If $(a,b,c)$ are the sides of a triangle and $x \ge 1$, what is the probability that $a+b > cx$?

Question

I am trying to generalize the triangle inequality in probabilistic terms. Assume that the vertices are uniformly distributed around the circumference of a fixed circle. In this question it was proved that if $(a,b,c)$ are the sides of a triangle than the probability that geometric mean of any two sides is greater than the third side i.e. $P\left(\sqrt{ab} > c\right) = \frac{2}{5}$. Since $a+b > 2\sqrt{ab}$, it implies that probability that $P(a+b > 2c) > \frac{2}{5}$. In general, we can ask:

Question: If $(a,b,c)$ are the sides of a triangle and $x \ge 1$, what is the probability that $a+b > cx$?

Equivalence with the Basel's Problem: The series in the accepted answer is actually the Lengendre Chi function $\chi_2\left(\frac1{x}\right)$; hence

$$ P(a+b > cx) = \frac{8}{\pi^2}\chi_2\left(\frac1{x}\right) $$

Taking $x = 1$ and applying the fact that the sum of the reciprocal of odd squares is $3/4$-th the sum over the reciprocal of the squares of natural numbers we find that the probability that the sum of any two sides of a triangle is greater than than the third side is $\displaystyle \frac{6\zeta(2)}{\pi^2}$. Since this probability must be $1$ and the proof does not require the value of $\zeta(2)$ to be known in advance, it unexpectedly shows that:

The triangle inequality equivalent to the Basel Problem $$ 1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6} $$

I have changed the title to reflect this remarkable connection.

Related question: If $(a,b,c)$ are the sides of a triangle, is it true that probability that $a+b > c^{\frac{3}{c}}$ is $\zeta(2)-1$?

Answer 1

I have a formula for the probability, which for large $x$ is roughly $$\frac4{\pi^2}\left(\frac2x+\frac2{9x^3}+\frac2{25x^5}+\cdots\right)$$ The series continues in the same form up to at least the $x^{-11}$ term. There is an integral form at the end of this answer.

Solution:

The angles in the triangle are all twice the corresponding angles at the centre of the circle. So the trio of angles $(\alpha,\beta,\gamma)$ is equally distributed over the flat region in 3space with vertices $(\pi,0, 0),(0,\pi,0),(0,0,\pi)$.
The area of a triangle with sides $a,b,c$ and angles $\alpha,\beta,\gamma$ is $$\Delta=\frac12ab\sin\gamma=\frac12ca\sin\beta=\frac12bc\sin\alpha$$
So for example $a=\frac{abc}{2\Delta}\sin\alpha$, and the ratio we want is $$\frac{a+b}c=\frac{\sin\alpha+\sin\beta}{\sin\gamma}\\=\frac{2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2}{\sin(\alpha+\beta)}\\ =\frac{\cos\frac{\alpha-\beta}2}{\cos\frac{\alpha+\beta}2}$$
The region of interest is $0\le\alpha,0\le \beta,\alpha+\beta\le\pi$. Change variables to $$\theta=\frac{\alpha+\beta}2,\phi=\frac{\alpha-\beta}2$$
The region is now $$0\le\theta\le\pi/2,\\-\theta\le\phi\le\theta$$ with area $\pi^2/4$.
The ratio of interest is greater than $x$ when $$\cos\theta\lt\frac{\cos\phi}x$$
For a given $\phi$, Theta ranges over an interval of length $$\pi/2-\arccos(\frac{\cos\phi}x)\\=\arcsin(\frac{\cos\phi}x)$$
Integrate that over the region to give $$\frac4{\pi^2}\int^{\pi/2}_{-\pi/2}\arcsin(\frac{\cos\phi}x)d\phi$$
I used a series expansion in arcsine $x+\frac{x^3}6+\frac{3x^5}{40}+\cdots$ to get the series at the top.

Answer 2

The posted code is wrong. I've refactored/bugfixed it slightly and the asymptotic value is in fact $8 / \pi^2$. The issue with sorting angles is that $a$ will be too small compared to $c$ (compared to if $a, b, c$ were generated from uniformly selected points and not ordered themselves).

from math import pi
import numpy as np
k = 2 * np.pi
n = 10**5
x = 1.0
max = 50
inc = 0.5
just run n trials for each value of x instead
while x < max:
    success = 0
    for _ in range(1, n):
        # remove the sort
        angles = np.random.uniform(0, k, 3)
    vertices_x = np.cos(angles)
    vertices_y = np.sin(angles)

    vertices_x = np.append(vertices_x, vertices_x[0])
    vertices_y = np.append(vertices_y, vertices_y[0])

    x_diff = np.diff(vertices_x)
    y_diff = np.diff(vertices_y)
    side_lengths = np.sqrt(x_diff ** 2 + y_diff ** 2)

    a = side_lengths[0]
    b = side_lengths[1]
    c = side_lengths[2]

    if a + b &gt; x * c:
        success += 1

print(f'x: {x:.4f}'[:-1], f'x * P(x): {x * success / n:.10f}'[:-1], f'8/pi^2: {8 / pi ** 2}')

x += 0.5

Sample output:

x: 1.000 x * P(x): 0.999990000 8/pi^2: 0.8105694691387022
x: 1.500 x * P(x): 0.861450000 8/pi^2: 0.8105694691387022
x: 2.000 x * P(x): 0.830600000 8/pi^2: 0.8105694691387022
x: 2.500 x * P(x): 0.830900000 8/pi^2: 0.8105694691387022
x: 3.000 x * P(x): 0.825840000 8/pi^2: 0.8105694691387022
x: 3.500 x * P(x): 0.817775000 8/pi^2: 0.8105694691387022
x: 4.000 x * P(x): 0.818160000 8/pi^2: 0.8105694691387022
x: 4.500 x * P(x): 0.814860000 8/pi^2: 0.8105694691387022
x: 5.000 x * P(x): 0.818700000 8/pi^2: 0.8105694691387022
x: 5.500 x * P(x): 0.818895000 8/pi^2: 0.8105694691387022
x: 6.000 x * P(x): 0.804540000 8/pi^2: 0.8105694691387022
x: 6.500 x * P(x): 0.800540000 8/pi^2: 0.8105694691387022
x: 7.000 x * P(x): 0.809200000 8/pi^2: 0.8105694691387022
x: 7.500 x * P(x): 0.809550000 8/pi^2: 0.8105694691387022
x: 8.000 x * P(x): 0.808240000 8/pi^2: 0.8105694691387022
x: 8.500 x * P(x): 0.808520000 8/pi^2: 0.8105694691387022
x: 9.000 x * P(x): 0.815310000 8/pi^2: 0.8105694691387022
x: 9.500 x * P(x): 0.797240000 8/pi^2: 0.8105694691387022
x: 10.000 x * P(x): 0.815900000 8/pi^2: 0.8105694691387022
x: 10.500 x * P(x): 0.802200000 8/pi^2: 0.8105694691387022
x: 11.000 x * P(x): 0.827090000 8/pi^2: 0.8105694691387022
x: 11.500 x * P(x): 0.813280000 8/pi^2: 0.8105694691387022
x: 12.000 x * P(x): 0.818640000 8/pi^2: 0.8105694691387022
x: 12.500 x * P(x): 0.813875000 8/pi^2: 0.8105694691387022
x: 13.000 x * P(x): 0.815750000 8/pi^2: 0.8105694691387022
x: 13.500 x * P(x): 0.806895000 8/pi^2: 0.8105694691387022
x: 14.000 x * P(x): 0.832300000 8/pi^2: 0.8105694691387022
x: 14.500 x * P(x): 0.802865000 8/pi^2: 0.8105694691387022
x: 15.000 x * P(x): 0.823050000 8/pi^2: 0.8105694691387022
x: 15.500 x * P(x): 0.805225000 8/pi^2: 0.8105694691387022
x: 16.000 x * P(x): 0.816480000 8/pi^2: 0.8105694691387022
x: 16.500 x * P(x): 0.805035000 8/pi^2: 0.8105694691387022
x: 17.000 x * P(x): 0.801380000 8/pi^2: 0.8105694691387022
```

Answer 3

I make a connection with a famous problem the disk triangle picking see

Dtp

Note that the probability decrease as the graph of the Op .