Difference between revisions of "2017 AIME II Problems/Problem 12"
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− | + | Circle <math>C_0</math> has radius <math>1</math>, and the point <math>A_0</math> is a point on the circle. Circle <math>C_1</math> has radius <math>r<1</math> and is internally tangent to <math>C_0</math> at point <math>A_0</math>. Point <math>A_1</math> lies on circle <math>C_1</math> so that <math>A_1</math> is located <math>90^{\circ}</math> counterclockwise from <math>A_0</math> on <math>C_1</math>. Circle <math>C_2</math> has radius <math>r^2</math> and is internally tangent to <math>C_1</math> at point <math>A_1</math>. In this way a sequence of circles <math>C_1,C_2,C_3,\ldots</math> and a sequence of points on the circles <math>A_1,A_2,A_3,\ldots</math> are constructed, where circle <math>C_n</math> has radius <math>r^n</math> and is internally tangent to circle <math>C_{n-1}</math> at point <math>A_{n-1}</math>, and point <math>A_n</math> lies on <math>C_n</math> <math>90^{\circ}</math> counterclockwise from point <math>A_{n-1}</math>, as shown in the figure below. There is one point <math>B</math> inside all of these circles. When <math>r = \frac{11}{60}</math>, the distance from the center <math>C_0</math> to <math>B</math> is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | |
<asy> | <asy> |
Revision as of 17:39, 23 March 2017
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
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
, we find that the distance from the origin 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.