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1996 AMC 12 第 27 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

27.

考虑两个实心球体：一个球心为 (0,0,212)(0,0,\frac{21}{2})，半径为 66；另一个球心为 (0,0,1)(0,0,1)，半径为 92\frac92。两个球体的交集中有多少个坐标全为整数的点 (x,y,z)(x,y,z)（格点）？

Consider two solid spherical balls, one centered at (0,0,212)(0,0,\frac{21}{2}) with radius 6,6, and the other centered at (0,0,1)(0,0,1) with radius 92.\frac92. How many points (x,y,z)(x,y,z) with only integer coordinates (lattice points) are there in the intersection of the balls?

77

99

1111

1313

1515

答案：D
知识点：lattice pointsspheresinequalities
难度评级：2100
小提示：

先取两个球体中 zz 坐标可能整数范围的交集

First intersect the possible integer ranges for the zz-coordinate in the two balls

大提示：

在唯一可能的高度，把两个球面不等式都化为对 x2+y2x^2+y^2 的限制

At the only possible height, reduce both sphere inequalities to a bound on x2+y2x^2+y^2

解答：

第一个球体允许的整数高度为 55 到 1616，第二个允许的整数高度为 −3-3 到 55，所以交集中的格点必须满足 z=5z=5。在这个高度，两个限制为 x2+y2≤36−(5−212)2=234,x2+y2≤814−(5−1)2=174。 \begin{aligned} x^2+y^2 &\le36-\left(5-\frac{21}{2}\right)^2\\ &=\frac{23}{4},\\ x^2+y^2 &\le\frac{81}{4}-(5-1)^2\\ &=\frac{17}{4} \end{aligned}\text{。}因此 x2+y2x^2+y^2 可以是 0,1,20,1,2 或 44。它们分别给出 1,4,41,4,4 和 44 个有序整数对，共有 1313 个点。正确答案是 D。

The first ball permits integer heights 55 through 16,16, while the second permits −3-3 through 5,5, so an intersection lattice point must have z=5.z=5. At that height the two bounds are x2+y2≤36−(5−212)2=234,x2+y2≤814−(5−1)2=174. \begin{aligned} x^2+y^2 &\le36-\left(5-\frac{21}{2}\right)^2\\ &=\frac{23}{4},\\ x^2+y^2 &\le\frac{81}{4}-(5-1)^2\\ &=\frac{17}{4}. \end{aligned} Thus x2+y2x^2+y^2 can be 0,1,2,0,1,2, or 4.4. These give 1,4,4,1,4,4, and 44 ordered integer pairs, respectively, for 1313 points. The correct answer is D.

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