Let $M^{\prime}$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_{2}$ and $M^{\prime} M$ is a diameter of $\omega_{2}$. It suffices to prove that $M^{\prime} A$ is perpendicular to $L M$, or equivalently, to $A K$. To see this, let $S$ be the point of intersection of $M^{\prime} A$ with $L M$. We will then have $\angle M^{\prime} S M=90^{\circ}$ which shows that $S$ belongs on $\omega_{2}$ as $M^{\prime} M$ is a diameter of $\omega_{2}$. We also have that $S$ belongs on $\omega_{1}$ as $A L$ is diameter of $\omega_{1}$.

Since $T$ and $C$ are the midpoints of $M^{\prime} M$ and $K M$ respectively, then $T C$ is parallel to $M^{\prime} K$ and so $M^{\prime} K$ is perpendicular to $A B$. Since $K A=K B$, then $K M^{\prime}$ is the perpendicular bisector of $A B$. But then the triangles $K B M^{\prime}$ and $K A M^{\prime}$ are equal, showing that $\angle M^{\prime} A K=\angle M^{\prime} B K=\angle M^{\prime} B M=90^{\circ}$ as required.
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Alternative Solution by Proposers. Since $C A=C L$, then $L$ belongs on $\omega_{1}$. Let $S$ be the other point of intersection of $\omega_{1}$ with the line $L M$. We need to show that $S$ belongs on $\omega_{2}$. Since $T B=T M$ ( $T$ is on the perpendicular bisector of $B M$ ) it is enough to show that $T S=T M$.

Let $N, T^{\prime}$ be points on the lines $A L$ and $L M$ respectively, such that $M N \perp L M$ and $T T^{\prime} \perp L M$. It is enough to prove that $T^{\prime}$ is the midpoint of $S M$. Since $A L$ is diameter of $\omega_{1}$ we have that $A S \perp L S$. Thus, it is enough to show that $T$ is the midpoint of $A N$. We have

$$A T=\frac{A N}{2} \Leftrightarrow A C-C T=\frac{A L-L N}{2} \Leftrightarrow 2 A C-2 C T=A L-L N \Leftrightarrow L N=2 C T$$

as $A L=2 A C$. So it suffices to prove that $L N=2 C T$.

Let $D$ be the midpoint of $B M$. Since $B K=K C=C M$, then $D$ is also the midpoint of $K C$. The triangles $L M N$ and $C T D$ are similar since they are right-angled with
$\angle T C D=\angle C A K=\angle M L N$. (AK=KC and $A K$ is parallel to $L M$.) So we have

$$\frac{L N}{C T}=\frac{L M}{C D}=\frac{A K}{C D}=\frac{C K}{C D}=2$$

as required.

## NUMBER THEORY