1996 AHSME Problems/Problem 27: Difference between revisions
| Line 5: | Line 5: | ||
<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 | ||
| Line 32: | Line 32: | ||
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
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
![\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]](http://latex.artofproblemsolving.com/0/5/d/05d38c791a9a4ff1eed17200cb427f9367e0230e.png)
![\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]](http://latex.artofproblemsolving.com/5/4/b/54b478de5cf20da9b1d5cc509351834c2c69d668.png)
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:
![\[x^2 + y^2 \le \frac{23}{4}\]](http://latex.artofproblemsolving.com/c/6/a/c6acb0118bd08e6796e64602521f8d40b48a41e3.png)
![\[x^2 + y^2 \le \frac{17}{4}\]](http://latex.artofproblemsolving.com/e/6/4/e64df4289bec71e1c500aa3408416be49f57a41d.png)
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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