Difference between revisions of "2015 AMC 10B Problems/Problem 12"
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The equation of the circle is <math>(x-5)^2+(y-5)^2=100</math>. Plugging in the given conditions we have <math>(x-5)^2+(-x-5)^2 \leq 100</math>. Expanding gives: <math>x^2-10x+25+x^2+10x+25\leq 100</math>, which simplifies to | The equation of the circle is <math>(x-5)^2+(y-5)^2=100</math>. Plugging in the given conditions we have <math>(x-5)^2+(-x-5)^2 \leq 100</math>. Expanding gives: <math>x^2-10x+25+x^2+10x+25\leq 100</math>, which simplifies to | ||
<math>x^2\leq 25</math> and therefore | <math>x^2\leq 25</math> and therefore | ||
| − | <math>x\leq 5</math> | + | <math>x\leq 5</math> and <math>x\geq -5</math>. So <math>x</math> ranges from <math>-5</math> to <math>5</math>, for a total of <math>\boxed{\mathbf{(A)}\ 11}</math> values. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2015|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:42, 4 February 2019
Problem
For how many integers 
is the point 
inside or on the circle of radius 10 centered at 
?
Solution
The equation of the circle is 
. Plugging in the given conditions we have 
. Expanding gives: 
, which simplifies to
and therefore
and 
. So ranges from 
to 
, for a total of 
values.
See Also
| 2015 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
