Difference between revisions of "1985 AIME Problems/Problem 9"
Alexlikemath (talk | contribs) |
Alexlikemath (talk | contribs) (Add solution 2 using Law of Cosines directly) |
||
Line 45: | Line 45: | ||
− | ==Solution 2 (Law of | + | ==Solution 2 (Law of Cosines)== |
<center><asy> | <center><asy> | ||
size(200); | size(200); | ||
Line 75: | Line 75: | ||
− | + | -Alexlikemath | |
Revision as of 00:25, 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,A1,18)); 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 |