7. Circles $\omega$ and $\Omega$ touch externally at point $F$, and their common external tangent touches circles $\omega$ and $\Omega$ at points $A$ and $B$, respectively. Line $\ell$ passes through point $B$, intersects circle $\Omega$ again at point $C$, and intersects $\omega$ at points $D$ and $E$ (point $D$ is between $C$ and $E$). The common tangent of the circles passing through point $F$ intersects lines $A B$ and $B E$ at points $P$ and $H$, respectively (point $H$ lies between points $P$ and $F$). It is known that $B C=60$, $D H=H C=2$. Find the length of segment $H P$ and the radii of both circles.
Answer: $H P=8 \sqrt{31}, r=6 \sqrt{\frac{93}{5}}, R=2 \sqrt{465}$.

Solution. We apply the theorem of the tangent and the secant three times:

$$
\begin{gathered}
H F^{2}=H C \cdot H B=2 \cdot 62 \Rightarrow H F=2 \sqrt{31} \\
H F^{2}=H D \cdot H E \Rightarrow H E=\frac{H F^{2}}{H D}=62 \\
B A^{2}=B D \cdot B E=64 \cdot 124=\Rightarrow B A=16 \sqrt{31}
\end{gathered}
$$

Since the segments of the tangents drawn from a point to a circle are equal to each other, $P F=P A=P B$, therefore, $P F=\frac{1}{2} A B=8 \sqrt{31}$. Thus, $P H=P F-F H=6 \sqrt{31}$. Let $\angle B P H=\gamma$. Then, by the cosine theorem for triangle $B P H$, we get $B H^{2}=B P^{2}+H P^{2}-2 B P \cdot P H \cdot \cos \gamma$, i.e., $62^{2}=64 \cdot 31+36 \cdot 31-2 \cdot 6 \cdot 8 \cdot 31 \cos \gamma$, from which $31=16+9-24 \cos \gamma, \cos \gamma=-\frac{1}{4}$.

Let $O$ and $Q$ be the centers, and $R$ and $r$ be the radii of circles $\Omega$ and $\omega$ respectively; since the circles are tangent, the point of tangency $F$ lies on the line of centers $O Q$, and $O Q \perp P H$. Angles $B$ and $F$ of quadrilateral $O B P F$ are right angles, so $\angle B O F=180^{\circ}-\angle B P F=180^{\circ}-\gamma$.

Consider the right trapezoid $A B O Q$. In it, $O Q=R+r, O B=R, A Q=r, A B=16 \sqrt{31}$, $\angle B O Q=180^{\circ}-\gamma$. Dropping a perpendicular $Q H$ from point $Q$ to the base $B O$, we get a right triangle $O H Q$, in which $Q H=A B=16 \sqrt{31}, O H=R-r$. By the Pythagorean theorem, we get
$(R+r)^{2}=(R-r)^{2}+(16 \sqrt{31})^{2}$; moreover, $\frac{1}{4}=\cos \left(180^{\circ}-\gamma\right)=\frac{R-r}{R+r}$. From the last equation, we get $R=\frac{5 r}{3}$, and from the first, it follows that $R r=64 \cdot 31$. Solving this system of equations, we find that $R=8 \sqrt{\frac{155}{3}}, r=8 \sqrt{\frac{93}{5}}$.

## Ticket 4