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 altitude of $\triangle AEG$ to $AG$ to that of $\triangle AFG$ to $AG$, which is equal to $\frac{EQ}{QF}$: $$\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}$$ $$\frac{EQ}{EF}=\frac{6}{13}$$

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

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 altitude of $\triangle APG$ to $AG$ to that of $\triangle AEG$ to $AG$, which is equal to $\frac{PQ}{QE}$: $$\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}$$ $$\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}$$ $$\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}$$ $$\frac{PE}{EF}=\frac{4}{7}$$ $$\frac{PQ}{PE} \cdot \frac{PE}{EF} = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}$$ $$\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}$$

Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)

We set the points $D(0, 0)$, $A(0, 4)$, $E(4, 4)$, $B(5, 4)$, $G(5, 1)$, $C(5, 0)$, and $F(2, 0)$. The equation of $\overline{AC}$ is $y=-\frac{4}{5}x+4$, the equation of $\overline{AG}$ is $y=-\frac{3}{5}x+4$, and the equation of $\overline{EF}$ is $y=2x-4$. Solving the system of equations for $\overline{AC}$ and $\overline{EF}$ to find point $P$, $y=-\frac{4}{5}x+4=2x-4 \longrightarrow \frac{14}{5}x=8 \longrightarrow x=\frac{20}{7}$ and $y=2x-4=\frac{12}{7}$. So the coordinate of point P is $P(\frac{20}{7}, \frac{12}{7})$. Next find point Q by solving the system of equations for $\overline{AG}$ and $\overline{EF}$ to get $Q(\frac{40}{13}, \frac{28}{13})$. Using the distance formula, $$PQ=\sqrt{\left(\frac{40}{13}-\frac{20}{7}\right)^{2}+\left(\frac{28}{13}-\frac{12}{7}\right)^{2}}=\sqrt{\left(\frac{20}{91}\right)^{2}+\left(\frac{40}{91}\right)^{2}}$$ $$=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}$$ Also using the distance formula, $$EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$$ Finally, $$\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}$$ ~JH. L

See Also

 2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 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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