AMC12 2013 B

AMC12 2013 B · Q20

AMC12 2013 B · Q20. It mainly tests Functions basics, Trigonometry (basic).

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 $\sin(2x)$?

对于 $135^\circ<x<180^\circ$，点 $P=(\cos x,\cos^2 x)$、$Q=(\cot x,\cot^2 x)$、$R=(\sin x,\sin^2 x)$、$S=(\tan x,\tan^2 x)$ 构成一个梯形的四个顶点。求 $\sin(2x)$。

(A) 2 - 2\sqrt{2} 2 - 2\sqrt{2}

(B) 3\sqrt{3} - 6 3\sqrt{3} - 6

(D) -\frac{3}{4} -\frac{3}{4}

(E) 1 - \sqrt{3} 1 - \sqrt{3}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Because $135^\circ<x<180^\circ$, it follows that $\cos x<0<\sin x$ and $|\sin x|<|\cos x|$. Thus $\tan x<0$, $\cot x<0$, and $$|\tan x|=\frac{|\sin x|}{|\cos x|}<1<\frac{|\cos x|}{|\sin x|}=|\cot x|.$$ Therefore $\cot x<\tan x$. Moreover, $\cot x=\frac{\cos x}{\sin x}<\cos x$. Thus the four vertices $P,Q,R,$ and $S$ are located on the parabola $y=x^2$ and $P$ and $S$ are in between $Q$ and $R$. If $AB$ and $CD$ are chords on the parabola $y=x^2$ such that the $x$-coordinates of $A$ and $B$ are less than the $x$-coordinates of $C$ and $D$, then the slope of $AB$ is less than the slope of $CD$. It follows that the two parallel sides of the trapezoid must be $\overline{QR}$ and $\overline{PS}$. Thus the slope of $\overline{QR}$ is equal to the slope of $\overline{PS}$. Thus, $$\cot x+\sin x=\tan x+\cos x.$$ Multiplying by $\sin x\cos x\ne 0$ and rearranging gives the equivalent identity $$(\cos x-\sin x)(\cos x+\sin x-\sin x\cos x)=0.$$ Because $\cos x-\sin x\ne 0$ in the required range, it follows that $\cos x+\sin x-\sin x\cos x=0$. Squaring and using the fact that $2\sin x\cos x=\sin(2x)$ gives $$1+\sin(2x)=\frac14\sin^2(2x).$$ Solving this quadratic equation in the variable $\sin(2x)$ gives $\sin(2x)=2\pm2\sqrt2$. Because $-1<\sin2x<1$, the only solution is $\sin(2x)=2-2\sqrt2$. There is indeed such a trapezoid for $x=180^\circ+\frac12\arcsin(2-2\sqrt2)\approx152.031^\circ$.

答案（A）：因为 $135^\circ<x<180^\circ$，可得 $\cos x<0<\sin x$ 且 $|\sin x|<|\cos x|$。因此 $\tan x<0$，$\cot x<0$，并且 $$|\tan x|=\frac{|\sin x|}{|\cos x|}<1<\frac{|\cos x|}{|\sin x|}=|\cot x|.$$ 所以 $\cot x<\tan x$。另外，$\cot x=\frac{\cos x}{\sin x}<\cos x$。因此四个顶点 $P,Q,R,S$ 都在抛物线 $y=x^2$ 上，且 $P$ 与 $S$ 位于 $Q$ 与 $R$ 之间。若 $AB$ 与 $CD$ 是抛物线 $y=x^2$ 上的弦，并且 $A,B$ 的 $x$ 坐标小于 $C,D$ 的 $x$ 坐标，则弦 $AB$ 的斜率小于弦 $CD$ 的斜率。由此可知该梯形的两条平行边必须是 $\overline{QR}$ 与 $\overline{PS}$。因此 $\overline{QR}$ 的斜率等于 $\overline{PS}$ 的斜率，于是 $$\cot x+\sin x=\tan x+\cos x.$$ 两边同乘 $\sin x\cos x\ne 0$ 并整理，得到等价恒等式 $$(\cos x-\sin x)(\cos x+\sin x-\sin x\cos x)=0.$$ 因为在所需范围内 $\cos x-\sin x\ne 0$，故 $\cos x+\sin x-\sin x\cos x=0$。两边平方并利用 $2\sin x\cos x=\sin(2x)$，得到 $$1+\sin(2x)=\frac14\sin^2(2x).$$ 把 $\sin(2x)$ 视为变量解此二次方程，得 $\sin(2x)=2\pm2\sqrt2$。由于 $-1<\sin2x<1$，唯一可行解为 $\sin(2x)=2-2\sqrt2$。确有这样的梯形，对应 $$x=180^\circ+\frac12\arcsin(2-2\sqrt2)\approx152.031^\circ.$$

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