If $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$ form a triangle, what is the probability that $(a,b,c)$ also form a triangle?

Question

Let $A_N$ be set of triplets $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$, $1\le a \le b \le c \le N$ such that $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$ are the sides of a triangle.

Let $B_N$ be the set of triplets $(a,b,c)$, $1\le a \le b \le c \le N$, $(\gcd(a,b), \gcd(b,c), \gcd(c,a)) \in A_N$ such that $(a,b,c)$ are the sides a triangle.

Let $A(N)$ and $B(N)$ be the number of elements in $A_N$ and $B_N$ respectively. Is the following conjecture true?

$$ \lim_{n \to \infty} \frac{A(N)}{B(N)} = \frac{1}{2} $$

Experimental evidence: Out the the first $1,012,270,546$ triplets where $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$ formed a triangle, there were $505,222,520$ triplets of $(a,b,c)$ which also formed a triangle, giving a ratio of $0.499098$. Increasing the range for the first $24,292,129,662$ triplets $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$ increased the ratio only slightly to $0.499697$.Thus experimental data indicates that the ratio is possibly $\displaystyle \frac{1}{2}$. Can this be proved?

Note: The converse is different i.e. if $(a,b,c)$ form a triangle then the probability that $(\gcd(a,b), \gcd(b,c), \gcd(c,a))$ also form a triangle is about $0.345$ (not sure if thi converges to a well known constant).

Code

import math as mt
a = 1
t = tg = ttg = 0

while True:
    b = 1
    while b <= a:
        c = 1
        while c <= b:
            ab = mt.gcd(a,b)
            bc = mt.gcd(b,c)
            ca = mt.gcd(c,a)
                
            if ab+bc>ca and bc+ca>ab and ca+ab>bc:
                tg = tg+1
                if a+b>c and b+c > a and c+a>b:
                    t = t + 1
            c = c + 1
        b = b + 1
    
    print(a,tg,t,t/tg)
    
    a = a + 1

Answer 1

Let $d,e,f = \gcd(a,b),\gcd(b,c),\gcd(c,a)$ respectively.

Note that the unconditional probability that $a,b,c$ form a triangle is already $1/2$; in the limit, this is the same as asking for the probability that $a,b,c$ form a triangle when drawn uniformly from $[0,1]$. Since we want to exclude the three disjoint outcomes $a>b+c, b>a+c, c>a+b$, we can do a bit of calculus to see that each of these three events has a probability of $1/6$ so the chance that none of them happen is $1-3/6 = 1/2$.

So all we want to show here is that the gcd event has no bearing on our prior beliefs about $a,b,c$ forming a triangle - ie, that after learning about $d$, $e$, and $f$, we still think the probability is $1/2$ like we did before.

To see this, note that as $M$ goes to infinity, the odds that $d,e,f<M$ approaches $1$. So almost all the probability mass is contained within $M^3$ different cases for $(d,e,f)$.

But for any given values of $d,e,f$, as we let $N$ go to infinity, learning that $a,b,c$ have those gcds doesn't tell us anything about their size relative to one another - when we zoom out the distribution looks uniform. This is because for any triple $(a,b,c)$ with $\gcd(a,b)=d,\gcd(b,c)=e,\gcd(c,a)=f$, we can add any multiple of $d\cdot e\cdot f$ to any of $a,b,c$ and get another valid triple. So when $N\gg def$ this gcd information doesn't tell us anything about the global structure.