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 $\displaystyle BC.$

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,