Difference between revisions of "2016 AMC 10B Problems/Problem 20"
(→Solution 2 added, by @adihaya) |
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==Solution 2: Bashing== | ==Solution 2: Bashing== | ||
− | + | <asy> /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | |
/* by adihaya */ | /* by adihaya */ | ||
import graph; size(13cm); | import graph; size(13cm); | ||
Line 33: | Line 33: | ||
/* dots and labels */ | /* dots and labels */ | ||
dot(O,blue); | dot(O,blue); | ||
− | label(" | + | label("$O$", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); |
dot(A,blue); | dot(A,blue); | ||
− | label(" | + | label("$A$", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); |
dot((5.,6.),blue); | dot((5.,6.),blue); | ||
− | label(" | + | label("$A'$", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); |
dot(B,xdxdff); | dot(B,xdxdff); | ||
− | label(" | + | label("$B$", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); |
− | label(" | + | label("$c$", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); |
dot(C,xdxdff); | dot(C,xdxdff); | ||
− | label(" | + | label("$C$", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); |
− | label(" | + | label("$d$", (3.477574142815031,8.107752774436745), NE * labelscalefactor); |
− | label(" | + | label("$a$", (7.255026033677397,9.404829628528034), NE * labelscalefactor); |
− | label(" | + | label("$b$", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); |
− | label(" | + | label("$e$", (4.615360856930201,9.404829628528034), NE * labelscalefactor); |
dot(D,linewidth(3.pt) + uuuuuu); | dot(D,linewidth(3.pt) + uuuuuu); | ||
/* Solution by adihaya */ | /* Solution by adihaya */ | ||
− | label(" | + | label("$D$", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); |
dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); | dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); | ||
− | label(" | + | label("$E$", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); |
− | label(" | + | label("$f$", (4.933941136882449,9.404829628528034), NE * labelscalefactor); |
dot(F,linewidth(3.pt) + uuuuuu); | dot(F,linewidth(3.pt) + uuuuuu); | ||
− | label(" | + | label("$\Large{(-4,-6)}$", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); |
− | label(" | + | label("$\Large{2\sqrt{13}}$", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); |
dot(G,linewidth(3.pt) + uuuuuu); | dot(G,linewidth(3.pt) + uuuuuu); | ||
− | label(" | + | label("$G$", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); |
− | label(" | + | label("$\Large{10}$", (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq); |
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
/* re-scale y/x */ | /* re-scale y/x */ | ||
currentpicture = yscale(0.9090909090909091) * currentpicture; | currentpicture = yscale(0.9090909090909091) * currentpicture; | ||
− | /* end of picture */ | + | /* end of picture */</asy> |
Using analytic geometry, we find that the center of dilation is at <math>(-4,-6)</math> and the coefficient/factor is <math>1.5</math>. Then, we see that the origin is <math>2\sqrt{13}</math> from the center, and will be <math>1.5 \times 2\sqrt{13} = 3\sqrt{13}</math> from it afterwards. | Using analytic geometry, we find that the center of dilation is at <math>(-4,-6)</math> and the coefficient/factor is <math>1.5</math>. Then, we see that the origin is <math>2\sqrt{13}</math> from the center, and will be <math>1.5 \times 2\sqrt{13} = 3\sqrt{13}</math> from it afterwards. | ||
Revision as of 15:04, 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 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: Bashing
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.