Difference between revisions of "1998 AHSME Problems/Problem 28"
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==Solution 2== | ==Solution 2== | ||
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Let <math>AC=2</math> and <math>AD=3</math>. By the Pythagorean Theorem, <math>CD=\sqrt{5}</math>. Let point <math>P</math> be on segment <math>CD</math> such that <math>AP</math> bisects <math>\angle CAD</math>. Thus, angles <math>CAP</math>, <math>PAD</math>, and <math>DAB</math> are congruent. Applying the angle bisector theorem on <math>ACD</math>, we get that <math>CP=\frac{2\sqrt{5}}{5}</math> and <math>PD=\frac{3\sqrt{5}}{5}</math>. Pythagorean Theorem gives <math>AP=\frac{\sqrt{5}\sqrt{24}}{5}</math>. | Let <math>AC=2</math> and <math>AD=3</math>. By the Pythagorean Theorem, <math>CD=\sqrt{5}</math>. Let point <math>P</math> be on segment <math>CD</math> such that <math>AP</math> bisects <math>\angle CAD</math>. Thus, angles <math>CAP</math>, <math>PAD</math>, and <math>DAB</math> are congruent. Applying the angle bisector theorem on <math>ACD</math>, we get that <math>CP=\frac{2\sqrt{5}}{5}</math> and <math>PD=\frac{3\sqrt{5}}{5}</math>. Pythagorean Theorem gives <math>AP=\frac{\sqrt{5}\sqrt{24}}{5}</math>. | ||
Let <math>DB=x</math>. By the Pythagorean Theorem, <math>AB=\sqrt{(x+\sqrt{5})^{2}+2^2}</math>. Applying the angle bisector theorem again on triangle <math>APB</math>, we have <cmath>\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\sqrt{5}}</cmath> | Let <math>DB=x</math>. By the Pythagorean Theorem, <math>AB=\sqrt{(x+\sqrt{5})^{2}+2^2}</math>. Applying the angle bisector theorem again on triangle <math>APB</math>, we have <cmath>\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\sqrt{5}}</cmath> | ||
− | The right side simplifies to<math>\frac{sqrt{24}}{3}</math>. Cross multiplying, squaring, and simplifying, we get a quadratic: <cmath>5x^2-6\sqrt{5}x-27=0</cmath> Solving this quadratic and taking the positive root gives <cmath>x=\frac{9\sqrt{5}}{5}</cmath> Finally, taking the desired ratio and canceling the roots gives < | + | The right side simplifies to<math>\frac{\sqrt{24}}{3}</math>. Cross multiplying, squaring, and simplifying, we get a quadratic: <cmath>5x^2-6\sqrt{5}x-27=0</cmath> Solving this quadratic and taking the positive root gives <cmath>x=\frac{9\sqrt{5}}{5}</cmath> Finally, taking the desired ratio and canceling the roots gives <math>\frac{CD}{BD}=\frac{5}{9}</math> The answer is <math>\fbox{(B) 14}</math>. |
== See also == | == See also == |
Revision as of 10:55, 2 August 2016
Contents
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 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 |
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