HMMT-Nov 2021 gen, Problem 9

$A B C D E$ is a cyclic convex pentagon, and $A C=B D=C E$. $A C$ and $B D$ intersect at $X$, and $B D$ and $C E$ intersect at $Y$. If $A X=6, X Y=4$, and $Y E=7$, then the area of pentagon $A B C D E$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are integers, $c$ is positive, $b$ is square-free, and $\operatorname{gcd}(a, c)=1$. Find $100 a+10 b+c$.

Solution: \section*{Solution:} Since $A C=B D, A B C D$ is an isosceles trapezoid. Similarly, $B C D E$ is also an isosceles trapezoid. Using this, we can now calculate that $C Y=D Y=D X-X Y=A X-X Y=2$, and similarly $B X=C X=3$. By applying Heron's formula we find that the area of triangle $C X Y$ is $\frac{3}{4} \sqrt{15}$. Now, note that \[ [A B C]=\frac{A C}{C X}[B X C]=3[B X C]=3 \frac{X Y}{B X}[C X Y]=\frac{9}{4}[C X Y] \] Similarly, $[C D E]=\frac{9}{4}[C X Y]$. Also, \[ [A C E]=\frac{C A \cdot C E}{C X \cdot C Y}[C X Y]=\frac{81}{6}[C X Y]=\frac{27}{2}[C X Y] \] Thus, $[A B C D E]=(9 / 4+9 / 4+27 / 2)[C X Y]=18[C X Y]=\frac{27}{2} \sqrt{15}$. $\fbox{2852}$.

Answer: 2852