Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(+ <asy> not to scale)
(→‎See also)
 
(3 intermediate revisions by 3 users not shown)
Line 5: Line 5:
 
size(120);
 
size(120);
 
defaultpen(0.7);
 
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);
 
draw(A--D--C--E--B--C--D--B--cycle);
 
label("\(A\)",A,S);
 
label("\(A\)",A,S);
Line 20: Line 20:
 
\qquad\mathrm{(E)}\ 18</math>
 
\qquad\mathrm{(E)}\ 18</math>
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
Let <math>\alpha = DBC</math>. Then the first condition tells us that
 +
<cmath>
 +
\tan^2 DBE = \tan(DBE - \alpha)\tan(DBE + \alpha) = \frac {\tan^2 DBE - \tan^2 \alpha}{1 - \tan ^2 DBE \tan^2 \alpha},
 +
</cmath>
 +
and multiplying out gives us <math>(\tan^4 DBE - 1) \tan^2 \alpha = 0</math>. Since <math>\tan\alpha \neq 0</math>, we have <math>\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}</math>.
 +
 
 +
The second condition tells us that <math>2\cot (45 - \alpha) = 1 + \cot \alpha</math>. Expanding, we have <math>1 + \cot \alpha = 2\left[\frac {\cot \alpha + 1}{\cot \alpha - 1}\right] \Longrightarrow (\cot \alpha - 3)(\cot \alpha + 1) = 0</math>. Evidently <math>\cot \alpha \neq - 1</math>, so we get <math>\cot \alpha = 3</math>.
 +
 
 +
Now <math>BD = 5\sqrt {2}</math> and <math>AC = \frac {2BD} {\cot \alpha} = \frac {10\sqrt {2}}{3}</math>. Thus, <math>[ABC] = \frac {1}{2} \cdot 5\sqrt {2} \cdot \frac {10\sqrt {2}}{3} = \frac {50}{3}\ \mathrm{(B)}</math>.
  
 
== See also ==
 
== See also ==
Line 27: Line 35:
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 
[[Category:Intermediate Trigonometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:59, 3 July 2013

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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_24&oldid=53519"