Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solutions)
 
(26 intermediate revisions by 17 users not shown)
Line 5: Line 5:
 
<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>
  
−
==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>. 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, then 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 equations, 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 equations, 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==
 
<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 */
Line 16: Line 15:
 
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 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.);  
 
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);  
+
pair O = (3.,0.), A = (2.,2.), B = (2.,1.), C = (4.203155585,5.592712848525), D = (5.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175);  
 
/* by adihaya */
 
/* by adihaya */
 
draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq);  
 
draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq);  
Line 68: Line 67:
 
Thus, it will move <math>3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}</math>.
 
Thus, it will move <math>3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}</math>.
  
−
===Solution 3: Logic and Geometry===
+
==Solution 3: Logic and Geometry==
  
 
Using the ratios of radii of the circles, <math>\frac{3}{2}</math>, we find that the scale factor is <math>1.5</math>. If the origin had not moved, this indicates that the center of the
 
Using the ratios of radii of the circles, <math>\frac{3}{2}</math>, we find that the scale factor is <math>1.5</math>. If the origin had not moved, this indicates that the center of the
−
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} = \boxed{\sqrt{13}}</math>.
  
−
===Solution 4: Simple and Practical===
+
==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 <math>r=2</math> to <math>r=3</math> would mean the origin point now transforms into the point <math>(-1,-1)</math>. Now apply the position shift: <math>3</math> to the right and <math>4</math> up. This gets you the point <math>(2,3)</math>. Now simply apply the Pythagorean theorem with the points <math>(0,0)</math> and <math>(2,3)</math> to find the requested distance.
 +
 
 +
==Solution 5: Using the Axes==
 +
 
 +
Before dilation, notice that the two axes are tangent to the circle with center <math>(2,2)</math>. Using this, we can draw new axes tangent to the radius 3 circle with center <math>(5,6)</math>, resulting in a "new origin" that is 3 units left and 3 units down from the center <math>(5,6)</math>, or <math>(2,3)</math>. Using the distance formula, the distance from <math>(0,0)</math> and <math>(2,3)</math> is <math>\boxed{\sqrt{13}}</math>.
 +
~Mightyeagle
 +
 
 +
==Solution 6: Geometry==
 +
 
 +
Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be <math>\frac{3}{2}</math>. The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point <math>(-4, -6)</math>. The distance from that point to the origin is <math>2\sqrt{13}</math>, and halving that to get the distance the origin moves gives <math>\boxed{\sqrt{13}}</math>. ~Meeshbot
  
 
==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}}

Latest revision as of 10:57, 28 February 2024

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 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, then 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 equations, 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 = (2.,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: $\sqrt{(6-3)^2 + (5-3)^2} = \boxed{\sqrt{13}}$.

Solution 4: Simple and Practical

Start with the size transformation. Transforming the circle from $r=2$ to $r=3$ would mean the origin point now transforms into the point $(-1,-1)$. Now apply the position shift: $3$ to the right and $4$ up. This gets you the point $(2,3)$. Now simply apply the Pythagorean theorem with the points $(0,0)$ and $(2,3)$ to find the requested distance.

Solution 5: Using the Axes

Before dilation, notice that the two axes are tangent to the circle with center $(2,2)$. Using this, we can draw new axes tangent to the radius 3 circle with center $(5,6)$, resulting in a "new origin" that is 3 units left and 3 units down from the center $(5,6)$, or $(2,3)$. Using the distance formula, the distance from $(0,0)$ and $(2,3)$ is $\boxed{\sqrt{13}}$. ~Mightyeagle

Solution 6: Geometry

Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\frac{3}{2}$. The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point $(-4, -6)$. The distance from that point to the origin is $2\sqrt{13}$, and halving that to get the distance the origin moves gives $\boxed{\sqrt{13}}$. ~Meeshbot

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_20&oldid=216284"