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]] | ||
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which yields finally <math>x = 277</math>. | which yields finally <math>x = 277</math>. | ||
| − | + | ==See Also== | |
| − | + | {{Mock AIME box|year=2006-2007|n=2|num-b=13|num-a=15}} | |
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Latest revision as of 10:53, 4 April 2012
Contents
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 
.
See Also
| Mock AIME 2 2006-2007 (Problems, Source) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
Problem Source
4everwise thought of this problem after reading the first chapter of Geometry Revisited.
