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1998 AHSME Problems/Problem 28

Revision as of 21:53, 20 December 2020 by Quantomaticguy (talk | contribs) (→‎Solution)

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

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 3

By the application of ratio lemma for CD/BD, we get CD/BD = 2cos3AcosA,where we let A = angleDAB. we already know cos2A hence the rest is easy

Solution 2

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 $\fbox{(B) 14}$.

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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