Sangaku Problem

Let the radius of the circle be $r$ and the smallest angle of the four congruent right triangle be $\theta$ . We note that $2r + 4r \cos \theta = 1 \cdots (1)$ (the blue line) and $\sin \theta + 2r = \cos \theta \cdots (2)$ (the short leg $+ 2r =$ the long leg of the right triangle.

From $(2): 2r = \cos \theta - \sin \theta$ . Substitutes in $(1)$ :

$\begin{aligned} (\cos \theta - \sin \theta) (1+2\cos \theta) & = 1 \\ \cos \theta - \sin \theta + 2 \cos^2 \theta - 2\sin \theta \cos \theta & = 1 \\ \cos \theta - \sin \theta & = 2\sin \theta \cos \theta - 2\cos^2 \theta + 1 \\ \cos \theta - \sin \theta & = \sin 2 \theta - \cos 2 \theta \\ \cos (\theta + 45^\circ) & = \sin (2\theta - 45^\circ) = \cos (135^\circ - 2\theta) \\ \theta + 45^\circ & = 135^\circ - 2\theta \\ 3 \theta & = 90^\circ \\ \implies \theta & = 30^\circ \end{aligned}$

From $(2): 2r = \dfrac {\sqrt 3}2 - \dfrac 12 \implies r = \dfrac {\sqrt 3 -1}4$ . Therefore $a+b+c = 3+1+4 = \boxed 8$ .

Without the circles, this is a classic proof of Pythagoras' theorem; in this vein, let the shorter sides of the right-angled triangles be $u>v$ .

The side of the smaller central square is then $u-v$ ; but, from the diagram, this is also $2r$ .

The inradius of a right-angled triangle with sides $x,y,z$ is $r=\frac{x+y-z}{2}$ ; so in this case $r=\frac{u+v-1}{2}$ . Hence $2r=u-v=u+v-1$

so that $v=\frac12$ .

Also, $u^2+v^2=1$ , so $u=\frac12 \sqrt3$ .

Finally, $r=\frac14 \left(\sqrt3-1\right)$ giving the answer $\boxed8$ .

Code I made which works on https://www.khanacademy.org/computer-programming/new/pjs I know, it's poorly commented, I am just throwing it here SHOULD NOTE I also have a ton of scratch-paper I used to work this out too, as well as wolframalpha for some of the trigonometric equations

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// Code by chase marangu 2020 Monday January 11th
// works on https://www.khanacademy.org/computer-programming/new/pjs
// th is an angle of the right triangle -change it see what happen!
// h is how zoomed in your view is. Also it is the hypotenouse of the right triangle
var th = 25;
var h = 200;

// center x and y of the circle inscribed in the first right triangle
var cy = cos(th)/(1+
     (1/(tan(th/2)) ) );
var cx = (cos(th)-cy);

// orient the view
scale(1, -1);
translate(0, -height);
translate(20, 200);

// coloring options
strokeWeight(2);
stroke(255, 0, 0);

// draw the four right triangles and their incircles
pushMatrix();
for (var i=0; i<4; ++i) {
    // draw the triangle
    line(0, 0, cos(th)*h, 0);
    line(0, 0, cos(th)*h, sin(th)*h);
    line(cos(th)*h, 0, cos(th)*h, sin(th)*h);
    // draw the circle inscribed in it
    ellipse(cx*h, cy*h, abs(cy)*2*h, abs(cy)*2*h);
    // translate and rotate in the for-loop to make the
    // four-right-triangles-in-a-square arrangement
    translate(cos(th)*h, sin(th)*h);
    rotate(-360/4);
}
popMatrix();

// draw the circle inscribed in the square
ellipse(
    ( cos(th)/2+sin(th)/2 )*h,
    -(cos(th)/2-sin(th)/2)*h,
    abs(cos(th)-sin(th))*h,
    abs(cos(th)-sin(th))*h);

There is a solution in this mathematical link