Difference between revisions of "1985 AIME Problems/Problem 9"
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== Problem == | == Problem == | ||
− | In a [[circle]], [[parallel]] [[chord]]s of | + | In a [[circle]], [[parallel]] [[chord]]s of lengths 2, 3, and 4 determine [[central angle]]s of <math>\alpha</math>, <math>\beta</math>, and <math>\alpha + \beta</math> [[radian]]s, respectively, where <math>\alpha + \beta < \pi</math>. If <math>\cos \alpha</math>, which is a [[positive]] [[rational number]], is expressed as a [[fraction]] in lowest terms, what is the sum of its numerator and denominator? |
− | == Solution == | + | |
+ | == Solution 1== <!-- Images obsoleted Image:1985_AIME-9.png, Image:1985_AIME-9a.png by asymptote --> | ||
+ | <center><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(" | ||
+ | </asy></center> | ||
+ | |||
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]]. | 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]]. | ||
+ | <center><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(" | ||
+ | label(" | ||
+ | </asy></center> | ||
− | {{ | + | This triangle has [[semiperimeter]] <math>\frac{2 + 3 + 4}{2}</math> so by [[Heron's formula]] it has [[area]] <math>K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}</math>. The area of a given triangle with sides of length <math>a, b, c</math> and circumradius of length <math>R</math> is also given by the formula <math>K = \frac{abc}{4R}</math>, so <math>\frac6R = \frac{3}{4}\sqrt{15}</math> and <math>R = \frac8{\sqrt{15}}</math>. |
− | This | + | Now, consider the triangle formed by two radii and the chord of length 2. This [[isosceles triangle]] has vertex angle <math>\alpha</math>, so by the [[Law of Cosines]], |
+ | |||
+ | <cmath>2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}</cmath> | ||
+ | and the answer is <math>17 + 32 = \boxed{049}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 2 (Law of Cosines)== | ||
+ | <center><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)); D(anglemark(A2,A3,A1,18)); | ||
+ | label(" | ||
+ | label(" | ||
+ | label(" | ||
+ | </asy></center> | ||
+ | |||
+ | 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 = 24\cos\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 | |
− | + | ==Solution 3 (trig)== | |
+ | Using the first diagram above, | ||
+ | <cmath>\sin \frac{\alpha}{2} = \frac{1}{r}</cmath> | ||
+ | <cmath>\sin \frac{\beta}{2} = \frac{1.5}{r}</cmath> | ||
+ | <cmath>\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}</cmath> | ||
+ | by the Pythagorean trig identities, | ||
+ | <cmath>\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}</cmath> | ||
+ | <cmath>\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}</cmath> | ||
+ | so by the composite sine identity | ||
+ | <cmath>\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}</cmath> | ||
+ | multiply both sides by <math>2r</math>, then subtract <math>\sqrt{4-\frac{9}{r^2}}</math> from both sides | ||
+ | squaring both sides, we get | ||
+ | <cmath>16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}</cmath> | ||
+ | <cmath>\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}</cmath> | ||
+ | <cmath>\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}</cmath> | ||
+ | plugging this back in, | ||
+ | <cmath>\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}</cmath> | ||
+ | so | ||
+ | <cmath>\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}</cmath> | ||
+ | and the answer is <math>17+32=\boxed{049}</math> | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1985|num-b=8|num-a=10}} | |
− | |||
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
− | |||
[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] |
Revision as of 15:20, 28 September 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 |