View source for Mock AIME 2 2006-2007 Problems/Problem 14

== Problem ==
In [[triangle]] <math>ABC</math>, <math>\displaystyle AB = 308</math> and <math>\displaystyle AC=35</math>.  Given that <math>\displaystyle AD</math>, <math>\displaystyle BE,</math> and <math>\displaystyle CF,</math> [[intersect]] at <math>\displaystyle P</math> and are an [[angle bisector]], [[median of a triangle | median]], and [[altitude]] of the triangle, respectively, compute the [[length]] of <math>\displaystyle BC.</math>

[[Image:Mock AIME 2 2007 Problem14.jpg]]

==Solution==
Let <math>BC = x</math>.

By the [[Angle Bisector Theorem]], <math>\frac{CD}{BD} = \frac{AC}{AB} = \frac{35}{308} = \frac{5}{44}</math>.

Let <math>CF = h</math>.  Then by the [[Pythagorean Theorem]], <math>h^2 + AF^2 = 35^2</math> and <math>h^2 + BF^2 = x^2</math>.  Subtracting the former [[equation]] from the latter to eliminate <math>h</math>, we have <math>BF^2 - AF^2 = x^2 - 35^2</math> so <math>(BF + AF)(BF - AF) = x^2 - 1225</math>.  Since <math>BF + AF = AB = 308</math>, <math>BF - AF = \frac{x^2 - 1225}{308}</math>.  We can solve these equations for <math>BF</math> and <math>AF</math> in terms of <math>x</math> to find that <math>BF = 154 + \frac{x^2 - 1225}{616} = </math> and <math>AF = 154 - \frac{x^2 - 1225}{616}</math>.

Now, by [[Ceva's Theorem]], <math>\frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BF}{FA} = 1</math>, so <math>1 \cdot \frac{5}{44} \cdot \frac{BF}{AF} = 1</math> and <math>5BF = 44AF</math>.  Plugging in the values we previously found,

<math>5\cdot 154 + \frac{5(x^2 - 1225)}{616} = 44\cdot 154 - \frac{44(x^2 - 1225)}{616}</math>

so

<math>\frac{49}{616}(x^2 - 1225) = 39\cdot 154</math>

and

<math>x^2 - 1225 = 75504</math>

which yields finally <math>x = 277</math>.

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*[[Mock AIME 2 2006-2007]]


== Problem Source ==
4everwise thought of this problem after reading the first chapter of Geometry Revisited.

[[Category:Intermediate Geometry Problems]]