Four is a crowd
Three points are chosen uniformly at random within a unit circle in the x y -plane.
A fourth point is chosen uniformly at random on the perimeter of a circle with radius r .
For r > > 1 , (say r ≈ 1 , 0 0 0 , 0 0 0 , 0 0 0 ), what does the probability p that the four points form a convex quadrilateral approach?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Oh darn. This is why I hate multiple-choice questions. :( I used the same logic, but I chose to take region 8 into account, making P = 0 . 5 + δ , i.e., not e x a c t l y 0 . 5 . But since the 4th point has essentially half of R 2 to choose from I will grudgingly accept your answer.
Nice problem, regardless. It is an interesting variation of Sylvester's Four-Point Problem , the difference being that in your problem 3 points are inside the planar region and one is forced outside. With Sylvester's the minimum of the range is 2 / 3 , so I find it intriguing how much the restriction on one of the points affects the range.
Some obvious follow-ups: (i) 1 point in a unit circle, 3 point outside; (ii) 2 points in a unit circle, 2 outside, and (iii) for R 3 , the probability that the convex hull of 4 points in a unit sphere and 1 point outside has 5 vertices.
Log in to reply
Yeah, I thought about that later, and changed it from p > 0 . 5 to p > 0 . 5 1 , but perhaps I should have changed it to something like this... Which of these is closest to the value? 0 , 4 1 , 2 1 , 4 3 , 1 . Do you suppose I should I change the options?
Interesting follow ups... I'll need to think about those for a bit...
Log in to reply
Oh, haha, I hadn't noticed you had 0 . 5 1 ; my brain immediately processed it as 0 . 5 . :p I think your options and answer are still o.k., but I've found an interesting situation. If we have the 3 points inside a circle of any finite radius R and the 4th point outside, then region 8 is always finite, (and the other regions still infinite). So would we anticipate the same 0 . 5 probability for a n y finite radius R ? If so, then there would be a "discontinuity" when we send R → ∞ , for then the probability would jump (I believe) to the Sylvester's disk probability of 1 − 1 2 π 2 3 5 ≈ 0 . 7 0 4 4 8 . Something seems weird about that, but weird things do happen at infinity so maybe there isn't any problem with this result.
Log in to reply
@Brian Charlesworth – Interesting, and definitely odd... I wonder what happens if we keep two of the three points inside the unit circle, and send R → ∞ for only one of the points? (I guess it becomes one of your follow up questions, but does it also "jump" from 0.5 to some other value?)
Log in to reply
@Geoff Pilling – My brain needs to be a bit fresher to think about this. For now, though, I was wondering if the correct option were changed from 0 . 5 to 0 . 4 9 ≤ p ≤ 0 . 5 1 whether that would be better at (i) covering all of [ 0 , 1 ] with the options given, and (ii) dealing with the potentially pesky δ -factor? Just a thought ....
Log in to reply
@Brian Charlesworth – Interesting idea... @Calvin Lin thoughts? Do you suppose that we should update the answer to give p a range from 0 . 4 9 to 0 . 5 1 ? See Michael's note above.
Maybe it's not "exactly" 0.5, but wouldn't that be a limit value? If the 4th point can be picked from the "entire" x-y plane, then region 8 becomes irrelevant. So, say, from an engineer's point of view, a value of 0.5 is more accurate and useful than " 0 . 4 9 ≤ p ≤ 0 . 5 1 "
Log in to reply
Yes, you're right, and from a mathematician's point of view a limiting value of e x a c t l y 0 . 5 is more consequential than something like [ 0 . 4 9 , 0 . 5 1 ] . But as you and Calvin have noted, this is all moot until the issue of how the points are distributed is dealt with. I think Geoff's updated wording might do the job for the 4th point, so now the distribution of the first 3 points needs to be clarified to something like "uniformly over r ∈ [ 0 , 1 ] and θ ∈ [ 0 , 2 π ] ".
Unfortunately, there is no uniform distribution on the entire x-y plane So, the question is somewhat moot.
Log in to reply
Would it be possible to first consider a finite area in the x-y plane such as a circle centered at the origin, and then ask for the probability as a limit as the circle is expanded indefinitely?
Log in to reply
Or perhaps the opposite? Choose three points uniformly at random in a unit circle, then let the radius of that circle get shrunk to zero with the coordinates of the points scaled accordingly? Then pick the fourth point uniformly at random on a (non-scaled) unit circle?
Log in to reply
@Geoff Pilling – Yes, something like that. Calvin is correct that for problems like this, you need to be careful about exactly specifying the distribution of so-called "randomly selected points". That's too often a pitfall that leads to contradictory results.
Log in to reply
@Michael Mendrin – Should we update the problem with a better phraseology?
How about if we say this?
"A fourth point is chosen uniformly at random on the perimeter of the unit circle, and then its coordinates are scaled together indefinitely. i.e. A value for θ is chosen uniformly at random on the interval [ 0 , 2 π ) and then the fourth point as chosen as ( r cos θ , r sin θ ) and we let r → ∞ .
Log in to reply
@Geoff Pilling – In that case, you have to be very careful, because the "infinite area" argument no longer holds true. In fact, that was my first iffy part of the problem, of considering these infinite areas to be equal. I do believe that areas 2, 4, 6 are going to be larger than areas 1, 3, 5, in any reasonably bounded region.
I would not be surprised if the answer works out to be 2/3 or even 3/4.
Log in to reply
@Calvin Lin – So would we then have to determine a general function p ( r ) , where r ≥ 1 is the distance of the 4th point from the origin, and then find r → ∞ lim p ( r ) ? If we extended this function so that r ≥ 0 then taking the (weighted) average from 0 to 1 we should get the Sylvester's result, which would be a good check of whatever function we come up with.
Log in to reply
@Brian Charlesworth – Right. It should be possible to look at a derivation of Sylvester's result, where it's obtained by summing conditionally on the max radius of the last point.
Log in to reply
@Calvin Lin – OK, I've updated the problem a bit... Hopefully the wording is a little more mathematically correct? I will plan to add a clarifying image to the solution as well when I get a chance...
@Calvin Lin – Without doing the math, I believe if you bound it, say, at R = 1,000,000,000 for the fourth point, and constrain the other three points to the unit circle, the sum total of the areas of 2,4,6 are extremely close to 1,3,5, within a fraction of a percentage...
However, lemme think about this all for a bit, and see if I can come up with a better way to rework the question... Perhaps something along the lines of what Brian suggests.
Log in to reply
@Geoff Pilling – Ah yes. I agree that at the limit, the areas do work out to be approximately equal.
For a non-rigorous argument, just shift the lines till they pass through the center of the circle, in which case these alternating 6 regions would have equal area by angles. Then confirm that the amount of shift is at most 1/r (or some other fraction) the area of the circle.
@Geoff Pilling – Perhaps it should also be mentioned that the first 3 points are chosen randomly and uniformly over r ∈ [ 0 , 1 ] and θ ∈ [ 0 , 2 π ] .
I was thinking of an alternative wording based on a previous problem of mine. It would go
"Suppose we have four concentric circles with respective radii 1 , 2 , 3 and 4 . Choose one point, uniformly and at random, on each of these circles. What is the probability p that the four points form a convex quadrilateral?"
Although this doesn't line up with the original intent of your problem, it might avoid the issues Calvin and Michael have mentioned. We could then see how the probability varies when we play around with the different radii, or put two points of one circle and single points on two different circles, etc..
Log in to reply
@Brian Charlesworth – Sounds good... Lemme think about this "new" version of the problem. Ultimately I'd like to have r 1 = r 2 = r 3 = 1 and r 4 = 1 , 0 0 0 , 0 0 0 , 0 0 0 :0)
Log in to reply
@Geoff Pilling – Yeah, that would align pretty well with your original intent. I'm curious to see what the final state of evolution of this problem ends up being. :)
Let the intersections of these three lines be the three initial points:
Since the fourth point is chosen very far from the origin ( r > > 1 ) , the spacing of the points, region 8 in the picture above, in the unit circle is negligible.
In order for the four points to form a convex quadrilateral, the fourth point needs to reside outside the angles opposite the interior angles of the triangle formed by the first three points, i.e. in regions 2 , 4 , or 6 in the figure above. Since the sum of the interior angles of the triangle formed by the first three points is 1 8 0 ∘ , the sum of the three opposite angles will be the same, which will therefore cover half of the x , y coordinate system.
Therefore, the probability will be:
P = 0 . 5
Note : For very large r (~ 1 billion) the unit circle starts to look a bit more like a single point, and the regions end up looking a bit like the picture below. Here you can see visually why the probability is 0.5. (i.e. The blue area will be equal to the white area)