Comments by "EebstertheGreat" (@EebstertheGreat) on "Heron’s formula: What is the hidden meaning of 1 + 2 + 3 = 1 x 2 x 3 ?" video.
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Hero's formula for the area of a triangle is one of those things introduced as a curiosity in a math textbook way back in middle school that I never really got a handle on. To a sixth grader, that is an impressively complex formula for an ancient to have discovered, and no proof was forthcoming. Through high school I saw it a couple more times, always in passing, a sort of neat oddity that seems compact but rarely gets used in practice. It was really neat to see an intuitive proof and motivation after all these years.
That said, it doesn't exactly seem useful. Even if you somehow do know the lengths of a triangle but not its angles, this formula is still not the fastest way to find the area. Typically, if you're doing this by hand, you will either have a table of square roots (for Hero's method) or of logs and logs of sines (for the law of sines method). That method is still faster, because you skip all the multiplication steps. If you want to compute the area of the triangle with a computer, you can use Newton's method to get the square root, and I assume Heron's formula really is faster. But the thing is, you basically never know all the side lengths of a triangle (and nothing else) before trying to find its area. Rather, you probably have coordinates, in which case the shoelace formula is by far the fastest.
So like, what is this formula actually good for? Is it just a novelty like the quartic formula? If it's never used, then no, I don't think it should be taught as part of a standard curriculum. The brief mentions in books for interested students are probably enough. There is so much I want to add to the math curriculum, and the curriculum is already packed as it is. It's hard to justify cramming in more random formulas to teach, prove, and memorize.
(BTW, although the phrase "Heron's formula" is seen pretty often in mathematical texts, in pretty much all other contexts in English, "Hero" is far more common than "Heron." Similarly, we say "Plato" rather than "Platon." The practice of Latinizing ancient Greek names is pretty standard in English. In classical Latin, the nominative singular would be "HERO," and the genitive singular would be "HERONIS." Since the Latin stem is still Heron-, the English adjective would be "Heronic" rather than "Heroic." Again, that's like the adjective "Platonic" rather than "Platoic. Other examples include "Pluto/Plutonic" and "Apollo/Apollonic." Admittedly, there are some exceptions, like the word "gnomon.")
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Well, if you know the x- and y-coordinates of each point, there is a much easier method. You have A = ½ |det([[x₁,y₁,1],[x₂,y₂,1],[x₃,y₃,1]])| = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁-y₂)| by the shoelace formula. No messing around with square roots or anything. You do have to take one absolute value in case the triangle is negatively oriented, but that's still much faster than taking a square root.
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If (1³+2³+3³)=(1+8+27)=36=∆, then √∆ = √36 = 6 = 1+2+3 = a+b+c, and
if (0³+2³+2³)=(0+8+8) = 16=∆, then √∆ = √16 = 4 = 0+2+2 = a+b+c, but
if (0³+1³+1³)=(0+1+1) = 2 = ∆, then √∆ = √2 = 1.41... ≠ 2 = 0+1+1 = a+b+c.
However, there are other solutions. In particular, a=b=c=3 gives
(3³+3³+3³)=(27+27+27)=81=∆, so √∆ = √81 = 9 = 3+3+3 = a+b+c,
a = 0, b=1, c=2 and other permutations give
(0³+1³+2³)=(0+1+8)=9=∆, so √∆ = √9 = 3 = 0+1+2 = a+b+c,
a=b=0, c=1 and other permutations give
(0³+0³+1³)=(0+0+1)=1=∆, so √∆ = √1 = 1 = 0+0+1 = a+b+c,
and a=b=c=0 gives
(0³+0³+0³)=(0+0+0)=0=∆, so √∆ = √0 = 0 = 0+0+0 = a+b+c.
So the complete solution set over the nonnegative integers is (a,b,c)∈{(0,0,0); (0,0,1); (0,1,0); (1,0,0); (0,1,2); (0,2,1); (1,0,2); (1,2,0); (2,0,1); (2,1,0); (0,2,2); (2,0,2); (2,2,0); (1,2,3); (1,3,2); (2,1,3); (2,3,1); (3,1,2); (3,2,1); (3,3,3)}.
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