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

(asy)
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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>?
  
  
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==Solution 1 (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 2 (Coordinate Geometry)==
+
 
 +
==Solution 1 (Coordinate Geometry)==
  
 
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:
 
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:
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dot(C1); dot(C2);
 
dot(C1); dot(C2);
 
dot((0,0)); dot((5,4));</asy>
 
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{20}{7} - \frac{40}{13} = \frac{20}{91}</math>.  Now we can find <math>\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \textbf{(D)}~\frac{10}{91}</math>
+
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 3 (Similar Triangles)==
+
==Solution 2 (Similar Triangles)==
 
<asy>
 
<asy>
 
  pair A1=(2,0),A2=(4,4);
 
  pair A1=(2,0),A2=(4,4);
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dot((20/7,12/7));
 
dot((20/7,12/7));
 
dot((3.07692307692,2.15384615384));
 
dot((3.07692307692,2.15384615384));
label("<math>Q</math>",(3.07692307692,2.15384615384),N);
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label("$Q$",(3.07692307692,2.15384615384),N);
label("<math>P</math>",(20/7,12/7),W);
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label("$P$",(20/7,12/7),W);
label("<math>A</math>",(0,4), NW);
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label("$A$",(0,4), NW);
label("<math>B</math>",(5,4), NE);
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label("$B$",(5,4), NE);
label("<math>C</math>",(5,0),SE);
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label("$C$",(5,0),SE);
label("<math>D</math>",(0,0),SW);
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label("$D$",(0,0),SW);
label("<math>F</math>",(2,0),S); label("<math>G</math>",(5,1),E);
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label("$F$",(2,0),S); label("$G$",(5,1),E);
label("<math>E</math>",(4,4),N);
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label("$E$",(4,4),N);
label("<math>H</math>",H,E);
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label("$H$",H,E);
 +
 
  
dot(A1); dot(A2);
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</asy>
dot(B1); dot(B2);
 
dot(C1); dot(C2);
 
dot(H);
 
dot((0,0)); dot((5,4));
 
<\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>  
 
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{EQ}{QF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}</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}{FE} = \dfrac{3}{7}.</cmath>  
 
and <cmath>\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{FE} = \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{\dfrac{10}{91}}.</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>.
  
 
==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 16:40, 24 December 2020

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 \[\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{FE} = \dfrac{3}{7}.\] 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 $\boxed{\textbf{(D)}~\frac{10}{91}}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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