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
Solution 1
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 .
Solution 2
Let be the intersection of and , or rather the incenter of triangle . Noting that and are right, we conclude that is a cyclic quadrilateral, so by Ptolemy's Theorem, Now let and be inradii to and respectively in the following picture, which is not to scale.
We know that . In triangle , we have Therefore, , and . Thus . Using a similar method, we can find that . Therefore, our Ptolemy's expression simplifies to where is the inradius of triangle . Thus, . Also, right triangles and tell us that and . But then , and this is equal to by the Angle Bisector Theorem. Therefore, solving this for and substituting yields . Similarly, . We now replace these in our Ptolemy's expression to get
We can also use mass points, assigning masses of , , and to points , , and , respectively. Then point has a mass of and point has a mass of , so and . This simplifies our expression further to
Then using the angle bisector formula, we find that and . Also, Heron's Formula tells us that so when we substitute this all in, we get
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.