2005 AIME II Problems/Problem 14
Problem
In triangle and Point is on with Point is on such that Given that where and are relatively prime positive integers, find
Solution
By the Law of Sines and since , we have
\frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC}
= \frac{\sin BAE \sin BAD}{\sin ADB \sin AEB} = \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2}
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Substituting our knowns, we have . The answer is .
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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