Difference between revisions of "1998 AHSME Problems/Problem 28"
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+ | == Problem == | ||
+ | In triangle <math>ABC</math>, angle <math>C</math> is a [[right angle]] and <math>CB > CA</math>. Point <math>D</math> is located on <math>\overline{BC}</math> so that angle <math>CAD</math> is twice angle <math>DAB</math>. If <math>AC/AD = 2/3</math>, then <math>CD/BD = m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | <math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math> | ||
+ | |||
+ | == Solution 1== | ||
+ | Let <math>\theta = \angle DAB</math>, so <math>2\theta = \angle CAD</math> and <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and | ||
+ | |||
+ | |||
+ | <center><math>\frac{BD}{CD} = \frac{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center> | ||
+ | |||
+ | |||
+ | Now, through the use of trigonometric identities, <math>\cos 2\theta = 2\cos^2 \theta - 1 = \frac{2}{\sec ^2 \theta} - 1 = \frac{1 - \tan^2 \theta}{1 + \tan ^2 \theta} = \frac{2}{3}</math>. Solving yields that <math>\tan^2 \theta = \frac 15</math>. Using the tangent addition identity, we find that <math>\tan 2\theta = \frac{2\tan \theta}{1 - \tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}</math>, and | ||
+ | |||
+ | |||
+ | <center><math>\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}</math></center> | ||
+ | |||
+ | |||
+ | 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 2 == | ||
+ | By the application of ratio lemma for <math>\frac{CD}{BD}</math>, we get <math>\frac{CD}{BD} = 2\cos{3A}\cos{A}</math>, where we let <math>A = \angle{DAB}</math>. We already know <math>\cos{2A}</math> hence the rest is easy | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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}}{\frac{3\sqrt{5}}{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 <math>\frac{CD}{BD}=\frac{5}{9}</math>. The answer is <math>\fbox{(B) 14}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>AC = 2</math>, <math>AD = 3</math>. <math>\cos \angle CAD = \frac23</math> | ||
+ | |||
+ | By the pythagorean theorem <math>CD = \sqrt{3^2-2^2} = \sqrt{5}</math> | ||
+ | |||
+ | <math>\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23</math> | ||
+ | |||
+ | <math>\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}</math> | ||
+ | |||
+ | By the Law of Sine, <math>\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}</math> | ||
+ | |||
+ | <math>\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}</math> | ||
+ | |||
+ | <math>8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)</math> | ||
+ | |||
+ | <math>5BD^2 - 6 BD \sqrt{5} -27=0</math> | ||
+ | |||
+ | As <math>BD>0</math>, <math>BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}</math> | ||
+ | |||
+ | <math>\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59</math>. | ||
+ | |||
+ | <math>5+9=\boxed{\textbf{(B) } 14}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1998|num-b=27|num-a=29}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 00:51, 3 October 2023
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 1
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
By the application of ratio lemma for , we get
, where we let
. We already know
hence the rest is easy
Solution 3
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
.
Solution 4
Let ,
.
By the pythagorean theorem
By the Law of Sine,
As ,
.
.
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.