# 1996 AHSME Problems/Problem 27

## Problem

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 $\frac{9}{2}$. How many points with only integer coordinates (lattice points) are there in the intersection of the balls?

$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

## Solution

The two equations of the balls are

$$x^2 + y^2 + (z - \frac{21}{2})^2 \le 36$$

$$x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}$$

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$, and the seocnd ball goes from $1 \pm 4.5$. The only integer value that $z$ can be is $z=5$.

Plugging that in to both equations, we get:

$$x^2 + y^2 \le \frac{23}{4}$$

$$x^2 + y^2 \le \frac{17}{4}$$

The second inequality implies the first inequality, so the only condition that matters is the second inequality.

From here, we do casework, noting that $|x|, |y| \le 3$:

For $x=\pm 2$, we must have $y=0$. This gives $2$ points.

For $x = \pm 1$, we can have $y\in \{-1, 0, 1\}$. This gives $2\cdot 3 = 6$ points.

For $x = 0$, we can have $y \in \{-2, -1, 0, 1, 2\}$. This gives $5$ points.

Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{D}$.