Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2015 AMC 10B Problems/Problem 12"
m ((I noticed the video link led to question 11, not 12, so I matched the link. ))
 
(8 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==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 <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>
 +
 +
==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 <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
 
<math>x^2\leq 25</math> and therefore
 
<math>x^2\leq 25</math> 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.
+
<math>x\leq 5</math> and <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> integer values.
 +
 
 +
Note by Williamgolly:
 +
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}}

Latest revision as of 22:53, 21 October 2024

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 $y=-x$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_12&oldid=230314"