Difference between revisions of "2015 AMC 10B Problems/Problem 12"
(→Solution) |
Ethanspoon (talk | contribs) m (→Solution=) |
||
Line 9: | Line 9: | ||
~Education, the Study of Everything= | ~Education, the Study of Everything= | ||
− | =Solution | + | =Solution= |
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 |
Revision as of 16:28, 15 August 2022
Contents
[hide]Problem
For how many integers is the point inside or on the circle of radius centered at ?
Video Solution 1
~Education, the Study of Everything=
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 integer values.
Note by Williamgolly: Alternatively, draw out the circle and see that these points must be on the line .
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.