1998 AHSME Problems/Problem 28
Revision as of 17:16, 8 August 2011 by Talkinaway (talk | contribs)
Problem
In triangle , angle
is a right angle and
. Point
is located on
so that angle
is twice angle
. If
, then
, where
and
are relatively prime positive integers. Find
.
Solution
Let , so
and
. Then, it is given that
and
![$\frac{BD}{CD} = \frac{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.$](http://latex.artofproblemsolving.com/7/f/9/7f9e8b19b091fcc2af041035e6e91d142adff013.png)
Now, through the use of trigonometric identities, . Solving yields that
. Using the tangent addition identity, we find that
, and
![$\frac{BD}{CD} = \frac{\tan 3\theta}{\tan 2\theta} - 1 = \frac{(3 - \tan^2 \theta)(1-\tan ^2 \theta)}{2(1 - 3\tan^2 \theta)} - 1 = \frac{(1 + \tan^2 \theta)^2}{2(1 - 3\tan^2 \theta)} = \frac{9}{5}$](http://latex.artofproblemsolving.com/7/8/5/785e8db7854afad3f4a10fe4e5419fc5764eb10a.png)
and . (This also may have been done on a calculator by finding
directly)
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |