Difference between revisions of "1985 AIME Problems/Problem 9"
Alexlikemath (talk | contribs) |
Alexlikemath (talk | contribs) |
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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,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1,10)); |
label("\(\alpha\)",(0.07,0.16),NE,fontsize(8)); | label("\(\alpha\)",(0.07,0.16),NE,fontsize(8)); | ||
label("\(\beta\)",(0.12,-0.16),NE,fontsize(8)); | label("\(\beta\)",(0.12,-0.16),NE,fontsize(8)); | ||
− | label("\(\alpha\)/2",(0. | + | label("\(\alpha\)/2",(0.82,-1.25),NE,fontsize(8)); |
</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 that <cmath>cos\frac{\alpha}{2} = 7/8</cmath> and using that <math>\cos 2\theta = 2\cos^2 \theta - 1 we get that | ||
+ | </math>\cos\alpha = 17/32<math>, | ||
+ | which gives an answer of </math>\boxed{049}<math> | ||
+ | |||
+ | — Alexlikemath | ||
==Solution 3 (trig)== | ==Solution 3 (trig)== | ||
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so by the composite sine identity | 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> | <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< | + | 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 | 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>16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}</cmath> | ||
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so | so | ||
<cmath>\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}</cmath> | <cmath>\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}</cmath> | ||
− | and the answer is <math>17+32=\boxed{049} | + | and the answer is </math>17+32=\boxed{049}$ |
== See also == | == See also == |
Revision as of 00:07, 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
![[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]](http://latex.artofproblemsolving.com/c/c/c/ccc44394730c0756402714001ccba571e69a1bf7.png)
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]](http://latex.artofproblemsolving.com/0/6/1/0618ad736b55ffe052dd977aed3aa0c72dca89d9.png)
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)
![[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,A1,10)); label("\(\alpha\)",(0.07,0.16),NE,fontsize(8)); label("\(\beta\)",(0.12,-0.16),NE,fontsize(8)); label("\(\alpha\)/2",(0.82,-1.25),NE,fontsize(8)); [/asy]](http://latex.artofproblemsolving.com/c/2/9/c29f2115e38d509bd58556ad568a3ead9e36b003.png)
It’s easy to see that the angle opposite the side 2 is , and using the Law of Cosines, we get:
Which, rearranges to:
And, that gets that
and using that
\cos\alpha = 17/32
\boxed{049}$— Alexlikemath
==Solution 3 (trig)==
Using the first diagram above,
<cmath>\sin \frac{\alpha}{2} = \frac{1}{r}</cmath>
<cmath>\sin \frac{\beta}{2} = \frac{1.5}{r}</cmath>
<cmath>\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}</cmath>
by the Pythagorean trig identities,
<cmath>\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}</cmath>
<cmath>\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}</cmath>
so by the composite sine identity
<cmath>\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}</cmath>
multiply both sides by$ (Error compiling LaTeX. Unknown error_msg)2r\sqrt{4-\frac{9}{r^2}}
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 |