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

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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>
 
<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>
  
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==Solution==
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==Solution 1==
  
 
It is easy to see that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math>
 
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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-solution by FRaelya
 
-solution by FRaelya
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 +
==Solution 2==
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 +
We shall make use of the pairs of similar triangles present in the problem, Ptolemy's Theorem, and Power of a Point.
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Let <math>Z</math> be the intersection of <math>AC</math> and <math>BD</math>. First, from <math>ABCD</math> being a cyclic quadrilateral, we have that <math>\triangle BCZ \sim \triangle AZD</math>, <math>\triangle BZA \sim \triangle CDZ</math>. Therefore, <math>\frac{2}{BZ} = \frac{8}{AZ}</math>, <math>\frac{6}{CZ} = \frac{3}{BZ}</math>, and <math>\frac{2}{CZ} = \frac{8}{DZ}</math>, so we have <math>BZ = \frac{1}{2}CZ</math>, <math>AZ = 2CZ</math>, and <math>DZ = 4CZ</math>. By Ptolemy's Theorem, <cmath>(AB)(CD) + (BC)(DA) = (AC)(BD) = (AZ + ZC)(BZ + ZD)</cmath> <cmath>\rightarrow 3 \cdot 6 + 2 \cdot 8 = 34 = \left(2CZ + ZC\right)\left(\frac{1}{2}CZ + 4CZ\right) = \frac{27}{2}CZ^2.</cmath> Thus, <math>CZ^2 = \frac{68}{27}</math>. Then, by Power of a Point, <math>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}</math>. So, <math>XG = \frac{9 \cdot 17}{16XC}</math>.
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Next, observe that <math>\triangle ACX \sim \triangle EFX</math>, so <math>\frac{XE}{XF} = \frac{AX}{XC}</math>. Also, <math>\triangle{AXD} \sim \triangle{EXY}</math>, so <math>\frac{8}{AX} = \frac{EY}{XY}</math>. We can compute <math>EY = \frac{128}{9}</math> after noticing that <math>XY = BD - BY - DX = BD - \frac{11}{36}BD - \frac{1}{4}BD = \frac{4}{9}BD</math> and that <math>\frac{8}{DX} = \frac{32}{BD} = \frac{EY}{XY} = \frac{EY}{\frac{4}{9}BD}</math>. So, <math>\frac{8}{AX} = \frac{128}{9XE}</math>. Then, <math>\frac{XE}{AX} = \frac{XF}{XC} = \frac{16}{9} \rightarrow XF = \frac{16}{9}XC</math>.
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Multiplying our equations for <math>XF</math> and <math>XG</math> yields that <math>XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{\textbf{(A)}\ 17}.</math>
  
 
==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 00:56, 12 July 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 1

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

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 $\frac{8}{AX} = \frac{EY}{XY}$. 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 $XF \cdot XG = \frac{9 \cdot 17}{16XC} \cdot \frac{16}{9}XC = \boxed{\textbf{(A)}\ 17}.$

See Also

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

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