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 $XF\cdot XG$ ?

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

Diagram

$[asy] size(8cm); real r = 4.01754; draw(circle((0,0), r)); pair C = r * dir(-30), B = r * dir(28.83-30), A = r * dir(72.68-30), D = r * dir(241.98-30); draw(A--B--C--D--cycle); draw(B--D); pair X = B * 1/4 + D * 3/4, Y = B * 25/36 + D * 11/36; label("A", A, N); label("B", B, NE); label ("C", C, E); label("D", D, S); label("Y", Y, N); label("X", X, N); pair G = X * 1.445 - C*0.445; label("G", G, NW); pair E = Y + (D - A) * 1.48; draw(Y--E); draw(A--E); label("E", E, S); pair F = E + (A - C) * 1.45; draw(C--F--E); label("F",F,NW); [/asy]$ ~raxu, put in by fuzimiao2013

Solution 1

Using the given ratios, note that $\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.$

By AA Similarity, $\triangle AXD \sim \triangle EXY$ with a ratio of $\frac{DX}{XY} = \frac{9}{16}$ and $\triangle ACX \sim \triangle EFX$ 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

Solution 2

We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point. Let $Z$ be the intersection of $AC$ and $BD$ . First, from $ABCD$ being a cyclic quadrilateral, we have that $\triangle BCZ \sim \triangle AZD$ , $\triangle BZA \sim \triangle CDZ$ . Therefore, $\frac{2}{BZ} = \frac{8}{AZ}$ , $\frac{6}{CZ} = \frac{3}{BZ}$ , and $\frac{2}{CZ} = \frac{8}{DZ}$ , so we have $BZ = \frac{1}{2}CZ$ , $AZ = 2CZ$ , and $DZ = 4CZ$ . By Ptolemy's Theorem, $(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)$ $\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.$ Thus, $CZ^2 = \frac{68}{27}$ . Then, by Power of a Point, $GX \cdot XC = BX \cdot XD = \frac{3}{4} \cdot \frac{1}{4}BD^2 = \frac{3}{16} \cdot \left(\frac{9}{2}CZ\right)^2 = \frac{9 \cdot 17}{16}$ . So, $XG = \frac{9 \cdot 17}{16XC}$ . Next, observe that $\triangle ACX \sim \triangle EFX$ , so $\frac{XE}{XF} = \frac{AX}{XC}$ . Also, $\triangle{AXD} \sim \triangle{EXY}$ , so $\frac{8}{AX} = \frac{EY}{XE}$ . We can compute $EY = \frac{128}{9}$ after noticing that $XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD$ and that $\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}$ . So, $\frac{8}{AX} = \frac{128}{9XE}$ . Then, $\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC$ .

Multiplying our equations for $XF$ and $XG$ yields that $XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{\textbf{(A)}\ 17}.$

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4 - AMBRIGGS

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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