2015 AMC 10B Problem 12

Attempt Problem 12 of the 2015 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

For how many integers xx is the point (x,−x)(x, -x) inside or on the circle of radius 1010 centered at (5,5)?(5, 5)?
11 11
12 12
13 13
14 14
15 15
Answer: A
Concepts:coordinate geometrycircleinequality
Small hint:
Substitute (x,−x)(x,-x) into the distance formula from (5,5)(5,5)
Big hint:
Solve 2x2+50≤1002x^2+50\le100
Solution:
The squared distance from (x,−x)(x,-x) to (5,5)(5,5) is (x−5)2+(−x−5)2=2x2+50. \begin{aligned} &(x-5)^2+(-x-5)^2 \\ &= 2x^2+50. \end{aligned}
Being inside or on the circle means 2x2+50≤1002x^2+50\le100, so x2≤25x^2\le25. Thus −5≤x≤5-5\le x\le5, giving 1111 integer values.
Thus, the correct answer is A.
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