# Difference between revisions of "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

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

Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Similarly, $\frac{DQ}{QB}=\frac{3}{2}$. Call the hypotonuse $l$. This means that ${DQ}=\frac{3l}{5}$. Applying similar triangles to ${ADP}$ and ${BEP}$, we see that $\frac{PD}{DB}=\frac{3}{1}$. Thus $DB=\frac{1}{4}$. Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{\textbf{(E) }20.}$