Difference between revisions of "1985 AIME Problems/Problem 9"
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and the answer is <math>17 + 32 = \boxed{049}</math>. | and the answer is <math>17 + 32 = \boxed{049}</math>. | ||
+ | == Video Solution by Pi Academy (Fast and Easy) == | ||
− | ==Solution 2 (Law of | + | https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D |
+ | |||
+ | ~ Pi Academy | ||
+ | |||
+ | ==Solution 2 (Law of Cosines)== | ||
<center><asy> | <center><asy> | ||
size(200); | size(200); | ||
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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 = | + | 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)== | ==Solution 3 (trig)== |
Latest revision as of 20:05, 8 December 2024
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
![[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
.
Video Solution by Pi Academy (Fast and Easy)
https://youtu.be/fHFD0TEfBnA?si=DRKVU_As7Rv0ou5D
~ Pi Academy
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,A1,18)); 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/4/4/8/448c27a4bb9a18d7576a79968b61141e098c460b.png)
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 |