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

(Solution 1: Algebraic)
(Solution 1: Algebraic)
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==Solutions==
 
==Solutions==
 
===Solution 1: Algebraic===
 
===Solution 1: Algebraic===
The center of dilation must lie on the line <math>A A'</math>, which can be expressed as <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 as <math>y = \dfrac{4x}{3} - \dfrac{2}{3}</math>. Note that the center of dilation must have an <math>x</math>-coordinate less than <math>2</math>: if the <math>x</math>-coordinate were otherwise, than the circle under the transformation would not have an increased <math>x</math>-coordinate in the coordinate plane. 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>. We do not have to include absolute value symbols because we know that the center of dilation has a lower <math>x</math>-coordinate, and hence a lower <math>y</math>-coordinate, from our reasoning above. Solving the two equation, 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: Geometric===
 
===Solution 2: Geometric===

Revision as of 07:07, 28 October 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?

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

Solutions

Solution 1: Algebraic

The center of dilation must lie on the line $A A'$, which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$. Note that the center of dilation must have an $x$-coordinate less than $2$: if the $x$-coordinate were otherwise, than the circle under the transformation would not have an increased $x$-coordinate in the coordinate plane. 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 $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$. We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$-coordinate, and hence a lower $y$-coordinate, from our reasoning above. Solving the two equation, 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 (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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