Difference between revisions of "1998 AHSME Problems/Problem 28"
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and <math>\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}</math>. (This also may have been done on a calculator by finding <math>\theta</math> directly) | and <math>\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}</math>. (This also may have been done on a calculator by finding <math>\theta</math> directly) | ||
== Solution 3 == | == Solution 3 == | ||
− | By the application of ratio lemma for CD/ | + | By the application of ratio lemma for <math>\frac{CD}{BD}</math>, we get <math>\frac{CD}{BD} = 2cos3AcosA</math>, where we let <math>A = \angle{DAB}</math>. We already know <math>cos2A</math> hence the rest is easy |
+ | (latex edit by aopspandy) | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:55, 22 May 2021
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)
Solution 3
By the application of ratio lemma for , we get
, where we let
. We already know
hence the rest is easy
(latex edit by aopspandy)
Solution 2
Let and
. By the Pythagorean Theorem,
. Let point
be on segment
such that
bisects
. Thus, angles
,
, and
are congruent. Applying the angle bisector theorem on
, we get that
and
. Pythagorean Theorem gives
.
Let . By the Pythagorean Theorem,
. Applying the angle bisector theorem again on triangle
, we have
The right side simplifies to
. Cross multiplying, squaring, and simplifying, we get a quadratic:
Solving this quadratic and taking the positive root gives
Finally, taking the desired ratio and canceling the roots gives
. The answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.