5.1. Let $C H, C L$ and $C M$ be altitude, bisector and a median of the triangle $A B C$ with points $H, L$ and $M$ lying on $A B$.

The ratio of areas of triangles $H M C$ and $A B C$ is $\frac{1}{4}$, while the corresponding ratio for triangles $L M C$ and $A B C$ is $1-\frac{\sqrt{3}}{2}$.

Determine the angles of the triangle $A B C$.

(Bulgaria)
5.1. Let $A B=c, A C=b$ and $B C=a$. Without loss of generality we may assume that $0<b<a$, so that the points $H$ and $L$ lie on the segment $A M$. From the conditions of the problem we have

## Figure 5.1.

$$
\begin{aligned}
& \frac{P(H M C)}{P(A B C)}=\frac{H M \cdot H C}{A B \cdot H C}=\frac{H M}{A B}=\frac{1}{4} \\
& \frac{P(L M C)}{P(A B C)}=\frac{L M \cdot H C}{A B \cdot H C}=\frac{L M}{A B}=1-\frac{\sqrt{3}}{2}
\end{aligned}
$$

It is well known that for bisectors we have the following relation:

$$
\frac{A L}{b}=\frac{c-A L}{a}
$$

so that

$$
A L=\frac{b c}{a+b}
$$

Therefore

$$
L M=\frac{c}{2}-A L=\frac{c(a-b)}{2(a+b)}
$$

As $L M / c=1-\sqrt{3} / 2$, we obtain that $a=\sqrt{3} b$. On the other hand, from the cosine theorem

$$
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha=b^{2}+c^{2}-2 c \cdot A H
$$

follows

$$
H M=\frac{c}{2}-A H=\frac{a^{2}-b^{2}}{2 c}=\frac{b^{2}}{c}
$$

and from this

$$
\frac{1}{4}=\frac{H M}{c}=\frac{b^{2}}{c^{2}}
$$

i.e. $c=2 b$. Therefore $\alpha=60 \mathrm{deg}, \beta=30 \mathrm{deg}, \gamma=90 \mathrm{deg}$.

## Alternate solution.

1) From $A H=H M$ we obtain $\angle H M C=\alpha$, and from this $C M=A C=b$.
2) As $L M=(1-\sqrt{3} / 2) c$, we have $L B=\frac{3-\sqrt{3}}{2} c$ and from $A L: L B=b: a$ we deduce

$$
a=\sqrt{3} b
$$

3) Now $H B=H M+M B=3 H M=3 A C \cos \alpha$ and $H B^{2}+H C^{2}=B C^{2}$, so that $9 \cos ^{2} \alpha+\sin ^{2} \alpha=3$, i.e. $\cos \alpha=1 / 2$ ( $\cos \alpha$ cannot be negative). Hence $\alpha=60$ deg .
4) We conclude that $A M=M C=M B$, and therefore $M$ is the center of the circumscribed circle, i.e. $\gamma=90 \mathrm{deg}$.