2016 AMC 10B Problems/Problem 19

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 (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 $P'Q' = DQ' - DP' = \frac{40}{13} - \frac{20}{7} = \frac{20}{91}$ . Now we can find $\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \boxed{\textbf{(D)}~\frac{10}{91}}$

Solution 2 (Similar Triangles)

$[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$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$",(4,4),N); label("$H$",H,E); [/asy]$

Extend $AG$ to intersect $CD$ at $H$ . Letting $x=\overline{HC}$ , we have that $\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.$

Then, notice that $\triangle{AEQ}\sim\triangle{HFQ}$ and $\triangle{AEP}\sim\triangle{CFP}$ . Thus, we see that $\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{AE}{HF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}$ and $\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{EF} = \dfrac{3}{7}.$ Thus, we see that $\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\textbf{(D)}~ \dfrac{10}{91}}.$

Solution 3 (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 4 (Area)

I will calculate $\frac{EP}{EF}$ using similar triangle, and $\frac{EQ}{EF}$ using ratio of area of $\triangle AEG$ to $\triangle AFG$ .

$[asy]pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1),B3=(4,4); pair C1=(5,0),C2=(0,4),C3=(2,0); draw(A1--A2); draw(B1--B2); draw(B2--B3); draw(C1--C2); draw(C2--C3); draw(A1--B2); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$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$",(4,4),N);[/asy]$

$\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}$

$[AEG]=\frac{1}{2} \cdot 4\cdot 3=6$ $[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7$ Because $\triangle AEG$ and $\triangle AFG$ share the same base $AG$ , the ratio $\frac{[AEG]}{[AFG]}$ is equal to the ratio of the altitudes of $\triangle AEG$ to $AG$ to that of $\triangle AFG$ to $AG$ , which is equal to $\frac{EQ}{QF}$ : $\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}$ $\frac{EQ}{EF}=\frac{6}{13}$

$\frac{PQ}{EF}=\frac{EP}{EF}-\frac{EQ}{EF}=\frac{4}{7}-\frac{6}{13}=\frac{10}{91}$ $\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}$

~isabelchen

Solution 5 (Area)

I will calculate $\frac{PQ}{QE}$ using the ratio of area of $\triangle APG$ to that of $\triangle AEG$ .

$[asy]pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1),B3=(20/7,12/7); pair C1=(5,0),C2=(0,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(A2--B2); draw(B2--B3); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$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$",(4,4),N);[/asy]$

$[ACG]=\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}$ $[CAB]=\frac{1}{2} \cdot 5 \cdot 4=10$ $\triangle AEP \sim \triangle CFP$ $\frac{CP}{AP}=\frac{CF}{AE}=\frac{3}{4}$ $\frac{CP}{AC}=\frac{3}{7}$ $\frac{[CPG]}{[CAB]}=\frac{CP}{CA} \cdot \frac{CG}{CB}=\frac{3 \cdot 1}{7 \cdot 4}=\frac{3}{28}$ $[CPG]=\frac{3}{28} \cdot [CAB]=\frac{3}{28} \cdot 10=\frac{15}{14}$ $[APG]=[ACG]-[CPG]=\frac{5}{2}-\frac{15}{14}=\frac{35-15}{14}=\frac{20}{14}=\frac{10}{7}$ $[AEG]=\frac{1}{2} \cdot 4 \cdot 3=6$ Because $\triangle APG$ and $\triangle AEG$ share the same base $AG$ , the ratio $\frac{[APG]}{[AEG]}$ is equal to the ratio of altitudes of $\triangle APG$ to $AG$ to that of $\triangle AEG$ to $AG$ , which is equal to $\frac{PQ}{QE}$ : $\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}$ $\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}$ $\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}$ $\frac{PE}{EF}=\frac{4}{7}$ $\frac{PQ}{PE} \cdot \frac{PE}{EF} = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}$ $\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}$

~isabelchen

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