)}{\\frac{1}{2}(\\cos0.5^\\circ-\\cos44.5^\\circ)}$ $=\\frac{\\cos45.5^\\circ-\\cos89.5^\\circ}{\\cos0.5^\\circ-\\cos44.5^\\circ}$ $=\\frac{2\\sin\\left(\\frac{45.5+89.5}{2}\\right)^\\circ\\sin\\left(\\frac{45.5-89.5}{2}\\right)^\\circ}{2\\sin\\left(\\frac{0.5+44.5}{2}\\right)^\\circ\\sin\\left(\\frac{0.5-44.5}{2}\\right)^\\circ}$ $=\\frac{2\\sin\\frac{135}{2}^\\circ\\sin(-22)^\\circ}{2\\sin\\frac{45}{2}^\\circ\\sin(-22)^\\circ}$ $=\\frac{\\sin\\frac{135}{2}^\\circ}{\\sin\\frac{45}{2}^\\circ}=\\tan\\frac{135}{2}^\\circ=\\frac{1+\\frac{\\sqrt{2}}{2}}{\\frac{\\sqrt{2}}{2}}=1+\\sqrt{2}\\approx 2.4142$ $\\Rightarrow 100x\\approx 241.42\\Rightarrow241 ~eevee9406",
"$x=\\frac{\\sum\\limits_{n=1}^{44} \\cos n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ}=\\frac{\\sum\\limits_{n=46}^{89} \\sin n^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ}$ $=\\frac{\\sum\\limits_{n=46}^{89} \\sin n^\\circ\\sin0.5^\\circ}{\\sum\\limits_{n=1}^{44} \\sin n^\\circ\\sin0.5^\\circ}$ $=\\frac{\\sum\\limits_{n=46}^{89}\\frac{1}{2}\\left(\\cos(n-0.5)^\\circ-\\cos(n+0.5)^\\circ\\right)}{\\sum\\limits_{n=1}^{44}\\frac{1}{2}\\left(\\cos(n-0.5)^\\circ-\\cos(n+0.5)^\\circ\\right)}$ $=\\frac{\\frac{1}{2}(\\cos45.5^\\circ-\\cos89.5^\\circ)}{\\frac{1}{2}(\\cos0.5^\\circ-\\cos44.5^\\circ)}$ $=\\frac{\\cos45.5^\\circ-\\cos89.5^\\circ}{\\cos0.5^\\circ-\\cos44.5^\\circ}$ $=\\frac{2\\sin\\left(\\frac{45.5+89.5}{2}\\right)^\\circ\\sin\\left(\\frac{45.5-89.5}{2}\\right)^\\circ}{2\\sin\\left(\\frac{0.5+44.5}{2}\\right)^\\circ\\sin\\left(\\frac{0.5-44.5}{2}\\right)^\\circ}$ $=\\frac{2\\sin\\frac{135}{2}^\\circ\\sin(-22)^\\circ}{2\\sin\\frac{45}{2}^\\circ\\sin(-22)^\\circ}$ $=\\frac{\\sin\\frac{135}{2}^\\circ}{\\sin\\frac{45}{2}^\\circ}=\\tan\\frac{135}{2}^\\circ=\\frac{1+\\frac{\\sqrt{2}}{2}}{\\frac{\\sqrt{2}}{2}}=1+\\sqrt{2}\\approx 2.4142$ $\\Rightarrow 100x\\approx 241.42\\Rightarrow241 ~eevee9406"
] |
1997-I-12
| 1,997 | 12 |
The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ , $b$ , $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ , $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$ .
| 58 | null |
[
"First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have $\\frac {a\\frac {ax + b}{cx + d} + b}{c\\frac {ax + b}{cx + d} + d} = x$, which reduces to \\[\\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} =\\frac {px + q}{rx + s} = x.\\] In order for this fraction to reduce to $x$, we must have $q = r = 0$ and $p = s\\not = 0$. From $c(a + d) = b(a + d) = 0$, we get $a = - d$ or $b = c = 0$. The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$, so $f(x) = \\frac {ax + b}{cx - a}$. The only value that is not in the range of this function is $\\frac {a}{c}$. To find $\\frac {a}{c}$, we use the two values of the function given to us. We get $2(97)a + b = 97^2c$ and $2(19)a + b = 19^2c$. Subtracting the second equation from the first will eliminate $b$, and this results in $2(97 - 19)a = (97^2 - 19^2)c$, so \\[\\frac {a}{c} = \\frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 .\\] Alternatively, we could have found out that $a = -d$ by using the fact that $f(f(-b/a))=-b/a$.",
"First, we note that $e = \\frac ac$ is the horizontal asymptote of the function, and since this is a linear function over a linear function, the unique number not in the range of $f$ will be $e$. $\\frac{ax+b}{cx+d} = \\frac{b-\\frac{cd}{a}}{cx+d} + \\frac{a}{c}$. Without loss of generality, let $c=1$, so the function becomes $\\frac{b- \\frac{d}{a}}{x+d} + e$. (Considering $\\infty$ as a limit) By the given, $f(f(\\infty)) = \\infty$. $\\lim_{x \\rightarrow \\infty} f(x) = e$, so $f(e) = \\infty$. $f(x) \\rightarrow \\infty$ as $x$ reaches the vertical asymptote, which is at $-\\frac{d}{c} = -d$. Hence $e = -d$. Substituting the givens, we get \\begin{align*} 19 &= \\frac{b - \\frac da}{19 - e} + e\\\\ 97 &= \\frac{b - \\frac da}{97 - e} + e\\\\ b - \\frac da &= (19 - e)^2 = (97 - e)^2\\\\ 19 - e &= \\pm (97 - e) \\end{align*} Clearly we can discard the positive root, so $e = 58$.",
"We first note (as before) that the number not in the range of \\[f(x) = \\frac{ax+b}{cx+ d} = \\frac{a}{c} + \\frac{b - ad/c}{cx+d}\\] is $a/c$, as $\\frac{b-ad/c}{cx+d}$ is evidently never 0 (otherwise, $f$ would be a constant function, violating the condition $f(19) \\neq f(97)$). We may represent the real number $x/y$ as $\\begin{pmatrix}x \\\\ y\\end{pmatrix}$, with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function $F(x) = \\frac{Ax + B}{Cx + D}$ as a matrix $\\begin{pmatrix} A & B\\\\ C& D \\end{pmatrix}$. Function composition and evaluation then become matrix multiplication. Now in general, \\[f^{-1} = \\begin{pmatrix} a & b\\\\ c&d \\end{pmatrix}^{-1} = \\frac{1}{\\det(f)} \\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix} .\\] In our problem $f^2(x) = x$. It follows that \\[\\begin{pmatrix} a & b \\\\ c& d \\end{pmatrix} = K \\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix} ,\\] for some nonzero real $K$. Since \\[\\frac{a}{d} = \\frac{b}{-b} = K,\\] it follows that $a = -d$. (In fact, this condition condition is equivalent to the condition that $f(f(x)) = x$ for all $x$ in the domain of $f$.) We next note that the function \\[g(x) = x - f(x) = \\frac{c x^2 + (d-a) x - b}{cx + d}\\] evaluates to 0 when $x$ equals 19 and 97. Therefore \\[\\frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \\frac{c(x-19)(x-97)}{cx+d}.\\] Thus $-19 - 97 = \\frac{d-a}{c} = -\\frac{2a}{c}$, so $a/c = (19+97)/2 = 58$, our answer.",
"Any number that is not in the domain of the inverse of $f(x)$ cannot be in the range of $f(x)$. Starting with $f(x) = \\frac{ax+b}{cx+d}$, we rearrange some things to get $x = \\frac{b-f(x)d}{f(x)c-a}$. Clearly, $\\frac{a}{c}$ is the number that is outside the range of $f(x)$. Since we are given $f(f(x))=x$, we have that \\[x = \\frac{a\\frac{ax+b}{cx+d}+b}{c\\frac{ax+b}{cx+d}+d} = \\frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \\frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}\\] \\[cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)\\] All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that $a = -d$. This solution follows in the same manner as the last paragraph of the first solution.",
"Since $f(f(x))$ is $x$, it must be symmetric across the line $y=x$. Also, since $f(19)=19$, it must touch the line $y=x$ at $(19,19)$ and $(97,97)$. $f$ a hyperbola that is a scaled and transformed version of $y=\\frac{1}{x}$. Write $f(x)= \\frac{ax+b}{cx+d}$ as $\\frac{y}{cx+d}+z$, and z is our desired answer $\\frac{a}{c}$. Take the basic hyperbola, $y=\\frac{1}{x}$. The distance between points $(1,1)$ and $(-1,-1)$ is $2\\sqrt{2}$, while the distance between $(19,19)$ and $(97,97)$ is $78\\sqrt{2}$, so it is $y=\\frac{1}{x}$ scaled by a factor of $39$. Then, we will need to shift it from $(-39,-39)$ to $(19,19)$, shifting up by $58$, or $z$, so our answer is $58.",
"From $f(f(x))=x$, it is obvious that $\\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\\frac{b}{d}$, $f(\\frac{b}{d})=0$ which means $a(\\frac{b}{d})+b=0$ so $a=-d$. Using $f(19)=19$, we have that $b=361c+38d$; on $f(97)=97$ we obtain $b=9409c+194d$. Solving for $d$ in terms of $c$ leads us to $d=-58c$, so the answer is $058. ~solution by mathleticguyyy",
"From $f(f(x))=x$, it is obvious that $\\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\\frac{b}{d}$, $f(\\frac{b}{d})=0$ which means $a(\\frac{b}{d})+b=0$ so $a=-d$. Using $f(19)=19$, we have that $b=361c+38d$; on $f(97)=97$ we obtain $b=9409c+194d$. Solving for $d$ in terms of $c$ leads us to $d=-58c$, so the answer is $058. ~solution by mathleticguyyy",
"Begin by finding the inverse function of $f(x)$, which turns out to be $f^{-1}(x)=\\frac{19d-b}{a-19c}$. Since $f(f(x))=x$, $f(x)=f^{-1}(x)$, so substituting 19 and 97 yields the system, $\\begin{array}{lcl} \\frac{19a+b}{19c+d} & = & \\frac{19d-b}{a-19c} \\\\ \\frac{97a+b}{97c+d} & = & \\frac{97d-b}{a-97c} \\end{array}$, and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get $116c=a-d$. Coincidentally, then $116c+d=a$, which is familiar because $f(116)=\\frac{116a+b}{116c+d}$, and since $116c+d=a$, $f(116)=\\frac{116a+b}{a}$. Also, $f(f(116))=\\frac{a(\\frac{116a+b}{a})+b}{c(\\frac{116a+b}{a})+d}=116$, due to $f(f(x))=x$. This simplifies to $\\frac{116a+2b}{c(\\frac{116a+b}{a})+d}=116$, $116a+2b=116(c(\\frac{116a+b}{a})+d)$, $116a+2b=116(c(116+\\frac{b}{a})+d)$, $116a+2b=116c(116+\\frac{b}{a})+116d$, and substituting $116c+d=a$ and simplifying, you get $2b=116c(\\frac{b}{a})$, then $\\frac{a}{c}=58$. Looking at $116c=a-d$ one more time, we get $116=\\frac{a}{c}+\\frac{-d}{c}$, and substituting, we get $\\frac{-d}{c}=58, and we are done.",
"Because there are no other special numbers other than $19$ and $97$, take the average to get $58 questions like these)",
"Because there are no other special numbers other than $19$ and $97$, take the average to get $58 questions like these)",
"By the function definition, $f(f(x))$, $f$ is its own inverse, so the only value not in the range of $f$ is the value not in the domain of $f$ (which is $-d/c$). Since $f(f(x))$, $f(f(0)=0$ (0 is a convenient value to use). $f(f(0))=f(f(\\tfrac{b}{d})=\\dfrac{a\\cdot\\tfrac{b}{d}+b}{c\\cdot\\tfrac{b}{d}+d}=\\dfrac{ab+bd}{bc+d^2}=0 \\Rightarrow ab+bd=0$. Then $ab+bd=b(a+d)=0$ and since $b$ is nonzero, $a=-d$. The answer we are searching for, $\\dfrac{-d}{c}$ (the only value not in the range of $f$), can now be expressed as $\\dfrac{a}{c}$. We are given $f(19)=19$ and $f(97)$, and they satisfy the equation $f(x)=x$, which simplifies to $\\dfrac{ax+b}{cx+d}=x\\Rightarrow x(cx+d)=ax+b\\Rightarrow cx^2+(d-a)x+b=0$. We have written this quadratic with roots $19$ and $97$. By Vieta, $\\dfrac{-(d-a)}{c}=\\dfrac{-(-a-a)}{c}=\\dfrac{2a}{c}=19+97$. So our answer is $\\dfrac{116}{2}=058. ~BakedPotato66",
"By the function definition, $f(f(x))$, $f$ is its own inverse, so the only value not in the range of $f$ is the value not in the domain of $f$ (which is $-d/c$). Since $f(f(x))$, $f(f(0)=0$ (0 is a convenient value to use). $f(f(0))=f(f(\\tfrac{b}{d})=\\dfrac{a\\cdot\\tfrac{b}{d}+b}{c\\cdot\\tfrac{b}{d}+d}=\\dfrac{ab+bd}{bc+d^2}=0 \\Rightarrow ab+bd=0$. Then $ab+bd=b(a+d)=0$ and since $b$ is nonzero, $a=-d$. The answer we are searching for, $\\dfrac{-d}{c}$ (the only value not in the range of $f$), can now be expressed as $\\dfrac{a}{c}$. We are given $f(19)=19$ and $f(97)$, and they satisfy the equation $f(x)=x$, which simplifies to $\\dfrac{ax+b}{cx+d}=x\\Rightarrow x(cx+d)=ax+b\\Rightarrow cx^2+(d-a)x+b=0$. We have written this quadratic with roots $19$ and $97$. By Vieta, $\\dfrac{-(d-a)}{c}=\\dfrac{-(-a-a)}{c}=\\dfrac{2a}{c}=19+97$. So our answer is $\\dfrac{116}{2}=058. ~BakedPotato66",
"Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$, we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$. The only point that is undefined under an inversion is the center of the circle, which we know is $\\frac{19+97}{2}=58.",
"Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$, we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$. The only point that is undefined under an inversion is the center of the circle, which we know is $\\frac{19+97}{2}=58.",
"First, consider the equation $\\frac{ax+b}{cx+d}=x$. This is a quadratic in terms of $x$, and we are given that $19,97$ are solutions to this equation. Rearranging yields $cx^2-(a-d)x-b=0$, so by Vieta’s Formulas we must have $\\cfrac{a-d}{c}=19+97=116$. Next, we consider the second condition. Note that for $m=-\\cfrac{b}{a},f(m)=0$, so $f(f(m))=f(0)$. We are also given that $f(f(m))=m$, so $f(0)=m$, which simplifies to $\\cfrac{b}{d}=-\\cfrac{b}{a}$ and $a=-d$. Substituting this into the equation above yields \\[\\cfrac{a+a}{c}=116\\] \\[\\cfrac{2a}{c}=116\\] \\[\\cfrac{a}{c}=58\\] This is the horizontal asymptote of the function, so the function does not include this $y$-value. Thus the answer is $58. ~eevee9406"
] |
1997-I-14
| 1,997 | 14 |
Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ . Let $m/n$ be the probability that $\sqrt{2+\sqrt{3}}\le |v+w|$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 582 | null |
[
"Solution 1 $z^{1997}=1=1(\\cos 0 + i \\sin 0)$ Define $\\theta = 2\\pi/1997$. By De Moivre's Theorem the roots are given by $z=\\cos (k\\theta) +i\\sin(k\\theta), \\qquad k \\in \\{0,1,\\ldots,1996\\}$ Now, let $v$ be the root corresponding to $m\\theta=2m\\pi/1997$, and let $w$ be the root corresponding to $n\\theta=2n\\pi/ 1997$. Then \\begin{align*} |v+w|^2 &= \\left(\\cos(m\\theta) + \\cos(n\\theta)\\right)^2 + \\left(\\sin(m\\theta) + \\sin(n\\theta)\\right)^2 \\\\ &= 2 + 2\\cos\\left(m\\theta\\right)\\cos\\left(n\\theta\\right) + 2\\sin\\left(m\\theta\\right)\\sin\\left(n\\theta\\right) \\end{align*} The cosine difference identity simplifies that to \\[|v+w|^2 = 2+2\\cos((m-n)\\theta)\\] We need $|v+w|^2 \\ge 2+\\sqrt{3}$, which simplifies to \\[\\cos((m-n)\\theta) \\ge \\frac{\\sqrt{3}}{2}.\\] Hence, $-\\frac{\\pi}{6} \\le (m-n) \\theta \\le \\frac{\\pi}{6},$ or alternatively, $|m - n| \\theta \\le \\frac{\\pi}{6}.$ Recall that $\\theta=\\frac{2\\pi}{1997}.$ Thus, \\[|m - n| \\le \\frac{\\pi}{6} \\cdot \\frac{1997}{2 \\pi} = \\left\\lfloor \\frac{1997}{12} \\right\\rfloor =166.\\] Therefore, $m$ and $n$ cannot be more than $166$ away from each other. This means that for a given value of $m$, there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$, and $166$ of them $< m$ ($m \\neq n$ since the problem says that $m$ and $n$ are distinct). Since $m$ and $n$ must be distinct, $n$ can have $1996$ total possible values for each value of $m$. Therefore, the probability is $\\frac{332}{1996}=\\frac{83}{499}$. The answer is then $499+83=582 ~SigmaPiE",
"$z^{1997}=1=1(\\cos 0 + i \\sin 0)$ Define $\\theta = 2\\pi/1997$. By De Moivre's Theorem the roots are given by $z=\\cos (k\\theta) +i\\sin(k\\theta), \\qquad k \\in \\{0,1,\\ldots,1996\\}$ Now, let $v$ be the root corresponding to $m\\theta=2m\\pi/1997$, and let $w$ be the root corresponding to $n\\theta=2n\\pi/ 1997$. Then \\begin{align*} |v+w|^2 &= \\left(\\cos(m\\theta) + \\cos(n\\theta)\\right)^2 + \\left(\\sin(m\\theta) + \\sin(n\\theta)\\right)^2 \\\\ &= 2 + 2\\cos\\left(m\\theta\\right)\\cos\\left(n\\theta\\right) + 2\\sin\\left(m\\theta\\right)\\sin\\left(n\\theta\\right) \\end{align*} The cosine difference identity simplifies that to \\[|v+w|^2 = 2+2\\cos((m-n)\\theta)\\] We need $|v+w|^2 \\ge 2+\\sqrt{3}$, which simplifies to \\[\\cos((m-n)\\theta) \\ge \\frac{\\sqrt{3}}{2}.\\] Hence, $-\\frac{\\pi}{6} \\le (m-n) \\theta \\le \\frac{\\pi}{6},$ or alternatively, $|m - n| \\theta \\le \\frac{\\pi}{6}.$ Recall that $\\theta=\\frac{2\\pi}{1997}.$ Thus, \\[|m - n| \\le \\frac{\\pi}{6} \\cdot \\frac{1997}{2 \\pi} = \\left\\lfloor \\frac{1997}{12} \\right\\rfloor =166.\\] Therefore, $m$ and $n$ cannot be more than $166$ away from each other. This means that for a given value of $m$, there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$, and $166$ of them $< m$ ($m \\neq n$ since the problem says that $m$ and $n$ are distinct). Since $m$ and $n$ must be distinct, $n$ can have $1996$ total possible values for each value of $m$. Therefore, the probability is $\\frac{332}{1996}=\\frac{83}{499}$. The answer is then $499+83=582 ~SigmaPiE",
"The solutions of the equation $z^{1997} = 1$ are the $1997$th roots of unity and are equal to $\\text{cis}(\\theta_k)$, where $\\theta_k = \\tfrac {2\\pi k}{1997}$ for $k = 0,1,\\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane. The quantity $|v+w|$ is unchanged upon rotation around the origin, so, WLOG, we can assume $v=1$ after rotating the axis till $v$ lies on the real axis. Let $w=\\text{cis}(\\theta_k)$. Since $w\\cdot \\overline{w}=|w|^2=1$ and $w+\\overline{w}=2\\text{Re}(w) = 2\\cos\\theta_k$, we have \\[|v + w|^2 = (1+w)(1+\\overline{w}) = 2+2\\cos\\theta_k\\] We want $|v + w|^2\\ge 2 + \\sqrt {3}.$ From what we just obtained, this is equivalent to \\[\\cos\\theta_k\\ge \\frac {\\sqrt {3}}2 \\qquad \\Leftrightarrow \\qquad -\\frac {\\pi}6\\le \\theta_k \\le \\frac {\\pi}6\\] which is satisfied by $k = 166,165,\\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$). So out of the $1996$ possible $k$, $332$ work. Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = 582 ~SigmaPiE",
"We can solve a geometrical interpretation of this problem. Without loss of generality, let $w = 1$. We are now looking for a point exactly one unit away from $w$ such that the point is at least $\\sqrt{2 + \\sqrt{3}}$ units away from the origin. Note that the \"boundary\" condition is when the point will be exactly $\\sqrt{2+\\sqrt{3}}$ units away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\\sqrt{2+\\sqrt{3}}$. The equations of these circles are $(x-1)^2 + y^2 = 1$ and $x^2 + y^2 = 2 + \\sqrt{3}$. Solving for $x$ yields $x = 1 + \\frac{\\sqrt{3}}{2}$. Clearly, this means that the real part of $v$ is greater than $\\frac{\\sqrt{3}}{2}$. Solving, we note that $332$ possible $v$s exist, meaning that $\\frac{m}{n} = \\frac{332}{1996} = \\frac{83}{499}$. Therefore, the answer is $83 + 499 = 582 ~SigmaPiE",
"Since $z^{1997}=1$, the roots will have magnitude $1$. Thus, the roots can be written as $\\cos(\\theta)+i\\sin(\\theta)$ and $\\cos(\\omega)+i\\sin(\\omega)$ for some angles $\\theta$ and $\\omega$. We rewrite the requirement as $\\sqrt{2+\\sqrt3}\\le|\\cos(\\theta)+\\cos(\\omega)+i\\sin(\\theta)+i\\sin(\\omega)|$, which can now be easily manipulated to $2+\\sqrt{3}\\le(\\cos(\\theta)+\\cos(\\omega))^2+(\\sin(\\theta)+\\sin(\\omega))^2$. WLOG, let $\\theta = 0$. Thus, our inequality becomes $2+\\sqrt{3}\\le(1+\\cos(\\omega))^2+(\\sin(\\omega))^2$, $2+\\sqrt{3}\\le2+2\\cos(\\omega)$, and finally $\\cos(\\omega)\\ge\\frac{\\sqrt{3}}{2}$. Obviously, $\\cos(\\frac{\\pi}{6})=\\frac{\\sqrt{3}}{2}$, and thus it follows that, on either side of a given point, $\\frac{1997}{12}\\approx166$ points will work. The probability is $\\frac{166\\times2}{1996} = \\frac{83}{499}$, and thus our requested sum is $582 ~SigmaPiE"
] |
1997-I-15
| 1,997 | 15 |
The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ , $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$ .
| 554 | null |
[
"Consider points on the complex plane $A (0,0),\\ B (11,0),\\ C (11,10),\\ D (0,10)$. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$, and the other two points $E$ and $F$ on $BC$ and $CD$, respectively. Let $E (11,a)$ and $F (b, 10)$. Since it's equilateral, then $E\\cdot\\text{cis}60^{\\circ} = F$, so $(11 + ai)\\left(\\frac {1}{2} + \\frac {\\sqrt {3}}{2}i\\right) = b + 10i$, and expanding we get $\\left(\\frac {11}{2} - \\frac {a\\sqrt {3}}{2}\\right) + \\left(\\frac {11\\sqrt {3}}{2} + \\frac {a}{2}\\right)i = b + 10i$. We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\\sqrt {3},22 - 10\\sqrt {3})$. Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\\sqrt{3}$. Using the area formula $\\frac{s^2\\sqrt{3}}{4}$, the area of the equilateral triangle is $\\frac{(884-440\\sqrt{3})\\sqrt{3}}{4} = 221\\sqrt{3} - 330$. Thus $p + q + r = 221 + 3 + 330 = 554.",
"Consider points on the complex plane $A (0,0),\\ B (11,0),\\ C (11,10),\\ D (0,10)$. Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$, and the other two points $E$ and $F$ on $BC$ and $CD$, respectively. Let $E (11,a)$ and $F (b, 10)$. Since it's equilateral, then $E\\cdot\\text{cis}60^{\\circ} = F$, so $(11 + ai)\\left(\\frac {1}{2} + \\frac {\\sqrt {3}}{2}i\\right) = b + 10i$, and expanding we get $\\left(\\frac {11}{2} - \\frac {a\\sqrt {3}}{2}\\right) + \\left(\\frac {11\\sqrt {3}}{2} + \\frac {a}{2}\\right)i = b + 10i$. We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\\sqrt {3},22 - 10\\sqrt {3})$. Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\\sqrt{3}$. Using the area formula $\\frac{s^2\\sqrt{3}}{4}$, the area of the equilateral triangle is $\\frac{(884-440\\sqrt{3})\\sqrt{3}}{4} = 221\\sqrt{3} - 330$. Thus $p + q + r = 221 + 3 + 330 = 554.",
"This is a trigonometric re-statement of the above. Let $x = \\angle EAB$; by alternate interior angles, $\\angle DFA=60+x$. Let $a = EB$ and the side of the equilateral triangle be $s$, so $s= \\sqrt{a^2+121}$ by the Pythagorean Theorem. Now $\\frac{10}{s} = \\sin(60+x)= \\sin {60} \\cos x+ \\cos {60} \\sin x = \\left(\\frac{\\sqrt{3}}2\\right)\\left(\\frac{11}s\\right)+\\left(\\frac 12\\right)\\left( \\frac as \\right)$. This reduces to $a=20-11\\sqrt{3}$. Thus, the area of the triangle is $\\frac{s^2\\sqrt{3}}{4} =(a^2+121)\\frac{\\sqrt{3}}{4}$, which yields the same answer as above.",
"Since $\\angle{BAD}=90$ and $\\angle{EAF}=60$, it follows that $\\angle{DAF}+\\angle{BAE}=90-60=30$. Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$. Let the image of $D$ be $D'$. Since angles are preserved under rotation, $\\angle{DAF}=\\angle{D'AE}$. It follows that $\\angle{D'AE}+\\angle{BAE}=\\angle{D'AB}=30$. Since $\\angle{ADF}=\\angle{ABE}=90$, it follows that quadrilateral $ABED'$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\\frac{s}{2}$. Let $O$ be its circumcenter. By Inscribed Angles, $\\angle{BOD'}=2\\angle{BAD'}=60$. By the definition of circle, $OB=OD'$. It follows that triangle $OBD'$ is equilateral. Therefore, $BD'=r=\\frac{s}{2}$. Applying the Law of Cosines to triangle $ABD'$, $\\frac{s}{2}=\\sqrt{10^2+11^2-(2)(10)(11)(\\cos{30})}$. Squaring and multiplying by $\\sqrt{3}$ yields $\\frac{s^2\\sqrt{3}}{4}=221\\sqrt{3}-330\\implies{p+q+r=221+3+330=554 -Solution by thecmd999",
"Clearly one vertex of the equilateral triangle is on a vertex of the rectangle, and the other two are lying on two other sides. Let $m$ be the side length of the triangle, and let the rectangle be partitioned into the equilateral triangle, a right triangle with sides 11, $y$, $m$, a right triangle with sides 10, $x$, $m$, and a right triangle with sides $11-x$, $10-y$, $m$. Simple area analysis nets \\[110=\\frac{\\sqrt{3}}{4}m^2+\\frac{11}{2}y+\\frac{10}{2}x+\\frac{(11-x)(10-y)}{2}\\implies 110-\\frac{\\sqrt{3}}{2}m^2=xy\\] By the Pythagorean Theorem, $11^2+y^2=m^2$ and $10^2+x^2=m^2$, so $x^2y^2=(m^2-10^2)(m^2-11^2)$. Thus, \\[(110-\\frac{\\sqrt{3}}{2}m^2)^2=(m^2-10^2)(m^2-11^2)\\] \\[110^2-110\\sqrt{3}m^2+\\frac34m^4=m^4-221m^2+110^2\\] Obviously $m^2\\neq0$ so we can divide by $m^2$ after cancellation: \\[-110\\sqrt{3}+221=\\frac14m^2\\] The area of the triangle is $\\frac{\\sqrt{3}}{4}m^2$, so the finish is simple. \\[\\frac{\\sqrt{3}}{4}m^2=221\\sqrt{3}-330\\implies p+q+r=221+3+330=554\\] - clarkculus",
"Clearly one vertex of the equilateral triangle is on a vertex of the rectangle, and the other two are lying on two other sides. Let $m$ be the side length of the triangle, and let the rectangle be partitioned into the equilateral triangle, a right triangle with sides 11, $y$, $m$, a right triangle with sides 10, $x$, $m$, and a right triangle with sides $11-x$, $10-y$, $m$. Simple area analysis nets \\[110=\\frac{\\sqrt{3}}{4}m^2+\\frac{11}{2}y+\\frac{10}{2}x+\\frac{(11-x)(10-y)}{2}\\implies 110-\\frac{\\sqrt{3}}{2}m^2=xy\\] By the Pythagorean Theorem, $11^2+y^2=m^2$ and $10^2+x^2=m^2$, so $x^2y^2=(m^2-10^2)(m^2-11^2)$. Thus, \\[(110-\\frac{\\sqrt{3}}{2}m^2)^2=(m^2-10^2)(m^2-11^2)\\] \\[110^2-110\\sqrt{3}m^2+\\frac34m^4=m^4-221m^2+110^2\\] Obviously $m^2\\neq0$ so we can divide by $m^2$ after cancellation: \\[-110\\sqrt{3}+221=\\frac14m^2\\] The area of the triangle is $\\frac{\\sqrt{3}}{4}m^2$, so the finish is simple. \\[\\frac{\\sqrt{3}}{4}m^2=221\\sqrt{3}-330\\implies p+q+r=221+3+330=554\\] - clarkculus"
] |
1998-I-1
| 1,998 | 1 |
For how many values of $k$ is $12^{12}$ the least common multiple of the positive integers $6^6$ and $8^8$ , and $k$ ?
| 25 | null |
[
"It is evident that $k$ has only 2s and 3s in its prime factorization, or $k = 2^a3^b$. $6^6 = 2^6\\cdot3^6$ $8^8 = 2^{24}$ $12^{12} = 2^{24}\\cdot3^{12}$ The LCM of any numbers an be found by writing out their factorizations and taking the greatest power for each factor. $[6^6,8^8] = 2^{24}3^6$. Therefore $12^{12} = 2^{24}\\cdot3^{12} = [2^{24}3^6,2^a3^b] = 2^{\\max(24,a)}3^{\\max(6,b)}$, and $b = 12$. Since $0 \\le a \\le 24$, there are $25.",
"We want the number of $k$ such that $\\operatorname{lcm}(6^6,8^8,k)=12^{12}$. Using $\\operatorname{lcm}$ properties, this is $\\operatorname{lcm}(\\operatorname{lcm}(2^6\\cdot 3^6,2^{24}),k)=2^{24} \\cdot 3^{12}$, or $\\operatorname{lcm}(2^{24}\\cdot 3^6,k)=2^{24}\\cdot 3^{12}$. At this point, we realize that $k=2^a\\cdot3^b$, as any other prime factors would be included in the $\\operatorname{lcm}$. Also, $b=12$ (or the power of $3$ in the $\\operatorname{lcm}$ wouldn't be $12$) and $0\\le a\\le 24$ (or the power of $2$ in the $\\operatorname{lcm}$ would be $a$ and not $24$). Therefore, $a$ can be any integer from $0$ to $24$, for a total of $25$ values of $a$ and $025."
] |
1998-I-2
| 1,998 | 2 |
Find the number of ordered pairs $(x,y)$ of positive integers that satisfy $x \le 2y \le 60$ and $y \le 2x \le 60$ .
| 480 | null |
[
"Solution 1 Pick's theorem states that: $A = I + \\frac B2 - 1$ The conditions give us four inequalities: $x \\le 30$, $y\\le 30$, $x \\le 2y$, $y \\le 2x$. These create a quadrilateral, whose area is $\\frac 12$ of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square. So $A = \\frac 12 \\cdot 30^2 = 450$. $B$ we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 lattice points, and the two diagonals each have 14 lattice points (for the top diagonal, every value of $x$ corresponds with an integer value of $y$ as $y = 2x$ and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders. $450 = I + \\frac {60}2 - 1$$I = 421$ Since the inequalities also include the equals case, we include the boundaries, which gives us $421 + 60 = 481$ ordered pairs. However, the question asks us for positive integers, so $(0,0)$ doesn't count; hence, the answer is $480$. Solution 2 First, note that all pairs of the form $(a,a)$, $1\\le a\\le30$ work. Now, considered the ordered pairs with $x < y$, so that $x < 2y$ is automatically satisfied. Since $x < y\\le 2x$, there are $2x - x = x$ possible values of $y$. Hence, given $x$, there are $x$ values of possible $y$ for which $x < y$ and the above conditions are satisfied. But $2y\\le60$, so this only works for $x\\le15$. Thus, there are $\\sum_{i=1}^{15} i=\\frac{(15)(16)}{2}$ ordered pairs. For $x > 15$, $y$ must follow $x < y\\le 30$. Hence, there are $30 - x$ possibilities for $y$, and there are $\\sum_{i=16}^{30}(30-i)=\\sum_{i=0}^{14}i=\\frac{(14)(15)}{2}$ ordered pairs. By symmetry, there are also $\\frac {(15)(16)}{2} + \\frac {(14)(15)}{2}$ ordered pairs with $x > y$ and the above criteria satisfied. Hence, the total is $\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+30=480.$ Solution 3 $y\\le2x\\le60$ Multiplying both sides by 2 yields: $2y\\le4x\\le120$ Then the two inequalities can be merged to form the following inequality: $x\\le2y\\le4x\\le120$ Additionally, we must ensure that $2y<60$ Therefore we must find pairs $(x,y)$ that satisfy the inequality above. A bit of trial and error and observing patterns leads to the answer $480$. It should be noted that the cases for $x\\le15$ and $x>15$ should be considered separately in order to ensure that $2y < 60$. Solution 4 - Unrigorous engineers induction solution We will try out small cases. By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480. -Alexlikemath Solution 5 - Counting Head on Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006",
"Pick's theorem states that: $A = I + \\frac B2 - 1$ The conditions give us four inequalities: $x \\le 30$, $y\\le 30$, $x \\le 2y$, $y \\le 2x$. These create a quadrilateral, whose area is $\\frac 12$ of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square. So $A = \\frac 12 \\cdot 30^2 = 450$. $B$ we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 lattice points, and the two diagonals each have 14 lattice points (for the top diagonal, every value of $x$ corresponds with an integer value of $y$ as $y = 2x$ and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders. $450 = I + \\frac {60}2 - 1$$I = 421$ Since the inequalities also include the equals case, we include the boundaries, which gives us $421 + 60 = 481$ ordered pairs. However, the question asks us for positive integers, so $(0,0)$ doesn't count; hence, the answer is $480$. Solution 2 First, note that all pairs of the form $(a,a)$, $1\\le a\\le30$ work. Now, considered the ordered pairs with $x < y$, so that $x < 2y$ is automatically satisfied. Since $x < y\\le 2x$, there are $2x - x = x$ possible values of $y$. Hence, given $x$, there are $x$ values of possible $y$ for which $x < y$ and the above conditions are satisfied. But $2y\\le60$, so this only works for $x\\le15$. Thus, there are $\\sum_{i=1}^{15} i=\\frac{(15)(16)}{2}$ ordered pairs. For $x > 15$, $y$ must follow $x < y\\le 30$. Hence, there are $30 - x$ possibilities for $y$, and there are $\\sum_{i=16}^{30}(30-i)=\\sum_{i=0}^{14}i=\\frac{(14)(15)}{2}$ ordered pairs. By symmetry, there are also $\\frac {(15)(16)}{2} + \\frac {(14)(15)}{2}$ ordered pairs with $x > y$ and the above criteria satisfied. Hence, the total is $\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+30=480.$ Solution 3 $y\\le2x\\le60$ Multiplying both sides by 2 yields: $2y\\le4x\\le120$ Then the two inequalities can be merged to form the following inequality: $x\\le2y\\le4x\\le120$ Additionally, we must ensure that $2y<60$ Therefore we must find pairs $(x,y)$ that satisfy the inequality above. A bit of trial and error and observing patterns leads to the answer $480$. It should be noted that the cases for $x\\le15$ and $x>15$ should be considered separately in order to ensure that $2y < 60$. Solution 4 - Unrigorous engineers induction solution We will try out small cases. By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480. -Alexlikemath Solution 5 - Counting Head on Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006",
"First, note that all pairs of the form $(a,a)$, $1\\le a\\le30$ work. Now, considered the ordered pairs with $x < y$, so that $x < 2y$ is automatically satisfied. Since $x < y\\le 2x$, there are $2x - x = x$ possible values of $y$. Hence, given $x$, there are $x$ values of possible $y$ for which $x < y$ and the above conditions are satisfied. But $2y\\le60$, so this only works for $x\\le15$. Thus, there are $\\sum_{i=1}^{15} i=\\frac{(15)(16)}{2}$ ordered pairs. For $x > 15$, $y$ must follow $x < y\\le 30$. Hence, there are $30 - x$ possibilities for $y$, and there are $\\sum_{i=16}^{30}(30-i)=\\sum_{i=0}^{14}i=\\frac{(14)(15)}{2}$ ordered pairs. By symmetry, there are also $\\frac {(15)(16)}{2} + \\frac {(14)(15)}{2}$ ordered pairs with $x > y$ and the above criteria satisfied. Hence, the total is $\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+\\frac{(15)(16)}{2}+\\frac{(14)(15)}{2}+30=480.$ Solution 3 $y\\le2x\\le60$ Multiplying both sides by 2 yields: $2y\\le4x\\le120$ Then the two inequalities can be merged to form the following inequality: $x\\le2y\\le4x\\le120$ Additionally, we must ensure that $2y<60$ Therefore we must find pairs $(x,y)$ that satisfy the inequality above. A bit of trial and error and observing patterns leads to the answer $480$. It should be noted that the cases for $x\\le15$ and $x>15$ should be considered separately in order to ensure that $2y < 60$. Solution 4 - Unrigorous engineers induction solution We will try out small cases. By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480. -Alexlikemath Solution 5 - Counting Head on Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006",
"$y\\le2x\\le60$ Multiplying both sides by 2 yields: $2y\\le4x\\le120$ Then the two inequalities can be merged to form the following inequality: $x\\le2y\\le4x\\le120$ Additionally, we must ensure that $2y<60$ Therefore we must find pairs $(x,y)$ that satisfy the inequality above. A bit of trial and error and observing patterns leads to the answer $480$. It should be noted that the cases for $x\\le15$ and $x>15$ should be considered separately in order to ensure that $2y < 60$. Solution 4 - Unrigorous engineers induction solution We will try out small cases. By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480. -Alexlikemath Solution 5 - Counting Head on Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006",
"We will try out small cases. By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480. -Alexlikemath Solution 5 - Counting Head on Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006",
"Notice $x$ and $y$ both must be equal or less than 30. The inequalities given have no complicated qualities. We can recompile them by understanding: Two times the larger integer will also be larger than the smaller integer; Two times the smaller integer is greater or equal to the greater integer if and only if the greater integer is less or equal to the double of the smaller integer. Knowing this, we create a chart. We will first solve without order, then multiply pairs by 2 at the end. $x$ can be 1-30, so we'll start with 1. The only possible value for $y$ is 1. For $x = 2$, $y$ can be 2-4. For $x = 3$, $y$ can be 3-6. There is an obvious pattern here. For every integer after 1, the possible values for $y$ will be numbers $x$-$2x$. This predictably ends at $x = 15$ because $y$ will reach 30. When $x = 16$, then the number of possible values of $y$ will begin to drop again, equaling the amount when $x = 14$. Then when we finally sum the group together, for $x = 2$ to $x = 14$ there are 104 pairs with 2 distinct values, and 13 values with congruent values. These will not be multiplied by 2 later on. $x = 16$ to $x = 28$ gives the same amount. Then $x = 15$ and $x = 29$ gives 16 and 2 values respectively, with two congruent values each. Finally, $x = 1$ and $x = 30$ give 3 and 1 respectively. Sum them together and you will get 480. -jackshi2006"
] |
1998-I-3
| 1,998 | 3 |
The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions. What is the area of the bounded region?
| 800 | null |
[
"The equation given can be rewritten as: $40|x| = - y^2 - 2xy + 400$ We can split the equation into a piecewise equation by breaking up the absolute value: $40x = -y^2 - 2xy + 400\\quad\\quad x\\ge 0$ $40x = y^2 + 2xy - 400 \\quad \\quad x < 0$ Factoring the first one: (alternatively, it is also possible to complete the square) $40x + 2xy = -y^2 + 400$ $2x(20 + y)= (20 - y)(20 + y)$ Hence, either $y = -20$, or $2x = 20 - y \\Longrightarrow y = -2x + 20$. Similarily, for the second one, we get $y = 20$ or $y = -2x - 20$. If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $800.",
"The equation can be rewritten as: $(x+y)^2=(|x|-20)^2$. Do casework as above."
] |
1998-I-4
| 1,998 | 4 |
Nine tiles are numbered $1, 2, 3, \cdots, 9,$ respectively. Each of three players randomly selects and keeps three of the tiles, and sums those three values. The probability that all three players obtain an odd sum is $m/n,$ where $m$ and $n$ are relatively prime positive integers . Find $m+n.$
| 17 | null |
[
"In order for a player to have an odd sum, he must have an odd number of odd tiles: that is, he can either have three odd tiles, or two even tiles and an odd tile. Thus, since there are $5$ odd tiles and $4$ even tiles, the only possibility is that one player gets $3$ odd tiles and the other two players get $2$ even tiles and $1$ odd tile. We count the number of ways this can happen. (We will count assuming that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.) $\\dbinom{5}{3} = 10$ choices for the tiles that he gets. The other two odd tiles can be distributed to the other two players in $2$ ways, and the even tiles can be distributed between them in $\\dbinom{4}{2} \\cdot \\dbinom{2}{2} = 6$ ways. This gives us a total of $10 \\cdot 2 \\cdot 6 = 120$ possibilities in which all three people get odd sums. In order to calculate the probability, we need to know the total number of possible distributions for the tiles. The first player needs three tiles which we can give him in $\\dbinom{9}{3} = 84$ ways, and the second player needs three of the remaining six, which we can give him in $\\dbinom{6}{3} = 20$ ways. Finally, the third player will simply take the remaining tiles in $1$ way. So, there are $\\dbinom{9}{3} \\cdot \\dbinom{6}{3} \\cdot 1 = 84 \\cdot 20 = 1680$ ways total to distribute the tiles. We must multiply the probability by 3, since any of the 3 players can have the 3 odd tiles.Thus, the total probability is $\\frac{360}{1680} = \\frac{3}{14},$ so the answer is $3 + 14 = 017. When we simplify it, we have: \\[\\frac{3}{14}\\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \\[017\\] -Pi_is_3.14",
"Let $O$ stand for an odd number and $E$ an even. Therefore, one person must pick $OOO$, the other person must pick $OEE$ and the last person must pick $OEE$. Since any permutation of the order of who is picking or change in the order of the even numbers (e.g. $EOE$ instead of $OEE$) doesn't change the probability, we just need to multiply the probability of one case by $\\binom{3}{2}^3 = 27$ as there are 27 such cases (by cases I mean ordered triples of ordered multisets $A, B, C$ such that one of them has 3 $O$'s and the other two have two $E$'s and an $O$ in them, respectively.) . Let's do the case $OOO$, $OEE$, $OEE$. $\\frac{5}{9} \\times \\frac{4}{8} \\times \\frac{3}{7} \\times \\frac{2}{6} \\times \\frac{4}{5} \\times \\frac{3}{4} \\times \\frac{1}{3} \\times \\frac{2}{2} \\times \\frac{1}{1} = \\frac{1}{126}$. We now multiply by 27 to get $\\frac{3}{14} \\implies m + n = 017. When we simplify it, we have: \\[\\frac{3}{14}\\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \\[017\\] -Pi_is_3.14",
"For this problem, let's think about parity. There are $5$ odd numbers from $1-9$ and there are four even numbers from $1-9$. Since this problem is asking for the probability that the each player gets an odd sum, we also have to calculate the total numeber of ways. In this case, there are only two ways to get an odd sum. Either have the sequence $OOO$ or $OEE$ where the letters $O$ and $E$ stand for odd and even respectively. Since we constrained to only $5$ odds, the only way to do the pairing is \\[OOO-OEE-OEE\\] There are of couse three ways to choose who gets the three odds. Once we have chosen who has gotten the three odds, we can actually reorder the sequence like this: \\[OEE-OEE-OOO\\] Now since we are choosing in groups, we can ignor order of these terms. For the first $OEE$, there are $5$ ways to choose which odd to use, and $\\dbinom{4}{2}$ or $6$ ways to choose the evens. Great, let's move on to the second $OEE$. There are $4$ odds left, so there are $4$ ways choose the odds. Since there are only two even numbers, they have to go here, so there is only $1$ way to choose the evens. The three odds will now fall in place. We can now multiply all the numbers since they are independent, and we have $5*6*4*3$ or $360$. Since this is a probability question, we have to ask ourselves how many ways are there to distribute the 9 tiles equally among 3 players. Fortunately for us, this is not hard as the first player has $\\dbinom{9}{3}$ options and the second player has $\\dbinom{6}{3}$. When we multiply these, we get $1680$. This is our denominator. When we make the fraction, we have $\\frac{360}{1680}$. When we simplify it, we have: \\[\\frac{3}{14}\\] We are asked to find the sum of the numerator and the denominator, so summing these, we have: \\[017\\] -Pi_is_3.14"
] |
1998-I-5
| 1,998 | 5 |
Given that $A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,$ find $|A_{19} + A_{20} + \cdots + A_{98}|.$
| 40 | null |
[
"Though the problem may appear to be quite daunting, it is actually not that difficult. $\\frac {k(k-1)}2$ always evaluates to an integer (triangular number), and the cosine of $n\\pi$ where $n \\in \\mathbb{Z}$ is 1 if $n$ is even and -1 if $n$ is odd. $\\frac {k(k-1)}2$ will be even if $4|k$ or $4|k-1$, and odd otherwise. So our sum looks something like: $\\left|\\sum_{i=19}^{98} A_i\\right| =- \\frac{19(18)}{2} + \\frac{20(19)}{2} + \\frac{21(20)}{2} - \\frac{22(21)}{2} - \\frac{23(22)}{2} + \\frac{24(23)}{2} \\cdots - \\frac{98(97)}{2}$ If we group the terms in pairs, we see that we need a formula for $-\\frac{(n)(n-1)}{2} + \\frac{(n+1)(n)}{2} = \\left(\\frac n2\\right)(n+1 - (n-1)) = n$. So the first two fractions add up to $19$, the next two to $-21$, and so forth. If we pair the terms again now, each pair adds up to $-2$. There are $\\frac{98-19+1}{2 \\cdot 2} = 20$ such pairs, so our answer is $|-2 \\cdot 20| = 040."
] |
1998-I-6
| 1,998 | 6 |
Let $ABCD$ be a parallelogram . Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$
| 308 | null |
[
"Solution 1 There are several similar triangles. $\\triangle PAQ\\sim \\triangle PDC$, so we can write the proportion: $\\frac{AQ}{CD} = \\frac{PQ}{PC} = \\frac{735}{112 + 735 + RC} = \\frac{735}{847 + RC}$ Also, $\\triangle BRQ\\sim DRC$, so: $\\frac{QR}{RC} = \\frac{QB}{CD} = \\frac{112}{RC} = \\frac{CD - AQ}{CD} = 1 - \\frac{AQ}{CD}$ $\\frac{AQ}{CD} = 1 - \\frac{112}{RC} = \\frac{RC - 112}{RC}$ Substituting, $\\frac{AQ}{CD} = \\frac{735}{847 + RC} = \\frac{RC - 112}{RC}$ $735RC = (RC + 847)(RC - 112)$ $0 = RC^2 - 112\\cdot847$ Thus, $RC = \\sqrt{112*847} = 308$. Solution 2 We have $\\triangle BRQ\\sim \\triangle DRC$ so $\\frac{112}{RC} = \\frac{BR}{DR}$. We also have $\\triangle BRC \\sim \\triangle DRP$ so $\\frac{ RC}{847} = \\frac {BR}{DR}$. Equating the two results gives $\\frac{112}{RC} = \\frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=308",
"There are several similar triangles. $\\triangle PAQ\\sim \\triangle PDC$, so we can write the proportion: $\\frac{AQ}{CD} = \\frac{PQ}{PC} = \\frac{735}{112 + 735 + RC} = \\frac{735}{847 + RC}$ Also, $\\triangle BRQ\\sim DRC$, so: $\\frac{QR}{RC} = \\frac{QB}{CD} = \\frac{112}{RC} = \\frac{CD - AQ}{CD} = 1 - \\frac{AQ}{CD}$ $\\frac{AQ}{CD} = 1 - \\frac{112}{RC} = \\frac{RC - 112}{RC}$ Substituting, $\\frac{AQ}{CD} = \\frac{735}{847 + RC} = \\frac{RC - 112}{RC}$ $735RC = (RC + 847)(RC - 112)$ $0 = RC^2 - 112\\cdot847$ Thus, $RC = \\sqrt{112*847} = 308$. Solution 2 We have $\\triangle BRQ\\sim \\triangle DRC$ so $\\frac{112}{RC} = \\frac{BR}{DR}$. We also have $\\triangle BRC \\sim \\triangle DRP$ so $\\frac{ RC}{847} = \\frac {BR}{DR}$. Equating the two results gives $\\frac{112}{RC} = \\frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=308",
"We have $\\triangle BRQ\\sim \\triangle DRC$ so $\\frac{112}{RC} = \\frac{BR}{DR}$. We also have $\\triangle BRC \\sim \\triangle DRP$ so $\\frac{ RC}{847} = \\frac {BR}{DR}$. Equating the two results gives $\\frac{112}{RC} = \\frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=308"
] |
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