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> 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.
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<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> integer values.
  
 
Note by Williamgolly:
 
Note by Williamgolly:

Revision as of 11:55, 18 July 2021

Problem

For how many integers $x$ is the point $(x, -x)$ inside or on the circle of radius $10$ centered at $(5, 5)$?

$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15$

Solution

The equation of the circle is $(x-5)^2+(y-5)^2=100$. Plugging in the given conditions we have $(x-5)^2+(-x-5)^2 \leq 100$. Expanding gives: $x^2-10x+25+x^2+10x+25\leq 100$, which simplifies to $x^2\leq 25$ and therefore $x\leq 5$ and $x\geq -5$. So $x$ ranges from $-5$ to $5$, for a total of $\boxed{\mathbf{(A)}\ 11}$ integer values.

Note by Williamgolly: Alternatively, draw out the circle and see that these points must be on the line $y=-x$.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
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

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