Difference between revisions of "2011 AIME II Problems/Problem 4"
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=== Solution 2 === | === Solution 2 === | ||
− | Assign [[mass points]] as follows: by Angle-Bisector Theorem, <math>BD / DC = 20/11</math>, so we assign <math>m(B) = 11, m(C) = 20, m(D) = 31</math>. Since <math>AM = MD</math>, then <math>m(A) = 31</math>, and <math>\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}</math>. So m+n=051 | + | Assign [[mass points]] as follows: by Angle-Bisector Theorem, <math>BD / DC = 20/11</math>, so we assign <math>m(B) = 11, m(C) = 20, m(D) = 31</math>. Since <math>AM = MD</math>, then <math>m(A) = 31</math>, and <math>\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}</math>. So $m+n=051 |
=== Solution 3 === | === Solution 3 === |
Revision as of 19:18, 11 September 2017
Problem 4
In triangle ,
. 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
.
Contents
Solutions
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
Assign mass points as follows: by Angle-Bisector Theorem, , so we assign
. Since
, then
, and
. So $m+n=051
Solution 3
By Menelaus' Theorem on with transversal
,
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.