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 |
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