Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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==Solution 3: Logic and Geometry== | ==Solution 3: Logic and Geometry== | ||
− | Using the ratios of the circles | + | Using the ratios of the circles <math>\frac{3}{2}</math>, we find that the scale factor is <math>1.5</math>. If the origin had not moved, this indicates that the center of the circle would be <math>(3,3)</math>, simply because of <math>(2 \cdot 1.5, 2 \cdot 1.5)</math>. 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 08:54, 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
Using the ratios of the circles , we find that the scale factor is . If the origin had not moved, this indicates that the center of the circle would be , simply because of . 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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