Two Inscribed Circles in a Right Triangle

Let the triangle be $ABC$ with $\angle A = \theta$ and the centers of the big and small circles be $O$ and $P$ respectively. Note that $\triangle ABC$ is a $3$ - $4$ - $5$ right triangle. Then we have:

$\begin{aligned} AB & = 10 \blue{\cot \frac \theta 2} + 10 & \small \blue{\text{Note that }\tan \frac \theta 2 = \frac {1-\cos \theta}{\sin \theta}} \\ & = 10 \cdot \blue 3 + 10 & \small \blue{= \frac {1-\frac 45}{\frac 35}= \frac 13} \\ & = 40 \end{aligned}$

Then $BC = \dfrac 34 \times 40 = 30$ . Let the radius of the small circle be $r$ . And we have:

$\begin{aligned} 10 + \sqrt{(10+r)^2-(10-r)^2} + r \cot \left(\frac {90^\circ - \theta}2\right) & = BC \\ 10 + \sqrt{40r} + r \left(\frac {1+\frac 13}{1-\frac 13} \right) & = 30 \\ 2\sqrt{10r} + 2r & = 20 \\ (\sqrt r)^2 + \sqrt{10} \cdot \sqrt r & = 10 \\ \left(\sqrt r + \frac {\sqrt{10}}2 \right)^2 & = 10 + \frac {10}4 = \frac {25}2 \\ \sqrt r & = \frac 5{\sqrt 2} - \frac {\sqrt{10}}2 = \frac {5\sqrt 2 - \sqrt{10}}2 \\ \implies r & = \frac {60-10\sqrt{20}}4 = 5(3-\sqrt 5) \end{aligned}$

Therefore the diameter of the small circle is $2r = 10(3-\sqrt 5)$ $\implies a + b + c = 10 +3+5 = \boxed{18}$ .

Let A and B be the lengths of the legs of the triangle and C be the length of the hypotenuse. Since the slope of the hypotenuse is -4/3, we see that B = 3 A/4. Hence, by the Pythagorean Theorem, C = Sqrt(A^2 + 9 (A^2)/4) = 5*A/4 Since the radius of the incircle is 10, we then have (using the formula for the length of the inradius):

10 = A (3 A/4) / (A + 3 A/4 + 5 A/4), or

A = 40, B = 3 A/4 = 30, C = 5 A/4 = 50.

If w is the angle opposite the side of length 40, we have tan(w) = 4/3. Now, draw the angle bisector of w, noting that this line passes through the centers of both circles. We note that:

tan(w) = 2 tan(w/2) / (1-tan(w/2) tan(w/2)) = 4/3.

Solving this, and rejecting the negative root, yields:

tan(w/2) = 1/2 (equation 1)

Consider the triangle formed by connecting the centers of the two circles. If r is the length of the radius of the smaller circle, the length of the hypotenuse of this triangle is 10+r, the length of one leg is 10-r. If x is the length of the length of the other leg, equation 1 implies:

tan(w/2) = (10-r)/x = 1/2, or

x = (20-2*r)

Now, the apply the Pythagorean Theorem to this triangle to get:

(20-2*r)^2 + (10-r)^2 = (10 + r)^2

or r = 5*(3-sqrt(5))

The diameter is then twice the radius, so that a = 10, b = 3, and c = 5. Hence a + b + c = 18