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Mock AIME 2 2006-2007/Problem 14

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Problem

In triangle ABC, \displaystyle AB = 308 and \displaystyle AC=35. Given that \displaystyle AD, \displaystyle BE, and \displaystyle CF, intersect at \displaystyle P and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of \displaystyle BC.

Image:Mock AIME 2 2007 Problem14.jpg

Solution

Let BC = x.

By the Angle Bisector Theorem, \frac{CD}{BD} = \frac{AC}{AB} = \frac{35}{308} = \frac{5}{44}.

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

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

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

so

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

and

x^2 - 1225 = 75504

which yields finally x = 277.



Problem Source

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

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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