 Touching Circles

Two circles touch each other externally at point $A$. Through point $B$ on their common tangent $A B$, two lines are drawn, one intersecting the first circle at points $M$ and $N$, and the other intersecting the second circle at points $P$ and $Q$. It is known that $A B=6, B M=9, B P=5$. Find the ratio of the areas of triangles $M N O$ and $P Q O$, where point $O$ is the intersection of lines $M P$ and $N Q$.
Prove that points $M, N, P$, and $Q$ lie on the same circle. From this, it follows that triangles $M N O$ and $Q P O$ are similar.

## Solution

Notice that point $N$ lies between $B$ and $M$, and point $P$ lies between $B$ and $Q$. By the tangent-secant theorem, $B N \cdot B M = A B^{2} = B P \cdot B Q$. Therefore, points $M, N, P$, and $Q$ lie on the same circle.

(Indeed, from the equality $B N \cdot B M = B P \cdot B Q$, it follows that $\frac{B N}{B P} = \frac{B Q}{B M}$. This means that triangles $B P N$ and $B M Q$ are similar by two sides and the angle between them. Therefore, $\angle B P N = \angle B M Q$. Then $\angle N M Q + \angle N P Q = 180^{\circ}$, which means that quadrilateral $M N P Q$ is cyclic.)

Chords $M P$ and $N Q$ intersect at point $O$. Therefore, triangle $M N O$ is similar to triangle $Q P O$ by two angles ($\angle O M N = \angle O Q P$ as inscribed angles subtending the same arc), and the similarity ratio $k$ is $\frac{M N}{P Q}$.

From the equalities $B N \cdot B M = A B^{2}$ and $B P \cdot B Q = A B^{2}$, we find that

$$
B N = \frac{A B^{2}}{B M} = \frac{36}{9} = 4 \text{ and } B Q = \frac{A B^{2}}{B P} = \frac{36}{5}
$$

Then

$$
M N = B M - B N = 9 - 4 = 5 \text{ and } P Q = B Q - B P = \frac{36}{5} - 5 = \frac{11}{5}.
$$

Therefore,

$$
\frac{S_{\triangle M N O}}{S_{\triangle P Q O}} = k^{2} = \left(\frac{M N}{P Q}\right)^{2} = \left(\frac{5}{\frac{11}{5}}\right)^{2} = \frac{625}{121}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-10.jpg?height=443&width=644&top_left_y=1&top_left_x=707)

## Answer

$\frac{625}{121}$.