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Difference between revisions of "2016 AMC 10A Problems/Problem 19"
Line 18: Line 18:
 
label("$C$", C, dir(-45));
 
label("$C$", C, dir(-45));
 
label("$D$", D, dir(-135));
 
label("$D$", D, dir(-135));
−
label("$P$", extension(A,(6,1),B,D),dir(-90));
+
label("$Q$", extension(A,(6,1),B,D),dir(-90));
−
label("$Q$", extension(A,(6,2),B,D), dir(90));
+
label("$P$", extension(A,(6,2),B,D), dir(90));
−
label("$X$", (6,1), dir(0));
+
label("$F$", (6,1), dir(0));
−
label("$Y$", (6,2), dir(0));
+
label("$E$", (6,2), dir(0));
 
</asy>
 
</asy>
  
−
As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{PD}=\frac{AD}{BX}=\frac{3}{2}.</math> Similarly, <math>\frac{DQ}{BQ}=3.</math> From this, it is not hard to find <math>r+s+t=\boxed{\textbf{(E) }20.}</math>
+
As <math>\triangle APD \sim \triangle XPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}.</math> From this, it is not hard to find <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math>
  
 
==See Also==
 
==See Also==

Revision as of 12:16, 4 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]

As $\triangle APD \sim \triangle XPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Similarly, $\frac{DQ}{QB}=\frac{3}{2}.$ From this, it is not hard to find $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ 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
2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 12 Problems and Solutions

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