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Difference between revisions of "2017 AMC 12A Problems/Problem 24"

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

Revision as of 15:17, 10 February 2017

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

See Also

2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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