Difference between revisions of "2009 AIME I Problems/Problem 5"
(→Solution) |
(→Solution) |
||
Line 11: | Line 11: | ||
Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent. | Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent. | ||
− | Therefore, triangle <math>AMK</math> is congruent to triangle <math>CPK</math> | + | Therefore, triangle <math>AMK</math> is congruent to triangle <math>CPK</math> because of SAS |
angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC | angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC |
Revision as of 23:31, 20 March 2009
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
Sorry, I fail to get the diagram up here, someone help me.
Since is the midpoint of
.
Thus, and the opposite angles are congruent.
Therefore, triangle is congruent to triangle
because of SAS
angle is congruent to
because of CPCTC
That shows is parallel to
(also
)
That makes triangle congruent to
Thus,
Now let apply angle bisector thm.
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 |