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Difference between revisions of "2017 AIME II Problems/Problem 12"
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==Problem==
 
==Problem==
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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,\cdots</math> and a sequence of points on the circles <math>A_1,A_2,A_3,\cdots</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>.
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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>.
  
 
<asy>
 
<asy>

Revision as of 17:37, 23 March 2017

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 $\frac{49}{61}$. $49+61=\boxed{110}$.

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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