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 2-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 $6$ , 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 $(9/2,0)$ .

Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent dome shapes on either end. Because the tops of the domes are at $(0,0,9/2)$ and $(0,0,11/2)$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $9/2$ and $11/2$ ). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.

Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles.

We know all three side lengths of this triangle: $9/2$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $19/2$ (the distance between the centers of circles $A$ and $B$ ). Using Heron's formula:

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