Difference between revisions of "2005 AIME II Problems/Problem 14"
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By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have | By the [[Law of Sines]] and since <math>\angle BAE = \angle CAD, \angle BAD = \angle CAE</math>, we have | ||
− | < | + | <cmath>\begin{align*} |
− | \frac{CD \cdot CE}{AC^2} &= \frac{\sin CAD}{\sin ADC} \cdot \frac{\sin CAE}{\sin AEC} | + | \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 BAD}{\sin ADB \sin AEB} \\ |
− | = \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} | + | &= \frac{\sin BAE}{\sin AEB} \cdot \frac{\sin BAD}{\sin ADB}\\ &= \frac{BE \cdot BD}{AB^2} |
\end{align*} | \end{align*} | ||
− | </ | + | </cmath> |
− | Substituting our knowns, we have <math>\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 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>. | + | Substituting our knowns, we have <math>\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 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>. |
== See also == | == See also == |
Revision as of 19:59, 13 March 2015
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
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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