Difference between revisions of "2016 AMC 10B Problems/Problem 20"
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However, take this solution with a grain of salt; generally, do not do this kind of guessing in the real AMC10 since leaving it blank gives you a guaranteed 1.5 points. ~peelybonehead | However, take this solution with a grain of salt; generally, do not do this kind of guessing in the real AMC10 since leaving it blank gives you a guaranteed 1.5 points. ~peelybonehead | ||
− | Well, according to Kai's theorem. we can conclude that 1+1=3 in this case, so we chose C~Kai Gao | + | Well, according to Kai's theorem. we can conclude that 1+1=3, 2+3=6 in this case, so we chose C~ Kai Gao and his goood teacher Nanson Wang |
==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 00:20, 7 July 2023
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: Answers
When you have no idea what the answer could possibly be, we can observe that C is the only non-integer number. Knowing AMC it's probably not that simple and thus ends in a non-integer number (this is usually not the usual case but since this is the only non-integer number it has an excuse). Thus we guess the answer is C and get it correct. -RealityWrites
lol ~nezha
However, take this solution with a grain of salt; generally, do not do this kind of guessing in the real AMC10 since leaving it blank gives you a guaranteed 1.5 points. ~peelybonehead
Well, according to Kai's theorem. we can conclude that 1+1=3, 2+3=6 in this case, so we chose C~ Kai Gao and his goood teacher Nanson Wang
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.