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I was trying to solve this problem from Dan.

What I did:

Firstly, I used $2tan^{-1}(p)$ as the mapping function to map a point $p$ on the real line to unit circle. This should map [$-\infty,\infty$] to [$-\pi,\pi$] on the circle. The chord length (for radius=$1$) is $2sin\left(\frac{arc\ length}{2}\right)$.

If I choose 3 points on the real line as $(-x,0),(0,0)$ and $(y,0)$, (where $x$ and $y$ are positive numbers) with the mid point to map to an arbitrary $0$ position on the circle, the chord lengths for the 3 sides forming a triangle (let's call them side1, side2 and side3) are $\frac{2x}{\sqrt{x^2+1}}$, $\frac{2y}{\sqrt{y^2+1}}$ and $\frac{2(x+y)}{\sqrt{x^2+1} \sqrt{y^2+1}}$.

If we pick side3 as $c$ for the inequality $ab < c$ (from Dan's question), this corresponds to, using the expressions for side1/2/3 above, $x+y \gt 2 x y$.

If we pick side2 as $c$ for the inequality $ab < c$ , this corresponds to, using the expressions for side1/2/3 above, $\frac{2x(x+y)}{x^2+1} \lt y$.

If we pick side1 as $c$ for the inequality $ab < c$ , this corresponds to, using the expressions for side1/2/3 above, $\frac{2y(x+y)}{y^2+1} \lt x$.

I tried plotting the union of the inequalities (assuming that the choice of labelling one of the sides, side1/2/3 as "$c$" has a probability of $\frac13$.

Plot

This seems to suggest that the probability is close to 1 as we expand the plot to positive quadrant of the real plane (becasue the only requirement is $x \gt 0$ and $y \gt 0$). The unshaded region should be negligible if we take the ratio of shaded to shaded plus unshaded, as we expand the square. Obviously this is wrong, given the number of proofs for the probability to be $\frac12$.

Question: Where am I going wrong in my approach?

Srini
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  • What is the probability density function that you're using? IE Are you saying that it's uniformly distributed on $ \mathbb{R}^2$? If yes, are there concerns about doing so? – Calvin Lin Apr 17 '24 at 15:52
  • I guess you are saying that due to nonlinear mapping of real line to the unit circle, uniform distribution on $R^2$ has to be undone with some arctan notion when calculating the probability using the terms in the triangle on the unit circle. I don't know how to do it, if that is the issue. – Srini Apr 17 '24 at 15:58
  • Right, you have to account for the probability density function, and so it's not a simple "geometric area comparison" as you seem to be suggesting. $ \quad$ You can deal with that using calculus, probability theory. – Calvin Lin Apr 17 '24 at 16:41
  • Thanks Calvin. Interestingly, Dan's neat result makes no assumption of the density function or the algorithm for picking the random points on the circle, sugesting that his approach is agnostic to those. However, I do see a point that in my noninear mapping case, it is relevant. I will do more work on this and post if I find anything that I can pursue further with this approach. – Srini Apr 17 '24 at 17:56
  • Dan specifically says "vertices of a triangle are three uniformly random points on a unit circle", which is an assumption on the density function. – Calvin Lin Apr 17 '24 at 18:46
  • Ah, sorry, I didn't read that carefully as I should have. Thanks – Srini Apr 17 '24 at 18:49
  • I tried to undo the $2 tan^{-1}(p)$ with $tan(\frac{()}{2})$ to see if I can then do a uniform area ratio in $R^2$. Most likely this is wrong too and I need to avoid the nonlinear map to start with. Anyway, here is the code for what I tried. NIntegrate[Boole[Tan[x/2]+Tan[y/2]>Tan[x/2]Tan[ y/2] || 2 Tanx/2/(Tan[x/2]^2+1) <Tan[y/2] || 2Tany/2/(Tan[y/2]^2+1) < Tan[x/2]],{x,0,10000},{y,0,10000},Method->{"GlobalAdaptive","MaxErrorIncreases"->10000},MaxRecursion->100]/10000^2 – Srini Apr 17 '24 at 19:37

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