Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

$A B C D E$ 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 $100 a+10 b+c$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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, $[A B C D E]=(9 / 4+9 / 4+27 / 2)[C X Y]=18[C X Y]=\frac{27}{2} \sqrt{15}$ .

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