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> | ||
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== See also == | == See also == |
Revision as of 19:18, 3 April 2011
Problem 4
In triangle ,
. The angle bisector of $\ang A$ (Error compiling LaTeX. Unknown error_msg) 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
.
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
.
Solution 3
By Menelaus' Theorem on with transversal
,
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 |