2017 AMC 12A Problems/Problem 24

Problem

Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on

$\overline{BD}$ such that $\frac{DX}{BD} = \frac{1}{4}$ and $\frac{BY}{BD} = \frac{11}{36}$ . Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $\overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $\overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XF\cdot XG$ ?

$\textbf{(A) }17\qquad\textbf{(B) }\frac{59 - 5\sqrt{2}}{3}\qquad\textbf{(C) }\frac{91 - 12\sqrt{3}}{4}\qquad\textbf{(D) }\frac{67 - 10\sqrt{2}}{3}\qquad\textbf{(E) }18$

Solution

It is easy to see that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$ First we note that $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}.$ Then $\triangle ACX \sim \triangle EFX$ with a ratio of $\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}$ , so $\frac{XF}{CX} = \frac{16}{9}.$ Now we find the length of $BD$ . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. $BD^2=3^2+8^2-48\cos\angle BAD=2^2+6^2-24\cos (180-\angle BAD)=2^2+6^2+24\cos\angle BAD$ $\rightarrow \cos\angle BAD = \frac{11}{24}$ $\rightarrow BD=\sqrt{51}$ By Power of a Point, $CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}$ . Thus $XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{\textbf{(A)}\ 17}.$

-solution by FRaelya

2017 AMC 12A (Problems • Answer Key • Resources)
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2017 AMC 12A Problems/Problem 24

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