Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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==Solution 3: Logic and Geometry== | ==Solution 3: Logic and Geometry== | ||
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+ | By using simple geometry, we find that the scale factor is 1.5. If the origin had not moved, this indicates that the center of the circle would be <math>(3,3)</math>, simply because of (2 <math>\cdot 1.5, (2 </math>\cdot 1.5). Since the origin has moved from <math>(3,3)</math> to <math>(5,6)</math>, we apply the distance formula and get: <math>\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{13}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2016|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 08:50, 13 October 2017
Contents
[hide]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 1: Algebraic
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.
Solution 2: Geometric
Using analytic geometry, we find that the center of dilation is at and the coefficient/factor is . Then, we see that the origin is from the center, and will be from it afterwards.
Thus, it will move .
Solution 3: Logic and Geometry
By using simple geometry, we find that the scale factor is 1.5. If the origin had not moved, this indicates that the center of the circle would be , simply because of (2 \cdot 1.5). Since the origin has moved from to , we apply the distance formula and get:
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 |
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