In an acute-angled triangle, two altitudes are equal to 3 and $2 \sqrt{2}$, and their point of intersection divides the third altitude in the ratio 5:1, counting from the vertex of the triangle. Find the area of the triangle.
Let $\alpha$ and $\beta$ be the acute angles of a triangle, from the vertices of which the given altitudes are drawn. Formulate a system of trigonometric equations in terms of $\alpha$ and $\beta$.

## Solution

Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes of triangle $ABC$, and $H$ be their point of intersection; $A A_{1}=3, B B_{1}=2 \sqrt{2}$.

Denote

$$
\angle C A B=\alpha, \angle A B C=\beta, H C_{1}=x
$$

Then

$$
C C_{1}=6 x, \angle A H C_{1}=\beta, \angle B H C_{1}=\alpha .
$$

From the right triangles $A A_{1} B$ and $B B_{1} A$, we find that

$$
A B=\frac{3}{\sin \beta}=\frac{2 \sqrt{2}}{\sin \alpha}
$$

Therefore, $\sin \beta=\frac{3 \sin \alpha}{2 \sqrt{2}}$. From the right triangles $B C_{1} C$ and $B C_{1} H$, we find that

$$
B C_{1}=\frac{6 x}{\tan \beta}=x \tan \alpha .
$$

Therefore

$$
\tan \alpha \tan \beta=6 \Rightarrow \sin \alpha \sin \beta=6 \cos \alpha \cos \beta \Rightarrow \sin ^{2} \alpha \sin ^{2} \beta=36 \cos ^{2} \alpha \cos ^{2} \beta
$$

Substitute into the last equation

$$
\sin ^{2} \beta=\frac{9 \sin ^{2} \alpha}{8}, \cos ^{2} \beta=1-\sin ^{2} \beta=1-\frac{9 \sin ^{2} \alpha}{8}, \cos ^{2} \alpha=1-\sin ^{2} \alpha
$$

We get the equation

$$
35 \sin ^{4} \alpha-68 \sin ^{2} \alpha+32=0
$$

from which we find that $\sin \alpha=\frac{2}{\sqrt{5}}$ ( $\alpha$ and $\beta$ are less than 90° ). Therefore,

$$
\cos \alpha=\frac{1}{\sqrt{5}}, \sin \beta=\frac{3}{\sqrt{10}}, \cos \beta=\frac{1}{\sqrt{10}}, A B=\frac{3}{\sin \beta}=\sqrt{10}
$$

From triangle $ABC$, we find by the sine rule:

$$
A C=\frac{A B \sin \beta}{\sin (\alpha+\beta)}=3 \sqrt{2}
$$

Therefore,

$$
S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \alpha=\frac{1}{2} \cdot \sqrt{10} \cdot 3 \sqrt{2} \cdot \frac{2}{\sqrt{5}}=6
$$

## Answer

6.