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

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==Problem==
 
==Problem==
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For how many integers <math>x</math> is the point <math>(x, -x)</math> inside or on the circle of radius 10 centered at <math>(5, 5)</math>?
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For how many integers <math>x</math> is the point <math>(x, -x)</math> inside or on the circle of radius <math>10</math> centered at <math>(5, 5)</math>?
  
 
<math>\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15</math>
 
<math>\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15</math>
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Note by Williamgolly:
 
Note by Williamgolly:
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Alternatively, draw out the circle and see that these points must be on the line y=-x
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Alternatively, draw out the circle and see that these points must be on the line <math>y=-x</math>.
  
 
==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 17:43, 13 July 2020

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 $\boxed{\mathbf{(A)}\ 11}$ 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 (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

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