# Problem 3.

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$, if $A B=\sqrt{13}$.
Answer: 12

Solution. Let $A B=c, B E=A D=2 a$. Since triangle $A B D$ is isosceles (the bisector is perpendicular to the base, $A B=B D=c, B C=2 c$), then by the formula for the length of the bisector (where $\angle A B C=\beta$) $2 a=\frac{4 c^{2}}{3 c} \cos \frac{\beta}{2} \Leftrightarrow \frac{3 a}{2}=c \cos \frac{\beta}{2}$. Consider triangle $A B F$ (where $F$ is the midpoint of segment AD and the intersection point of the bisector and the median). We have $a=c \sin \frac{\beta}{2}$, $c=\frac{a}{\sin \frac{\beta}{2}}$. Therefore, $\operatorname{tg} \frac{\beta}{2}=\frac{2}{3}$ and $\cos \frac{\beta}{2}=\frac{3}{\sqrt{13}}, \sin \frac{\beta}{2}=\frac{2}{\sqrt{13}}, \sin \beta=\frac{12}{13}$. The area of triangle $A B C$ is $S_{A B C}=c^{2} \sin \beta=\frac{12}{13} c^{2}$.

## B-2

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $A B=\sqrt{26}$.

Answer: 24

## B-3

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=4$.

Answer: 12

## B-4

In triangle $A B C$, the bisector $B E$ and the median $A D$ are equal and perpendicular. Find the area of triangle $A B C$ if $B E=A D=6$.

Answer: 27

## Lomonosov High School Mathematics Olympiad

Preliminary stage 2020/21 academic year for 9th grade

## B-1

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