Difference between revisions of "2017 AIME II Problems/Problem 12"
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− | Let <math>A_0</math> be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to r. Now accounting for rotation by <math>\frac{\pi}{2}</math> radians, we see that the common ratio is <math>ri</math>. Thus since our first term is <math>A_1=-r+ri</math>, the total sum (by geometric series formula) is <math>\frac{-r+ri}{1-ri}=\frac{-781+538i}{3721}</math>. We need the distance from <math>C_0=-1</math> so our distance is <math>|B-C_0|=\sqrt{\left(\frac{-781}{3721}-(-1)\right)^2+\left(\frac{538i}{3721}\right)^2}=\sqrt{\frac{2401}{3721}}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math> | + | Let <math>A_0</math> be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to <math>r</math>. Now accounting for rotation by <math>\frac{\pi}{2}</math> radians, we see that the common ratio is <math>ri</math>. Thus since our first term is <math>A_1=-r+ri</math>, the total sum (by geometric series formula) is <math>\frac{-r+ri}{1-ri}=\frac{-781+538i}{3721}</math>. We need the distance from <math>C_0=-1</math> so our distance is <math>|B-C_0|=\sqrt{\left(\frac{-781}{3721}-(-1)\right)^2+\left(\frac{538i}{3721}\right)^2}=\sqrt{\frac{2401}{3721}}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math> |
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=11|num-a=13}} | {{AIME box|year=2017|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:37, 1 December 2021
Contents
Problem
Circle has radius
, and the point
is a point on the circle. Circle
has radius
and is internally tangent to
at point
. Point
lies on circle
so that
is located
counterclockwise from
on
. Circle
has radius
and is internally tangent to
at point
. In this way a sequence of circles
and a sequence of points on the circles
are constructed, where circle
has radius
and is internally tangent to circle
at point
, and point
lies on
counterclockwise from point
, as shown in the figure below. There is one point
inside all of these circles. When
, the distance from the center
to
is
, where
and
are relatively prime positive integers. Find
.
Solution 1
Impose a coordinate system and let the center of be
and
be
. Therefore
,
,
,
, and so on, where the signs alternate in groups of
. The limit of all these points is point
. Using the geometric series formula on
and reducing the expression, we get
. The distance from
to the origin is
Let
, and the distance from the origin is
.
.
Solution 2
Let the center of circle be
. Note that
is a right triangle, with right angle at
. Also,
, or
. It is clear that
, so
. Our answer is
-william122
Solution 3
Note that there is an invariance, Consider the entire figure . Perform a
counterclockwise rotation, then scale by
with respect to
. It is easy to see that the new figure
, so
is invariant.
Using the invariance, Let . Then rotating and scaling,
. Equating, we find
. The distance is thus
. Our answer is
-Isogonal
Solution 4
Using the invariance again as in Solution 3, assume is
away from the origin. The locus of possible points is a circle with radius
. Consider the following diagram.
Let the distance from to
be
. As
is invariant,
. Then by Power of a Point,
. Solving,
. Our answer is
-Isogonal
Solution 5 (complex)
Let be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to
. Now accounting for rotation by
radians, we see that the common ratio is
. Thus since our first term is
, the total sum (by geometric series formula) is
. We need the distance from
so our distance is
. Our answer is
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.