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Difference between revisions of "2002 AIME II Problems/Problem 13"

m (Undo revision 241710 by Faliure167 (talk))
(Tag: Undo)
 
Line 157: Line 157:
 
</asy>
 
</asy>
 
Use the mass of point. Denoting the mass of <math>C=15,B=6,A=5,D=21,E=20</math>, we can see that the mass of <math>P</math> is <math>26</math>, hence we know that <math>\frac{BP}{PE}=\frac{10}{3}</math>, now we can find that <math>\frac{PQ}{AE}=\frac{10}{13}</math> which implies <math>PQ=\frac{30}{13}</math>, it is obvious that <math>\triangle{PQR}</math> is similar to <math>\triangle{ACB}</math> so we need to find the ration between PQ and AC, which is easy, it is <math>\frac{15}{26}</math>, so our final answer is <math>\left( \frac{15}{26} \right)^2= \frac{225}{676}</math> which is <math>\boxed{901}</math>. ~bluesoul
 
Use the mass of point. Denoting the mass of <math>C=15,B=6,A=5,D=21,E=20</math>, we can see that the mass of <math>P</math> is <math>26</math>, hence we know that <math>\frac{BP}{PE}=\frac{10}{3}</math>, now we can find that <math>\frac{PQ}{AE}=\frac{10}{13}</math> which implies <math>PQ=\frac{30}{13}</math>, it is obvious that <math>\triangle{PQR}</math> is similar to <math>\triangle{ACB}</math> so we need to find the ration between PQ and AC, which is easy, it is <math>\frac{15}{26}</math>, so our final answer is <math>\left( \frac{15}{26} \right)^2= \frac{225}{676}</math> which is <math>\boxed{901}</math>. ~bluesoul
 +
 +
==Solution 4(Ceva & Menelaus)==
 +
Construct <math>\overline{CP}</math> and extend it to line <math>\overline{AB}</math> at point <math>F</math>.
 +
 +
<asy>
 +
size(10cm);
 +
pair A,B,C,D,E,F,P,Q,R;
 +
A=(0,0);
 +
B=(8,0);
 +
C=(1.9375,3.4994);
 +
D=(3.6696,2.4996);
 +
E=(1.4531,2.6246);
 +
F=(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(F);
 +
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("$F$",F,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--F);
 +
</asy>
 +
 +
Using Ceva's Theorem on triangle <math>ABC</math> and point <math>P</math>, we get <cmath>\frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}=1</cmath>
 +
Thus, <math>\frac{AF}{BF}=\frac{6}{5}</math>.
 +
Then, using this info, we apply Menelaus on triangle <math>ACF</math> and line <math>BE</math>, obtaining <cmath>\frac{AE}{BE} \times \frac{CP}{FP} \times \frac{FB}{AB}=1</cmath>
 +
Simplifying and substituting, we find that <math>\frac{CP}{FP}=\frac{11}{15}</math>.
 +
Alternatively, <math>\frac{FP}{CF}=\frac{15}{26}</math>, which is also the ratio between the heights of the desired triangles.
 +
Finishing, <math>\left( \frac{15}{26} \right)^2= \frac{225}{676}</math> achieving the final answer of <math>\boxed{901}</math>.
 +
~faliure167
  
 
== See also ==
 
== See also ==

Latest revision as of 23:10, 1 February 2025

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 $m+n=\boxed{901}$. Note we can just use mass points to get $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$.

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

  • Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)

Solution 3

[asy] size(10cm); pair A,B,C,D,E,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); 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("$P$",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); [/asy] Use the mass of point. Denoting the mass of $C=15,B=6,A=5,D=21,E=20$, we can see that the mass of $P$ is $26$, hence we know that $\frac{BP}{PE}=\frac{10}{3}$, now we can find that $\frac{PQ}{AE}=\frac{10}{13}$ which implies $PQ=\frac{30}{13}$, it is obvious that $\triangle{PQR}$ is similar to $\triangle{ACB}$ so we need to find the ration between PQ and AC, which is easy, it is $\frac{15}{26}$, so our final answer is $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ which is $\boxed{901}$. ~bluesoul

Solution 4(Ceva & Menelaus)

Construct $\overline{CP}$ and extend it to line $\overline{AB}$ at point $F$.

[asy] size(10cm); pair A,B,C,D,E,F,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); F=(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(F); 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("$F$",F,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--F); [/asy]

Using Ceva's Theorem on triangle $ABC$ and point $P$, we get \[\frac{AF}{BF} \times \frac{BD}{CD} \times \frac{CE}{AE}=1\] Thus, $\frac{AF}{BF}=\frac{6}{5}$. Then, using this info, we apply Menelaus on triangle $ACF$ and line $BE$, obtaining \[\frac{AE}{BE} \times \frac{CP}{FP} \times \frac{FB}{AB}=1\] Simplifying and substituting, we find that $\frac{CP}{FP}=\frac{11}{15}$. Alternatively, $\frac{FP}{CF}=\frac{15}{26}$, which is also the ratio between the heights of the desired triangles. Finishing, $\left( \frac{15}{26} \right)^2= \frac{225}{676}$ achieving the final answer of $\boxed{901}$. ~faliure167

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

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

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