Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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<math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | <math>\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> | ||
− | ==Solution== | + | ==Solution 1: Algebraic== |
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== | ||
+ | [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("<math>O</math>", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); | ||
+ | dot(A,blue); | ||
+ | label("<math>A</math>", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); | ||
+ | dot((5.,6.),blue); | ||
+ | label("<math>A'</math>", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); | ||
+ | dot(B,xdxdff); | ||
+ | label("<math>B</math>", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); | ||
+ | label("<math>c</math>", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); | ||
+ | dot(C,xdxdff); | ||
+ | label("<math>C</math>", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); | ||
+ | label("<math>d</math>", (3.477574142815031,8.107752774436745), NE * labelscalefactor); | ||
+ | label("<math>a</math>", (7.255026033677397,9.404829628528034), NE * labelscalefactor); | ||
+ | label("<math>b</math>", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); | ||
+ | label("<math>e</math>", (4.615360856930201,9.404829628528034), NE * labelscalefactor); | ||
+ | dot(D,linewidth(3.pt) + uuuuuu); | ||
+ | /* Solution by adihaya */ | ||
+ | label("<math>D</math>", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); | ||
+ | dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); | ||
+ | label("<math>E</math>", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); | ||
+ | label("<math>f</math>", (4.933941136882449,9.404829628528034), NE * labelscalefactor); | ||
+ | dot(F,linewidth(3.pt) + uuuuuu); | ||
+ | label("<math>\Large{(-4,-6)}</math>", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); | ||
+ | label("<math>\Large{2\sqrt{13}}</math>", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); | ||
+ | dot(G,linewidth(3.pt) + uuuuuu); | ||
+ | label("<math>G</math>", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); | ||
+ | label("<math>\Large{10}</math>", (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 <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. | ||
+ | |||
+ | Thus, it will move <math>3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}</math>. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 15:04, 21 February 2016 (EST) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2016|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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
[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("", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); dot(A,blue); label("", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); dot((5.,6.),blue); label("", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); dot(B,xdxdff); label("", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); label("", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); dot(C,xdxdff); label("", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); label("", (3.477574142815031,8.107752774436745), NE * labelscalefactor); label("", (7.255026033677397,9.404829628528034), NE * labelscalefactor); label("", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); label("", (4.615360856930201,9.404829628528034), NE * labelscalefactor); dot(D,linewidth(3.pt) + uuuuuu); /* Solution by adihaya */ label("", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); label("", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); label("", (4.933941136882449,9.404829628528034), NE * labelscalefactor); dot(F,linewidth(3.pt) + uuuuuu); label("", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); label("", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); dot(G,linewidth(3.pt) + uuuuuu); label("", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); label("", (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 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.