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Difference between revisions of "1996 AHSME Problems/Problem 27"
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<math> \text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15 </math>
 
<math> \text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15 </math>
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 +
==Solution==
 +
 +
The two equations of the balls are
 +
 +
<cmath>x^2 + y^2 + (z - \frac{21}{2})^2 \le 36</cmath>
 +
 +
<cmath>x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}</cmath>
 +
 +
Note that along the <math>z</math> axis, the first ball goes from <math>10.5 \pm 6</math>, and the seocnd ball goes from <math>1 \pm 4.5</math>.  The only integer value that <math>z</math> can be is <math>z=5</math>.
 +
 +
Plugging that in to both equations, we get:
 +
 +
<cmath>x^2 + y^2 \le \frac{23}{4}</cmath>
 +
 +
<cmath>x^2 + y^2 \le \frac{17}{4}</cmath>
 +
 +
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
 +
 +
From here, we do casework, noting that <math>|x|, |y| \le 3</math>:
 +
 +
For <math>x=\pm 2</math>, we must have <math>y=0</math>.  This gives <math>2</math> points.
 +
 +
For <math>x = \pm 1</math>, we can have <math>y\in \{-1, 0, 1\}</math>.  This gives <math>2\cdot 3 = 6</math> points.
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 +
For <math>x = 0</math>, we can have <math>y \in {-2, -1, 0, 1, 2\}</math>.  This gives <math>5</math> points.
 +
 +
Thus, there are <math>\boxed{13}</math> possible points, giving answer <math>\boxed{D}</math>.
  
 
==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=26|num-a=28}}
 
{{AHSME box|year=1996|num-b=26|num-a=28}}

Revision as of 17:35, 20 August 2011

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 $y \in {-2, -1, 0, 1, 2\}$ (Error compiling LaTeX. Unknown error_msg). This gives $5$ points.

Thus, there are $\boxed{13}$ possible points, giving answer $\boxed{D}$.

See also

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
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