A Simple Sangaku

Plot this on the Cartesian plane with $A(0,0), B(1,0), C(1,1), D(0,1), E(1/2, 1)$ .

The equation of the straight line $EC$ is $\frac{y - 1}{x-1/2} = \frac{1- 0}{1/2 -1}$ . And the equation of the striaght line $AC$ is $y = x$ .

Solving these two equations simultaneously gives the intersection point $F(2/3, 2/3)$ .

So we have all the 3 coordinates of the triangles, by Pythagoras theorem, the lengths of this triangle is $1, \frac{\sqrt{5}} 3, \frac23 \sqrt2$ .

Using the formula for the area of a triangle, $A = rs$ , where $r$ is the radius of the inscribed circle and $s$ is the semiperimeter of the triangles gives $r = \frac{1 + \sqrt2 + \sqrt5 - \sqrt{10}}6.$

The answer is $2 + 5 + 10 + 6 = \boxed{23}$ .

Let the center of the circle be $O$ . Then $O$ is also the incenter of the green triangle. Therefore, $OA$ bisects $\angle CAB$ $\implies \angle OAB = \frac \pi 8$ and $OB$ bisects $\angle EBA = \theta$ $\implies \angle OBA = \frac \theta 2$ . Let the radius of the circle be $r$ . Then

$\begin{aligned} \frac r\blue{\tan \frac \pi 8} + \frac r \red{\tan \frac \theta 2} & = 1 & \small \blue{\text{Since }\frac {2\tan \frac \pi8}{1-\tan^2 \frac \pi 8}=\tan \frac \pi 4 = 1} \\ \frac r\blue{\sqrt 2-1} + \frac {\red 2r}\red{\sqrt 5-1} & = 1 & \small \red{\text{and }\frac {2\tan \frac \theta 2}{1-\tan^2 \frac \theta 2}=\tan \theta = 2} \end{aligned}$

$\begin{aligned} \implies r & = \frac 1{\frac 1{\sqrt 2-1} + \frac 2{\sqrt 5-1}} \\ & = \frac 1{\sqrt 2+1+\frac {\sqrt 5+1}2} \\ & = \frac 2{2\sqrt 2+\sqrt 5+3} \\ & = \frac {2(2\sqrt 2+\sqrt 5-3)}{8+5+4\sqrt{10}-9} \\ & = \frac {2\sqrt 2+\sqrt 5-3}{2+2\sqrt{10}} \\ & = \frac {(2\sqrt 2+\sqrt 5-3)(\sqrt{10}-1)}{2(\sqrt{10}+1)(\sqrt{10}-1)} \\ & = \frac {4\sqrt 5+5\sqrt 2 - 3\sqrt{10}-2\sqrt 2-\sqrt 5+3}{2(9)} \\ & = \frac {3\sqrt 5+3\sqrt 2 - 3\sqrt{10}+3}{2(9)} \\ & = \frac {1 + \sqrt 5+\sqrt 2 - \sqrt{10}}6 \end{aligned}$

Therefore, $a+b+c+d = 5+2+10+6 = \boxed{23}$ .