Difference between revisions of "2015 AMC 10B Problems/Problem 12"

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Revision as of 20:55, 4 March 2015

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 $x\leq 5$ or $x\geq -5$. So $x$ ranges from $-5$ to $5$, for a total of $\boxed{\textbf{(A)} 11}$ values.

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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