[ Ratio of areas of similar triangles ] Law of Sines

In an acute-angled triangle $A B C$, heights $A P$ and $C Q$ are dropped from vertices $A$ and $C$ to sides $B C$ and $A B$. It is known that the area of triangle $A B C$ is 18, the area of triangle $B P Q$ is 2, and $P Q=2 \sqrt{2}$. Find the radius of the circumcircle of triangle $A B C$.
Triangles $B P Q$ and $B A C$ are similar with a similarity coefficient of $\frac{1}{3}$.

## Solution

Triangles $B P Q$ and $B A C$ are similar by two angles. Since the ratio of their areas is $\frac{2}{18}=\frac{1}{9}$, the similarity coefficient is $\frac{1}{3}$. Therefore, $A C=3 P Q=6 \sqrt{2}$.

On the other hand, the similarity coefficient is equal to $\frac{B P}{A B}=\cos \angle B$. Therefore, $\cos \angle B=\frac{1}{3}$. Then $\sin \angle B=\frac{2 \sqrt{2}}{3}$. If $R$ is the radius of the circumscribed circle of triangle $A B C$, then

$$
R=\frac{A C}{2 \sin \angle B}=\frac{6 \sqrt{2}}{2 \cdot \frac{2 \sqrt{2}}{3}}=\frac{9}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-38.jpg?height=409&width=412&top_left_y=1648&top_left_x=823)

## Answer

$\frac{9}{2}$.