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

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Quadrilateral
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==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>.
 +
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>?
 +
 
 +
<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>
 +
 
 +
==Diagram==
 +
 
 +
<asy>
 +
size(8cm);
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real r = 4.01754;
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draw(circle((0,0), r));
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pair C = r * dir(-30), B = r * dir(28.83-30), A = r * dir(72.68-30), D = r * dir(241.98-30);
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draw(A--B--C--D--cycle); draw(B--D);
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pair X = B * 1/4 + D * 3/4, Y = B * 25/36 + D * 11/36;
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label("A", A, N); label("B", B, NE); label ("C", C, E); label("D", D, S);
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label("Y", Y, N); label("X", X, N);
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pair G = X * 1.445 - C*0.445;
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label("G", G, NW);
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pair E = Y + (D - A) * 1.48;
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draw(Y--E);
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draw(A--E);
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label("E", E, S);
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pair F = E + (A - C)  * 1.45;
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draw(C--F--E);
 +
label("F",F,NW);
 +
</asy>
 +
~raxu, put in by fuzimiao2013
 +
 
 +
==Solution 1==
 +
 
 +
Using the given ratios, note that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math>
 +
 
 +
By AA Similarity, <math>\triangle AXD \sim \triangle EXY</math> with a ratio of <math>\frac{DX}{XY} = \frac{9}{16}</math> and <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{16}{9}</math>.
 +
 
 +
Now 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>
 +
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{\textbf{(A)}\ 17}.</math>
 +
 
 +
-solution by FRaelya
 +
 
 +
==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.
 +
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>.
 +
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}{XE}</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>
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==Solution 3==
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Denote <math>P</math> to be the intersection between line <math>AE</math> and circle <math>O</math>. Note that <math>\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE</math>, making <math>\angle GFE + \angle GDE = 180</math>. Thus, <math>PEFG</math> is a cyclic quadrilateral. Using [[Power of a Point]] on <math>X</math> gives <math>XP \cdot XE = XG \cdot XF</math>.
 +
 
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Since <math>\triangle ADX \sim \triangle EYX</math> and <math>\triangle ACX \sim \triangle EFX</math>, <math>AX/XE = XD/YX = 9/16</math>. Using Power of a Point on <math>X</math> again, <math>(AX)(PX) = (BX)(DX)</math>. Plugging in <math>AX=9/16 XE</math> gives:
 +
<cmath>\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)</cmath>
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By [[Law of Cosines]], we can find <math>BD = \sqrt{51}</math>, as in Solution 1. Now, <math>BX = 3/4 (\sqrt{51})</math> and <math>DX = 1/4 (\sqrt{51})</math>, making <math>\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}</math>. This gives us <math>FX \cdot GX = \boxed{\textbf{(A)}\ 17}</math> as a result.
 +
 
 +
-Solution by sml1809
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 +
===Note===
 +
 
 +
You could have also got the relation <math>XP \cdot XE = XG \cdot XF</math> as follows: From the similarities, <math>AX/XE = CX/XF = 9/16</math>. PoP on <math>X</math> gives <math>(AX)(PX) = (CX)(GX)</math>. Plugging in <math>AX = 9/16 XE</math> and <math>CX=9/16 FX</math> gives
 +
<cmath>9/16 (XE)(PX) = 9/16(FX)(GX),</cmath>
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implying that <math>(XP)(XE) = (XG)(XP)</math>.
 +
 
 +
~sml1809
 +
 
 +
==Solution 4==
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 +
<math>\because</math> <math>AC \parallel EF</math>, <math>\quad \therefore</math> <math>\triangle ACX \sim \triangle EFX</math>, <math>\quad \frac{XF}{XC} = \frac{XE}{XA}</math>
 +
 
 +
By Power of a Point, <math>XG \cdot XC = XD \cdot XB</math>
 +
 
 +
By multiplying the <math>2</math> equations we get <math>XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB</math>
 +
 
 +
<math>\because</math> <math>YE \parallel AD</math>, <math>\quad \therefore</math> <math>\triangle EYX \sim \triangle ADX</math>, <math>\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}</math>
 +
 
 +
By substitution, <math>XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}</math>
 +
 
 +
Let <math>a = AB</math>, <math>b = BC</math>, <math>c = CD</math>, <math>d = AD</math>, <math>p = AC</math>, and <math>q = BD</math>
 +
 
 +
By Ptolemy's theorem, <math>p \cdot q = a \cdot c + b \cdot d</math>
 +
 
 +
<cmath>[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A</cmath>
 +
 
 +
<cmath>[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B</cmath>
 +
 
 +
<cmath>[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A</cmath>
 +
 
 +
<cmath>[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B</cmath>
 +
 
 +
<cmath>\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}</cmath>
 +
 
 +
<cmath>\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd</cmath>
 +
 
 +
<cmath>BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51</cmath>
 +
 
 +
<cmath>XF \cdot XG = \frac{51}{3} = \boxed{\textbf{(A) } 17}</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Video Solution by Richard Rusczyk==
 +
https://www.youtube.com/watch?v=JdERP0d0W64&list=PLyhPcpM8aMvLZmuDnM-0vrFniLpo7Orbp&index=4
 +
- AMBRIGGS
 +
 
 +
==See Also==
 +
{{AMC12 box|year=2017|ab=A|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Latest revision as of 02:24, 7 October 2023

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

Solution 3

Denote $P$ to be the intersection between line $AE$ and circle $O$. Note that $\angle GFE = \angle ACG = \angle APG = 180 - \angle GPE$, making $\angle GFE + \angle GDE = 180$. Thus, $PEFG$ is a cyclic quadrilateral. Using Power of a Point on $X$ gives $XP \cdot XE = XG \cdot XF$.

Since $\triangle ADX \sim \triangle EYX$ and $\triangle ACX \sim \triangle EFX$, $AX/XE = XD/YX = 9/16$. Using Power of a Point on $X$ again, $(AX)(PX) = (BX)(DX)$. Plugging in $AX=9/16 XE$ gives: \[\dfrac{9}{16}(XE)(PX) = (BX)(DX) = \dfrac{9}{16}(FX)(GX)\] By Law of Cosines, we can find $BD = \sqrt{51}$, as in Solution 1. Now, $BX = 3/4 (\sqrt{51})$ and $DX = 1/4 (\sqrt{51})$, making $\dfrac{9}{16}(FX)(GX) = \left( \dfrac{\sqrt{51}}{4}\right)\left( \dfrac{3\sqrt{51}}{4}\right) = \dfrac{3(51)}{16}$. This gives us $FX \cdot GX = \boxed{\textbf{(A)}\ 17}$ as a result.

-Solution by sml1809

Note

You could have also got the relation $XP \cdot XE = XG \cdot XF$ as follows: From the similarities, $AX/XE = CX/XF = 9/16$. PoP on $X$ gives $(AX)(PX) = (CX)(GX)$. Plugging in $AX = 9/16 XE$ and $CX=9/16 FX$ gives \[9/16 (XE)(PX) = 9/16(FX)(GX),\] implying that $(XP)(XE) = (XG)(XP)$.

~sml1809

Solution 4

$\because$ $AC \parallel EF$, $\quad \therefore$ $\triangle ACX \sim \triangle EFX$, $\quad \frac{XF}{XC} = \frac{XE}{XA}$

By Power of a Point, $XG \cdot XC = XD \cdot XB$

By multiplying the $2$ equations we get $XF \cdot XG = \frac{XE}{XA} \cdot XD \cdot XB$

$\because$ $YE \parallel AD$, $\quad \therefore$ $\triangle EYX \sim \triangle ADX$, $\quad \frac{XD}{XY} = \frac{XA}{XE}, \quad XD \cdot XE = XA \cdot XY, \quad XD = \frac{XA \cdot XY}{XE}$

By substitution, $XF \cdot XG = \frac{XE}{XA} \cdot \frac{XA \cdot XY}{XE} \cdot XB = XY \cdot XB = \frac{4}{9} BD \cdot \frac{3}{4} BD = \frac{BD^2}{3}$

Let $a = AB$, $b = BC$, $c = CD$, $d = AD$, $p = AC$, and $q = BD$

By Ptolemy's theorem, $p \cdot q = a \cdot c + b \cdot d$

\[[ABD] = \frac12 \cdot ad \cdot \sin A, \quad [BCD] = \frac12 \cdot bc \cdot \sin C = \frac12 \cdot bc \cdot \sin A\]

\[[ABC] = \frac12 \cdot ab \cdot \sin B, \quad [ACD] = \frac12 \cdot cd \cdot \sin D = \frac12 \cdot cd \cdot \sin B\]

\[[ABCD] = [ABD] + [BCD] = \frac12 \cdot ad \cdot \sin A + \frac12 \cdot bc \cdot \sin A = \frac12 (ad + bc) \sin A\]

\[[ABCD] = [ABC] + [ACD] = \frac12 \cdot ab \cdot \sin B + \frac12 \cdot cd \cdot \sin B = \frac12 (ab + cd) \sin B\]

\[\frac{ab + cd}{ad + bc} = \frac{ \sin A }{ \sin B} = \frac{ \frac{q}{2R} }{ \frac{p}{2R} } = \frac{q}{p}, \quad p = \frac{q(ad + bc)}{ab + cd}\]

\[\frac{q(ad + bc)}{ab + cd} \cdot q = ac + bd\]

\[BD^2 = q^2 = \frac{ (ac + bd)(ab + cd) }{ad + bc} = \frac{(3 \cdot 6 + 2 \cdot 8)(3 \cdot 2 + 6 \cdot 8)}{3 \cdot 8 + 2 \cdot 6} = 51\]

\[XF \cdot XG = \frac{51}{3} = \boxed{\textbf{(A) } 17}\]

~isabelchen

Video Solution by Richard Rusczyk

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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