[  Pythagorean Theorem (direct and inverse)  Two tangents drawn from one point  [ Area of a triangle (through semiperimeter and radius of inscribed or exscribed circle

Find the radii of the inscribed and exscribed circles of a triangle with sides 5, 12, and 13.

#
If $a$ and $b$ are the legs of a right triangle, and $c$ is the hypotenuse, then the sought radii are $\frac{a+b-c}{2}, \frac{a+b+c}{2}, \frac{a+c-b}{2}$, and $\frac{b+c-a}{2}$.

## Solution

First method. Let the circle with center $O$ and radius $r$ be inscribed in the right triangle $ABC$, where $BC=a$, $AC=b$ are the legs, and $AB=c$ is the hypotenuse. If the circle touches the segments $BC$, $AC$, and $AB$ at points $K$, $L$, and $M$ respectively, then $OKCL$ is a square with side length $r$. Therefore, $BM = BK = BC - CK = a - r$, $AM = AL = AC - CL = b - r$, and since $AM + BM = AB$, we have $a - r + b - r = c$. From this, we find that

$$
r = \frac{a + b - c}{2} = \frac{5 + 12 - 13}{2} = 2
$$

Let the circle with center $O_a$ and radius $r_a$ touch the leg $BC$ at point $P$, and the extensions of the leg $AC$ and the hypotenuse $AB$ at points $Q$ and $T$ respectively. Then $O_aPCQ$ is a square with side length $r_a$, so

$$
r_a = OP = CQ = AQ - AC = p - b = \frac{a + b + c}{2} - b = \frac{a + c - b}{2} = \frac{5 + 13 - 12}{2} = 3
$$

where $p$ is the semiperimeter of the triangle.

Similarly, we find the radii of the other circles.

Second method. Let $a$, $b$, and $c$ be the sides of an arbitrary triangle, $S$ its area, $p$ its semiperimeter, $r$ the radius of the inscribed circle, and $r_a$, $r_b$, and $r_c$ the radii of the excircles touching sides $a$, $b$, and $c$ respectively. Then

$$
r = \frac{S}{p}, \quad r_a = \frac{S}{p - a}
$$

In our case,

$$
\begin{gathered}
r = \frac{S}{p} = \frac{\frac{1}{2} \cdot 5 \cdot 12}{15} = 2, \quad r_a = \frac{S}{p - a} = \frac{30}{15 - 5} = 3 \\
r_b = \frac{S}{p - b} = \frac{30}{15 - 12} = 10, \quad r_c = \frac{S}{p - c} = \frac{30}{15 - 13} = 15
\end{gathered}
$$

## Answer

$2, 15, 3, 10$.