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Difference between revisions of "1985 AIME Problems/Problem 9"
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==Solution 2== (trig)
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==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

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 1

[asy] 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*expi(pi/3); D(CR(O,r)); D(O--rotate(a/2)*A--rotate(-a/2)*A--cycle); D(O--rotate(b/2)*A--rotate(-b/2)*A--cycle); D(O--rotate((a+b)/2)*A--rotate(-(a+b)/2)*A--cycle); MP("2",(rotate(a/2)*A+rotate(-a/2)*A)/2,NE); MP("3",(rotate(b/2)*A+rotate(-b/2)*A)/2,NE); MP("4",(rotate((a+b)/2)*A+rotate(-(a+b)/2)*A)/2,NE); D(anglemark(rotate(-(a+b)/2)*A,O,rotate((a+b)/2)*A,5)); label("\(\alpha+\beta\)",(0.08,0.08),NE,fontsize(8)); [/asy]

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.

[asy] 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*expi(pi/3), A1 = rotate(a/2)*A, A2 = rotate(-a/2)*A, A3 = rotate(-a/2-b)*A; D(CR(O,r)); D(O--A1--A2--cycle); D(O--A2--A3--cycle); D(O--A1--A3--cycle); MP("2",(A1+A2)/2,NE); MP("3",(A2+A3)/2,E); MP("4",(A1+A3)/2,E); D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); label("\(\alpha\)",(0.07,0.16),NE,fontsize(8)); label("\(\beta\)",(0.12,-0.16),NE,fontsize(8)); [/asy]

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 $17 + 32 = \boxed{049}$.


Solution 2 (trig)

Using the first diagram above, \[\sin \frac{\alpha}{2} = \frac{1}{r}\] \[\sin \frac{\beta}{2} = \frac{1.5}{r}\] \[\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}\] by the Pythagorean trig identities, \[\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}\] \[\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}\] so by the composite sine identity \[\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}\] multiply both sides by $2r$, then subtract $\sqrt{4-\frac{9}{r^2}}$ from both sides squaring both sides, we get \[16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}\] \[\Longrightarrow 16+4=9+8\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}\] \[\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}\] plugging this back in, \[\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}\] so \[\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}\] and the answer is $17+32=\boxed{049}$

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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