Difference between revisions of "2013 AMC 12B Problems/Problem 20"
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Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>: | Now, we perform some algebraic manipulation to find <math>\sin (2x)</math>: | ||
− | < | + | <math> |
\sin x+\cot x = \cos x+\tan x \\ | \sin x+\cot x = \cos x+\tan x \\ | ||
\sin x - \cos x = \tan x - \cot x = (\sin x - \cos x) (\sin x + \cos x) / (\sin x \cos x) \\ | \sin x - \cos x = \tan x - \cot x = (\sin x - \cos x) (\sin x + \cos x) / (\sin x \cos x) \\ | ||
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(\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\ | (\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\ | ||
(\sin x \cos x)^2 - 2\sin x \cos x - 1 =0 | (\sin x \cos x)^2 - 2\sin x \cos x - 1 =0 | ||
− | </ | + | </math> |
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>. | 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>. |
Revision as of 00:00, 6 April 2015
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. )
Solution
Let be (not respectively). Then we have four points , and a pair of lines each connecting two points must be parallel (as we are dealing with a trapezoid). WLOG, take the line connecting the first two points and the line connecting the last two points to be parallel, so that , or .
Now, we must find how to match up to so that the above equation has a solution. On the interval , we have , and so the sum of the largest and the smallest is equal to the sum of the other two, namely, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.