# sangaku - a geometrical puzzle

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

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. $$\square$$

• This one's the most elegant of the lot, IMHO. :) Sep 13, 2010 at 1:49
• 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, 2010 at 20:47
• @Blue Your 'curiosity' is the easiest way to recognize that the triangles are 30-60-90 triangles, as in my answer. Apr 22, 2014 at 16:02
• You could have added this as another answer I believe. Anyways thanks for the solution, I needed it. :D
– user983206
Apr 2 at 1:14

Here is my attempt at a purely geometric approach:

Arrange a set of equal-size circles as in image 2, and position the top and bottom circles so that their centers form one side of a square. From the four-fold symmetry, it is seen that the left-most circle is in the center of the square, and since the altitude to the equilateral triangle is tangent to each of the two circles and is perpendicular to the side of the triangle in image 3, it follows that the left and right circles, and the two central circles just above these, are coincident with those from the original diagram in image 1 (and of course the images have been aligned to convince you of this as well). Thus all that remains is to measure $r$; in image 4 we note that the side of the triangle, which has length $1$, is also length $(2+2\sqrt3)r$ (the $\sqrt3$ measurements come from this being an equilateral triangle); from this we get

$$r=\frac1{2(1+\sqrt3)}=\frac{\sqrt3-1}4.$$

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

• +1 for the solution and the adequate picture. Just a curiosity: The angles of the triangles are $\pi /6,\pi /3,\pi /2$. Sep 12, 2010 at 20:56
• i.e. a 30-60-90 triangle, but it isn't that obvious until you actually go through the derivation. :) Sep 12, 2010 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 :-( Sep 13, 2010 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:

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

In the picture above, let the red lines have lengths $$r$$, and the blue lines have lengths $$x$$. Then the yellow lines have lengths $$1-x$$.

From Pythagoras: $$(x+r)^2+(1-x+r)^2=1^2$$ Which simplifies to: $$r+r^2=x-x^2$$ Considering the area of the outer square as the sum of 4 rectangles and the inner square: $$4\cdot\frac{1}{2}\cdot(x+r)(1-x+r)+(2r)^2=1^2$$ Which simplifies to: $$2(x-x^2+r+r^2)+4r^2=1$$ Subbing in the previous expression: $$2(r+r^2+r+r^2)+4r^2=1$$ Which simplifies to: $$8r^2+4r-1=0$$ Solving for the positive root: $$r=\frac{\sqrt{3}-1}{4}$$

Great Sangaku! Here's a quick and easy solution:

Let $r$ be the radius of the circles, let $L$ be the side length of the square, and let $a$ and $b$ be the remaining side lengths of the triangles where $a > b$.

Then we can see that $$a = b + 2r$$ The radius of the incircle of a right triangle is $$\frac{a + b - L}{2} = r$$

From these two equations, we easily obtain $b = \frac{L}{2}$.

The Pythagorean Theorem on $a, b = \frac{L}{2},$ and $L$ gives

$$a^{2} + (\frac{L}{2})^{2} = L^{2}$$

meaning $a = \frac{\sqrt{3}}{2}L$. Plugging our values for a and b into the top equation gives

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