2011 AIME II Problems/Problem 4
Contents
Problem 4
In triangle ,
and
. The angle bisector of
intersects
at point
, and point
is the midpoint of
. Let
be the point of the intersection of
and
. The ratio of
to
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let
be on
such that
. It follows that
, so
by the Angle Bisector Theorem. Similarly, we see by the Midline Theorem that
. Thus,
and
.
Solution 2 (mass points)
Assign mass points as follows: by Angle-Bisector Theorem, , so we assign
. Since
, then
, and
, so
.
Solution 3
By Menelaus' Theorem on with transversal
,
So
.
Solution 4
We will use barycentric coordinates. Let ,
,
. By the Angle Bisector Theorem,
. Since
is the midpoint of
,
. Therefore, the equation for line BM is
. Let
. Using the equation for
, we get
Therefore,
so the answer is
.
Solution 5
Let . Then by the Angle Bisector Theorem,
. By the Ratio Lemma, we have that
Notice that
since their bases have the same length and they share a height. By the sin area formula, we have that
Simplifying, we get that
Plugging this into what we got from the Ratio Lemma, we have that
Solution 6 (quick Menelaus)
First, we will find . By Menelaus on
and the line
, we have
This implies that
. Then, by Menelaus on
and line
, we have
Therefore,
The answer is
. -brainiacmaniac31
Solution 7 (Visual)
vladimir.shelomovskii@gmail.com, vvsss
Solution 8 (Cheese)
Assume is a right triangle at
. Line
and
. These two lines intersect at
which have coordinates
and thus
has coordinates
. Thus, the line
. When
,
has
coordinate equal to
=
which equals
giving an answer of
Video Solution by OmegaLearn
https://youtu.be/Gjt25jRiFns?t=314
~ pi_is_3.14
See also
2011 AIME II (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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