36. [25] The Marathon. Let $\omega$ denote the incircle of triangle $A B C$. The segments $B C, C A$, and $A B$ are tangent to $\omega$ at $D, E$, and $F$, respectively. Point $P$ lies on $E F$ such that segment $P D$ is perpendicular to $B C$. The line $A P$ intersects $B C$ at $Q$. The circles $\omega_{1}$ and $\omega_{2}$ pass through $B$ and $C$, respectively, and are tangent to $A Q$ at $Q$; the former meets $A B$ again at $X$, and the latter meets $A C$ again at $Y$. The line $X Y$ intersects $B C$ at $Z$. Given that $A B=15, B C=14$, and $C A=13$, find $\lfloor X Z \cdot Y Z\rfloor$.
Answer: 101. Construct $D^{\prime}$ diametrically opposed to $D$, so that $\angle D F D^{\prime}$ and $\angle D E D^{\prime}$ are right, and note that $P$ lies on $D D^{\prime}$. By standard angle chasing, $m \angle F D D^{\prime}=\beta$ (half angle $B$ ) and $m \angle D^{\prime} D E=\gamma$. Thus, $m \angle D D^{\prime} F=90^{\circ}-\beta$ and $m \angle E D^{\prime} D=90^{\circ}-\gamma$. Then by the law of sines, $D E: E D^{\prime}: D^{\prime} F$ : $F D=\cos (\gamma): \sin (\gamma): \sin (\beta): \sin (\gamma)$. Now using $\triangle D E P \sim \triangle F D^{\prime} P$ and $\triangle D F P \sim \triangle E D^{\prime} P$, we have
$$
\frac{E P}{P F}=\frac{E D \cdot E D^{\prime}}{F D \cdot F D^{\prime}}=\frac{\sin (\gamma) \cos (\gamma)}{\sin (\beta) \sin (\beta)}=\frac{c}{b} .
$$

Let the dilation centered at $A$ sending $E$ to $C$ map $P$ and $F$ to $P^{\prime}$ and $F^{\prime}$, respectively. Note that $A F^{\prime}=A C$ as $A E$ and $A F$ are equal tangents, and $C P^{\prime}: P^{\prime} F^{\prime}=E P: P F=c: b$ by similarity. Then by Menelaus' theorem,
$$
1=\frac{B Q}{Q C} \frac{C P^{\prime}}{P^{\prime} F^{\prime}} \frac{F^{\prime} A}{A B}=\frac{B Q}{Q C} \frac{c}{b} \frac{b}{c},
$$
so that $B Q=Q C$ and $A Q$ is actually a median. So, $A Q^{2}=\frac{1}{4}\left(2 b^{2}+2 c^{2}-a^{2}\right)=148$. Now by Power of a Point, $A B \cdot A X=A Q^{2}=A C \cdot A Y$, so $A X=148 / 15$ and $A Y=148 / 13$. Moreover, $B X C Y$ is cyclic as $\triangle A B C \sim \triangle A Y X$. Thus, $X Z \cdot Y Z=B Z \cdot C Z$, and it suffices to compute $B Z / C Z$. Menelaus once more gives
$$
1=\frac{B Z}{Z C} \frac{C Y}{Y A} \frac{A X}{X B},
$$
whence, $B Z / C Z=(A Y / A X)(B X / C Y)=(15 / 13)((77 \cdot 13) /(21 \cdot 15))=11 / 3$. We write $C Z=3 d$ and $B Z=11 d$. Because $A X<A B$ and $A Y<A C, Z$ does not lie on segment $B C$. Given the configuration information, $B C=8 d=14$, so $d=7 / 4$, and finally $\lfloor B Z \cdot C Z\rfloor=\left\lfloor 33 d^{2}\right\rfloor=\left\lfloor\frac{1617}{16}\right\rfloor=101$.