2009 AIME I Problems/Problem 5
Contents
Problem
Triangle has
and
. Points
and
are located on
and
respectively so that
, and
is the angle bisector of angle
. Let
be the point of intersection of
and
, and let
be the point on line
for which
is the midpoint of
. If
, find
.
Solution 1
![[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2)); label("$72$",(L+P)/2,(-1,0));label("$300$",(B+C)/2,(1,1)); [/asy]](http://latex.artofproblemsolving.com/7/d/a/7da806e0aa0782795b140c81cd725875ad3dfc7c.png)
Since is the midpoint of
and
, quadrilateral
is a parallelogram, which implies
and
is similar to
Thus,
Now lets apply the angle bisector theorem.
Solution 2
![[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$y$",(P+C)/2,(-1,0));label("$x$",(A+K)/2,(0,2));label("$x$",(K+C)/2,(0,2)); label("$2y/5$",(L+P)/2,(-1,0));label("$180$",(B+C)/2,(1,1)); [/asy]](http://latex.artofproblemsolving.com/3/c/2/3c2e0d82fd6a211b32332c5caa82c904b872634c.png)
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem:
So, we can weight
as
and
as
and
as
. Since
is the midpoint of
and
, the weight of
is equal to the weight of
, which equals
.
Also, since the weight of
is
and
is
, we can weight
as
.
\[\n\] (Error compiling LaTeX. Unknown error_msg)
By the definition of mass points,
\[\frac{LP}{CP}=\frac{2}{5}\impliesLP=\frac{2}{5}CP\] (Error compiling LaTeX. Unknown error_msg)
By vertical angles, angle angle
.
Also, it is given that
and
.
\[\n\] (Error compiling LaTeX. Unknown error_msg)
By the SAS congruence, triangle = triangle
. So,
=
= 180.
Since ,
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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