Difference between revisions of "2009 AIME I Problems/Problem 5"
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Also, it is given that <math>AK=CK</math> and <math>PK=MK</math>. | Also, it is given that <math>AK=CK</math> and <math>PK=MK</math>. | ||
− | By the SAS congruence, triangle <math>MKA</math> = triangle <math>PKC</math>. So, <math>MA</math> = <math>CP</math> = 180. | + | By the SAS congruence, triangle <math>MKA</math> = triangle <math>PKC</math>. So, <math>MA</math> = <math>CP</math> = <math>180</math>. |
− | Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}180 = \boxed{072}</math> | + | Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}(180) = \boxed{072}</math> |
== See also == | == See also == |
Revision as of 18:11, 1 June 2016
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
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
.
By the definition of mass points,
By vertical angles, angle
angle
.
Also, it is given that
and
.
By the SAS congruence, triangle = triangle
. So,
=
=
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.