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2016 AMC 10B Problems/Problem 19

Revision as of 16:00, 21 February 2016 by Electroredstoner (talk | contribs) (Provided a method using coordinate geometry for obtaining the solution.)

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 2 (Coordinate Geometry)

First, we will define point $D$ as the origin. Then, we will find the equations of the following three lines: $AG$, $AC$, and $EF$. The slopes of these lines are $-\frac{3}{5}$, $-\frac{4}{5}$, and $2$, respectively. Next, we will find the equations of $AG$, $AC$, and $EF$. They are as follows: \[AG = f(x) = -\frac{3}{5}x + 4\] \[AC = g(x) = -\frac{4}{5}x + 4\] \[EF = h(x) = 2x - 4\] After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\frac{PQ}{EF} = \frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. [asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); 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); label("$P'$", (20/7,0),SSW); label("$Q'$", (40/13,0),SSE); label("$E'$", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy] Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\frac{20}{7}$ and $\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \frac{20}{7} - \frac{40}{13} = \frac{20}{91}$. Now we can find $\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \textbf{(D)}~\frac{10}{91}$

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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