Timeline for Dissection proof of Heron's formula?
Current License: CC BY-SA 4.0
14 events
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| Oct 13 at 19:51 | comment | added | Timothy Chow | @MichaelHardy Matt Parker's YouTube channel is entitled "Stand-up Maths", and while he does try to be accurate, he's not above a bit of overstatement for rhetorical effect. In particular, I don't think he's claiming that the formula is an "arbitrary hodge-podge"; it's true, after all! I also think Parker would appreciate the street-fighting mathematics character of your observations, and regard it as a first step toward removing the opacity, but maybe not more than that. | |
| Oct 13 at 19:39 | comment | added | Michael Hardy | @TimothyChow : I wonder if you understood what I meant by "does it show that Matt Parker's pessimistic assessment is mistaken?". Note what Matt Parker wrote: "opaque formula, and I feel like you just chuck in the side-lengths, turn a series of arbitrary mathematical handles, and out pops the answer. It works, but it's not satisfying." My argument is not a proof, maybe even at an "intuitive" level but Matt Parker seems to think the formula is an arbitrary hodge-podge, and it seems to me that what I wrote should at least show that it's better than that. | |
| Oct 11 at 9:56 | comment | added | Emil Jeřábek | ... for some polynomial $g$, and $g$ has to be a symmetric homogeneous linear polynomial, thus the only possibility is that it is a scalar multiple of $a+b+c$. | |
| Oct 11 at 9:53 | comment | added | Emil Jeřábek | The argument can be made rigorous under the assumption that the square of the area is a polynomial $f(a,b,c)$. It, indeed, has to be homogeneous of degree $4$, and divisible by the square-free polynomials $a+b-c$, $a+c-b$, $b+c-a$ as it vanishes on $\{(a,b,c)\in\mathbb R^3:a,b,c\ge0,a+b=c\}$, and this set is Zariski dense in $\{(a,b,c)\in\mathbb C^3:a+b=c\}$ (this is intuitively clear, and easy to prove). The argument for the $a+b+c$ factor in the answer is thouroughly nonsensical, but it can be circumvented as follows: we already know $f(a,b,c)=(s-a)(s-b)(s-c)g(a,b,c)$ ... | |
| Oct 11 at 3:08 | comment | added | Timothy Chow | @MichaelHardy In my opinion, your argument at best has the form of a sanity check, similar to the way dimensional analysis of certain formulas in physics provides a sanity check. In my mind, a sanity check is only a small step toward an explanation. For comparison, we know from elementary Newtonian mechanics that kinetic energy is $\frac{1}{2} mv^2$, so by dimensional analysis we see that $E = mc^2$ also makes sense. But if that's the only justification we have for $E = mc^2$ then the formula remains mysterious. | |
| Oct 11 at 1:19 | comment | added | Michael Hardy | Although this argument is not a complete proof that this particular function is the one, does it show that Matt Parker's pessimistic assessment is mistaken? | |
| Oct 10 at 23:06 | comment | added | Bruno Le Floch | The area is (half) the product of a side by the height, so the question of why we get square roots boils down to why the height is a square root. This can be tracked down by applying Pythagoras to both right triangles made by the height. | |
| Oct 10 at 21:21 | comment | added | Michael Hardy | @EmilJeřábek : I think if I ponder this for a few decades, I can probably come up with a cogent reason why it ought to be the simplest thing that meets some simple desiderata, and your thing involving fifth powers is not simple enough. | |
| Oct 10 at 20:35 | comment | added | Emil Jeřábek | Hmm. One could argue that it should be in an iterated quadratic extension of $\mathbb Q(a,b,c)$ as it is constructible with a ruler and a compass. In that case, consider e.g. $\sqrt[4]{(a^5+b^5+c^5)(s-a)(s-b)(s-c)}$. | |
| Oct 10 at 20:26 | comment | added | Emil Jeřábek | $(s-a)\sqrt{(s-b)(s-c)}$ can be ruled out as it is not symmetric, but one can consider e.g. $((s-a)(s-b)(s-c))^{2/3}$ instead. | |
| Oct 10 at 20:21 | comment | added | Emil Jeřábek | Also, the reasoning does not explain why we should expect a square root of a polynomial of degree $4$, rather than, say, a cube root of a polynomial of degree $6$, or even a more complicated algebraic function. | |
| Oct 10 at 20:19 | comment | added | Emil Jeřábek | The argument for the $a+b+c$ factor is faulty, as polynomials that vanish at $(0,0,0)$ are not necessarily multiples of $a+b+c$; indeed, the other three factors already make the expression $0$ when $a=b=c=0$. The constraints you list would be satisfied e.g. by the wrong formula $(s-a)\sqrt{(s-b)(s-c)}$ (scaled by a suitable constant). | |
| Oct 10 at 20:11 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Oct 10 at 19:59 | history | answered | Michael Hardy | CC BY-SA 4.0 |