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 18:39, 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 |
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