sangaku - a geometrical puzzle

Question

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

(read about the origin of sangaku)

Answer 1

Edit. (Almost a duodecade later!) Combining and streamlining my previous arguments (now consigned to the Edit History) ...

The figure shows that the right triangle's hypotenuse (ie, $2r+2t$) is twice the length of its short leg (ie, $r+t$), hence we have a $30^\circ$-$60^\circ$-$90^\circ$ triangle.

The inradius can then be calculated from standard formulas, or we can continue to leverage the figure and write

$$2r = \underbrace{(3r+t)}_{\text{long leg}}-\underbrace{(r+t)}_{\text{short leg}} \qquad\to\qquad r = \frac{s}{2}\left(\sqrt{3}-1\right)$$ where $s$ is half the length of the side of the square. $\square$

Answer 2

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 $r = (-1+\sqrt{3})/4$.

Answer 3

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:

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