Difference between revisions of "2017 AIME II Problems/Problem 12"
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− | draw(Circle((0, | + | draw(Circle((0,0),125)); |
draw(Circle((25,0),100)); | draw(Circle((25,0),100)); | ||
draw(Circle((25,20),80)); | draw(Circle((25,20),80)); | ||
− | draw(Circle(( | + | draw(Circle((9,20),64)); |
dot((125,0)); | dot((125,0)); | ||
− | label("$A_0$",( | + | label("$A_0$",(125,0),E); |
dot((25,100)); | dot((25,100)); | ||
label("$A_1$",(25,100),SE); | label("$A_1$",(25,100),SE); |
Revision as of 02:29, 26 February 2018
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
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.