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: Bashing==
+
==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 $2$ 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 $3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}$. — @adihaya (talk) 15:04, 21 February 2016 (EST)

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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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