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}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}</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)}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}</math> |
==Solution== | ==Solution== |
Latest revision as of 00:54, 28 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.