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== | + | ==Solution 1== |
The two equations of the balls are | The two equations of the balls are | ||
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Thus, there are <math>\boxed{13}</math> possible points, giving answer <math>\boxed{D}</math>. | Thus, there are <math>\boxed{13}</math> possible points, giving answer <math>\boxed{D}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=26|num-a=28}} | {{AHSME box|year=1996|num-b=26|num-a=28}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:56, 30 December 2018
Contents
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 1
The two equations of the balls are
Note that along the axis, the first ball goes from , and the second 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 . This gives points.
Thus, there are possible points, giving answer .
Solution 2
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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