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2005 AIME II Problems/Problem 14

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Problem

In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$

Solution

pointpen = black; pathpen = black + linewidth(0.7); pen f = fontsize(10);pair C = (0,0), B=(15,0), A=IP(CR(B,13), CR(C,14)), ...

By the Law of Sines and since $\angle BAE = \angle CAD, \angle BAD = \angle CAE$ , we have

$\begin{align*} \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} = \frac{\sin BAE \s...$

Substituting our knowns, we have $\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrightarrow BE = \frac{13^2 \c...$ . The answer is $q = \boxed{463}$ .

See also

2005 AIME II (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_II_Problems/Problem_14"

Category: Intermediate Geometry Problems

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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