Point in a circle

A circle of radius 1 is randomly chosen strictly inside a circle of radius 2. A point is chosen, again at random, inside the circle with radius 1. What is the probability this point is closer to the center of the circle with radius 2 than the circumference of the circle with radius 2?

#Geometry #MathProblem #Math

Easy Math Editor

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Comments

Edited to match the corrected question...

Let $R$ be the distance of the centre $O_2$ of the circle $C$ of radius $1$ from the centre $O_1$ of the circle of radius $2$ . Since $O_2$ can be anywhere inside the circle of radius $1$ with centre $O_1$ (shown dotted in the diagram) we have $\mathbb{P}[R \le r] \,=\, \tfrac{\pi r^2}{\pi} \,=\, r^2$ for any $0 \le r \le 1$ and hence $R$ has the probability distribution function $f(r) \; = \; \left\{ \begin{array}{ll} 2r & 0 \le r \le 1 \\ 0 & \mbox{o.w.} \end{array} \right.$

alternate<em>text alternatetext

Given that $R = r$ , the probability that the randomly chosen point in the circle $C$ is closer to the centre of the original circle than its circumference is the probability that the randomly chosen point lies inside the region common to $C$ and the (dotted) circle of centre $O_1$ and radius $1$ , which is $\begin{array}{rcl} \mathbb{P}[b > a | R = r] & = & \tfrac{2}{\pi}(\theta - \sin\theta\cos\theta) \\ & = & \tfrac{1}{\pi}\left[2\cos^{-1}(\tfrac{r}{2}) - r\sqrt{1 - \tfrac14r^2}\right] \end{array}$ and hence $\begin{array}{rcl} \mathbb{P}[b > a] & = & \displaystyle\int_0^1 \mathbb{P}[b > a|R=r]\,2r\,dr \; = \; \frac{2}{\pi}\int_0^1\left(2r\cos^{-1}(\tfrac{r}{2}) - r^2\sqrt{1 - \tfrac14r^2}\right)\,dr \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac12\pi}^{\frac13\pi}\left(4u\cos u - 4\cos^2u\sin u\right)(-2\sin u)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac13\pi}^{\frac12\pi}\left(4u\sin2u - 2\sin^22u\right)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac23\pi}^\pi \left(v\sin v - \sin^2v\right)\,dv \\ & = & \displaystyle\frac{2}{\pi}\left[ -v\cos v + \sin v - \tfrac12v + \tfrac14\sin2v\right]_{\frac23\pi}^\pi \\ & = & \frac{2}{\pi} \times \frac{1}{8}(4\pi - 3\sqrt{3}) \; = \; 1 - \frac{3\sqrt{3}}{4\pi} \end{array}$

Mark Hennings - 7 years, 7 months ago

Sorry, little typo there. Fixed. That would mean the probability is $\dfrac{4\pi-3\sqrt{3}}{4\pi}$ , correct? I never thought this problem would actually be that hard.

Daniel Liu - 7 years, 7 months ago

Hi,

I tried out a simulation, choosing the coordinates $(r_1\, \cos \theta+r_2\, \cos \alpha ,r_1\, \sin \theta+r_2\, \sin \alpha )$ at random, where $0\le r_1, r_2 \le 1$ and $0\le \theta, \alpha \le 2\,\pi$ .

The distance from the center of the bigger circle, $a=\sqrt{r_1^2+r_2^2+2\, r_1\, r_2\, \cos {(\theta-\alpha)}}$ and $b = 2 -a$ , simulating for 1000000 trials, I got the average $b > a$ about 0.777, which is more than 0.5865.

Can't see any mistake, though.

gopinath no - 7 years, 7 months ago

Did you assume that $r_1$ and $r_2$ were uniform on $[0,1]$ ?

Mark Hennings - 7 years, 7 months ago

@Mark Hennings – Yes, I did..

gopinath no - 7 years, 7 months ago

@Gopinath No – The pdf of $r_1$ and $r_2$ is $2x$ for $0\le x \le 1$ . Look at the beginning of my solution.

Mark Hennings - 7 years, 7 months ago

@Mark Hennings – I saw, but I thought it was required only for deriving the answer analytically.

$r_1$ is the distance from the center of the circle with radius 2 centered at $(0,0)$ , and get the coordinates $(r_1\cos \theta, r_1 \sin \theta )$ , which is the center of the smaller circle. Then select $(r_2\cos \alpha, r_2 \sin \alpha )$ , which is the point within the smaller circle. Then I did my simulation as in the earlier post.

Is there something wrong with this thinking? Aren't the variables independent?

gopinath no - 7 years, 7 months ago

@Gopinath No – If you are going to simulate this, you must adjust your simulation so that the chances of choosing $r_1$ and $r_2$ correct. You can ignore $\theta$ . Provided that you choose $r_1$ and $r_2$ independently from the correct pdf, and choose $\alpha$ randomly and uniformly in $[0,2\pi)$ , all should be well. It will not work if you choose $r_1$ and $r_2$ randomly from a uniform distribution.

To model $r_1$ and $r_2$ correctly, choose $r_1^2$ and $r_2^2$ from a uniform distribution on $[0,1]$ . If you do that, your simulation should agree with my theoretical result (my one did).

Mark Hennings - 7 years, 7 months ago

@Mark Hennings – Oh, I see! It's agreeing now.

Thanks for your help, much appreciated.

gopinath no - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$