2015 AMC 10B Problems/Problem 12: Difference between revisions

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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://www.youtube.com/watch?v=BeD8xOvfzE0

~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 $y=-x$ .

Solution(analytical)

Drawing out the diagram, it becomes apparent that all of the successive points are on the line $y=-x$ . Since the lowest point of the circle is ( $5$ , $5-10=-5$ ), and the point most to the left is ( $5-10=-5$ , $5$ ), we see that these $2$ points and the origin are collinear, meaning every point in between the $2$ extreme points of the circle works, inclusive. We have $-5 \le x \le 5$ , so counting them or using $5-(-5)+1$ , we have $\boxed{\mathbf{(A)}\ 11}$ integer values.

-mrcosmic

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

Revision as of 13:39, 11 November 2025 view source Mrcosmic (talk \| contribs) editor 5 edits →Solution ← Older edit		Latest revision as of 13:40, 11 November 2025 view source Mrcosmic (talk \| contribs) editor 5 edits →Solution
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	==Solution==		==Solution(analytical)==
	Drawing out the diagram, it becomes apparent that all of the successive points are on the line <imath>y=-x</imath>. Since the lowest point of the circle is (<imath>5</imath>, <imath>5-10=-5</imath>), and the point most to the left is (<imath>5-10=-5</imath>, <imath>5</imath>), we see that these <imath>2</imath> points and the origin are collinear, meaning every point in between the <imath>2</imath> extreme points of the circle works, inclusive. We have <imath>-5 \le x \le 5</imath>, so counting them or using <imath>5-(-5)+1</imath>, we have <imath>\boxed{\mathbf{(A)}\ 11}</imath> integer values.		Drawing out the diagram, it becomes apparent that all of the successive points are on the line <imath>y=-x</imath>. Since the lowest point of the circle is (<imath>5</imath>, <imath>5-10=-5</imath>), and the point most to the left is (<imath>5-10=-5</imath>, <imath>5</imath>), we see that these <imath>2</imath> points and the origin are collinear, meaning every point in between the <imath>2</imath> extreme points of the circle works, inclusive. We have <imath>-5 \le x \le 5</imath>, so counting them or using <imath>5-(-5)+1</imath>, we have <imath>\boxed{\mathbf{(A)}\ 11}</imath> integer values.

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