2016 AMC 10A Problems/Problem 19

Revision as of 08:46, 4 February 2016 by Benq (talk | contribs)

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("$P$", extension(A,(6,1),B,D),dir(-90)); label("$Q$", extension(A,(6,2),B,D), dir(90)); label("$X$", (6,1), dir(0)); label("$Y$", (6,2), dir(0)); [/asy]

As $\triangle APD \sim \triangle XPB,$ $\frac{DP}{PD}=\frac{AD}{BX}=\frac{3}{2}.$ Similarly, $\frac{DQ}{BQ}=3.$ From this, it is not hard to find $r+s+t=\boxed{\textbf{(E) }20.}$

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png