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2018 AMC 12A 第 22 题

先试着解答 2018 AMC 12A 第 22 题，然后核对你的答案与精心整理的解答，解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2018 AMC 12A 解答，或核对答案。

所有题目均经美国数学协会（MAA）官方合法授权使用。

22.

方程 z2=4+415 iz^2 = 4 + 4\sqrt{15}\,i 和 z2=2+23 iz^2 = 2 + 2\sqrt{3}\,i 的解（其中 i=−1i = \sqrt{-1}）在复平面中构成一个平行四边形的顶点。这个平行四边形的面积可写成 pq−rsp\sqrt{q} - r\sqrt{s}，其中 pp、qq、rr 和 ss 是正整数，且 qq 和 ss 都不被任何质数的平方整除。p+q+r+sp + q + r + s 是多少？

The solutions to the equations z2=4+415 iz^2 = 4 + 4\sqrt{15}\,i and z2=2+23 i,z^2 = 2 + 2\sqrt{3}\,i, where i=−1,i = \sqrt{-1}, form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form pq−rs,p\sqrt{q} - r\sqrt{s}, where p,p, q,q, r,r, and ss are positive integers and neither qq nor ss is divisible by the square of any prime number. What is p+q+r+s?p + q + r + s?

2020

2121

2222

2323

2424

答案：A
知识点：复数鞋带公式
难度评级：2270
小提示：

写 z=a+biz = a + bi，通过比较实部和虚部解 z2=4+415 iz^2 = 4 + 4\sqrt{15}\,i；另一个方程也同样处理

Solve z2=4+415 iz^2 = 4 + 4\sqrt{15}\,i by writing z=a+biz = a + bi and matching real and imaginary parts; do the same for the other equation

大提示：

四个顶点是 ±z1\pm z_1 和 ±z2\pm z_2；对它们的坐标使用鞋带公式

The four vertices are ±z1\pm z_1 and ±z2;\pm z_2; apply the shoelace formula to their coordinates

解答：

写 z=a+biz = a + bi，由 (a+bi)2=4+415 i(a + bi)^2 = 4 + 4\sqrt{15}\,i 得 a2−b2=4a^2 - b^2 = 4 和 2ab=4152ab = 4\sqrt{15}。因此 a4−4a2−60=0a^4 - 4a^2 - 60 = 0，即 (a2−10)(a2+6)=0(a^2 - 10)(a^2 + 6) = 0，得 a=±10a = \pm\sqrt{10}，b=±6b = \pm\sqrt{6}。第一个方程给出的顶点为 ±(10+6 i)\pm(\sqrt{10} + \sqrt{6}\,i)。对 z2=2+23 iz^2 = 2 + 2\sqrt{3}\,i 用同样方法得到 ±(3+i)\pm(\sqrt{3} + i)。

对 (10,6)(\sqrt{10}, \sqrt{6})，(3,1)(\sqrt{3}, 1)，(−10,−6)(-\sqrt{10}, -\sqrt{6})，(−3,−1)(-\sqrt{3}, -1) 使用鞋带公式，面积为 62−2106\sqrt{2} - 2\sqrt{10}。因此 p+q+r+sp + q + r + s =6+2+2+10= 6 + 2 + 2 + 10 =20= 20。

所以正确答案是 A。

Writing z=a+biz = a + bi with (a+bi)2=4+415 i(a + bi)^2 = 4 + 4\sqrt{15}\,i gives a2−b2=4a^2 - b^2 = 4 and 2ab=415.2ab = 4\sqrt{15}. Then a4−4a2−60=0,a^4 - 4a^2 - 60 = 0, so (a2−10)(a2+6)=0,(a^2 - 10)(a^2 + 6) = 0, yielding a=±10,a = \pm\sqrt{10}, b=±6.b = \pm\sqrt{6}. The vertices from the first equation are ±(10+6 i).\pm(\sqrt{10} + \sqrt{6}\,i). The same method on z2=2+23 iz^2 = 2 + 2\sqrt{3}\,i gives ±(3+i).\pm(\sqrt{3} + i).

Applying the shoelace formula to (10,6),(\sqrt{10}, \sqrt{6}), (3,1),(\sqrt{3}, 1), (−10,−6),(-\sqrt{10}, -\sqrt{6}), (−3,−1)(-\sqrt{3}, -1) gives area 62−210.6\sqrt{2} - 2\sqrt{10}. Thus p+q+r+sp + q + r + s =6+2+2+10= 6 + 2 + 2 + 10 =20.= 20.

Thus, the correct answer is A.

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