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 $D E=18$, $D H=H C=3$. Find the length of segment $H P$ and the radii of both circles.
Answer: $H P=3 \sqrt{7}, r=2 \sqrt{21}, R=6 \sqrt{21}$.

Solution. We apply the tangent-secant theorem three times:

$$
\begin{gathered}
H F^{2}=H D \cdot H E=3 \cdot 21 \Rightarrow H F=3 \sqrt{7} \\
H F^{2}=H C \cdot H B \Rightarrow H B=\frac{H F^{2}}{H C}=21 \\
B A^{2}=B D \cdot B E=24 \cdot 42=\Rightarrow B A=12 \sqrt{7}
\end{gathered}
$$

Since the segments of tangents drawn from a point to a circle are equal, $P F=P A=P B$, hence $P F=\frac{1}{2} A B=6 \sqrt{7}$. Therefore, $P H=P F-F H=3 \sqrt{7}$. Let $\angle B P H=\gamma$. Then, by the cosine rule for triangle $B P H$, we have $B H^{2}=B P^{2}+H P^{2}-2 B P \cdot P H \cdot \cos \gamma$, i.e., $21^{2}=36 \cdot 7+9 \cdot 7-2 \cdot 3 \cdot 6 \cdot 7 \cos \gamma$, from which $7=4+1-4 \cos \gamma, \cos \gamma=-\frac{1}{2}$.

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, $O Q=R+r, O B=R, A Q=r, A B=12 \sqrt{7}$, $\angle B O Q=180^{\circ}-\gamma$. Dropping a perpendicular $Q H$ from point $Q$ to the base $B O$, we get a right triangle $O H Q$, in which $Q H=A B=12 \sqrt{7}, O H=R-r$. By the Pythagorean theorem, we have $(R+r)^{2}=(R-r)^{2}+(12 \sqrt{7})^{2} ;$ also, $\frac{1}{2}=\cos \left(180^{\circ}-\gamma\right)=\frac{R-r}{R+r}$. From the last equation, we get $R=3 r$, and from the first, we get $R r=252$. Solving this system of equations, we find that $R=6 \sqrt{21}$, $r=2 \sqrt{21}$.

A problem is considered fully solved (and the maximum number of points is awarded) only if all necessary transformations and logical steps are fully explained in the solution text; the answers obtained must be simplified.

The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.

A full and justified solution to the problem is awarded the full number of points (indicated in parentheses after the problem number). Some partial progress is evaluated according to the instructions. In other cases, the score is at the discretion of the checker.