Problem 1439

Let $\left(c_{1}\right)$ and $\left(c_{2}\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\left(c_{1}\right)$ and $(c)$ meet the common tangent to $\left(c_{2}\right)$ and $(c)$ at $Q$. Then the point $Q$ is the radical center of the circles $\left(c_{1}\right),\left(c_{2}\right)$ and $(c)$, and the line $K L$ passes through $Q$.

We have $m(\widehat{Q L B})=m(\widehat{Q B K})=m(\widehat{Q B A})=\frac{1}{2} m(\overparen{B A})=m(\widehat{Q O B})$. So, the quadrilateral $O B Q L$ is cyclic. We conclude that $m(\widehat{Q L O})=90^{\circ}$ and the points $O, B, Q, A$ and $L$ are cocyclic on a circle $(k)$.

In the sequel, we will denote $\mathcal{P}_{\omega}(X)$ the power of the point $X$ with respect of the circle $\omega$. The first continuation.

From $M O^{2}-O P^{2}=\mathcal{P}_{c}(M)=M A \cdot M B=\mathcal{P}_{k}(M)=M L \cdot M O=(M O-O L) \cdot M O=$ $M O^{2}-O L \cdot M O$ follows that $O P^{2}=O L \cdot O M$. Since $P L \perp O M$, this shows that the triangle $M P O$ is right at point $P$. Thus, the line $M P$ is tangent to the circle (c).

The second continuation.

Let $R \in(c)$ be so that $B R \perp M O$. The triangle $L B R$ is isosceles with $L B=L R$, so $\widehat{O L R} \equiv \widehat{O L B} \equiv \widehat{O Q B} \equiv \widehat{O Q A} \equiv \widehat{M L A}$. We conclude that the points $A, L$ and $R$ are collinear.

Now $m(\widehat{A M R})+m(\widehat{A O R})=m(\widehat{A M R})+2 m(\widehat{A B R})=m(\widehat{A M R})+m(\widehat{A B R})+m(\widehat{M R B})=$ $180^{\circ}$, since the triangle $M B R$ is isosceles. So, the quadrilateral $M A O R$ is cyclic.

This yields $L M \cdot L O=-\mathcal{P}_{(\text {MAOR })}(L)=L A \cdot L R=-\mathcal{P}_{c}(L)=L P^{2}$, which as above, shows that $O P \perp P M$.

The third continuation.

$\widehat{K L A} \equiv \widehat{K A Q} \equiv \widehat{K L B}$ and $m(\widehat{M L K})=90^{\circ}$ show that $[L K$ and $[L M$ are the internal and external bisectors of the angle $\widehat{A L B}$, so $(M, K)$ and $(A, B)$ are harmonic conjugates. So, $L K$ is the polar line of $M$ in the circle $(c)$.

## Chapter 4

## Number Theory