4. Chords $AB$ and $CD$ of circle $\Gamma$ with center $O$ have a length of 5. 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 6.5, and point $T$ is the center of the circle inscribed in triangle $ACP$. Find the length of segment $PT$ and the area of triangle $ACP$.
Answer: $A P=5, P T=\frac{3 \sqrt{41}-13}{2}, S_{\triangle A P C}=\frac{1000}{41}$.

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), so $P N=P H, \angle O P N=\angle O P H$ (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 $A N=H D=\frac{5}{2}$; 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 $\frac{y}{y+5}=\frac{1}{2}$, $y=5$.

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 $O H^{2}=O D^{2}-D H^{2}=\frac{169}{4}-\frac{25}{4}=$ 36. By the Pythagorean theorem for triangle $P H O: P O^{2}=P H^{2}+O H^{2}=\left(\frac{15}{2}\right)^{2}+36=\frac{369}{4}$. Therefore, $P T=P O-O T=\frac{3 \sqrt{41}}{2}-\frac{13}{2}$.

Let $\angle C P O=\beta$. Then $\sin \beta=\frac{O H}{O P}=6: \frac{3 \sqrt{41}}{2}=\frac{4}{\sqrt{41}}, \cos \beta=\sqrt{1-\sin ^{2} \beta}=\frac{5}{\sqrt{41}}, \sin 2 \beta=$ $2 \cos \beta \sin \beta=\frac{40}{41}, S_{\triangle A P C}=\frac{1}{2} P A \cdot P C \sin 2 \beta=\frac{1}{2} \cdot 5 \cdot 10 \cdot \frac{40}{41}=\frac{1000}{41}$.