Mock AIME 2 2006-2007 Problems/Problem 14

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Problem

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

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 $x = 277$.

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.

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