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

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==Problem==
==Problem==
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>?
For how many integers <imath>x</imath> is the point <imath>(x, -x)</imath> inside or on the circle of radius <imath>10</imath> centered at <imath>(5, 5)</imath>?


<math>\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15</math>
<imath>\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15</imath>
 
==Video Solution==
https://www.youtube.com/watch?v=BeD8xOvfzE0
 
~Education, the Study of Everything=


==Solution==
==Solution==
The equation of the circle is <math>(x-5)^2+(y-5)^2=100</math>. Plugging in the given conditions we have <math>(x-5)^2+(-x-5)^2 \leq 100</math>. Expanding gives: <math>x^2-10x+25+x^2+10x+25\leq 100</math>, which simplifies to
The equation of the circle is <imath>(x-5)^2+(y-5)^2=100</imath>. Plugging in the given conditions we have <imath>(x-5)^2+(-x-5)^2 \leq 100</imath>. Expanding gives: <imath>x^2-10x+25+x^2+10x+25\leq 100</imath>, which simplifies to
<math>x^2\leq 25</math> and therefore
<imath>x^2\leq 25</imath> and therefore
<math>x\leq 5</math> or <math>x\geq -5</math>. So <math>x</math> ranges from <math>-5</math> to <math>5</math>, for a total of <math>\boxed{\mathbf{(A)}\ 11}</math> values.
<imath>x\leq 5</imath> and <imath>x\geq -5</imath>. So <imath>x</imath> ranges from <imath>-5</imath> to <imath>5</imath>, for a total of <imath>\boxed{\mathbf{(A)}\ 11}</imath> integer values.
 
Note by Williamgolly:
Alternatively, draw out the circle and see that these points must be on the line <imath>y=-x</imath>.
 
 
==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.
 
-mrcosmic


==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}}

Latest revision as of 13:40, 11 November 2025

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

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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