2016 AMC 10A Problems/Problem 19

Problem

In rectangle $ABCD,$ $AB=6$ and $BC=3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE=EF=FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$ ?

$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$

Solution 1 (Similar Triangles)

$[asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]$

Use similar triangles. Our goal is to put the ratio in terms of ${BD}$ . Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Similarly, $\frac{DQ}{QB}=\frac{3}{2}$ . This means that ${DQ}=\frac{3\cdot BD}{5}$ . As $\triangle ADP$ and $\triangle BEP$ are similar, we see that $\frac{PD}{PB}=\frac{3}{1}$ . Thus $PB=\frac{BD}{4}$ . Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{\textbf{(E) }20.}$

Solution 2(Coordinate Bash)

We can set coordinates for the points. $D=(0,0), C=(6,0), B=(6,3),$ and $A=(0,3)$ . The line $BD$ 's equation is $y = \frac{1}{2}x$ , line $AE$ 's equation is $y = -\frac{1}{6}x + 3$ , and line $AF$ 's equation is $y = -\frac{1}{3}x + 3$ . Adding the equations of lines $BD$ and $AE$ , we find that the coordinates of $P$ are $\left(\frac{9}{2},\frac{9}{4}\right)$ . Furthermore we find that the coordinates of $Q$ are $\left(\frac{18}{5}, \frac{9}{5}\right)$ . Using the Pythagorean Theorem, we get that the length of $QD$ is $\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}$ , and the length of $DP$ is $\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.$ $PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.$ The length of $DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}$ . Then $BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.$ The ratio $BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.$ Then $r, s,$ and $t$ is $5, 3,$ and $12$ , respectively. The problem tells us to find $r + s + t$ , so $5 + 3 + 12 = \boxed{\textbf{(E) }20}$ ~ minor LaTeX edits by dolphin7

Solution 3

Extend $AF$ to meet $CD$ at point $T$ . Since $FC=1$ and $BF=2$ , $TC=3$ by similar triangles $\triangle TFC$ and $\triangle AFB$ . It follows that $\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}$ . Now, using similar triangles $\triangle BEP$ and $\triangle DAP$ , $\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}$ . WLOG let $BP=1$ . Solving for $PQ, QD$ gives $PQ=\frac{3}{5}$ and $QD=\frac{12}{5}$ . So our desired ratio is $5:3:12$ and $5+3+12=\boxed{\textbf{(E) } 20}$ .

Solution 4 (Mass Points)

Draw line segment $AC$ , and call the intersection between $AC$ and $BD$ point $K$ . In $\delta ABC$ , observe that $BE:EC=1:2$ and $AK:KC=1:1$ . Using mass points, find that $BP:PK=1:1$ . Again utilizing $\delta ABC$ , observe that $BF:FC=2:1$ and $AK:KC=1:1$ . Use mass points to find that $BQ:QK=4:1$ . Now, draw a line segment with points $B$ , $P$ , $Q$ , and $K$ ordered from left to right. Set the values $BP=x$ , $PK=x$ , $BQ=4y$ and $QK=y$ . Setting both sides segment $BK$ equal, we get $y= \frac{2}{5}x$ . Plugging in and solving gives $QK= \frac{2}{5}x$ , $PQ=\frac{3}{5}x$ , $BP=x$ . The question asks for $BP:PQ:QD$ , so we add $2x$ to $QK$ and multiply the ratio by $5$ to create integers. This creates $5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12$ . This sums up to $3+5+12=\boxed{\textbf{(E) }20}$

Solution 5 (Easy Coord Bash)

We set coordinates for the points. Let $A=(0,3), B=(6,3), C=(6,0)$ and $D=(0,0)$ . Then the equation of line $AE$ is $y = -\frac{1}{6}x + 3,$ the equation of line $AF$ is $y = -\frac{1}{3}x + 3,$ and the equation of line $BD$ is $y = \frac{1}{2}x$ . We find that the x-coordinate of point $P$ is $\frac 9 2$ by solving $-\frac{1}{6}x + 3=\frac{1}{2}x.$ Similarly we find that the x-coordinate of point $Q$ is $\frac {18} 5$ by solving $-\frac{1}{3}x + 3=\frac{1}{2}x.$ It follows that $BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.$ Hence $r,s,t=5,3,12$ and $r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.$ ~ Solution by dolphin7

Video Solution

https://www.youtube.com/watch?v=aG9JiBMd0ag

Video Solution 2

https://youtu.be/us3e2jMjWnw

~IceMatrix

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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