2017 AIME II Problems/Problem 12
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 .
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 . .
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
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
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
Solution 5 (complex)
Let be the origin. Now note that the ratio of lengths of consecutive line segments is constant and equal to . 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
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