In triangle $ABC$, where $AB = BC = 3$, $\angle ABC = \arccos \frac{1}{9}$, the median $AD$ and the angle bisector $CE$ intersect at point $M$. A line through point $M$ parallel to $AC$ intersects sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Find $PM$ and the radius of the circle inscribed in triangle $PQB$.

Answer

$\frac{16}{11}, \frac{14 \sqrt{5}}{55}$.

Circle $\sigma$ with center at point $O$ on side $AC$ of triangle $ABC$ touches sides $AB$ and $BC$ at points $D$ and $E$ respectively. It is known that $AD = 2CE$, and the angle $DOE$ is $\operatorname{arcctg} \frac{1}{3}$. Find the angles of triangle $ABC$ and the ratio of its area to the area of the circle bounded by circle $\sigma$.
Let $\angle A B C=\beta, \angle B A C=\alpha, \angle A C B=\gamma, O D=O E=R, C E=x, B D=B E=y$. Then $A D=2 x$,

$$
\beta=\angle A B C=\pi-\angle D O E=\pi-\operatorname{arcctg} \frac{1}{3}
$$

$\operatorname{tg} \beta=\operatorname{tg}\left(\pi-\operatorname{arcctg} \frac{1}{3}\right)=-\operatorname{tg}\left(\operatorname{arcctg} \frac{1}{3}\right)=-\operatorname{tg} \operatorname{arctg} 3=-3$.

$\sin \beta=\frac{1}{\sqrt{1+\alpha+g^{2} \beta}}=\frac{1}{\sqrt{1+\frac{1}{9}}}=\frac{3}{\sqrt{10}}$.

From the right triangles $A O D$ and $C O E$ we find that

$$
\operatorname{tg} \alpha=\operatorname{tg} \angle D A O=\frac{O D}{A D}=\frac{R}{2 x}, \operatorname{tg} \gamma=\operatorname{tg} \angle E C O=\frac{O E}{C E}=\frac{R I}{x}
$$

Then

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-54.jpg?height=134&width=686&top_left_y=1096&top_left_x=692)

Then

$$
-3=\operatorname{tg} \beta=\operatorname{tg}(\pi-\alpha-\gamma)=-\operatorname{tg}(\alpha+\gamma)=-\frac{3 H \pi x}{2 x^{2}-r^{2}},
$$

or $R 2+R x-2 x 2=0$, from which $R=x$. Therefore,

$$
\operatorname{tg} \gamma=\frac{R P}{x}=1, \gamma=\frac{\pi}{4}, \operatorname{tg} \alpha=\frac{\frac{R}{2 x}}{2 x}=\frac{1}{2}, \alpha=\operatorname{arctg} \frac{1}{2}=\operatorname{arcctg} 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-54.jpg?height=120&width=1927&top_left_y=1575&top_left_x=0)
From the right triangle $B O E$ we find that

$$
y=B E=O E \operatorname{ctg} \angle O B E=R \operatorname{ctg} \frac{\frac{9}{2}=\frac{R}{2}}{\operatorname{tg}}=\frac{3 x}{1+\sqrt{10}}=\frac{x(\sqrt{10}-1)}{3} .
$$

Then

$$
\begin{gathered}
B C=B E+E C=y+x=\frac{x(\sqrt{10}-1)}{3}+x=\frac{x(2+\sqrt{10})}{3}, \\
A B=B D+A D=y+2 x=\frac{x(\sqrt{10}-1)}{3}+2 x=\frac{x(5+\sqrt{10})}{3}, \\
S_{\triangle A B C}=\frac{1}{2} A B \cdot B C \sin \beta=\frac{1}{2} \cdot \frac{x(2+\sqrt{10})}{3} \cdot \frac{x(5+\sqrt{10})}{3} \cdot \frac{3}{\sqrt{10}}=\frac{x^{2}(2 \sqrt{10}+7)}{6},
\end{gathered}
$$

and since the area of the circle with radius $R=x$ is $\pi x 2$, the ratio of the area of triangle $A B C$ to the area of the circle is $\frac{2 \sqrt{10}+7}{6 \pi}$

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-55.jpg?height=654&width=1741&top_left_y=-1&top_left_x=204)

## Answer

$\angle A B C=\pi-\operatorname{arcctg} \frac{\frac{1}{3}}{3}, \angle A C B=\frac{\pi}{4}, \angle B A C=\operatorname{arcctg} 2, \frac{2 \sqrt{10}+7}{6 \pi}$.

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