8,9  In an isosceles triangle \(ABC\), where \(AB = BC\) and angle \(B\) is \(\frac{\pi}{4}\), a perpendicular \(AD\) is dropped onto side \(BC\). In triangles \(ABD\) and \(ADC\), semicircles are inscribed such that their diameters lie on \(BD\) and \(AD\), respectively. Find the ratio of the areas of the constructed semicircles.}
Let a semicircle of radius $R$ with center at point $Q$ on side $BC$ of triangle $ABC$ touch the side $AB$ at point $P$, and a circle of radius $r$ with center $O$ on segment $AD$ touch the base $AC$. From the isosceles right triangles $BQP$ and $ABD$, we find that

$$
\begin{gathered}
B Q=P Q \sqrt{2}=R \sqrt{2}, B D=B Q+Q D=R \sqrt{2}+R, A B=B D \sqrt{2}=2 R+R \sqrt{2}, \\
\text { therefore } \\
C D=B C-B D=A B-B D=(2 R+R \sqrt{2})-(R \sqrt{2}+R)=R .
\end{gathered}
$$

The center of the circle inscribed in an angle lies on the angle bisector, so

$$
\left.\angle O C D=\frac{1}{2} \angle A C B=\frac{1}{2} \frac{\pi}{2}-\frac{\pi}{8}\right)=\frac{3 \pi}{16}
$$

From the right triangle $OCD$, we find that

$$
\frac{r}{R}=\frac{O D}{D C}=\operatorname{tg} \angle O C D=\operatorname{tg} \frac{3 \pi}{16}
$$

Therefore, the ratio of the areas of the circles is

$$
\frac{\pi r^{2}}{\pi l^{2}}=\operatorname{tg} 2 \frac{3 \pi}{16}
$$

## Answer

$\operatorname{tg} 2 \frac{3 \pi}{16}$.

Submit a comment