AMC12 2006 B

AMC12 2006 B · Q24

AMC12 2006 B · Q24. It mainly tests Quadratic equations, Coordinate geometry.

Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \le x \le \frac{\pi}{2}$ and $0 \le y \le \frac{\pi}{2}$. What is the area of the subset of $S$ for which \[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\]

设 $S$ 为坐标平面中所有点 $(x,y)$ 的集合，使得 $0 \le x \le \frac{\pi}{2}$ 且 $0 \le y \le \frac{\pi}{2}$。满足 \[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\] 的 $S$ 的子集面积是多少？

(A) $\frac{\pi^2}{9}$ $\frac{\pi^2}{9}$

(B) $\frac{\pi^2}{8}$ $\frac{\pi^2}{8}$

(D) $\frac{3\pi^2}{16}$ $\frac{3\pi^2}{16}$

(E) $\frac{2\pi^2}{9}$ $\frac{2\pi^2}{9}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

We start out by solving the equality first. \begin{align*} \sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\ \sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\ &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\ \sin x &= \sin (y \pm \frac{\pi}{3}) \end{align*} We end up with three lines that matter: $x = y + \frac\pi3$, $x = y - \frac\pi3$, and $x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y$. We plot these lines below. Note that by testing the point $(\pi/6,\pi/6)$, we can see that we want the area of the pentagon. We can calculate that by calculating the area of the square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.) \begin{align*} A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\ &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\ &= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\text{(C)}\ \frac{\pi^2}{6}} \end{align*}

我们先解等式。 \begin{align*} \sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\ \sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\ &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\ \sin x &= \sin (y \pm \frac{\pi}{3}) \end{align*} 最终得到三条关键直线：$x = y + \frac\pi3$、$x = y - \frac\pi3$、以及 $x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y$。将这些直线画出。注意通过测试点 $(\pi/6,\pi/6)$，可知我们要的是五边形的面积。可以先求正方形面积再减去三个三角形的面积。（也可以通过把左下角的等腰三角形与以其斜边为长边的矩形面积相加来做。） \begin{align*} A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\ &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\ &= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\text{(C)}\ \frac{\pi^2}{6}} \end{align*}

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