2013 AMC 12B Problems/Problem 20

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For $135^\circ < x < 180^\circ$, points $P=(\cos x, \cos^2 x), Q=(\cot x, \cot^2 x), R=(\sin x, \sin^2 x)$ and $S =(\tan x, \tan^2 x)$ are the vertices of a trapezoid. What is $\sin(2x)$?

$\textbf{(A)}\ 2-2\sqrt{2}\qquad\textbf{(B)}\3\sqrt{3}-6\qquad\textbf{(C)}\ 3\sqrt{2}-5\qquad\textbf{(D}}\ -\frac{3}{4}\qquad\textbf{(E)}\ 1-\sqrt{3}$ (Error compiling LaTeX. )


Let $f,g,h,j$ be $\sin, \cos, \tan, \cot$ (not respectively). Then we have four points $(f,f^2),(g,g^2),(h,h^2),(j,j^2)$, 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 $\frac{g^2-f^2}{g-f} = \frac{j^2-h^2}{j-h}$, or $g+f = j+h$.

Now, we must find how to match up $\sin, \cos, \tan, \cot$ to $f,g,h,j$ so that the above equation has a solution. The three possible ways are $\sin x+\cos x = \tan x+\cot x$, $\sin x+\tan x = \cos x+\cot x$, and $\sin x+\cot x = \cos x+\tan x$. On the interval $135^\circ < x < 180^\circ$, we have $0<\sin x<\frac{1}{\sqrt{2}},-1< \cos x<-\frac{1}{\sqrt{2}},-1< \tan x<0,-\infty< \cot x<-1$, so the first two of those possible ways do not work because the LHS and the RHS have non-overlapping ranges. Thus, it must be the last possible way, $\sin x+\cot x = \cos x+\tan x$.

Now, we perform some algebraic manipulation to find $\sin (2x)$:

\[\sin x+\cot x = \cos x+\tan x \\ \sin x - \cos x = \tan x - \cot x = (\sin x - \cos x) (\sin x + \cos x) / (\sin x \cos x) \\ \sin x \cos x = \sin x + \cos x \\ (\sin x \cos x)^2 = (\sin x + \cos x)^2 \\ (\sin x \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x \\ (\sin x \cos x)^2 - 2\sin x \cos x - 1 =0\]

Solve the quadratic to find $\sin x \cos x = \frac{2 - 2\sqrt{2}}{2}$, so that $\sin(2x) = 2 \sin x \cos x = \boxed{\textbf{(A)} \ 2 - 2\sqrt{2}}$.

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions
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