Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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==Problem== | ==Problem== | ||
| − | A | + | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius <math>2</math> centered at <math>A(2,2)</math> to the circle of radius <math>3</math> centered at <math>A’(5,6)</math>. What distance does the origin <math>O(0,0)</math>, move under this transformation? |
<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 16:02, 21 February 2016
Problem
A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius 
centered at 
to the circle of radius 
centered at 
. What distance does the origin 
, move under this transformation?
Solution
The center of dilation must lie on the line 
, which can be expressed 
. Also, the ratio of dilation must be equal to 
, which is the ratio of the radii of the circles. Thus, we are looking for a point 
such that 
(for the 
-coordinates), and 
. Solving these, we get 
and 
. This means that any point 
on the plane will dilate to the point 
, which means that the point 
dilates to 
. Thus, the origin moves 
units.
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
