Difference between revisions of "2013 AMC 12B Problems/Problem 20"
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==Solution== | ==Solution== | ||
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+ | Let <math>f,g,h,j</math> be <math>\sin, \cos, \tan, \cot</math> (not respectively). Then we have four points <math>(f,f^2),(g,g^2),(h,h^2),(j,j^2)</math>, and the two lines connecting two pairs of them must be parallel (as we are dealing with a trapezoid). WLOG, we take the line connecting the first two and the line connecting the last two to be parallel, so that <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>. | ||
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+ | Now, we must find how to match up <math>\sin, \cos, \tan, \cot</math> to <math>f,g,h,j</math> so that the above equation has a solution. The three possible ways are <math>\sin x+\cos x = \tan x+\cot x</math>, <math>\sin x+\tan x = \cos x+\cot x</math>, and <math>\sin x+\cot x = \cos x+\tan x</math>. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1</math>, so the first two of those possible ways do not work because the LHS and the RHS have different ranges. Thus, it must be the last possible way, <math>\sin x+\cot x = \cos x+\tan x</math>. | ||
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+ | Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>: | ||
+ | |||
+ | <cmath> | ||
+ | \sin x+\cot x = \cos x+\tan x \\ | ||
+ | (\sin x+\cot x)(\sin x \cos x ) = (\cos x+\tan x)(\sin x \cos x ) \\ | ||
+ | \sin^2 x \cos x+\cos^2 x = \cos^2 x \sin x+\sin^2 x \\ | ||
+ | \sin^2 x \cos x -\cos^2 x \sin x = \sin^2 x - \cos^2 x \\ | ||
+ | \sin x \cos x (\sin x -\cos x) = (\sin x - \cos x)(\sin x + \cos x) \\ | ||
+ | \sin x \cos x = \sin x + \cos x \\ | ||
+ | (\sin x \cos x)^2 = (\sin x + \cos x)^2 \\ | ||
+ | (\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\ | ||
+ | (\sin x \cos x)^2 = 1 + 2\sin x \cos x \\ | ||
+ | (\sin x \cos x)^2 - 2\sin x \cos x - 1 =0 | ||
+ | </cmath> | ||
+ | |||
+ | Solve the quadratic to find <math>\sin x \cos x = \frac{2 - 2\sqrt{2}}{2}</math>, so that <math>\sin(2x) = 2 \sin x \cos x = \boxed{\textbf{(A)} \ 2 - 2\sqrt{2}}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2013|ab=B|num-b=19|num-a=21}} |
Revision as of 20:32, 22 February 2013
Problem
For , points and are the vertices of a trapezoid. What is ?
$\textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D}}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}$ (Error compiling LaTeX. Unknown error_msg)
Solution
Let be (not respectively). Then we have four points , and the two lines connecting two pairs of them must be parallel (as we are dealing with a trapezoid). WLOG, we take the line connecting the first two and the line connecting the last two to be parallel, so that , or .
Now, we must find how to match up to so that the above equation has a solution. The three possible ways are , , and . On the interval , we have , so the first two of those possible ways do not work because the LHS and the RHS have different ranges. Thus, it must be the last possible way, .
Now, we perform some algebraic manipulation to find :
Solve the quadratic to find , so that .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |