6. [7 points] An acute triangle $ABC$ is inscribed in a circle $\omega$ with center $O$. The circle passing through points $A, O$, and $C$ intersects the segment $BC$ at point $P$. The tangents to $\omega$ drawn through points $A$ and $C$ intersect at point $T$. Segment $TP$ intersects side $AC$ at point $K$. It is known that the areas of triangles $APK$ and $CPK$ are 15 and 13, respectively.

a) Find the area of triangle $ABC$.

b) Suppose it is additionally known that $\angle ABC = \operatorname{arctg} \frac{4}{7}$. Find $AC$.
Answer: a) $S_{A B C}=\frac{784}{13}$; b) $A C=\frac{14}{\sqrt{3}}$.

Solution. Since lines $T C$ and $T A$ are tangents to $\omega$, they are perpendicular to the radii drawn to the points of tangency, and $\angle O C T=\angle O A T=90^{\circ}$. It follows that points $A$ and $C$ lie on a circle with diameter $O T$ (let's call this circle $\Omega$). Point $P$ also lies on this circle, as it lies on the circle passing through points $A, O, C$. Let $\angle A B C=\beta$. Then, by the property of the angle between a chord and a tangent, we get that $\angle T A C=\beta$. Further, $\angle T P C=\angle T A C=\beta$ (angles inscribed in circle $\Omega$). From the fact that $\angle T P C=\angle A B C$, it follows that $A B \| P T$.

Since triangles $A P K$ and $C P K$ have a common height drawn from vertex $P$, their areas are in the ratio of their bases, i.e., $C K: A K=S_{\triangle C P K}: S_{\triangle A P K}=13: 15$. Triangles $A B C$ and $K P C$ are similar because $P K \| A B$, and the similarity coefficient $k$ is $\frac{A C}{C K}=\frac{A K+K C}{C K}=1+\frac{A K}{C K}=\frac{28}{13}$. Therefore, $S_{A B C}=$ $k^{2} \cdot S_{\triangle C P K}=\left(\frac{28}{13}\right)^{2} \cdot 13=\frac{784}{13}$.

b) Since $\angle A B C$ is acute, then $\angle A O C=2 \angle A B C=2 \beta$ (the central angle is twice the inscribed angle), $\angle A P C=\angle A O C=2 \beta$ (inscribed angles in $\Omega$ subtending the same arc). Therefore, $P K$ is the angle bisector of triangle $A C P$. The angle bisector of a triangle divides the opposite side proportionally to the other two sides, so $C P: A P=C K: A K=13: 15$. Let $C P=13 y$; then $A P=15 y$.

From the additional condition $\beta=\operatorname{arctg} \frac{4}{7}$. Therefore,

$$
\begin{aligned}
& \cos 2 \beta=\frac{\cos 2 \beta}{1}=\frac{\cos ^{2} \beta-\sin ^{2} \beta}{\cos ^{2} \beta+\sin ^{2} \beta}=\frac{1-\operatorname{tg}^{2} \beta}{1+\operatorname{tg}^{2} \beta}=\frac{1-\left(\frac{4}{7}\right)^{2}}{1+\left(\frac{4}{7}\right)^{2}}=\frac{33}{65} \\
& \sin 2 \beta=\sqrt{1-\cos ^{2} 2 \beta}=\sqrt{\left(1+\frac{33}{65}\right)\left(1-\frac{33}{65}\right)}=\sqrt{\frac{98 \cdot 32}{65^{2}}}=\frac{56}{65}
\end{aligned}
$$

The area of triangle $A C P$ is $\frac{1}{2} \cdot C P \cdot A P \sin 2 \beta=\frac{1}{2} \cdot 13 y \cdot 15 y \cdot \frac{56}{65}=84 y^{2}$, from which we get $84 y^{2}=28$, $y^{2}=\frac{1}{3}$. By the cosine rule in triangle $A P C$, we find that $A C^{2}=(13 y)^{2}+(15 y)^{2}-2 \cdot 13 y \cdot 15 y \cdot \cos 2 \beta=394 y^{2}+390 y^{2} \cdot \frac{33}{65}=196 y^{2}=\frac{196}{3}$, from which we finally get $A C=\frac{14}{\sqrt{3}}$.

## VARIANT 24