Difference between revisions of "1985 AIME Problems/Problem 9"
Line 46: | Line 46: | ||
− | ==Solution 2 | + | ==Solution 2 (trig)== |
Using the first diagram above, | Using the first diagram above, | ||
<cmath>\sin \frac{\alpha}{2} = \frac{1}{r}</cmath> | <cmath>\sin \frac{\alpha}{2} = \frac{1}{r}</cmath> |
Revision as of 13:01, 16 June 2018
Contents
[hide]Problem
In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of , , and radians, respectively, where . If , which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?
Solution 1
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 so by Heron's formula it has area . The area of a given triangle with sides of length and circumradius of length is also given by the formula , so and .
Now, consider the triangle formed by two radii and the chord of length 2. This isosceles triangle has vertex angle , so by the Law of Cosines,
and the answer is .
Solution 2 (trig)
Using the first diagram above, by the Pythagorean trig identities, so by the composite sine identity multiply both sides by , then subtract from both sides squaring both sides, we get plugging this back in, so and the answer is
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 |