1996 AHSME Problems/Problem 27

Problem

Consider two solid spherical balls, one centered at with radius , and the other centered at with radius . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?

Solution 1

The two equations of the balls are

Note that along the axis, the first ball goes from , and the second ball goes from . The only integer value that can be is .

Plugging that in to both equations, we get:

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

From here, we do casework, noting that :

For , we must have . This gives points.

For , we can have . This gives points.

For , we can have . This gives points.

Thus, there are possible points, giving answer .

Solution 2 (minimal 3D geometry)

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 while the y-axis remains the same).

The spheres now become circles with centers at and . They have radii and , respectively. Let circle be the circle centered on and circle be the one centered on .

The point on circle closest to the center of circle is . The point on circle B closest to the center of circle is .

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 and , respectively, the lattice points inside the area of intersection must have z-value (because is the only integer between and ). 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 and and third point at the first quadrant intersection of the circles. Let's call that altitude .

A=(1,0);
B=(10.5,0);
C=(5,2.2);
draw(A--B--C--A);
label("$A$",A,W);
<\asy><\center>

<center><asy>
size(8cm);
pair A, B, C, D, E, EE;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
EE = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--EE,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,EE,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,NE);
label("$4$",(A+D)/2,S);
label("$\frac{27}{8}$",(D+EE)/2,S);
label("$\frac{15}{8}$",(E+EE)/2,W);
 (Error making remote request. Unknown error_msg)

We know all three side lengths of this triangle: (the radius of circle ), (the radius of circle ), and (the distance between the centers of circles and ). We can now find the area of the triangle using Heron's formula:

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

Remember, the altitude 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 of the circle in a 2-dimensional plane is , a little greater than . We know that because , and .

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

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.