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AMC数学竞赛真题2016年B 21

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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 21What is the area of the region enclosed by the graph of the equation

Solution 1WLOG 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 , which implies that and , and multiply by at the end.

We can rearrange the equation to get , which describes a circle with center and radius It´s clear we now want to find the union of four circles with overlap.

There are several ways to find the area, but note that if you connect to its other three respective points in the other three quadrants, you get a square of area , along with four half-circles of diameter , for a total area of which is .

Solution 2Another 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 To make this as easy as possible, we can make both and positive. Simplifying the equation for and being positive, we get the equation

Using the complete the square method, we get

Therefore, the origin of this section of the shape is at

Using the equation we can also see that the radius has a length of .

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 so using special right triangles, we see that the area of the triangle is . The semicircle has the area of .

But this is only case. There are cases in total so we have to multiply by .

After multiplying, our answer is:

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