1996 AHSME Problems/Problem 27

Problem

Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ 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 1

The two equations of the balls are

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

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

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the second 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 $\boxed{D}$ .

Solution 2

Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a two-dimensional plane (the previous z-axis is the new x-axis and the y-axis remains the same).

The spheres now become circles with centers at (1,0) and (21/2,0). They have radii 9/2 and 4, respectively. Let circle A be the circle centered on (1,0) and circle B be the one centered on (21/2,0).

The point on circle A closest to the center of circle B is (11/2,0). The point on circle B closest to the center of circle A is (13/2,0).

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1996 AHSME Problems/Problem 27

Contents

Problem

Solution 1

Solution 2

See also