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Difference between revisions of "2017 AMC 12A Problems/Problem 24"
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==Problem==
 
==Problem==
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Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on <math>\overline{BD}</math> such that <math>\frac{DX}{BD} = \frac{1}{4}</math> and <math>\frac{BY}{BD} = \frac{11}{36}</math>.
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Quadrilateral <math>ABCD</math> is inscribed in circle <math>O</math> and has side lengths <math>AB=3, BC=2, CD=6</math>, and <math>DA=8</math>. Let <math>X</math> and <math>Y</math> be points on  
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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>.
 
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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==Solution==
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It is easy to see that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}</math>.
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First we note that <math>\triangle AXD \sim \triangle EXY</math> with a ratio of <math>\frac{DX}{XY} = \frac{9}{16}</math>. Then <math>\triangle ACX \sim \triangle EFX</math> with a ratio of <math>\frac{AX}{XE} = \frac{DX}{XY} = \frac{9}{16}</math>, so <math>\frac{XF}{CX} = \frac{9}{16}</math>.
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Then we find the length of <math>BD</math>. Because the quadrilateral is cyclic, we can simply use the Law of Cosines. <cmath>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</cmath><cmath>\rightarrow \cos\angle BAD = \frac{11}{24}</cmath><cmath>\rightarrow BD=\sqrt{51}</cmath>
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By Power of a Point, <math>CX\cdot XG = DX\cdot XB = \frac{\sqrt{51}}{4} \frac{3\sqrt{51}}{4}</math>. Thus <math>XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{17} (A)</math>.
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-solution by FRaelya
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:38, 8 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 $XF\cdot XG$?

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 $\frac{XF}{CX} = \frac{9}{16}$. Then 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 $XF\cdot XG = \frac{XF}{CX} CX\cdot XG = \frac{51}{3} = \boxed{17} (A)$.

-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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