1996 AHSME Problems/Problem 27

Revision as of 21:06, 30 December 2018 by Imbadatmath23 (talk | contribs) (Solution 2)

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 $(13/2,0)$.

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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