Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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The center of dilation must lie on the line <math>A A'</math>, which can be expressed <math>y = \dfrac{4x}{3} - \dfrac{2}{3}</math>. Also, the ratio of dilation must be equal to <math>\dfrac{3}{2}</math>, which is the ratio of the radii of the circles. Thus, we are looking for a point <math>(x,y)</math> such that <math>\dfrac{3}{2} \left( 2 - x \right) = 5 - x</math> (for the <math>x</math>-coordinates), and <math>\dfrac{3}{2} \left( 2 - y \right) = 6 - y</math>. Solving these, we get <math>x = -4</math> and <math>y = - 6</math>. This means that any point <math>(a,b)</math> on the plane will dilate to the point <math>\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)</math>, which means that the point <math>(0,0)</math> dilates to <math>\left( 6 - 4, 9 - 6 \right) = (2,3)</math>. Thus, the origin moves <math>\sqrt{2^2 + 3^2} = \boxed{\sqrt{13}}</math> units. | The center of dilation must lie on the line <math>A A'</math>, which can be expressed <math>y = \dfrac{4x}{3} - \dfrac{2}{3}</math>. Also, the ratio of dilation must be equal to <math>\dfrac{3}{2}</math>, which is the ratio of the radii of the circles. Thus, we are looking for a point <math>(x,y)</math> such that <math>\dfrac{3}{2} \left( 2 - x \right) = 5 - x</math> (for the <math>x</math>-coordinates), and <math>\dfrac{3}{2} \left( 2 - y \right) = 6 - y</math>. Solving these, we get <math>x = -4</math> and <math>y = - 6</math>. This means that any point <math>(a,b)</math> on the plane will dilate to the point <math>\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)</math>, which means that the point <math>(0,0)</math> dilates to <math>\left( 6 - 4, 9 - 6 \right) = (2,3)</math>. Thus, the origin moves <math>\sqrt{2^2 + 3^2} = \boxed{\sqrt{13}}</math> units. | ||
− | ==Solution 2: | + | ==Solution 2: Geometric== |
<asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | <asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
/* by adihaya */ | /* by adihaya */ |
Revision as of 11:54, 30 October 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 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 . — @adihaya (talk) 15:04, 21 February 2016 (EST)
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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