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> | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | 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 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
The two equations of the balls are
Note that along the axis, the first ball goes from , and the seocnd 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 $y \in {-2, -1, 0, 1, 2\}$ (Error compiling LaTeX. ! Missing } inserted.). This gives points.
Thus, there are possible points, giving answer .
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 |