Difference between revisions of "2003 AIME II Problems/Problem 11"
I_like_pie (talk | contribs) |
m |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math> | ||
== Solution == | == Solution == | ||
Line 5: | Line 6: | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2003|n=II|num-b=10|num-a=12}} | |
− | |||
− | |||
− | |||
− |
Revision as of 14:40, 21 November 2007
Problem
Triangle is a right triangle with
and right angle at
Point
is the midpoint of
and
is on the same side of line
as
so that
Given that the area of triangle
may be expressed as
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
2003 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |