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

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==Solution 2==
 
==Solution 2==
 
Let the center of circle <math>C_i</math> be <math>O_i</math>. Note that <math>O_0BO_1</math> is a right triangle, with right angle at <math>B</math>. Also, <math>O_1B=\frac{11}{60}O_0B</math>, or <math>O_0B = \frac{60}{61}O_0O_1</math>. It is clear that <math>O_0O_1=1-r=\frac{49}{60}</math>, so <math>O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
 
Let the center of circle <math>C_i</math> be <math>O_i</math>. Note that <math>O_0BO_1</math> is a right triangle, with right angle at <math>B</math>. Also, <math>O_1B=\frac{11}{60}O_0B</math>, or <math>O_0B = \frac{60}{61}O_0O_1</math>. It is clear that <math>O_0O_1=1-r=\frac{49}{60}</math>, so <math>O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}</math>. Our answer is <math>49+61=\boxed{110}</math>
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-william122
 
-william122
  

Revision as of 08:52, 27 July 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 $\frac{49}{61}$. $49+61=\boxed{110}$.

Solution 2

Let the center of circle $C_i$ be $O_i$. Note that $O_0BO_1$ is a right triangle, with right angle at $B$. Also, $O_1B=\frac{11}{60}O_0B$, or $O_0B = \frac{60}{61}O_0O_1$. It is clear that $O_0O_1=1-r=\frac{49}{60}$, so $O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}$. Our answer is $49+61=\boxed{110}$

-william122

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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