HMMO

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:
A sphere is centered at a point with integer coordinates and passes through the three points $(2,0,0)$ , $(0,4,0)$ , $(0,0,6)$ , but not the origin $(0,0,0)$ . If $r$ is the smallest possible radius of the sphere, compute $r^{2}$ .

Solution

Solution: Let $(x, y, z)$ be the center of the sphere. By the given condition, we have
$(x-2)^{2} + y^{2} + z^{2} = x^{2} + (y-4)^{2} + z^{2} = x^{2} + y^{2} + (z-6)^{2}.$
Subtracting $x^{2} + y^{2} + z^{2}$ yields
$x^{2} - (x-2)^{2} = y^{2} - (y-4)^{2} = z^{2} - (z-6)^{2},$
or
$4(x-1) = 8(y-2) = 12(z-3)$
Therefore $(x-1, y-2, z-3)$ must be $(6t, 3t, 2t)$ for some integer $t$ . Checking small values of $t$ yields that the possibilities for $(x, y, z)$ closest to $(2,0,0)$ are $(-5,-1,1)$ , $(1,2,3)$ , and $(7,5,5)$ . The second yields a sphere that passes through the origin and is thus forbidden. The other two yield $r^{2} = 51$ and $r^{2} = 75$ , so 51 is the answer.

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