Difference between revisions of "1985 AIME Problems/Problem 9"
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MP("3",(A2+A3)/2,E); | MP("3",(A2+A3)/2,E); | ||
MP("4",(A1+A3)/2,E); | MP("4",(A1+A3)/2,E); | ||
− | D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1, | + | D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1,18)); |
label(" | label(" | ||
label(" | label(" | ||
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</asy></center> | </asy></center> | ||
− | It’s easy to see that the angle opposite the side 2 is <math>\frac{\alpha}{2}</math>, and using the [[Law of Cosines]], we get: <cmath>2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}</cmath> Which, rearranges to: <cmath>21 = 24cos\frac{\alpha}{2}</cmath> And, that gets | + | It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is <math>\frac{\alpha}{2}</math>, and using the [[Law of Cosines]], we get: |
− | < | + | <cmath>2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}</cmath> |
− | + | Which, rearranges to: | |
+ | <cmath>21 = 24cos\frac{\alpha}{2}</cmath> | ||
+ | And, that gets us: | ||
+ | <cmath>cos\frac{\alpha}{2} = 7/8</cmath> | ||
+ | Using <math>\cos 2\theta = 2\cos^2 \theta - 1</math>, we get that: | ||
+ | <cmath>\cos\alpha = 17/32</cmath> | ||
+ | Which gives an answer of <math>\boxed{049}</math> | ||
+ | |||
— Alexlikemath | — Alexlikemath | ||
+ | |||
==Solution 3 (trig)== | ==Solution 3 (trig)== |
Revision as of 00:11, 1 May 2019
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 (Law of cosines)
It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is , and using the Law of Cosines, we get: Which, rearranges to: And, that gets us: Using , we get that: Which gives an answer of
— Alexlikemath
Solution 3 (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 |