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Let $(a,b,c)$ be the sides of a triangle inscribed inside a unit circle such that the vertices of the triangle are distributed uniformly on the circumference. The regular triangle inequality states that the sum of any two sides is greater than the third side. But what happens if we take the sum of the reciprocal of any two sides? Is it greater than the third side? It turns out that the reciprocal triangle inequality $\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}$ is not true in general however, experimental data shows an interesting observation that the probability

$$ P\left(\frac{1}{a} + \frac{1}{b} \ge \frac{1}{c}\right) = \frac{4}{5} $$

Can this be proved or disproved? Note that this is equivalent proving or disproving

$$ P\left(\frac{1}{\sin x} + \frac{1}{\sin y} \ge \frac{1}{|\sin (x+y)|} \right) = \frac{4}{5} $$

where $0 \le x,y \le \pi$.

Related question: Probability that the geometric mean of any two sides of a triangle is greater than the third side is $\displaystyle \frac{2}{5}$.

Nilotpal Sinha
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  • For clarity: of course, the usual triangle inequality holds that any two sides sum to more than the third. Is that what you want here or do you mean "for a specific choice of $c$"? – lulu Mar 09 '24 at 14:51
  • @lulu I meant for any two sides i.e. The reciprocal triangle inequality conjecture says that in any triangle the probability that the sum of the reciprocals of any two sides is greater than than the third side is $4/5$. I have now mentioend this specifically in the post. – Nilotpal Sinha Mar 09 '24 at 14:54
  • Are $x,y$ uniformly distributed? – Maximilian Janisch Mar 09 '24 at 14:56
  • But then it's not obvious (to me) that your trigonometric version is equivalent. Wouldn't you need to have three inequalities? Are you claiming that one of these implies the other two? – lulu Mar 09 '24 at 14:57
  • Of course it would suffice to consider the labelling $(a,b,c)$ for which $c$ was the shortest side. – lulu Mar 09 '24 at 15:03
  • @lulu Note that the probability $4/5$ in the unordered case where $c$ is not necessarily the smallest side is equivalent to the probability $2/5$ in ordered case where $c$ is the smallest side as explained in the comments and the solution of this related problem which also explains that it is easier to solve for the unordered case and then use it to derive the probability the ordered case. https://math.stackexchange.com/questions/4836901/if-a-b-c-are-the-sides-of-a-triangle-what-is-the-probability-that-acb2 – Nilotpal Sinha Mar 09 '24 at 15:37
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    @MaximilianJanisch Since the vertices are uniformly distributed on the circumference of the unit circle, it follows that $x,y$ will have to too when we shift from cartesian to polar coordinates as shown in the related question added to the post. – Nilotpal Sinha Mar 09 '24 at 15:51
  • In the OP, there should be absolute value signs around the $\sin(x+y)$. – Dan Mar 10 '24 at 11:53
  • @Dan Thanks. Updated – Nilotpal Sinha Mar 10 '24 at 11:54
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    @NilotpalSinha I've added a geometrical implication to my answer. – Dan Jun 22 '25 at 22:29

1 Answers1

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Assume that the circle is centred at the origin, and the vertices of the triangle are:
$A(\cos(-2Y),\sin(-2Y))$ where $0\le Y\le\pi$
$B(\cos(2X),\sin(2X))$ where $0\le X\le\pi$
$C(1,0)$

Let:
$a=BC=2\sin X$
$b=AC=2\sin Y$
$c=AB=\left|2\sin\left(\frac{2\pi-2X-2Y}{2}\right)\right|=|2\sin(X+Y)|$

$P\left[\frac1a+\frac1b\ge\frac1c\right]=1-P\left[\frac1a+\frac1b<\frac1c\right]$

$P\left[\frac1a+\frac1b<\frac1c\right]=P\left[\frac{1}{\sin X}+\frac{1}{\sin Y}<\frac{1}{|\sin(X+Y)|}\right]$

This last probability is the ratio of the area of the shaded region to the area of the square in the graph below.

enter image description here

Rotate these regions $45^\circ$ clockwise about the origin, then shrink by factor $\frac{1}{\sqrt2}$, by letting $X=x-y$ and $Y=x+y$.

enter image description here

Using symmetry, we only need to consider the left half of the blue "diamond". Note that in the left half, $0<x<\pi/2$, so $|\sin(2x)|=\sin(2x)$.

$P\left[\frac{1}{\sin X}+\frac{1}{\sin Y}<\frac{1}{|\sin(X+Y)|}\right]$
$=P\left[\frac{1}{\sin(x-y)}+\frac{1}{\sin(x+y)}<\frac{1}{\sin(2x)}\right]$

Now we want to express the inequality as $-f(x)<y<f(x)$ for some $f(x)$.

$=P\left[\sin(2x)(\sin(x+y)+\sin(x-y))<\sin(x-y)\sin(x+y)\right]$
$=P\left[2\sin(2x)(\sin x)(\cos y)<(\sin^2x)(\cos^2y)-(\cos^2x)(\sin^2y)\right]$
$=P\left[2\sin(2x)(\sin x)(\cos y)<(\sin^2x)(\cos^2y)-(\cos^2x)(1-\cos^2y)\right]$
$=P\left[\cos^2y-2\sin(2x)(\sin x)(\cos y)-\cos^2x>0\right]$

Solving the quadratic in $\cos y$ gives

$=P\left[\cos y>(\cos x)\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right]$
$=P\left[-f(x)<y<f(x)\right]$

where $\color{red}{f(x)=\arccos\left((\cos x)\left(2\sin^2x+\sqrt{1+4\sin^4x}\right)\right)}$

Note that $f(\arccos(1/4))=0$ and $f(\pi/2)=\pi/2$.

So we have $P\left[\frac1a+\frac1b\ge\frac1c\right]=1-\dfrac{\int_{\arccos(1/4)}^{\pi/2}f(x)\mathrm dx}{\frac12 \left(\frac{\pi}{2}\right)^2}$

Here it is shown that $\int_{\arccos(1/4)}^{\pi/2}f(x)\mathrm dx=\dfrac{\pi^2}{40}$.

$\therefore P\left[\frac1a+\frac1b\ge\frac1c\right]=1-\dfrac15=\dfrac45$.

A nice geometrical implication

This result is equivalent to:

A triangle's vertices are three uniformly random points on a circle. Call the altitudes $p,q,r$. The probability that line segments of lengths $p,q,r$ can form a triangle is $2/5$.

Explanation:

The altitude perpendicular to base $a$ is $\dfrac{2S}{a}$ where $S$ is the area of the triangle. Same for $b$ and $c$.

If the three altitudes cannot form a triangle, then at least one of the following must be true:

  • $\dfrac{2S}{a}>\dfrac{2S}{b}+\dfrac{2S}{c}$
  • $\dfrac{2S}{b}>\dfrac{2S}{a}+\dfrac{2S}{c}$
  • $\dfrac{2S}{c}>\dfrac{2S}{a}+\dfrac{2S}{b}$

The $2S$'s cancel. By my answer to the OP, each one is true with probability $1/5$, and they are mutually exclusive, so the probability that at least one is true is $3/5$. So the probability that the three altitudes can form a triangle is $2/5$.

Dan
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