Difference between revisions of "2004 AMC 12B Problems/Problem 24"
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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>. | 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 = 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>. | + | 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 == | ||
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[[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 ,
, and
is an altitude. Point
is on the extension of
such that
. The values of
,
, and
form a geometric progression, and the values of
form an arithmetic progression. What is the area of
?
![[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]](http://latex.artofproblemsolving.com/c/5/a/c5ae6e04f0710cb08f0e10ffc47a0f257844910e.png)
Solution
Let . Then the first condition tells us that
and multiplying out gives us
. Since
, we have
.
The second condition tells us that . Expanding, we have
. Evidently
, so we get
.
Now and
. Thus,
.
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 |
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