4. From the base $BC$ of the isosceles triangle $ABC$, the height $AH$ is dropped. On the side $AB$, a point $P$ is marked such that $CP = BC$. The segment $CP$ intersects the height $AH$ at point $Q$. It turns out that the area of triangle $BHQ$ is 4 times smaller than the area of triangle $APQ$. Find the angles of triangle $ABC$.
Answer: $30^{\circ}, 75^{\circ}, 75^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-28.jpg?height=482&width=820&top_left_y=553&top_left_x=698)

Solution 1. Since triangle $A B C$ is isosceles, its height $A H$ is also the bisector and median, hence $B H=C H$ and $\angle C A H=\angle B A H$. Let $N$ be the midpoint of segment $B P$. Then $N H$ is the midline of triangle $P B C$ and, therefore, $N H \| P C$. Triangle $B C P$ is isosceles, so segment $C N$ is its height. By the given condition

$$
4=\frac{S_{A P Q}}{S_{B H Q}}=\frac{S_{A P Q}}{S_{C H Q}}=\frac{A Q \cdot P Q}{Q H \cdot Q C} \Longrightarrow \frac{P Q}{Q C}=4 \cdot \frac{Q H}{A Q}=4 \cdot \frac{P N}{A P}
$$

Since $A Q$ is the bisector of triangle $A P C$, we get

$$
\frac{A P}{A B}=\frac{A P}{A C}=\frac{P Q}{Q C}=4 \cdot \frac{P N}{A P}
$$

Let $x=\frac{A N}{A B}$. Then

$$
\frac{P N}{A B}=\frac{B N}{A B}=1-x, \quad \frac{A P}{A B}=1-2 \cdot \frac{P N}{A B}=2 x-1
$$

and from the previous equality

$$
(2 x-1)^{2}=\left(\frac{A P}{A B}\right)^{2}=4 \cdot \frac{P N}{A P} \cdot \frac{A P}{A B}=4 \cdot \frac{P N}{A B}=4-4 x \Longleftrightarrow x=\frac{\sqrt{3}}{2}
$$

Since $x=\frac{A N}{A C}=\cos \angle B A C$, angle $B A C$ is $30^{\circ}$, and angles $A B C$ and $A C B$ are $75^{\circ}$.

Solution 2. Note that by the two legs, right triangles $A H B$ and $A H C$, as well as $Q H B$ and $Q H C$, are equal. Then $\angle B A H=\angle C A H$; denote the common value of these angles by $\alpha$. Triangles $B C P$ and $B A C$ are isosceles, so

$$
\angle C P B=\angle P B C=\angle A C B=90^{\circ}-\alpha \quad \text { and } \quad \angle A P C=90^{\circ}+\alpha .
$$

Triangles $A Q B$ and $A Q C$ are equal by two sides and an angle, hence

$$
\angle P B Q=\angle A C Q=180^{\circ}-\angle A P C-\angle P A C=90^{\circ}-3 \alpha \quad \text { and } \quad \angle Q B H=\angle Q C H=\angle A C B-\angle A C Q=2 \alpha
$$

By the formula for the area of a triangle

$$
\frac{1}{2} A Q \cdot P Q \cdot \sin \angle A Q P=S_{A P Q}=4 S_{B H Q}=4 S_{C H Q}=2 C Q \cdot H Q \cdot \sin \angle C Q H=2 C Q \cdot H Q \cdot \sin \angle A Q P
$$

Thus

$$
4 \cdot \frac{C Q}{A Q}=\frac{P Q}{H Q}=\frac{P Q}{B Q \sin \angle Q B H}=\frac{P Q}{B Q \sin 2 \alpha}
$$

By the sine rule for triangles $C Q A$ and $P B Q$

$$
\frac{C Q}{A Q}=\frac{\sin \alpha}{\sin \left(90^{\circ}-3 \alpha\right)}=\frac{\sin \alpha}{\cos 3 \alpha} \quad \text { and } \quad \frac{P Q}{B Q}=\frac{\sin \left(90^{\circ}-3 \alpha\right)}{\sin \left(90^{\circ}-\alpha\right)}=\frac{\cos 3 \alpha}{\cos \alpha} \text {. }
$$

Substituting these relations into the previous equality, we get

$$
4 \cdot \frac{\sin \alpha}{\cos 3 \alpha}=4 \cdot \frac{C Q}{A Q}=\frac{P Q}{B Q \sin 2 \alpha}=\frac{\cos 3 \alpha}{\sin 2 \alpha \cdot \cos \alpha}
$$

Therefore,

$$
\cos ^{2} 3 \alpha=4 \sin \alpha \cos \alpha \sin 2 \alpha=2 \sin ^{2} 2 \alpha
$$

from which

$$
\cos 3 \alpha=\sqrt{2} \sin 2 \alpha \Longleftrightarrow 1-4 \sin ^{2} \alpha=2 \sqrt{2} \sin \alpha \Longleftrightarrow \sin \alpha=\frac{\sqrt{3}-1}{2 \sqrt{2}} \Longleftrightarrow \alpha=15^{\circ}
$$