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Difference between revisions of "1985 AIME Problems/Problem 9"

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* [[1985 AIME Problems/Problem 10 | Next problem]]
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 14:33, 6 May 2007

Problem

In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution

All chords of a given length in a given circle subtend the same arc and therefore the same central angle. Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle.

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This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$. The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$, so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$.

Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle $\alpha$, so by the Law of Cosines,

$2^2 = R^2 + R^2 - 2R^2\cos \alpha$

and so $\cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}$ and the answer is $17 + 32 = 049$.

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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