Protasov V.Yu.

Given a triangle $ABC$ with an area of 1. A perpendicular $BM$ is dropped from vertex $B$ to the angle bisector of angle $C$. Find the area of triangle $AMC$.

#
The first method. Draw a line through point $B$ parallel to $AC$ until it intersects the bisector of angle $C$ at point $N$ (see figure). Since

$\angle B N C=\angle A C N=\angle B C N$, triangle $B C N$ is isosceles and $B M$ is its median. Therefore, $S_{A M C}=1 / 2$ $S_{A N C}=1 / 2 S_{A B C}=1 / 2$

![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-21.jpg?height=1358&width=1098&top_left_y=1322&top_left_x=480)

The second method. Since $S_{A M C}=1 / 2 A C \cdot C M \sin \angle C / 2$ and $C M=B C \cos \angle C / 2$, then $S_{A M C}=1 / 4 A C \cdot B C \sin \angle C=1 / 2 S_{A B C}=$ $1 / 2$.

Find the radii of the circumcircle, incircle, and excircles of a triangle with sides $13, 14, 15$.

## Solution

Let the sides $A B, A C$, and $B C$ of triangle $A B C$ be 13, 14, and 15 respectively (Fig. 1), $R$ and $r$ be the radii of the circumcircle and incircle respectively, $r_{a}, r_{b}$, and $r_{c}$ be the radii of the excircles touching sides $B C, A C$, and $A B$ respectively, $S$ be the area of triangle $A B C$, and $p$ be the semiperimeter. By the cosine rule,

$$
\cos \angle B A C=\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C}=\frac{169+196-225}{2 \cdot 13 \cdot 14}=\frac{5}{13}.
$$

Then

$$
\begin{gathered}
\sin \angle B A C=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13} \\
S=\frac{1}{2} A B \cdot A C \sin \angle B A C=\frac{1}{2} \cdot 13 \cdot 14 \cdot \frac{12}{13}=84
\end{gathered}
$$

Therefore,

$$
r=\frac{S}{p}=\frac{84}{21}=4
$$

By the sine rule,

$$
R=\frac{B C}{2 \sin \angle B A C}=\frac{15}{2 \cdot \frac{12}{13}}=\frac{65}{8}
$$

Let $O$ be the center of the excircle touching side $B C$ at point $M$ and the extensions of sides $A C$ and $A B$ at points $K$ and $L$ respectively (Fig. 2). Then

$$
\begin{gathered}
A K=A L, C K=C M, B L=B M, 2 p=A C+B C+A B=A C+(C M+B M)+A L= \\
=(A C+C M)+(A B+B M)=(A C+C K)+(A B+B L)=A K+A B
\end{gathered}
$$

Thus, $A K=A L=p$, so

$$
\begin{gathered}
S=S_{\triangle A O K}+S_{\triangle A O L}-S_{K C B L O}=S_{\triangle A O K}+S_{\triangle A O L}-2 S_{B O C}= \\
=\frac{1}{2} A K \cdot O K+\frac{1}{2} A L \cdot O L-2 \cdot \frac{1}{2} B C \cdot O M=\frac{1}{2} p r_{a}+\frac{1}{2} p r_{a}-B C \cdot r_{a}=(p-B C) r_{a}
\end{gathered}
$$

Therefore,

$$
r_{a}=\frac{S}{p-B C}=\frac{84}{21-15}=\frac{84}{6}=14
$$

Similarly, we find that

$$
r_{b}=\frac{S}{p-A C}=\frac{84}{21-14}=\frac{84}{7}=12, r_{c}=\frac{S}{p-A B}=\frac{84}{21-13}=\frac{84}{8}=\frac{21}{2}
$$

## Answer

$\frac{65}{8}, 4, 14, 12, \frac{21}{2}$.