Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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Before dilation, notice that the two axes are tangent to the circle with center <math>(2,2)</math>. Using this, we can draw new axes tangent to the radius 3 circle with center <math>(5,6)</math>, resulting in a "new origin" that is 3 units left and 3 units down from the center <math>(5,6)</math>, or <math>(2,3)</math>. Using the distance formula, the distance from <math>(0,0)</math> and <math>(2,3)</math> is <math>\boxed{\sqrt{13}}</math>. | Before dilation, notice that the two axes are tangent to the circle with center <math>(2,2)</math>. Using this, we can draw new axes tangent to the radius 3 circle with center <math>(5,6)</math>, resulting in a "new origin" that is 3 units left and 3 units down from the center <math>(5,6)</math>, or <math>(2,3)</math>. Using the distance formula, the distance from <math>(0,0)</math> and <math>(2,3)</math> is <math>\boxed{\sqrt{13}}</math>. | ||
~Mightyeagle | ~Mightyeagle | ||
+ | |||
+ | ==Solution 6: Answers== | ||
==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 16:14, 13 October 2021
Contents
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 as . Note that the center of dilation must have an -coordinate less than ; if the -coordinate were otherwise, then the circle under the transformation would not have an increased -coordinate in the coordinate plane. 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 . We do not have to include absolute value symbols because we know that the center of dilation has a lower -coordinate, and hence a lower -coordinate, from our reasoning above. Solving the two equations, 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 radii 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 center has moved from to , we apply the distance formula and get: .
Solution 4: Simple and Practical
Start with the size transformation. Transforming the circle from to would mean the origin point now transforms into the point . Now apply the position shift: to the right and up. This gets you the point . Now simply apply the Pythagorean theorem with the points and to find the requested distance.
Solution 5: Using the Axes
Before dilation, notice that the two axes are tangent to the circle with center . Using this, we can draw new axes tangent to the radius 3 circle with center , resulting in a "new origin" that is 3 units left and 3 units down from the center , or . Using the distance formula, the distance from and is . ~Mightyeagle
Solution 6: Answers
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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