Difference between revisions of "2004 AMC 12B Problems/Problem 24"

Revision as of 09:21, 29 April 2008

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

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2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2004 AMC 12B Problems/Problem 24"

Revision as of 09:21, 29 April 2008

Problem

Solution

See also

@@ Line 5: / Line 5: @@
 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 = (4/3*5*2^0.5,0);
+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);