Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (Not the same problem)
Line 28: Line 28:
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}
−
{{AMC12 box|year=2016|ab=A|num-b=12|num-a=14}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:24, 5 February 2016

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 $\frac{PD}{PB}=\frac{3}{1}$. Thus $PB=\frac{1}{4}l$. Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{\textbf{(E) }20.}$

See Also

2016 AMC 10A (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_19&oldid=75543"