1998 AMC 12 Problem 28

Attempt Problem 28 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

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28.

In triangle ABC,ABC, angle CC is a right angle and CB>CA.CB\gt CA. Point DD is located on BC‾\overline{BC} so that angle CADCAD is twice angle DAB.DAB. If ACAD=23,\frac{AC}{AD}=\frac{2}{3}, then CDBD=mn,\frac{CD}{BD}=\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m+n.
1010
1414
1818
2222
2626
Answer: B
Concepts:trigonometryangle multiplication
Small hint:
Let ∠DAB=α\angle DAB=\alpha, so ∠CAB=3α\angle CAB=3\alpha and cos⁡2α=ACAD\cos2\alpha=\frac{AC}{AD}
Big hint:
Set AC=1AC=1 and express CDCD and CBCB using tan⁡2α\tan2\alpha and tan⁡3α\tan3\alpha
Solution:
Let ∠DAB=α.\angle DAB=\alpha. Since ACAD=cos⁡2α=23,\frac{AC}{AD}=\cos2\alpha=\frac{2}{3}, the identity for cos⁡2α\cos2\alpha gives tan⁡α=15.\tan\alpha=\frac{1}{\sqrt5}. Taking AC=1,AC=1, CD=tan⁡2α=52,CB=tan⁡3α=755. \begin{aligned} CD&=\tan2\alpha=\frac{\sqrt5}{2},\\ CB&=\tan3\alpha=\frac{7\sqrt5}{5}. \end{aligned} Thus BD=CB−CD=9510,BD=CB-CD=\frac{9\sqrt5}{10}, and CDBD=59.\frac{CD}{BD}=\frac{5}{9}. Therefore m+n=14,m+n=14, so B is correct.
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