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>. | ||
− | == 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 18:42, 12 September 2015
Contents
Problem
In triangle point
is on
with
and
point
is on
with
and
and
and
intersect at
Points
and
lie on
so that
is parallel to
and
is parallel to
It is given that the ratio of the area of triangle
to the area of triangle
is
where
and
are relatively prime positive integers. Find
.
Solution 1
Let be the intersection of
and
.
Since and
,
and
. So
, and thus,
.
Using mass points:
WLOG, let .
Then:
.
.
.
.
Thus, . Therefore,
, and
.
Solution 2
First draw and extend it so that it meets with
at point
.
[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,WSW); label("
",B,ESE); label("
",C,NNW); label("
",D,NE); label("
",E,WNW); label("
",X,SSE); label("
",P,NNE); label("
",Q,SSW); label("
",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
By Ceva's, That means that
Now we apply mass points. Assume WLOG that . That means that
Notice now that is similar to
. Therefore,
Also, is similar to
. Therefore,
Because is similar to
,
.
As a result, .
Therefore,
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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