Question

For $135^\circ \lt x \lt 180^\circ,$ 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 $\sin(2x)?$

$2 - 2\sqrt2$

$3\sqrt3 - 6$

$3\sqrt2 - 5$

$-\dfrac{3}{4}$

$1 - \sqrt3$

Solution:

Each point $(t, t^2)$ lies on $y = t^2,$ and the chord through parameters $t_1, t_2$ has slope $t_1 + t_2.$ For $135^\circ \lt x \lt 180^\circ,$ both $\cos x$ and $\tan x$ lie between $\cot x$ and $\sin x,$ so $P$ and $S$ sit between $Q$ and $R$ and the parallel sides are $QR$ and $PS.$ Equal slopes give $\cot x + \sin x = \tan x + \cos x.$ Multiplying by $\sin x\cos x$ and simplifying yields $\cos x + \sin x - \sin x\cos x = 0.$ Squaring and using $2\sin x\cos x = \sin 2x$ gives $1 + \sin 2x = \tfrac14\sin^2 2x,$ whose only root in $(-1, 1)$ is $\sin 2x = 2 - 2\sqrt2.$ Thus, the correct answer is A.

2013 AMC 12B Problem 20

Solution:

Problem 20 in Other Years