Difference between revisions of "2011 AIME II Problems/Problem 4"
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=== Solution 3 === | === Solution 3 === | ||
− | By [[Menelaus' Theorem]] on <math>\triangle ACD</math> with [[transversal]] <math>PB</math>, <cmath>1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.</cmath> | + | By [[Menelaus' Theorem]] on <math>\triangle ACD</math> with [[transversal]] <math>PB</math>, <cmath>1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}. </cmath> So <math>m+n = \boxed{051}</math>. |
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===Solution 4=== | ===Solution 4=== | ||
We will use barycentric coordinates. Let <math>A = (1, 0, 0)</math>, <math>B = (0, 1, 0)</math>, <math>C = (0, 0, 1)</math>. By the [[Angle Bisector Theorem]], <math>D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)</math>. Since <math>M</math> is the midpoint of <math>AD</math>, <math>M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)</math>. Therefore, the equation for line BM is <math>20x = 31z</math>. Let <math>P = (x, 0, 1-x)</math>. Using the equation for <math>BM</math>, we get <cmath>20x = 31(1-x)</cmath> | We will use barycentric coordinates. Let <math>A = (1, 0, 0)</math>, <math>B = (0, 1, 0)</math>, <math>C = (0, 0, 1)</math>. By the [[Angle Bisector Theorem]], <math>D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)</math>. Since <math>M</math> is the midpoint of <math>AD</math>, <math>M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)</math>. Therefore, the equation for line BM is <math>20x = 31z</math>. Let <math>P = (x, 0, 1-x)</math>. Using the equation for <math>BM</math>, we get <cmath>20x = 31(1-x)</cmath> |
Revision as of 19:19, 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
. $$ (Error compiling LaTeX. Unknown error_msg) 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
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.