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Difference between revisions of "2002 AIME II Problems/Problem 13"
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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>.
 
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>.
  
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== Solution ==
+
== Solution 1 ==
 
Let <math>X</math> be the intersection of <math>\overline{CP}</math> and <math>\overline{AB}</math>.  
 
Let <math>X</math> be the intersection of <math>\overline{CP}</math> and <math>\overline{AB}</math>.  
  
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<math>W_P=W_C+W_X=15+11=26</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>.  
+
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>.
 +
 
 +
== Solution 2 ==
 +
First draw <math>\overline{CP}</math> and extend it so that it meets with <math>\overline{AB}</math> at point <math>X</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("<math>A</math>",A,WSW); label("<math>B</math>",B,ESE); label("<math>C</math>",C,NNW); label("<math>D</math>",D,NE); label("<math>E</math>",E,WNW); label("<math>X</math>",X,SSE); label("<math>P</math>",P,NNE); label("<math>Q</math>",Q,SSW); label("<math>R</math>",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]
 +
 
 +
We have that <math>[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}</math>
 +
 
 +
By Ceva's, <cmath>3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}</cmath> That means that <cmath> \frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}</cmath>
 +
 
 +
Now we apply mass points. Assume WLOG that <math>W_{A}=1</math>. That means that
 +
 
 +
<cmath>W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}</cmath>
 +
 
 +
Notice now that <math>\triangle{PBQ}</math> is similar to <math>\triangle{EBA}</math>. Therefore,
 +
 
 +
<cmath>\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}</cmath>
 +
 
 +
Also, <math>\triangle{PRA}</math> is similar to <math>\triangle{DBA}</math>. Therefore,
 +
 
 +
<cmath>\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}</cmath>
 +
 
 +
Because <math>\triangle{PQR}</math> is similar to <math>\triangle{CAB}</math>, <math>\angle{C}=\angle{P}</math>.
 +
 
 +
As a result, <math>[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}</math>.
 +
 
 +
Therefore, <cmath>\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}</cmath>
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:42, 12 September 2015

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 1

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 $m+n=\boxed{901}$.

Solution 2

First draw $\overline{CP}$ and extend it so that it meets with $\overline{AB}$ at point $X$.

[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]

We have that $[ABC]=\frac{1}{2}\cdot AC \cdot BC\sin{C}=\frac{1}{2}\cdot 4\cdot {7}\sin{C}=14\sin{C}$

By Ceva's, \[3\cdot{\frac{2}{5}}\cdot{\frac{BX}{AX}}=1\implies BX=\frac{5\cdot AX}{6}\] That means that \[\frac{11\cdot {AX}}{6}=8\implies AX=\frac{48}{11} \ \text{and} \ BX=\frac{40}{11}\]

Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that

\[W_{C}=3, W_{B}=\frac{6}{5}, W_{X}=\frac{11}{5}, W_{D}=\frac{21}{5}, W_{E}=4, W_{P}=\frac{26}{5}\]

Notice now that $\triangle{PBQ}$ is similar to $\triangle{EBA}$. Therefore,

\[\frac{PQ}{EA}=\frac{PB}{EB}\implies \frac{PQ}{3}=\frac{10}{13}\implies PQ=\frac{30}{13}\]

Also, $\triangle{PRA}$ is similar to $\triangle{DBA}$. Therefore,

\[\frac{PA}{DA}=\frac{PR}{DB}\implies \frac{21}{26}=\frac{PR}{5}\implies PR=\frac{105}{26}\]

Because $\triangle{PQR}$ is similar to $\triangle{CAB}$, $\angle{C}=\angle{P}$.

As a result, $[PQR]=\frac{1}{2}\cdot PQ \cdot PR \sin{C}=\frac{1}{2}\cdot \frac{30}{13}\cdot \frac{105}{26}\sin{P}=\frac{1575}{338}\sin{C}$.

Therefore, \[\frac{[PQR]}{[ABC]}=\frac{\frac{1575}{338}\sin{C}}{14\sin{C}}=\frac{225}{676}\implies 225+676=\boxed{901}\]

See also

2002 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions

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