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AMC数学竞赛真题2016年B 21

2018-12-05 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10数学竞赛真题以及官方解答吧：

Problem 21

What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$

$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$

Solution 1

WLOG note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that $x, y \ge 0$ , which implies that $|x|=x$ and $|y|=y$ , and multiply by $4$ at the end.

We can rearrange the equation to get $x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2$ , which describes a circle with center $(\tfrac12, \tfrac12)$ and radius $\tfrac{\sqrt2}{2}.$ It's clear we now want to find the union of four circles with overlap.

$AMC数学竞赛$

There are several ways to find the area, but note that if you connect $(0, 1)$ to its other three respective points in the other three quadrants, you get a square of area $2$ , along with four half-circles of diameter $\sqrt{2}$ , for a total area of $2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2$ which is $\boxed{\textbf{(B)}}$ .

Solution 2

Another way to solve this problem is using cases. Though this may seem tedious, we only have to do one case. The equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible, we can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive, we get the equation $x^{2} +y^{2} -x-y = 0.$

Using the complete the square method, we get $\left(x-\frac{1}{2}\right)^{2} + \left(y-\frac{1}{2}\right)^{2}=\frac{1}{2}$

Therefore, the origin of this section of the shape is at $\left(\frac{1}{2}, \frac{1}{2}\right).$

Using the equation we can also see that the radius has a length of $\frac{\sqrt{2}}{2}$ .

With this shape we see that this shape can be cut into a right triangle and a semicircle. The length of the hypotenuse of the triangle is $\sqrt{2}$ so using special right triangles, we see that the area of the triangle is $\frac{1}{2}$ . The semicircle has the area of $\frac{1}{4}\pi$ .