6
$\begingroup$

Imagine a circle $C_1$ that has radius $a$. We then chose at random, and independently, three points from the interior of that circle. These three points, if non-collinear, uniquely determine another circle $C_2$; $C_2$ may or may not be totally contained within $C_1$. What is the probability that $C_2$ lies totally inside $C_1$?

This is from Paul Nahin's book 'Inside Interesting Integrals' (pg. 25 in second edition)

Below is the full analytic solution given in the book. However, it goes on to discuss that performing a Monte Carlo simulation gives a value in the range $0.39992$ to $0.400972$ while our answer is $0.418879$; alternatively, the answer seems to be $0.4$ which is $\frac{2\cdot3}{15}$, while our answer is $\frac{2\cdot \pi}{15}$.

So where exactly did we make a mistake? Here are my thoughts-

When we are considering all those $C_2$ inside $C_1$, there will clearly be overlap between the cases. Say case 1 is circle of radius $\frac{a}{3}$ centered at $\frac{a}{2}$, and case 2 is same radius but centered at $\frac{a}{4}$, these will have some intersection area along radical axis. Here he seems to have assumed the two cases to be disjoint, but are they really? We call it as 'probability of all 3 being in circumference band of $C_2$, but in reality ofcourse it is sum over its own continuum of cases- probability of all 3 being in this little arc of the $\Delta(x)$ band of C2 + probability of all 3 being in the next little arc of C2 + and so on. NOW my point is- that little intersection area along radical axis, it will appear in this expansion in both cases, so we are in fact overcounting it.

It seems insignificant since we're thinking of only that particular pair, but any particular intersection area square in the plane will be counted infinitely many times (there’s an infinitude of circles of just the right appropriately diff radius and centers that intersect at the exact same place). In other words, number of times an intersection area is overcounted is proportional to number of pairs of circles, but number of times it is supposed to be counted is proportional to just the number of circles.

img1img2img3

$\endgroup$
7
  • $\begingroup$ Maybe I'm thick or still half-asleep, but it looks like there is something smelly in the proof. First the bands are defined around $C_2$ after the three points are chosen (with the extra condition that $C_2$ lies inside $C_1$), and then out of a hat he computes the cube of this probability. I can't see how that can be correct. And the integral of these infinitesimal probabilities becomes the probability that $C_2$ lies inside $C_1$, which was the assumption from the beginning. Weird. I'd like to see events and conditioning clearly defined here. $\endgroup$ Jul 8, 2023 at 8:11
  • $\begingroup$ @Jean-ClaudeArbaut I suppose you're right, but I think if we wanted to be systematic we could start from considering the set of all circles contained inside with centers on the horizontal, then creating a band around it, and then the cube would be the probability of choosing all 3 points from that band (wrt the sample space of the whole disc) $\endgroup$
    – Amadeus
    Jul 8, 2023 at 9:02
  • $\begingroup$ I can also confirm your estimation of $0.4$ by Monte Carlo simulation. The random selection of points matters very much in this kind of problem: I sampled from the square $[−1,1]^2$ and used rejection, so the distribution is uniform. Mersenne Twister for the RNG. With $10^{10}$ samples I get $0.399991776$ as the required probability. $\endgroup$ Jul 8, 2023 at 9:27
  • $\begingroup$ To answer your last comment: a band around what? If it's around a given circle, we already know the points. If it's around the union of those circles inside $C_1$, it will tell you nothing about the probability of having $C_2$ inside $C_1$ (having all three points inside this union doesn't guarantee anything). The more I read it, the more I'm convinced it's all nonsense. And the simulation doesn't leave much doubt anyway. $\endgroup$ Jul 8, 2023 at 9:39
  • 2
    $\begingroup$ Here is a proof that the probability is $2/5$. $\endgroup$
    – Dan
    Jul 10, 2023 at 0:15

1 Answer 1

4
$\begingroup$

Frankly, I don't understand the logic behind the computation shown in that book. Instead, I'll try to give a more plausible solution.

Take as $C_1$ the unit circle centred at $O=(0,0)$. WLOG we can take the first point on the positive $x$-axis: $A=(x_1,0)$, with $0\le x_1\le1$, and the second point $B=(x_2,y_2)$ inside the upper half of $C_1$, so that $-1\le x_2\le 1$ and $0\le y_2\le1$.

If we want circle $C_2$ to be inside $C_1$, then the third point $C$ must lie in the grey area shown in figure below. Let's see why.

The centre $Q$ of $C_2$ lies on the perpendicular bisector of $AB$ and its distance from $C_1$ must be greater than its distance from either $A$ or $B$ (which is the radius of $C_2$). But the distance of $Q$ from $C_1$ is $1-OQ$, hence the inequality $1-OQ\ge AQ$ must hold for $C_2$ to be inside $C_1$.

That inequality can be rewritten as $AQ+OQ\le1$, i.e. point $Q$ must lie inside the ellipse $E_1$ with foci $A$ and $O$ and unit major axis (dashed in figure below), which has equation $$ {(2x-x_1)^2}+{4y^2\over1-x_1^2}=1. $$ Combining that with the equation of the perpendicular bisector of $AB$: $$ y - {y_2\over2} = {x1 - x2\over y_2}\left(x - {x1 + x2\over2}\right) $$ we can compute its intersection points $P_1$ and $P_2$ with the ellipse as a function of $x_1$, $x_2$, $y_2$. The centre $Q$ of $C_2$ must lie on segment $P_1P_2$ and that happens if point $C$ lies inside the circle through $A$ with centre $P_1$ but outside the circle through $A$ with centre $P_2$, or vice-versa. That is $C$ must lie inside the grey region in figure below.

The probability that $C_2$ lies inside $C_1$ (given $A$ and $B$) is then the same as the probability that point $C$ lies in the grey region, that is the area of that region divided by $\pi$. But the area of that region can be computed as a function of $x_1$, $x_2$ and $y_2$. If $M$ is the midpoint of $AB$ and: $$ d_1=P_1M,\quad d_2=P_2M,\quad l=AM, $$ then the area can be found as the sum of the areas of the two circles, minus double the areas of the two circular segments in figure below: $$ \text{area}_1= (d_1^2 + d_2^2 + 2 l^2)\pi - 2 (d_1^2 + l^2)\arctan{l\over d_1} + 2 ld_1 - 2 (d_2^2 + l^2)\arctan{l\over d_2} + 2 ld_2. $$ EDIT.

The above result works if $M$ is between $P_1$ and $P_2$. If $P_1$ is between $M$ and $P_2$ it must be modified to: $$ \text{area}_2= (d_1^2 + d_2^2 + 2 l^2)\pi - 2 (d_1^2 + l^2)(\pi-\arctan{l\over d_1}) - 2 ld_1 - 2 (d_2^2 + l^2)\arctan{l\over d_2} + 2 ld_2. $$ The limiting case $P_1=M$ occurs when $M$ lies on ellipse $e_1$. In that case point $B$ lies on the homothetic of ellipse $e_1$ with respect to centre $A$ and ratio $2$, i.e. on an ellipse $e_2$ with $A$, $-A$ as foci and major axis $2$, whose equation is then: $$ y^2=(1-x^2)(1-x_1^2). $$ Hence we must use area$_1$ for $0\le y_2<\sqrt{(1-x_2^2)(1-x_1^2)}$ and area$_2$ for $\sqrt{(1-x_2^2)(1-x_1^2)}\le y_2\le\sqrt{1-x_1^2}$

The requested probability $p$ can then be found averaging over the positions of $A$ and $B$: $$ p={2\over \pi}\int\limits_0^1\int\limits_{-1}^1 \kern{-30pt}\int\limits_0^{\kern30pt\sqrt{(1-x_2^2)(1-x_1^2)}} \kern{-30pt}{\text{area}_1\over\pi}\,dy_2\,dx_2\,dx_1 +{2\over \pi}\int\limits_0^1\int\limits_{-1}^1 \kern{-30pt}\int\limits_{\kern30pt\sqrt{(1-x_2^2)(1-x_1^2)}} ^{\kern20pt\sqrt{1-x_2^2}} \kern{-30pt}{\text{area}_2\over\pi}\,dy_2\,dx_2\,dx_1. $$ I found expressions for the area with Mathematica (too long to be written here) but there's no way to get an exact result for the integrals. Resorting to numerical integration I found $p\approx0.40$, but couldn't achieve a greater precision.

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .