Problem

Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$ . Therefore $A_1=(1-r,r)$ , $A_2=(1-r-r^2,r-r^2)$ , $A_3=(1-r-r^2+r^3,r-r^2-r^3)$ , $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$ , and so on, where the signs alternate in groups of $2$ . The limit of all these points is point $B$ . Using the geometric series formula on $B$ and reducing the expression, we get $B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)$ . The distance from $B$ to the origin is $\sqrt{\left(\frac{1-r}{r^2+1}\right)^2+\left(\frac{r-r^2}{r^2+1}\right)^2}=\frac{1-r}{\sqrt{r^2+1}}.$ Let $r=\frac{11}{60}$ , and the distance from the origin is $\frac{49}{61}$ . $49+61=\boxed{110}$ .

$[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label("$A_0$",(125,0),E); dot((25,100)); label("$A_1$",(25,100),SE); dot((-55,20)); label("$A_2$",(-55,20),E); [/asy]$

Solution

Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$ . Therefore $A_1=(1-r,r)$ , $A_2=(1-r-r^2,r-r^2)$ , $A_3=(1-r-r^2+r^3,r-r^2-r^3)$ , $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$ , and so on, where the signs alternate in groups of $2$ . The limit of all these points is point $B$ . Using the geometric series formula on $B$ and reducing the expression, we get $B=(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1})$ . The distance from $B$ to the origin is $\sqrt{(\frac{1-r}{r^2+1})^2+(\frac{r-r^2}{r^2+1}))^2}=\frac{1-r}{\sqrt{r^2+1}}.$ Let $r=\frac{11}{60}$ , and the distance from the origin is $\frac{49}{61}$ . $49+61=\boxed{110}$ .

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

Page

Toolbox

Search

2017 AIME II Problems/Problem 12

Problem

Solution

See Also