In triangle $ABC$, $\angle C = 90^\circ$, $\angle A = 40^\circ$, and $AC = 10$. Find the radius of the incircle of triangle $ABC$.

Answer 1

Let $I$ be the incenter of triangle $ABC$, and let $M$ be the midpoint of $\overline{AC}$, as shown below.

[asy]
unitsize(1 cm);

pair A, B, C, I, M;

C = (0,0);
A = (8,0);
B = intersectionpoint(arc(C,10,0,180),arc(A,24,0,180));
I = incenter(A,B,C);
M = (A + C)/2;

draw(A--B--C--cycle);
draw(A--I--C);
draw(C--M);

label("$A$", A, ENE);
label("$B$", B, NW);
label("$C$", C, SW);
label("$I$", I, S);
label("$M$", M, S);
[/asy]

We draw the incircle of triangle $ABC$ and locate its center $I$.

[asy]
unitsize(1 cm);

filter grayscale;
pair A, B, C, I, M;

C = (0,0);
A = (8,0);
B = intersectionpoint(arc(C,10,0,180),arc(A,24,0,180));
I = incenter(A,B,C);
M = (A + C)/2;

draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(A--I--C);
draw(C--M);

label("$A$", A, ENE);
label("$B$", B, NW);
label("$C$", C, SW);
label("$I$", I, SE);
label("$M$", M, S);
label("$10/2 = 5$", (A + M)/2, S);
[/asy]

Let $r$ be the inradius of triangle $ABC$, and let $E$ be one of the points of tangency of the incircle of triangle $ABC$ with $\overline{AC}$. Then $IE = r$, so by the Pythagorean Theorem on triangle $IMC$, we have $r^2 + (5 - r)^2 = I^2$. Since $I$ lies on the angle bisector of $\angle C$, we have
$$\frac{AC}{IC}=\frac{AE}{EI}.$$In other words,
$$\frac{10}{IC}=\frac{5}{r}.$$Then $\frac{IC}{5} = \frac{r}{10}$. Since $\angle ABC = 50^\circ$, $\angle IBC = 90^\circ - 50^\circ = 40^\circ$, so $\angle ICB = 180^\circ - 90^\circ - 40^\circ = 50^\circ$. Therefore, triangle $IBC$ is isosceles, and $IC = IB$.

Hence, $\frac{IC + IB}{5} = \frac{2 IC}{5} = \frac{r}{10}$, so $2 IC r = 5 \cdot 10 = 50$. But $r^2 + (5 - r)^2 = I^2 = 50$. Hence,
$$\begin{array}{rl}2ICr& =2IC\cdot 5\cdot \sqrt{r^2+(5-r{)}^2}\\ & =2IC\cdot 5\cdot \sqrt{50}\\ & =2\cdot 5IC\cdot 5\\ & =100IC,\end{array}$$so $100 IC = 50$, and $IC = \frac{1}{2}$. Therefore, $r = IC \cdot \frac{r}{10} = \boxed{\frac{1}{20}}$.