1996 AMC 12 Problem 27

Question

Consider two solid spherical balls, one centered at

(0,0,\frac{21}{2})

with radius

6,

and the other centered at

(0,0,1)

with radius

\frac92.

How many points

(x,y,z)

with only integer coordinates (lattice points) are there in the intersection of the balls?

7

9

11

13

15

First intersect the possible integer ranges for the

z

-coordinate in the two balls

At the only possible height, reduce both sphere inequalities to a bound on

x^2+y^2

The first ball permits integer heights

5

through

16,

while the second permits

-3

through

5,

so an intersection lattice point must have

z=5.

At that height the two bounds are

\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

x^2+y^2

can be

0,1,2,

or

4.

These give

1,4,4,

and

4

ordered integer pairs, respectively, for

13

points. The correct answer is D.

Problem 27 in other years