1996 AMC 12 Problem 27

Attempt Problem 27 of the 1996 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 12 solutions, or check the answer key.

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27.

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

Answer: D
Concepts:lattice pointsspheresinequalities
Difficulty rating: 2100
Small Hint:

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

Big Hint:

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

Solution:

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+y236(5212)2=234,x2+y2814(51)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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