# sangaku - a geometrical puzzle

Find the radius of the circles if the size of the larger square is 1x1.
Enjoy!

(read about the origin of sangaku)

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## 3 Answers

Here's a nicer approach ...

Given two segments emanating from a point and ending at points of tangency with a circle, we know that those segments are congruent. In the figure, the three vertices of the right triangle give rise to three fundamental lengths; note that, since the "green" angle is a right angle, the lengths of the tangent segments are equal to the radius of the circle.

Pythagoras tells us that

$$(b+r)^2 + ( r + a )^2 = ( a + b )^2$$

But we also have that $a+b=1$ (the side of the square), and that $b=a+2r$ (via tangent circles on the "outside" of the right triangle). From these relations, we find that $a=1/2-r$ and $b=1/2+r$.

Therefore,

$$(1/2+2r)^2 + ( 1/2 )^2 = ( 1 )^2$$ $$1/4+2r+4r^2+1/4=1$$ $$8r^2+4r-1=0$$

The roots are $(-1\pm\sqrt{3})/4$, and we select the positive value: $r = (-1+\sqrt{3})/4$.

As Américo noted: The sides of the triangle have lengths $r+a=1/2$, $3r+a=\sqrt{3}/2$, and $1$, so that we have a 30-60-90 triangle.

(I like that there's but the one extraneous value this time, rather than the three in my first attempt. Is there yet a better approach that yields the answer directly, with no extraneous values?

Edit. There is. Immediately after we have deduced that $a=1/2-r$, we know that the short leg of the triangle has length $1/2$, so that its longer leg is $\sqrt{3}/2$. Since that longer leg is also $a+3r=1/2+2r$, we have that $r=(-1+\sqrt{3})/4$.)

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This one's the most elegant of the lot, IMHO. :) – J. M. Sep 13 '10 at 1:49
accepted as answer for indeed looking nicer than your other solution. – stevenvh Sep 13 '10 at 13:52

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Another curiosity: The lines that bisect the square vertically and horizontally are tangent to the outer circles (since $a+r=b-r=1/2$). – Blue Sep 13 '10 at 20:47

Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.

The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that

$(1)\hspace{1.0in}4r\cos\theta=1-2r$

Looking at the four right triangles, we see that the center circle's diameter is equal to the difference in the lengths of the legs; since the hypotenuse has length $1$, we have

$(2)\hspace{1.0in}2r = \cos\theta - \sin\theta$

From here, we simply need to eliminate $\theta$.

Multiplying (2) through by $4r$ and substituting in from (1) ...

$$8 r^2 = 4r\cos\theta - 4r \sin\theta = 1 - 2r - 4r \sin\theta$$ $$4r \sin\theta = 1 - 2r - 8 r^2$$

Therefore,

$$\begin{eqnarray}16r^2 &=& (4r \cos\theta)^2 + (4 r \sin\theta)^2 \\ &=& ( 1 - 2r )^2 + ( 1 - 2r - 8 r^2 )^2 \\ &=& 2 - 8 r - 8 r^2 + 32r^3 + 64 r^4 \end{eqnarray}$$

so that

$$0 = 32 r^4 + 16 r^3 - 12 r^2 - 4 r + 1 = (2r+1)(2r-1)(8 r^2 + 4 r - 1)$$

The roots of the polynomial are $\pm1/2$ and $(-1\pm\sqrt{3})/4$. We can eliminate three of them from consideration to conclude that $r = (-1+\sqrt{3})/4$.

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+1 for the solution and the adequate picture. Just a curiosity: The angles of the triangles are $\pi /6,\pi /3,\pi /2$. – Américo Tavares Sep 12 '10 at 20:56
i.e. a 30-60-90 triangle, but it isn't that obvious until you actually go through the derivation. :) – J. M. Sep 12 '10 at 21:54
This is the approach I followed too, but I had some trouble finding a second equation to get rid of the theta :-( – stevenvh Sep 13 '10 at 13:54

Somewhat related to Don's solution: From the figure, we see that the four triangles are 1: congruent, and 2: right triangles. The hypotenuse of one triangle has length 1, and if we let $\theta$ be the smaller of the two angles of the right triangle, and use $r$ to denote the radius of one circle, then the Pythagorean relation is

$$\cos^2\;\theta+(\cos\;\theta-2r)^2=1$$

This can now be solved as a simultaneous equation with any of the other two equations Don obtained, or we can use another equation, the expression for the inradius $r$:

$$r^2=\frac{(s-1)(s-\cos\;\theta)(s+2r-\cos\;\theta)}{s}$$

where $s=\frac{1+\cos\;\theta+(\cos\;\theta-2r)}{2}$ is the semiperimeter.

If we eliminate $\cos\;\theta$ and solve the two equations here for $r$, we find that the roots of the resulting quartic equation are

$$r=\frac{\pm 1\pm\sqrt{3}}{4}$$

If we carry out Don's approach as well, we find that only one positive value of $r$ is consistent with both systems, and thus has to be the correct answer:

$$r=\frac{-1+\sqrt{3}}{4}$$

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+1 for the elegant solution. – Américo Tavares Sep 12 '10 at 20:58
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