Difference between revisions of "2002 AIME II Problems/Problem 13"
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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 = 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>. | ||
+ | |||
+ | <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>. Note we can just use mass points to get <math>\left( \frac{15}{26} \right)^2= \frac{225}{676}</math> which is <math>\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("$A$",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 <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> | ||
+ | |||
+ | *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 <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 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2002|n=II|num-b=12|num-a=14}} | |
+ | |||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:06, 24 August 2024
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
. Note we can just use mass points to get
which is
.
Solution 2
First draw and extend it so that it meets with
at point
.
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,
- 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
Use the mass of point. Denoting the mass of
, we can see that the mass of
is
, hence we know that
, now we can find that
which implies
, it is obvious that
is similar to
so we need to find the ration between PQ and AC, which is easy, it is
, so our final answer is
which is
. ~bluesoul
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.