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

Revision as of 15:43, 21 February 2016

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 $\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \textbf{(D)}~\frac{10}{91}$

Solution 3 (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);

dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot(H); dot((0,0)); dot((5,4));[/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 $\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{EQ}{QF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}$ 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 $\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\dfrac{10}{91}}.$

@@ Line 75: / Line 75: @@
 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>
+==Solution 3 (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("<math>Q</math>",(3.07692307692,2.15384615384),N);
+label("<math>P</math>",(20/7,12/7),W);
+label("<math>A</math>",(0,4), NW);
+label("<math>B</math>",(5,4), NE);
+label("<math>C</math>",(5,0),SE);
+label("<math>D</math>",(0,0),SW);
+label("<math>F</math>",(2,0),S); label("<math>G</math>",(5,1),E);
+label("<math>E</math>",(4,4),N);
+label("<math>H</math>",H,E);
+dot(A1); dot(A2);
+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>
+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>
+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>
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=18|num-a=20}}
 {{MAA Notice}}

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

Page

Toolbox

Search