A green circle is tangent to a red circle and a black circle. The three circles have equal radii. Their centres are collinear and distinct.
Random point A is chosen on the red circle. Random points B and C are chosen on the green circle. (All random points are independent and uniformly distributed.)
Which line is more likely to intersect the black circle: $\overleftrightarrow{AB}$ or $\overleftrightarrow{BC}$?
Update: Now posted on MSE.
Self-answering.
My method uses complicated integration, but I suspect there is a more elegant method.
I'm not going to show all the details. I'm just posting this to give an idea of how tedious this method is - which should motivate the search for a more elegant method.
$\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$ are equally likely to intersect the black circle.
Define angle $x$ as shown.
If $\overleftrightarrow{BC}$ intersects the black circle, then the arc on which $C$ can lie has a central angle of $$f(x)=4\arcsin\left(\frac{1}{\sqrt{5-4\cos x}}\right)$$ (The derivation uses the sine rule, cosine rule and angle chasing.)
Here is the graph of $y=f(x)$.
The probability that $\overleftrightarrow{BC}$ intersects the black circle is the average value of $\frac{f(x)}{2\pi}$ for $0\le x\le 2\pi$, that is, $$\frac{1}{2\pi}\int_0^{2\pi}\frac{f(x)}{2\pi}dx$$ which turns out to equal
$$\dfrac34-\dfrac{2}{\pi^2}\left(\dfrac38\ln^22+\operatorname{Li}_2\left(-\sqrt2\right)+3\operatorname{Li}_2\left(\dfrac{1}{\sqrt2}\right)\right)=0.3871287106\dots$$ (Wolfram numerical calculation)
Define $x$ in the same way as in the previous diagram, but add the red circle on the left.
If $\overleftrightarrow{AB}$ intersects the black circle, then the arcs on which $A$ can lie have a total central angle of:
$g(x)=\begin{cases}2\pi&0\le x\le\frac{\pi}{6},\frac{11\pi}{6}\le x\le 2\pi\\2\arcsin\left(\frac{3-8\sqrt{\left(\sin^{2}x\right)\left(1-\cos x\right)}}{5-4\cos x}\right)+\pi&\frac{\pi}{6}\le x\le\frac{5\pi}{6},\frac{7\pi}{6}\le x\le\frac{11\pi}{6}\\2\arcsin\left(\frac{3-8\sqrt{\left(\sin^{2}x\right)\left(1-\cos x\right)}}{5-4\cos x}\right)+2\arcsin\left(\frac{3+8\sqrt{\left(\sin^{2}x\right)\left(1-\cos x\right)}}{5-4\cos x}\right)&\frac{5\pi}{6}\le x\le\frac{7\pi}{6}\end{cases}$
Here is the graph of $y=g(x)$ in red (a sort of batman curve), and the graph of $y=f(x)$ in green.
The probability that $\overleftrightarrow{AB}$ intersects the black circle is the average value of $\frac{g(x)}{2\pi}$ for $0\le x\le 2\pi$, that is, $$\frac{1}{2\pi}\int_0^{2\pi}\frac{g(x)}{2\pi}dx$$ which also turns out to equal $$\dfrac34-\dfrac{2}{\pi^2}\left(\dfrac38\ln^22+\operatorname{Li}_2\left(-\sqrt2\right)+3\operatorname{Li}_2\left(\dfrac{1}{\sqrt2}\right)\right)=0.3871287106\dots$$ (Wolfram numerical calculation)
I did a simulation sampling 1M random points a,b,c. Determined the equation for the line, calculated the points in the range of the black circle and checked if any fell in the circle (x^2 +y^2)<= r^2.
I got: a b: 0.386588, b c: 0.387247, n= 1000000
edit:
with n = 10M the numbers are a b: 0.3870692 b c: 0.3871753. The p-value from a z-test is 0.626, indicating the two distribution have the same mean. The error is 0.00015 using the 99.99% confidence interval
Editing: I’m working this one out on paper to make it clearer. I think this answer as it stands is in the right direction but has two flaws.
To me there seems to be an obvious symmetry here - the hard part is just figuring out exactly what the symmetry is! First I thought about the target circle being a reflection of the 1-point circle (A) but I didn't get anywhere.
However, considering the possible space of points covered, I found an intuitive solution I think. Consider the ‘shadow’ of the target circle - the region of the plane where a line passing through the middle circle must hit the target circle. Now consider the areas swept out by possible lines in either situation. With BC, the whole plane can be covered. With AB, I think again the whole plane can be covered (except maybe a 0-measure set the vertical line tangent to both circles A and B). So then (I'll update this assumption in the morning if someone reminds me) I'm PRETTY sure that the shadow for AB and the shadow for BC is the same (because both lines pass through B I think?) so looking at the shadow divided by swept area we find that the two probabilities are equal. I guess infinitesimally less for BC.
Okay, I’m awake and wrote this up:
This solution is based on thinking of the circles as being on the plane. First, let's name the circle A (which has the singleton point). B (which has two points), and T for target (the one the lines hit). Ultimately, we will equate the probability of a line hitting T with the fraction of the plane that T occludes. More precisely I think: we are considering the plane as covered by lines. The 'range' of B would be the entire plane: obviously, for any point X on the plane we can pick a line that passes through that point X and two points on B (B and C I believe in the question). The 'shadow' cast by T is the region of the plane which cannot be reached by such a line except by passing through T. If B were emitting (non bending) rays of light in the directions prescribed by (line BC) and T were opaque, then we'd see the shadow clearly. Then, we take the probability of hitting T to be the area of the shadow divided by the area of the range.
Concretely, this is:
I just read Nuclear Hoagie's answer a second ago to make sure mine doesn't overlap, and I think this is where we part ways. There need not be a 1-1 correspondence between lines of B and lines of A, because we're normalizing. This obviates the concern over the fact that each line through B can be achieved in two ways (by swapping points B and C). It doesn't matter - the areas and division erase this factor of 2. We're not concerned with comparative counts, we're concerned with ratios of areas.
Similarly, A has a 'range' defined by the allowed lines: the change is that these lines must pass through point A on circle A and at least one point B on circle B, rather than two points on the same circle (as for circle B). However, it is intuitively clear that still the entire plane is in the range of A, except for the vertical line which is tangent to both circle A and circle B. So, the denominator of our probability is the same (because this single line of area-measure 0 doesn't make a difference). In this sense, I suppose we are essentially integrating without actually integrating - we're totting up infinitesimal areas and tossing out sets of measure 0.
Moreover, it is clear that ALMOST EVERY (and I mean this in a way similar to the measure-theoretic sense) line allowed on A also could be a line of B; it intersects circle B twice. Actually this may not be strictly true in terms of measure, but it doesn't really matter. We only care about the shadow. This means we can restrict our attention to the lines which hit A, B, and T.
First we observe that if a line is permitted on B and hits T, it's part of the shadow of T for light source B (which we described above). This means that if a line (like the bulk of the lines permitted on A) hits two points on circle B and then intersects T, it's part of the shadow for B. Of course, because it's permitted on A and intersects T, it also is part of the shadow for A. So then these shadows are starting to look very similar, except if there are lines that are allowed on A but NOT on B, which still hit T.
So to pause and recap here, we've established the probability for B and removed the concern over 1-1 correspondences by normalizing using area division. We've also established that, again NOT worrying about 1-1 correspondence or 'layers' of covering the plane, the denominator for the probability fraction for A is the same as that for B. So if we can then establish that the numerators (the shadows) are the same (up to differences of measure 0) then we know the two probabilities are the same.
There are a few exceptions to our previous discussion. These are the lines that start on A, hit only 1 point on B (as required), and go on to hit T. We can pretty easily say that these are only the two tangents which are tangent to all three circles. This is a finite set of measure-0 lines, which contributes nothing to the area of course. So the shadow size is also the same. Then dividing, we get the same probability.
So the answer is as suggested by the brute integration (which I was frankly never going to do myself): they're equal.
Edit: Dan raises the potential objection that this argument might seem to imply the probabilities remain equal when the red circle (A) is shifted to the right. I contend that my argument would no longer apply. The reasoning:
My original argument depends on the ratio of shadow to total possible area being equal for A and B. In the original problem, for both A and B, it is only possible to cover the plane with lines twice. That is, for any given direction or angle of line, we could get it two ways. For lines on B, we could get (for example) a line using B at the southwest and C at the northeast for 45 degrees. Or, we could switch B and C and get the exact same line. I established that almost every line allowed on A also passes through two points on B, which means we can pick two identical lines again. The tangents to B that hit A can likewise pass through two points on A so we can again pick two identical lines. So for lines allowed on A, we can cover (almost) the entire plane twice, just like B. The vertical tangent to A and B can never be reached, but that’s infinitesimally small as an area: the plane is covered twice by A or B, just the same. Remember, we’re ignoring T here.
So the denominator would be the same.
HOWEVER: when Dan poses this shifting of A, we lose the equivalency of the denominators. This is because as long as even a tiny stretch of the perimeter of A is contained within the ring of B, we suddenly re-cover the plane an uncountably infinite number of times. For every point of A inside B, we can pick a point on B to get a line in whatever direction we want. In fact. We can pick two such points! So then the sizes of the numerator and denominator get all out of whack and we can’t equate them or keep track of them at all. This means there’s no conclusion about any situation other than when A and B are tangent at exactly one point.
They are
equally likely.
Here's why:
For any placement of point A, there are at most two placements of point B that would make line AB intersect the green circle only once (forming the two tangents), while all other placements of B would make line AB intersect the green circle twice - this means for any choice of line AB, we almost surely have a corresponding colinear point C which lies on the green circle, forming line ABC. If ABC intersects the black circle, both AB and BC do. Since there is a 0% chance of placing B such that we cannot draw line ABC, we find that for every AB that goes through the black circle there is a BC that does the same (actually two, if we swap B and C), and for every BC that goes through the black circle there is an AB that does the same (again two, if we move A to the other side of the circle).
We now need to show that C is still uniformly distributed when selecting it in this way. Note that almost surely, when uniformly selecting A and uniformly selecting B, there is a colinear C on the green circle. For each of these configurations, there is also a 1:1 corresponding configuration with B and C swapped. C therefore has the same distribution as B, which was explicitly uniform - if C wasn't uniform, it would imply there were placements for C without a corresponding B, or vice versa.
