2016 AMC 10B Problems/Problem 20

Problem

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1: Algebraic

The center of dilation must lie on the line $A A'$ , which can be expressed $y = \dfrac{4x}{3} - \dfrac{2}{3}$ . Also, the ratio of dilation must be equal to $\dfrac{3}{2}$ , which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\dfrac{3}{2} \left( 2 - x \right) = 5 - x$ (for the $x$ -coordinates), and $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$ . Solving these, we get $x = -4$ and $y = - 6$ . This means that any point $(a,b)$ on the plane will dilate to the point $\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)$ , which means that the point $(0,0)$ dilates to $\left( 6 - 4, 9 - 6 \right) = (2,3)$ . Thus, the origin moves $\sqrt{2^2 + 3^2} = \boxed{\sqrt{13}}$ units.

Solution 2: Geometric

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ /* by adihaya */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7., xmax = 9., ymin = -7., ymax = 9.6; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O = (0.,0.), A = (2.,2.), B = (2.,0.), C = (3.209655293155585,3.5927128026548525), D = (2.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175); /* by adihaya */ draw((2.482722020656878,0.)--(2.4827220206568783,0.4827220206568779)--(2.,0.48272202065687797)--B--cycle, qqwuqq); draw((5.482722020656878,0.)--(5.482722020656878,0.48272202065687797)--(5.,0.48272202065687797)--(5.,0.)--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(shift(A) * scale(2., 2.)*unitcircle); draw(shift((5.,6.)) * scale(3.000060969351735, 3.000060969351735)*unitcircle); draw((xmin, 1.3333333333333333*xmin-0.6666666666666666)--(xmax, 1.3333333333333333*xmax-0.6666666666666666)); /* line */ draw((2.,ymin)--(2.,ymax)); /* line */ draw((5.,ymin)--(5.,ymax)); /* line */ draw((xmin, 1.6666869897839114*xmin + 0.6666260204321771)--(xmax, 1.6666869897839114*xmax + 0.6666260204321771)); /* line */ draw(O--F, qqzzff); draw(F--A, ffwwqq); /* dots and labels */ dot(O,blue); label("$O$", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); dot(A,blue); label("$A$", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); dot((5.,6.),blue); label("$A'$", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); dot(B,xdxdff); label("$B$", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); label("$c$", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); dot(C,xdxdff); label("$C$", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); label("$d$", (3.477574142815031,8.107752774436745), NE * labelscalefactor); label("$a$", (7.255026033677397,9.404829628528034), NE * labelscalefactor); label("$b$", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); label("$e$", (4.615360856930201,9.404829628528034), NE * labelscalefactor); dot(D,linewidth(3.pt) + uuuuuu); /* Solution by adihaya */ label("$D$", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); label("$E$", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); label("$f$", (4.933941136882449,9.404829628528034), NE * labelscalefactor); dot(F,linewidth(3.pt) + uuuuuu); label("$\Large{(-4,-6)}$", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); label("$\Large{2\sqrt{13}}$", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); dot(G,linewidth(3.pt) + uuuuuu); label("$G$", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); label("$\Large{10}$", (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(0.9090909090909091) * currentpicture; /* end of picture */[/asy]$ Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$ . Then, we see that the origin is $2\sqrt{13}$ from the center, and will be $1.5 \times 2\sqrt{13} = 3\sqrt{13}$ from it afterwards.

Thus, it will move $3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}$ .

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 $(3,3)$ , simply because of (2 $\cdot 1.5, (2$ \cdot 1.5). Since the origin has moved from $(3,3)$ to $(5,6)$ , we apply the distance formula and get: $\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{13}$

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2016 AMC 10B Problems/Problem 20

Contents

Problem

Solution 1: Algebraic

Solution 2: Geometric

Solution 3: Logic and Geometry

See Also