Difference between revisions of "2017 AIME II Problems/Problem 12"
(→Problem) |
The turtle (talk | contribs) |
||
Line 16: | Line 16: | ||
==Solution== | ==Solution== | ||
− | Impose a coordinate system and let the center of <math>C_0</math> be <math>(0,0)</math> and <math>A_0</math> be <math>(1,0)</math>. Therefore <math>A_1=(1-r,r)</math>, <math>A_2=(1-r-r^2,r-r^2)</math>, <math>A_3=(1-r-r^2+r^3,r-r^2-r^3)</math>, <math>A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)</math>, and so on, where the signs alternate in groups of <math>2</math>. The limit of all these points is point <math>B</math>. Using the geometric series formula on <math>B</math> and reducing the expression, we get <math>B=(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1})</math>. The distance from <math>B</math> to the origin is <math>\sqrt{(\frac{1-r}{r^2+1})^2+(\frac{r-r^2}{r^2+1} | + | Impose a coordinate system and let the center of <math>C_0</math> be <math>(0,0)</math> and <math>A_0</math> be <math>(1,0)</math>. Therefore <math>A_1=(1-r,r)</math>, <math>A_2=(1-r-r^2,r-r^2)</math>, <math>A_3=(1-r-r^2+r^3,r-r^2-r^3)</math>, <math>A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)</math>, and so on, where the signs alternate in groups of <math>2</math>. The limit of all these points is point <math>B</math>. Using the geometric series formula on <math>B</math> and reducing the expression, we get <math>B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)</math>. The distance from <math>B</math> to the origin is <math>\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.</math> Let <math>r=\frac{11}{60}</math>, and the distance from the origin is <math>\frac{49}{61}</math>. <math>49+61=\boxed{110}</math>. |
− | |||
=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 19:28, 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 , and 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.