2004 AMC 12B Problems/Problem 24

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Problem

In $\triangle ABC$, $AB = BC$, and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $BE = 10$. The values of $\tan \angle CBE$, $\tan \angle DBE$, and $\tan \angle ABE$ form a geometric progression, and the values of $\cot \angle DBE,$ $\cot \angle CBE,$ $\cot \angle DBC$ form an arithmetic progression. What is the area of $\triangle ABC$?

[asy] size(120); defaultpen(0.7); pair A = (0,0), D = (5*2^.5/3,0), C = (10*2^.5/3,0), B = (5*2^.5/3,5*2^.5), E = (13*2^.5/3,0); draw(A--D--C--E--B--C--D--B--cycle); label("\(A\)",A,S); label("\(B\)",B,N); label("\(C\)",C,S); label("\(D\)",D,S); label("\(E\)",E,S); [/asy]

$\mathrm{(A)}\ 16 \qquad\mathrm{(B)}\ \frac {50}3 \qquad\mathrm{(C)}\ 10\sqrt{3} \qquad\mathrm{(D)}\ 8\sqrt{5} \qquad\mathrm{(E)}\ 18$

Solution

Let $\alpha = DBC$. Then the first condition tells us that \[\tan^2 DBE = \tan(DBE - \alpha)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},\] and multiplying out gives us $(\tan^4 DBE - 1) \tan^2 \alpha = 0$. Since $\tan\alpha \neq 0$, we have $\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}$.

The second condition tells us that $2\cot (45 - \alpha) = 1 + \cot \alpha$. Expanding, we have $1 + \cot \alpha = 2\left[\frac {\cot \alpha + 1}{\cot \alpha - 1}\right] \Longrightarrow (\cot \alpha - 3)(\cot \alpha + 1) = 0$. Evidently $\cot \alpha \neq - 1$, so we get $\cot \alpha = 3$.

Now $BD = 5\sqrt {2}$ and $AC = \frac {2BD} {\cot \alpha} = \frac {10\sqrt {2}}{3}$. Thus, $[ABC] = \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} = \frac {50}{3}\ \mathrm{(B)}$.

See also

2004 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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
All AMC 12 Problems and Solutions

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