# AMC数学竞赛真题2016年B 19

2018-12-03 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

## Problem 19

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

## Solution 3 (Similar Triangles)

$[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("Q",(3.07692307692,2.15384615384),N); label("P",(20/7,12/7),W); label("A",(0,4), NW); label("B",(5,4), NE); label("C",(5,0),SE); label("D",(0,0),SW); label("F",(2,0),S); label("G",(5,1),E); label("E",(4,4),N); label("H",H,E); [/asy]$

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 $$\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\textbf{(D)}~ \dfrac{10}{91}}.$$