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Difference between revisions of "2017 AIME II Problems/Problem 12"

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==Solution==
 
==Solution==
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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=(\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}))^2}=\frac{1-r}{\sqrt{r^2+1}}.</math> Let <math>r=\frac{11}{60}</math>, we find that the distance from the origin is <math>\frac{49}{61} \implies 49+61=\boxed{110}</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>.
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=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 $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\ldots$ and a sequence of points on the circles $A_1,A_2,A_3,\ldots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[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 $B=\left(\frac{1-r}{r^2+1},\frac{r-r^2}{r^2+1}\right)$. 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 $\frac{49}{61}$. $49+61=\boxed{110}$.

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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