1998 AHSME Problems/Problem 28

Problem

In triangle $ABC$ , angle $C$ is a right angle and $CB > CA$ . Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$ . If $AC/AD = 2/3$ , then $CD/BD = m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26$

Solution 1

Let $\theta = \angle DAB$ , so $2\theta = \angle CAD$ and $3 \theta = \angle CAB$ . Then, it is given that $\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}$ and

$\frac{BD}{CD} = \frac{AC(\tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.$

Now, through the use of trigonometric identities, $\cos 2\theta = 2\cos^2 \theta - 1 = \frac{2}{\sec ^2 \theta} - 1 = \frac{1 - \tan^2 \theta}{1 + \tan ^2 \theta} = \frac{2}{3}$ . Solving yields that $\tan^2 \theta = \frac 15$ . Using the tangent addition identity, we find that $\tan 2\theta = \frac{2\tan \theta}{1 - \tan ^2 \theta},\ \tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$ , and

$\frac{BD}{CD} = \frac{\tan 3\theta}{\tan 2\theta} - 1 = \frac{(3 - \tan^2 \theta)(1-\tan ^2 \theta)}{2(1 - 3\tan^2 \theta)} - 1 = \frac{(1 + \tan^2 \theta)^2}{2(1 - 3\tan^2 \theta)} = \frac{9}{5}$

and $\frac{CD}{BD} = \frac{5}{9} \Longrightarrow m+n = 14 \Longrightarrow \mathbf{(B)}$ . (This also may have been done on a calculator by finding $\theta$ directly)

Solution 2

By the application of ratio lemma for $\frac{CD}{BD}$ , we get $\frac{CD}{BD} = 2\cos{3A}\cos{A}$ , where we let $A = \angle{DAB}$ . We already know $\cos{2A}$ hence the rest is easy

Solution 3

Let $AC=2$ and $AD=3$ . By the Pythagorean Theorem, $CD=\sqrt{5}$ . Let point $P$ be on segment $CD$ such that $AP$ bisects $\angle CAD$ . Thus, angles $CAP$ , $PAD$ , and $DAB$ are congruent. Applying the angle bisector theorem on $ACD$ , we get that $CP=\frac{2\sqrt{5}}{5}$ and $PD=\frac{3\sqrt{5}}{5}$ . Pythagorean Theorem gives $AP=\frac{\sqrt{5}\sqrt{24}}{5}$ .

Let $DB=x$ . By the Pythagorean Theorem, $AB=\sqrt{(x+\sqrt{5})^{2}+2^2}$ . Applying the angle bisector theorem again on triangle $APB$ , we have $\frac{\sqrt{(x+\sqrt{5})^{2}+2^2}}{x}=\frac{\frac{\sqrt{5}\sqrt{24}}{5}}{\frac{3\sqrt{5}}{5}}$ The right side simplifies to $\frac{\sqrt{24}}{3}$ . Cross multiplying, squaring, and simplifying, we get a quadratic: $5x^2-6\sqrt{5}x-27=0$ Solving this quadratic and taking the positive root gives $x=\frac{9\sqrt{5}}{5}$ Finally, taking the desired ratio and canceling the roots gives $\frac{CD}{BD}=\frac{5}{9}$ . The answer is $\fbox{(B) 14}$ .

Solution 4

Let $AC = 2$ , $AD = 3$ . $\cos \angle CAD = \frac23$

By the pythagorean theorem $CD = \sqrt{3^2-2^2} = \sqrt{5}$

$\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23$

$\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}$

By the Law of Sine, $\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}$

$\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}$

$8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)$

$5BD^2 - 6 BD \sqrt{5} -27=0$

As $BD>0$ , $BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}$

$\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59$ .

$5+9=\boxed{\textbf{(B) } 14}$ .

~isabelchen

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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