Difference between revisions of "1996 AHSME Problems/Problem 27"

Revision as of 13:10, 5 July 2013

Problem

Consider two solid spherical balls, one centered at $(0, 0,\frac{21}{2})$ 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

The two equations of the balls are

$x^2 + y^2 + (z - \frac{21}{2})^2 \le 36$

$x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}$

Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the seocnd 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 $\boxed{D}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

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 ==See also==
 {{AHSME box|year=1996|num-b=26|num-a=28}}
+{{MAA Notice}}

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Difference between revisions of "1996 AHSME Problems/Problem 27"

Revision as of 13:10, 5 July 2013

Problem

Solution

See also