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

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(→‎Solution 1 (Coordinate Geometry))
 
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==Problem==
 
==Problem==
−
Rectangle <math>ABCD</math> has <math>AB=5</math> and <math>BC=4</math>. Point <math>E</math> lies on <math>\overline{AB}</math> so that <math>EB=1</math>, point <math>G</math> lies on <math>\overline{BC}</math> so that <math>CG=1</math>. and point <math>F</math> lies on <math>\overline{CD}</math> so that <math>DF=2</math>. Segments <math>\overline{AG}</math> and <math>\overline{AC}</math> intersect <math>\overline{EF}</math> at <math>Q</math> and <math>P</math>, respectively. What is the value of <math>\frac{PQ}{EF}</math>?
+
Rectangle <math>ABCD</math> has <math>AB=5</math> and <math>BC=4</math>. Point <math>E</math> lies on <math>\overline{AB}</math> so that <math>EB=1</math>, point <math>G</math> lies on <math>\overline{BC}</math> so that <math>CG=1</math>, and point <math>F</math> lies on <math>\overline{CD}</math> so that <math>DF=2</math>. Segments <math>\overline{AG}</math> and <math>\overline{AC}</math> intersect <math>\overline{EF}</math> at <math>Q</math> and <math>P</math>, respectively. What is the value of <math>\frac{PQ}{EF}</math>?
  
  
Line 28: Line 28:
  
  
−
==Solution==
 
−
<math>\textbf{(D)}~\frac{10}{91}</math>
 
  
−
solution by ngeorge
+
 
 +
 
 +
==Solution 1 (Coordbash)==
 +
 
 +
First, we will define point <math>D</math> as the origin.  Then, we will find the equations of the following three lines: <math>AG</math>, <math>AC</math>, and <math>EF</math>.  The slopes of these lines are <math>-\frac{3}{5}</math>, <math>-\frac{4}{5}</math>, and <math>2</math>, respectively.  Next, we will find the equations of <math>AG</math>, <math>AC</math>, and <math>EF</math>.  They are as follows:
 +
<cmath>AG = f(x) = -\frac{3}{5}x + 4</cmath>
 +
<cmath>AC = g(x) = -\frac{4}{5}x + 4</cmath>
 +
<cmath>EF = h(x) = 2x - 4</cmath>
 +
After drawing in altitudes to <math>DC</math> from <math>P</math>, <math>Q</math>, and <math>E</math>, we see that <math>\frac{PQ}{EF} = \frac{P'Q'}{E'F}</math> because of similar triangles, and so we only need to find the x-coordinates of <math>P</math> and <math>Q</math>.
 +
<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 <math>AC</math> and <math>EF</math>, and <math>AG</math> and <math>EF</math> gives the x-coordinates of <math>P</math> and <math>Q</math> to be <math>\frac{20}{7}</math> and <math>\frac{40}{13}</math>.  This means that <math>P'Q' = DQ' - DP' = \frac{40}{13} - \frac{20}{7} = \frac{20}{91}</math>.  Now we can find <math>\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \boxed{\textbf{(D)}~\frac{10}{91}}</math>
 +
 
 +
==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 <math>AG</math> to intersect <math>CD</math> at <math>H</math>. Letting <math>x=\overline{HC}</math>, we have that <cmath>\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}.</cmath>
 +
 
 +
Then, notice that <math>\triangle{AEQ}\sim\triangle{HFQ}</math> and <math>\triangle{AEP}\sim\triangle{CFP}</math>. Thus, we see that <cmath>\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}</cmath>
 +
and <cmath>\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{EF} = \dfrac{3}{7}.</cmath>
 +
Thus, we see that <cmath>\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}}.</cmath>
 +
 
 +
==Solution 3 (Answer Choices)==
 +
 
 +
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>.
 +
 
 +
==Solution 4 (Area)==
 +
 
 +
I will calculate <math>\frac{EP}{EF}</math> using similar triangle, and <math>\frac{EQ}{EF}</math> using ratio of area of <math>\triangle AEG</math> to <math>\triangle AFG</math>.
 +
 
 +
<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>
 +
 
 +
<cmath>\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}</cmath>
 +
 
 +
<cmath>[AEG]=\frac{1}{2} \cdot 4\cdot 3=6</cmath>
 +
<cmath>[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7</cmath>
 +
Because <math>\triangle AEG</math> and <math>\triangle AFG</math> share the same base <math>AG</math>, the ratio <math>\frac{[AEG]}{[AFG]}</math> is equal to the ratio of the altitude of <math>\triangle AEG</math> to <math>AG</math> to that of <math>\triangle AFG</math> to <math>AG</math>, which is equal to <math>\frac{EQ}{QF}</math>:
 +
<cmath>\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}</cmath>
 +
<cmath>\frac{EQ}{EF}=\frac{6}{13}</cmath>
 +
 
 +
<cmath>\frac{PQ}{EF}=\frac{EP}{EF}-\frac{EQ}{EF}=\frac{4}{7}-\frac{6}{13}=\frac{10}{91}</cmath>
 +
<cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 5 (Area)==
 +
 
 +
I will calculate <math>\frac{PQ}{QE}</math> using the ratio of area of <math>\triangle APG</math> to that of <math>\triangle AEG</math>.
 +
 
 +
<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>
 +
 
 +
<cmath>[ACG]=\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}</cmath>
 +
<cmath>[CAB]=\frac{1}{2} \cdot 5 \cdot 4=10</cmath>
 +
<cmath>\triangle AEP \sim \triangle CFP</cmath>
 +
<cmath>\frac{CP}{AP}=\frac{CF}{AE}=\frac{3}{4}</cmath>
 +
<cmath>\frac{CP}{AC}=\frac{3}{7}</cmath>
 +
<cmath>\frac{[CPG]}{[CAB]}=\frac{CP}{CA} \cdot \frac{CG}{CB}=\frac{3 \cdot 1}{7 \cdot 4}=\frac{3}{28}</cmath>
 +
<cmath>[CPG]=\frac{3}{28} \cdot [CAB]=\frac{3}{28} \cdot 10=\frac{15}{14}</cmath>
 +
<cmath>[APG]=[ACG]-[CPG]=\frac{5}{2}-\frac{15}{14}=\frac{35-15}{14}=\frac{20}{14}=\frac{10}{7}</cmath>
 +
<cmath>[AEG]=\frac{1}{2} \cdot 4 \cdot 3=6</cmath>
 +
Because <math>\triangle APG</math> and <math>\triangle AEG</math> share the same base <math>AG</math>, the ratio <math>\frac{[APG]}{[AEG]}</math> is equal to the ratio of altitude of <math>\triangle APG</math> to <math>AG</math> to that of <math>\triangle AEG</math> to <math>AG</math>, which is equal to <math>\frac{PQ}{QE}</math>:
 +
<cmath>\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}</cmath>
 +
<cmath>\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}</cmath>
 +
<cmath>\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}</cmath>
 +
<cmath>\frac{PE}{EF}=\frac{4}{7}</cmath>
 +
<cmath>\frac{PQ}{PE} \cdot \frac{PE}{EF}  = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}</cmath>
 +
<cmath>\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 6 (Coordinate Bash, not as efficient as Solution 1 but it works)==
 +
We set the points
 +
<math>D(0, 0)</math>, <math>A(0, 4)</math>, <math>E(4, 4)</math>, <math>B(5, 4)</math>, <math>G(5, 1)</math>, <math>C(5, 0)</math>, and <math>F(2, 0)</math>.
 +
The equation of <math>\overline{AC}</math> is <math>y=-\frac{4}{5}x+4</math>, the equation of <math>\overline{AG}</math> is <math>y=-\frac{3}{5}x+4</math>, and the equation of <math>\overline{EF}</math> is <math>y=2x-4</math>. Solving the system of equations for <math>\overline{AC}</math> and <math>\overline{EF}</math> to find point <math>P</math>, <math>y=-\frac{4}{5}x+4=2x-4 \longrightarrow \frac{14}{5}x=8 \longrightarrow x=\frac{20}{7}</math> and <math>y=2x-4=\frac{12}{7}</math>. So the coordinate of point P is <math>P(\frac{20}{7}, \frac{12}{7})</math>. Next find point Q by solving the system of equations for <math>\overline{AG}</math> and <math>\overline{EF}</math> to get <math>Q(\frac{40}{13}, \frac{28}{13})</math>. Using the distance formula, <cmath>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}}</cmath>
 +
<cmath>=\sqrt{\frac{400}{8281}+\frac{1600}{8281}}=\sqrt{\frac{2000}{8281}}=\frac{20\sqrt{5}}{91}</cmath> Also using the distance formula, <cmath>EF=\sqrt{\left(4-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}</cmath> Finally, <cmath>\frac{PQ}{EF}=\frac{\frac{20\sqrt{5}}{91}}{2\sqrt{5}}=\frac{10}{91} \Longrightarrow \boxed{\textbf{(D)}~\frac{10}{91}}</cmath>
 +
~JH. L
 +
 
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/GrCtzL0S-Uo?t=911
 +
 
 +
~ pi_is_3.14
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 20:16, 1 November 2024

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 (Coordbash)

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 $\frac{EQ}{QF}$: \[\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}\] \[\frac{EQ}{EF}=\frac{6}{13}\]

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

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

~isabelchen

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


Video Solution by OmegaLearn

https://youtu.be/GrCtzL0S-Uo?t=911

~ pi_is_3.14

See Also

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

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