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On a circle, choose three uniformly random points $A,B,C$.

Triangle $T_1$ has vertices $A,B,C$. The side lengths of $T_1$ are, in random order, $a,b,c$.

Triangle $T_2$ is formed by drawing tangents to the circle at $A,B,C$. ($T_2$ may or may not lie inside the circle.) The side lengths of $T_2$ are, in random order, $d,e,f$.

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Is the following conjecture true:

The distribution of ratios of $a,b,c$ is the same as the distribution of ratios of $d,e,f$.

So for example, if my conjecture is true, then we would have the following probability equations:

  • $P(a^2+b^2<c^2)=P(d^2+e^2<f^2)$
  • $P(a^3+b^3<c^3)=P(d^3+e^3<f^3)$
  • $P(ab<c^2)=P(de<f^2)$
  • $P\left(\frac1a+\frac1b<\frac1c\right)=P\left(\frac1d+\frac1e<\frac1f\right)$

(My conjecture, if true, would not imply that $P(ab<c)=P(de<f)$, because these inequalities are not dimensionally homogeneous.)

Basis for my conjecture

I found that $P(ab<c^2)=\frac35$. Then I wondered what $P(de<f^2)$ equals, and a simulation suggests that it also equals $\frac35$.

I found that $P\left(\frac1a+\frac1b<\frac1c\right)=\frac15$. Then I wondered what $P\left(\frac1d+\frac1e<\frac1f\right)$ equals, and a simulation suggests that it also equals $\frac15$.

I tried some other probabilities, and simulations suggest that each probability has the same value for the two triangles.

So I suspect that the two triangles have the same distribution of ratios of side lengths. But I don't know why.

Dan
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1 Answers1

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The angles in $T1$ are half the corresponding angles at the centre of the circle, so they are unuformly distributed on the triangle $\alpha+\beta+\gamma=\pi$.
The angles in $T2$ are either $\pi-2\alpha,\pi-2\beta,\pi-2\gamma$ or $2\alpha,2\beta,2\gamma-\pi$. I haven't worked out the details, but expect they are also uniformly distributed.

Empy2
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  • But what about the side lengths? I don't think the side lengths of $T_2$ are uniformly distributed. They can be arbitrarily large. – Dan Apr 04 '24 at 15:08
  • $T1$ and $T2$ are equally likely to be $1^\circ,29^\circ,150^\circ$, so are equally likely to have that specific ratio of lengths. – Empy2 Apr 04 '24 at 15:13
  • I don't think the central angles are uniformly distributed. Let $\theta$ be a central angle. $P(0^\circ<\theta<1^\circ)$ is much greater than $P(359^\circ<\theta<360^\circ)$. – Dan Apr 04 '24 at 16:55
  • I mean evenly spread on the triangle with vertices $(360^\circ,0^\circ,0^\circ),(0^\circ,360^\circ,0^\circ),(0^\circ,0^\circ,360^\circ)$ – Empy2 Apr 04 '24 at 21:11
  • That is, for example $32^\circ,146^\circ,182^\circ$ is just as likely as $61^\circ,200^\circ,99^\circ$ – Empy2 Apr 05 '24 at 06:52