7. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 6. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$. Line $PO$ intersects segment $AC$ at point $L$, such that $AL: LC = 1: 2$.

a) Find $AP$.

b) Suppose additionally that the radius of circle $\Gamma$ is 4. Let $T$ be the center of the circle inscribed in triangle $ACP$. Find the length of segment $PT$ and the area of triangle $ACP$.
Answer: $A P=6, P T=2 \sqrt{22}-4, S_{\triangle A P C}=\frac{81 \sqrt{7}}{11}$.

Solution. a) Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles $O P N$ and $O P H$ are equal by the leg and hypotenuse ( $O P$ is common), therefore $P N=P H, \angle O P N=\angle O P H$ (the latter means that $P O$ is the bisector of angle $B P C$). Since a diameter perpendicular to a chord bisects it, points $H$ and $N$ are the midpoints of $C D$ and $A B$ respectively, so $A N=H D=3$; from this it follows that $A P=D P$. Let $A P=D P=y$. Since $P L$ is the bisector of triangle $A P C, \frac{P A}{P C}=\frac{A L}{L C}$, hence $\frac{y}{y+6}=\frac{1}{2}, y=6$.

b) Triangles $O P A$ and $O P D$ are equal by two sides and the angle between them ( $P O$ is common, $A P=D P$, $\angle A P O=\angle D P O$), so $\angle A O P=\angle D O P$. This means that segment $O P$ bisects arc $A D$. Denote the intersection of segment $O P$ with the circle as $J$. Then $C J$ is the bisector of angle $A C D$ (angles $A C J$ and $D C J$ are inscribed, subtending equal arcs). Moreover, $P J$ is the bisector of angle $A P C$, so $J$ is the intersection of the bisectors of triangle $A P C$, i.e., the center of its inscribed circle. Thus, points $J$ and $T$ coincide.

By the Pythagorean theorem for triangle $D H O$ we get $O H^{2}=O D^{2}-D H^{2}=16-9=7$. By the Pythagorean theorem for triangle $P H O: P O^{2}=P H^{2}+O H^{2}=9^{2}+7=88$. Therefore, $P T=P O-O T=$ $\sqrt{88}-4$.

Let $\angle C P O=\beta$. Then $\sin \beta=\frac{O H}{O P}=\sqrt{7}: \sqrt{88}=\frac{\sqrt{7}}{2 \sqrt{22}}, \cos \beta=\sqrt{1-\sin ^{2} \beta}=\frac{9}{2 \sqrt{22}}, \sin 2 \beta=$ $2 \cos \beta \sin \beta=\frac{9 \sqrt{7}}{44}, S_{\triangle A P C}=\frac{1}{2} P A \cdot P C \sin 2 \beta=\frac{1}{2} \cdot 6 \cdot 12 \cdot \frac{9 \sqrt{7}}{44}=\frac{81 \sqrt{7}}{11}$.

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