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Difference between revisions of "2016 AMC 10B Problems/Problem 20"

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circle would be <math>(3,3)</math>, simply because of <math>(2 \cdot 1.5,  2 \cdot 1.5)</math>. Since the center 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>.
 
circle would be <math>(3,3)</math>, simply because of <math>(2 \cdot 1.5,  2 \cdot 1.5)</math>. Since the center 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>.
  
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==Solution 4: Simple and Practical==
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===Solution 4: Simple and Practical===
  
 
Start with the size transformation. Transforming the circle from radius 2 to radius 3 would mean the origin point now transforms into the point (-1,-1). Now apply the position shift; 3 to the right and 4 up which gets you the pt (2,3). Now simply do Pythagorean theorem with pts (0,0) and (2,3) to find the distance traveled.
 
Start with the size transformation. Transforming the circle from radius 2 to radius 3 would mean the origin point now transforms into the point (-1,-1). Now apply the position shift; 3 to the right and 4 up which gets you the pt (2,3). Now simply do Pythagorean theorem with pts (0,0) and (2,3) to find the distance traveled.

Revision as of 18:30, 20 January 2019

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(0.49019607843137253,50.49019607843137253,1.); pen uuuuuu = rgb(0.666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O = (3.,0.), A = (6.,2.), B = (2.,1.), C = (4.203155585,5.592712848525), D = (5.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175); /* by adihaya */ draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq); draw((5.482722020656878,0.)--(7.4827220878,1.48277797)--(5.,0.48272687797)--(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

Using the ratios of radii of the circles, $\frac{3}{2}$, 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 center has moved from $(3,3)$ to $(5,6)$, we apply the distance formula and get: $\sqrt{(6-3)^2 + (5-3)^2} = \sqrt{13}$.

Solution 4: Simple and Practical

Start with the size transformation. Transforming the circle from radius 2 to radius 3 would mean the origin point now transforms into the point (-1,-1). Now apply the position shift; 3 to the right and 4 up which gets you the pt (2,3). Now simply do Pythagorean theorem with pts (0,0) and (2,3) to find the distance traveled.

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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