10. Let $A B C$ be a triangle such that $A B=6, B C=5, A C=7$. Let the tangents to the circumcircle of $A B C$ at $B$ and $C$ meet at $X$. Let $Z$ be a point on the circumcircle of $A B C$. Let $Y$ be the foot of the perpendicular from $X$ to $C Z$. Let $K$ be the intersection of the circumcircle of $B C Y$ with line $A B$. Given that $Y$ is on the interior of segment $C Z$ and $Y Z=3 C Y$, compute $A K$.
Answer:
$\frac{147}{10}$
Let $\omega_{1}$ denote the circumcircle of $A B C$ and $\omega_{2}$ denote the circle centered at $X$ through $B$ and $C$. Let $\omega_{2}$ intersect $A B, A C$ again at $B^{\prime}, C^{\prime}$. The (signed) power of $Y$ with respect to $\omega_{1}$ is $-C Y \cdot Y Z$. The power of $Y$ with respect to $\omega_{2}$ is $X Y^{2}-C X^{2}=-C Y^{2}$. Thus the ratio of the powers of $Y$ with respect to the two circles is $3: 1$. The circumcircle of $B C Y$ passes through the intersection points of $\omega_{1}$ and $\omega_{2}$ ( $B$ and $C$ ) and thus contains exactly the set of points such that the ratio of their powers with respect to $\omega_{1}$ and $\omega_{2}$ is $3: 1$ (this fact can be verified in a variety of ways). We conclude that $K$ must be the point on line $A B$ such that $\frac{K B^{\prime}}{K A}=3$. It now suffices to compute $A B^{\prime}$. Note $A B^{\prime}=\frac{A C \cdot B^{\prime} C^{\prime}}{B C}$ by similar triangles. Also an angle chase gives that $B^{\prime} X C^{\prime}$ are collinear. We compute
$$
B X=\frac{B C}{2 \cos A}=\frac{5}{2 \cdot \frac{5}{7}}=\frac{7}{2}
$$
and thus $B^{\prime} C^{\prime}=7$ so $A B^{\prime}=\frac{49}{5}$ and $A K=\frac{3}{2} A B^{\prime}=\frac{147}{10}$.