Difference between revisions of "2013 AMC 12B Problems/Problem 20"
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For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? | For <math>135^\circ < x < 180^\circ</math>, points <math>P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)</math> and <math>S =(\tan x, \tan^2 x)</math> are the vertices of a trapezoid. What is <math>\sin(2x)</math>? | ||
− | <math> \textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D | + | <math>\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}</math> |
==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 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 <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>. | 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 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 <math>\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}</math>, or <math>g+f = j+h</math>. | ||
− | 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 | + | 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. On the interval <math>135^\circ < x < 180^\circ</math>, we have <math>\cot x <-1<\cos x<0<\sin x</math>, and <math>\cot x <-1<\tan x<0<\sin x</math> so the sum of the largest and the smallest is equal to the sum of the other two, namely, <math>\sin x+\cot x = \cos x+\tan x</math>. |
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>. | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:54, 27 September 2021
Problem
For , points and are the vertices of a trapezoid. What is ?
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 |
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