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 6 and 4, respectively.

a) Find the area of triangle $ABC$.

b) Suppose it is additionally known that $\angle ABC = \operatorname{arctg} \frac{7}{5}$. Find $AC$.
Answer: a) $S_{A B C}=25 ;$ b) $A C=\frac{50}{\sqrt{42}}$.

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 $\angle O C T=\angle O A T=90^{\circ}$. Therefore, 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 $C P K$ share a common height from vertex $P$, their areas are proportional to their bases, i.e., $C K: A K=S_{\triangle C P K}: S_{\triangle A P K}=4: 6=2: 3$. Triangles $A B C$ and $K P C$ are similar because $P K \| A B$, and the similarity coefficient $k$ is $\frac{A C}{C K}=\frac{A K+K C}{C K}=1+\frac{A K}{C K}=\frac{5}{2}$. Therefore, $S_{A B C}=$ $k^{2} \cdot S_{\triangle C P K}=\left(\frac{5}{2}\right)^{2} \cdot 4=25$.

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 $C P: A P=C K: A K=2: 3$. Let $C P=2 y$; then $A P=3 y$.

From the additional condition $\beta=\operatorname{arctg} \frac{7}{5}$. Therefore,

$$
\begin{aligned}
& \cos 2 \beta=\frac{\cos 2 \beta}{1}=\frac{\cos ^{2} \beta-\sin ^{2} \beta}{\cos ^{2} \beta+\sin ^{2} \beta}=\frac{1-\operatorname{tg}^{2} \beta}{1+\operatorname{tg}^{2} \beta}=\frac{1-\left(\frac{7}{5}\right)^{2}}{1+\left(\frac{7}{5}\right)^{2}}=-\frac{12}{37} \\
& \sin 2 \beta=\sqrt{1-\cos ^{2} 2 \beta}=\sqrt{\left(1+\frac{12}{37}\right)\left(1-\frac{12}{37}\right)}=\sqrt{\frac{49 \cdot 25}{37^{2}}}=\frac{35}{37}
\end{aligned}
$$

The area of triangle $A C P$ is $\frac{1}{2} \cdot C P \cdot A P \sin 2 \beta=\frac{1}{2} \cdot 2 y \cdot 3 y \cdot \frac{35}{37}=\frac{105}{37} y^{2}$, from which we get $\frac{105 y^{2}}{37}=10$, $y^{2}=\frac{74}{21}$. By the cosine rule in triangle $A P C$, we find that $A C^{2}=(2 y)^{2}+(3 y)^{2}-2 \cdot 2 y \cdot 3 y \cdot \cos 2 \beta=$ $13 y^{2}+12 y^{2} \cdot \frac{12}{37}=\frac{625 y^{2}}{37}=\frac{625 \cdot 74}{37 \cdot 21}$, from which we finally get $A C=\frac{25 \sqrt{2}}{\sqrt{21}}$.

## VARIANT 18