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2002 AIME II Problems/Problem 13: Difference between revisions

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== Problem ==
== Problem ==
In triangle <math>ABC,</math> point <math>D</math> is on <math>\overline{BC}</math> with <math>CD = 2</math> and <math>DB = 5,</math> point <math>E</math> is on <math>\overline{AC}</math> with <math>CE = 1</math> and <math>EA = 32,</math> <math>AB = 8,</math> and <math>\overline{AD}</math> and <math>\overline{BE}</math> intersect at <math>P.</math> Points <math>Q</math> and <math>R</math> lie on <math>\overline{AB}</math> so that <math>\overline{PQ}</math> is parallel to <math>\overline{CA}</math> and <math>\overline{PR}</math> is parallel to <math>\overline{CB}.</math> It is given that the ratio of the area of triangle <math>PQR</math> to the area of triangle <math>ABC</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
In triangle <math>ABC,</math> point <math>D</math> is on <math>\overline{BC}</math> with <math>CD = 2</math> and <math>DB = 5,</math> point <math>E</math> is on <math>\overline{AC}</math> with <math>CE = 1</math> and <math>EA = 3,</math> <math>AB = 8,</math> and <math>\overline{AD}</math> and <math>\overline{BE}</math> intersect at <math>P.</math> Points <math>Q</math> and <math>R</math> lie on <math>\overline{AB}</math> so that <math>\overline{PQ}</math> is parallel to <math>\overline{CA}</math> and <math>\overline{PR}</math> is parallel to <math>\overline{CB}.</math> It is given that the ratio of the area of triangle <math>PQR</math> to the area of triangle <math>ABC</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.


== Solution ==
== Solution ==
{{solution}}
Let <math>X</math> be the intersection of <math>\overline{CP}</math> and <math>\overline{AB}</math>.
 
<asy>
size(10cm);
pair A,B,C,D,E,X,P,Q,R;
A=(0,0);
B=(8,0);
C=(1.9375,3.4994);
D=(3.6696,2.4996);
E=(1.4531,2.6246);
X=(4.3636,0);
P=(2.9639,2.0189);
Q=(1.8462,0);
R=(6.4615,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(X);
dot(P);
dot(Q);
dot(R);
label("$A$",A,WSW);
label("$B$",B,ESE);
label("$C$",C,NNW);
label("$D$",D,NE);
label("$E$",E,WNW);
label("$X$",X,SSE);
label("$P$",P,NNE);
label("$Q$",Q,SSW);
label("$R$",R,SE);
draw(A--B--C--cycle);
draw(P--Q--R--cycle);
draw(A--D);
draw(B--E);
draw(C--X);
</asy>
 
Since <math>\overline{PQ} \parallel \overline{CA}</math> and <math>\overline{PR} \parallel \overline{CB}</math>, <math>\angle CAB = \angle PQR</math> and <math>\angle CBA = \angle PRQ</math>. So <math>\Delta ABC \sim \Delta QRP</math>, and thus, <math>\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2</math>.
 
Using [[mass points]]:
 
[[WLOG]], let <math>W_C=15</math>.
 
Then:
 
<math>W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5</math>.
 
<math>W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6</math>.
 
<math>W_X=W_A+W_B=5+6=11</math>.
 
<math>W_P=W_C+W_X=15+11=26</math>.
 
Thus, <math>\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}</math>. Therefore, <math>\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}</math>, and <math>m+n=\boxed{901}</math>.


== See also ==
== See also ==
{{AIME box|year=2002|n=II|num-b=12|num-a=14}}
{{AIME box|year=2002|n=II|num-b=12|num-a=14}}

Revision as of 22:43, 24 June 2008

Problem

In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD = 2$ and $DB = 5,$ point $E$ is on $\overline{AC}$ with $CE = 1$ and $EA = 3,$ $AB = 8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Let $X$ be the intersection of $\overline{CP}$ and $\overline{AB}$.

[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label("$A$",A,WSW); label("$B$",B,ESE); label("$C$",C,NNW); label("$D$",D,NE); label("$E$",E,WNW); label("$X$",X,SSE); label("$P$",P,NNE); label("$Q$",Q,SSW); label("$R$",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]

Since $\overline{PQ} \parallel \overline{CA}$ and $\overline{PR} \parallel \overline{CB}$, $\angle CAB = \angle PQR$ and $\angle CBA = \angle PRQ$. So $\Delta ABC \sim \Delta QRP$, and thus, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left(\frac{PX}{CX}\right)^2$.

Using mass points:

WLOG, let $W_C=15$.

Then:

$W_A = \left(\frac{CE}{AE}\right)W_C = \frac{1}{3}\cdot15=5$.

$W_B = \left(\frac{CD}{BD}\right)W_C = \frac{2}{5}\cdot15=6$.

$W_X=W_A+W_B=5+6=11$.

$W_P=W_C+W_X=15+11=26$.

Thus, $\frac{PX}{CX}=\frac{W_C}{W_P}=\frac{15}{26}$. Therefore, $\frac{[\Delta PQR]}{[\Delta ABC]} = \left( \frac{15}{26} \right)^2 = \frac{225}{676}$, and $m+n=\boxed{901}$.

See also

2002 AIME II (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
All AIME Problems and Solutions
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