Second pair of triangles with equal sum of integer Torricelli-Ferma distanses

Since a Torricelli-Fermat point $P$ has the property that $\angle APB = \angle APC = \angle BPC = 120°$ , then by the law of cosines:

$AB^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos 120° = PA^2 + PA \cdot PB + PB^2$

$AC^2 = PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos 120° = PA^2 + PA \cdot PC + PC^2$

$BC^2 = PB^2 + PC^2 - 2 \cdot PB \cdot PC \cdot \cos 120° = PB^2 + PB \cdot PC + PC^2$

The below Python program finds all numbers $a$ and $b$ between $0 < a < b < 15000$ such that $a^2 + ab + b^2$ is a perfect square, then uses that list to find integer sets of $(PA, PB, PC)$ that give integer triangle sides, and finally displays all $h = PA + PB + PC$ with more than one possible $(PA, PB, PC)$ set. The second $h$ value computes to $\boxed{20280}$ .

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# Second Pair of Triangles with Equal Sum of Integer Torricelli-Fermat Distances
# Python

# set up variables
# f[a][b] is a list of numbers such that a ** 2 + a * b + b ** 2 is a perfect square.
#   For example, f[5] = [3, 16] because 5 ** 2 + 5 * 3 + 3 ** 2 is a perfect square and
#   5 ** 2 + 5 * 16 + 16 ** 2 is a perfect square.
# g[h][(PA, PB, PC)] is a list of numbers such that PA and PB, PB and PC, and PA and PC are in f and
#   h = PA + PB + PC. For example, g[16464] = [(4095, 5544, 6825), (4224, 5200, 7040)]
import math
fmax = 15000
gmax = 3 * fmax
f = []
for a in range(0, fmax):
    f.append([])
g = []
for a in range(0, gmax):
    g.append([])

# function to determine if x is a perfect square
def isSquare(x):
    s = int(math.sqrt(x))
    return (s ** 2 == x)

# populate f
for a in range(1, fmax):
    for b in range(a + 1, fmax):
        if isSquare(a ** 2 + a * b + b ** 2):
            f[a].append(b)
            f[b].append(a)

# populate g
for a in range(0, fmax):
    for b in range(0, len(f[a])):
        for c in range(0, len(f[f[a][b]])):
            for d in range(0, len(f[f[f[a][b]][c]])):
                PA = a
                PB = f[a][b]
                PC = f[f[a][b]][c]
                P4 = f[f[f[a][b]][c]][d]
                if PA == P4 and PA < PB and PB < PC:
                    h = PA + PB + PC
                    g[h].append((PA, PB, PC))

# display results
for a in range(0, gmax):
    if len(g[a]) > 1:
        print("h = " + str(a) + " for (PA, PB, PC) = " + str(g[a])[1:-1])

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h = 16464 for (PA, PB, PC) = (4095, 5544, 6825), (4224, 5200, 7040)
h = 20280 for (PA, PB, PC) = (1305, 4968, 14007), (6307, 6765, 7208)
h = 32928 for (PA, PB, PC) = (8190, 11088, 13650), (8448, 10400, 14080)