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 1
Extend and such that they intersect line at points and , respectively. Since is the angle bisector of angle B, and is perpendicular to , so , M is the midpoint of . For the same reason, ,N is the midpoint of . Hence . But , so .
Solution 2
[There seem to be some mislabeled points going on here but the idea is sound.] Let be the incenter of . Now, since and , we have is a cyclic quadrilateral. Consequently, . Since , we have that . Letting be the point of contact of the incircle of with side , we have thus
Solution 3 (Bash)
Project onto and as and . and are both in-radii of so we get right triangles with legs (the in-radius length) and . Since is the hypotenuse for the 4 triangles ( and ), are con-cyclic on a circle we shall denote as which is also the circumcircle of and . To find , we can use the Law of Cosines on where is the center of . Now, the circumradius can be found with Pythagorean Theorem with or : . To find , we can use the formula and by Heron's, . To find , we can find since . . Thus, and since , we have . Plugging this into our Law of Cosines formula gives . To find , we use LoC on . Our formula now becomes . After simplifying, we get .
--lucasxia01
Solution 4
Because , is cyclic.
Ptolemy on CMIN:
by angle addition formula.
.
Let be where the incircle touches , then . , for a final answer of .
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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