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

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In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?
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== Problem ==
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In rectangle <math>ABCD,</math> <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ratio <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s,</math> and <math>t</math> is <math>1.</math> What is <math>r+s+t</math>?
  
 
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math>
 
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math>
  
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== Solution ==
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 +
<asy>
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size(6cm);
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pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);
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draw(A--B--C--D--cycle);
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draw(B--D);
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draw(A--(6,2));
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draw(A--(6,1));
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label("$A$", A, dir(135));
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label("$B$", B, dir(45));
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label("$C$", C, dir(-45));
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label("$D$", D, dir(-135));
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label("$P$", extension(A,(6,1),B,D),dir(-90));
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label("$Q$", extension(A,(6,2),B,D), dir(90));
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label("$X$", (6,1), dir(0));
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label("$Y$", (6,2), dir(0));
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</asy>
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:27, 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("$P$", 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]

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

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