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Difference between revisions of "2016 AMC 10B Problems/Problem 19"

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==Solution 1 (Answer Choices)==
 
==Solution 1 (Answer Choices)==
  
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Since the opposite sides of a rectangle are parallel and <math>\angle{APE}</math> <math>=</math> <math>\angle{CPF}</math> due to vertical angles, <math>\triangle{APE}</math> <math>\sim</math> <math>\triangle{CPF}</math>. Furthermore, the ratio between the side lengths of the two triangles is <math>\frac{AE}{FC}</math> or <math>\frac{4}{3}</math>. Labeling <math>EP</math> <math>=</math> <math>4x</math> and <math>FP</math> <math>=</math> <math>3x</math>, we see that <math>EF</math> turns out to be equal to <math>7x</math>. Since the denominator of <math>\frac{PQ}{EF} must now be a multiple of 7, the only possible solution in the answer choices is </math>\boxed{\textbf{(D)}~\frac{10}{91}}$.
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Since the opposite sides of a rectangle are parallel and <math>\angle{APE}</math> <math>=</math> <math>\angle{CPF}</math> due to vertical angles, <math>\triangle{APE}</math> <math>\sim</math> <math>\triangle{CPF}</math>. Furthermore, the ratio between the side lengths of the two triangles is <math>\frac{AE}{FC}</math> <math>=</math> <math>\frac{4}{3}</math>. Labeling <math>EP</math> <math>=</math> <math>4x</math> and <math>FP</math> <math>=</math> <math>3x</math>, we see that <math>EF</math> turns out to be equal to <math>7x</math>. Since the denominator of <math>\frac{PQ}{EF}</math> must now be a multiple of 7, the only possible solution in the answer choices is <math>\boxed{\textbf{(D)}~\frac{10}{91}}</math>.
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(Solution by akaashp11)
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:50, 21 February 2016

Problem

Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$. and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?


[asy]pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N);[/asy]

$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$


Solution 1 (Answer Choices)

Since the opposite sides of a rectangle are parallel and $\angle{APE}$ $=$ $\angle{CPF}$ due to vertical angles, $\triangle{APE}$ $\sim$ $\triangle{CPF}$. Furthermore, the ratio between the side lengths of the two triangles is $\frac{AE}{FC}$ $=$ $\frac{4}{3}$. Labeling $EP$ $=$ $4x$ and $FP$ $=$ $3x$, we see that $EF$ turns out to be equal to $7x$. Since the denominator of $\frac{PQ}{EF}$ must now be a multiple of 7, the only possible solution in the answer choices is $\boxed{\textbf{(D)}~\frac{10}{91}}$.

(Solution by akaashp11)

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 10 Problems and Solutions

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