Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
m
Line 10: Line 10:
 
==Solution 1==
 
==Solution 1==
  
−
It is easy to see that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math>
+
Using the given ratios, note that <math>\frac{XY}{BD} = 1 - \frac{1}{4} - \frac{11}{36} = \frac{4}{9}.</math>
−
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{16}{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>
 
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>
 
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>

Revision as of 10:01, 30 December 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

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_24&oldid=89278"