2017 AIME II Problems/Problem 12
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 r. 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 |
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