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 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, | ||
− | + | ==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 */ | ||
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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 = ( | + | 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); | ||
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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== | |
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} = \boxed{\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== | |
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. | 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 \boxed{\sqrt{13}}. | + | 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 | ~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 11:57, 28 February 2024
Contents
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 as
. Note that the center of dilation must have an
-coordinate less than
; if the
-coordinate were otherwise, then the circle under the transformation would not have an increased
-coordinate in the coordinate plane. 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
. We do not have to include absolute value symbols because we know that the center of dilation has a lower
-coordinate, and hence a lower
-coordinate, from our reasoning above. Solving the two equations, 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: Geometric
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 .
Solution 3: Logic and Geometry
Using the ratios of radii of the circles, , we find that the scale factor is
. If the origin had not moved, this indicates that the center of the
circle would be
, simply because of
. Since the center has moved from
to
, we apply the distance formula and get:
.
Solution 4: Simple and Practical
Start with the size transformation. Transforming the circle from to
would mean the origin point now transforms into the point
. Now apply the position shift:
to the right and
up. This gets you the point
. Now simply apply the Pythagorean theorem with the points
and
to find the requested distance.
Solution 5: Using the Axes
Before dilation, notice that the two axes are tangent to the circle with center . Using this, we can draw new axes tangent to the radius 3 circle with center
, resulting in a "new origin" that is 3 units left and 3 units down from the center
, or
. Using the distance formula, the distance from
and
is
.
~Mightyeagle
Solution 6: Geometry
Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be . 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
. The distance from that point to the origin is
, and halving that to get the distance the origin moves gives
. ~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.