2011 AIME I Problems/Problem 4
Problem 4
In triangle , , and . The angle bisector of angle intersects at point , and the angle bisector of angle intersects at point . Let and be the feet of the perpendiculars from to and , respectively. Find .
Solution
Extend such that it intersects lines and at points and , respectively.
Lemma 1: are midpoints of and
Proof: Consider the reflection of the vertex over the line , and let this point be . Since , we have that is the image of after reflection over , and from the definition of reflection . Then it is easily seen that since is an angle bisector, that , so lies on . Similarly, if we define to be the reflection of over , then we find that lies on . Then we can now see that , with a homothety of ratio taking the first triangle to the second. Then this same homothety takes everything on the line to everything on the line . So since lie on , this homothety also takes to so they are midpoints, as desired.
Lemma 2: are isosceles triangles
Proof: To show that is isosceles, note that , with similarity ratio of . So it suffices to show that triangle is isosceles. But this follows quickly from Lemma 1, since is both an altitude and an angle bisector of . is isosceles by the same reasoning.
Since is a midline, it then follows that and . Since and are both isosceles, we have that and . Since is a midline, . We want to find , which is just .
Substituting the values of , we have that the answer is .
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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