1985 AIME Problems/Problem 9

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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

size(200); pointpen = black; pathpen = black + linewidth(0.8);real r = 8/15^0.5, a = 57.91, b = 93.135;pair O = (0,0), A = r*...

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.

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,

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

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

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1985 AIME Problems/Problem 9

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Problem

Solution

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