Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1985 AIME Problems/Problem 9"

(added solution)
(→‎See also)
Line 15: Line 15:
  
 
== See also ==
 
== See also ==
−
* [[1985 AIME Problems/Problem 8 | Previous problem]]
+
{{AIME box|year=1985|num-b=8|num-a=10}}
−
* [[1985 AIME Problems/Problem 10 | Next problem]]
+
* [[AIME Problems and Solutions]]
−
* [[1985 AIME Problems]]
+
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
  
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]

Revision as of 13:33, 6 May 2007

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

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.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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,

$2^2 = R^2 + R^2 - 2R^2\cos \alpha$

and so $\cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}$ and the answer is $17 + 32 = 049$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1985_AIME_Problems/Problem_9&oldid=14922"