Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 14"
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== Problem == | == Problem == | ||
| − | In [[triangle]] <math>ABC</math>, <math> | + | In [[triangle]] <math>ABC</math>, <math>AB = 308</math> and <math>AC=35</math>. Given that <math>AD</math>, <math>BE,</math> and <math>CF,</math> [[intersect]] at <math>P</math> and are an [[angle bisector]], [[median of a triangle | median]], and [[altitude]] of the triangle, respectively, compute the [[length]] of <math>BC.</math> |
[[Image:Mock AIME 2 2007 Problem14.jpg]] | [[Image:Mock AIME 2 2007 Problem14.jpg]] | ||
| Line 27: | Line 27: | ||
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| − | *[[Mock AIME 2 2006-2007/Problem 13 | Previous Problem]] | + | *[[Mock AIME 2 2006-2007 Problems/Problem 13 | Previous Problem]] |
| − | *[[Mock AIME 2 2006-2007/Problem 15 | Next Problem]] | + | *[[Mock AIME 2 2006-2007 Problems/Problem 15 | Next Problem]] |
*[[Mock AIME 2 2006-2007]] | *[[Mock AIME 2 2006-2007]] | ||
Revision as of 14:32, 3 April 2012
Problem
In triangle 
, 
and 
. Given that 
, 
and 
intersect at 
and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of 
Solution
Let 
.
By the Angle Bisector Theorem, 
.
Let 
. Then by the Pythagorean Theorem, 
and 
. Subtracting the former equation from the latter to eliminate 
, we have 
so 
. Since 
, 
. We can solve these equations for 
and 
in terms of 
to find that 
and 
.
Now, by Ceva's Theorem, 
, so 
and 
. Plugging in the values we previously found,
so
and
which yields finally 
.
Problem Source
4everwise thought of this problem after reading the first chapter of Geometry Revisited.
