Difference between revisions of "2017 AIME II Problems/Problem 12"
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-Isogonal | -Isogonal | ||
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+ | ==Solution 5== | ||
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+ | We place this on the complex plane with <math>C_0=0</math>. Notice that <math>B</math> is the center of <math>C_k</math> as <math>k</math> approaches <math>\infty</math>. We have <math>C_1=C_0+1-\frac{11}{60}=C_0+\frac{49}{60}</math>. Similarly, <math>C_2=C_1+\frac{11}{60}\cdot\frac{49}{60}i</math>, <math>C_3=C_2+\left(\frac{11}{60}\right)^2\cdot\frac{49}{60}i^2</math>, and so on. Therefore, <math>B=\frac{49}{60}+\frac{49}{60}\cdot\frac{11}{60}i+\frac{49}{60}\cdot\left(\frac{11}{60}i\right)^2+\ldots</math> | ||
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+ | It follows that <math>B=\frac{\frac{49}{60}}{1-\frac{11}{60}i}=\frac{49}{60-11i}</math>. We seek <math>|B|</math>, which is <math>\frac{49}{|60-11i|}=\frac{49}{61}</math>, and the answer is <math>49+61=\boxed{110}</math>. | ||
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+ | -TheUltimate123 | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=11|num-a=13}} | {{AIME box|year=2017|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:07, 4 March 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
Solution 5
We place this on the complex plane with . Notice that is the center of as approaches . We have . Similarly, , , and so on. Therefore,
It follows that . We seek , which is , and the answer is .
-TheUltimate123
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.