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

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<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>
  
==Video Solution 1==
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==Video Solution==
 
https://youtu.be/RZDFs3qrw7Y
 
https://youtu.be/RZDFs3qrw7Y
  

Revision as of 19:41, 25 August 2023

Problem

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

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

Video Solution

https://youtu.be/RZDFs3qrw7Y

~Education, the Study of Everything=

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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