Difference between revisions of "2017 AIME II Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
Let the center of circle <math>C_i</math> be <math>O_i</math>. Note that <math>O_0BO_1</math> is a right triangle, with right angle at <math>B</math>. Also, <math>O_1B=\frac{11}{60}O_0B</math>, or <math>O_0B = \frac{60}{61}O_0O_1</math>. It is clear that <math>O_0O_1=1-r=\frac{49}{60}</math>, so <math>O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math> | Let the center of circle <math>C_i</math> be <math>O_i</math>. Note that <math>O_0BO_1</math> is a right triangle, with right angle at <math>B</math>. Also, <math>O_1B=\frac{11}{60}O_0B</math>, or <math>O_0B = \frac{60}{61}O_0O_1</math>. It is clear that <math>O_0O_1=1-r=\frac{49}{60}</math>, so <math>O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math> | ||
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-william122 | -william122 | ||
Revision as of 09:52, 27 July 2017
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
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
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.