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 $(\frac{21}{2},0)$ . They have radii $\frac{9}{2}$ and $6$ , respectively. Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$ .

The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$ . The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$ .

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 $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$ ). 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. Let's call that altitude $h$ .

We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ). We can now find the area of the triangle using Heron's formula:

$s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20$

$A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}$

Using the area of the triangle, we can find that altitude $h$ from the x-axis:

$\sqrt{110}=\frac{h*\frac{19}{2}}{2}$

$h=\frac{4*\sqrt{110}}{19}$

Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.

Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.

The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$ , a little greater than $2$ . We know that $h>2$ because $4\sqrt{110}>4*10$ , and $\frac{40}{19}>2$ .

Finally, looking at a circle with radius slightly larger than 2, we see that there are 13 lattice points within it.

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

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