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Tue, 04 Apr 2006
Hero's formula
this mathematician's best known work is the formula for the area of a triangle in terms of the lengths of it's sides. who is this and when did they live?This is the kind of question that always trips me up in Jeopardy. (And doesn't that first sentence sounds just like a Jeopardy question? "I'll take triangles for $600, Alex. . .") The ears hear "mathematician", "triangle", "formula", and the mouth says "Who was Pythagoras!" without any conscious intervention. But the answer here is almost certainly Hero of Alexandria. Hero's formula, or Heron's formula, has always seemed to me like something of an oddity. It doesn't look like any other formula in geometry. Usually to get anything useful from a triangle you need to involve the angles, and express the results in terms of trigonometric functions. This will be obvious if you pause to remember what the phrase "trigonometric function" means. The few rare cases in which you can avoid trigonometry (there's that word again) usually involve special cases, such as right triangles. But Hero's formula expresses the area of a triangle in terms of the lengths of its sides, with no angles and no trigonometric functions. If we were trying to discover such a formula, we might proceed trigonometrically, and then hope we could somehow eliminate the trigonometry by the end. So we might proceed as follows: The magnitude of the cross product a×b is the area of the parallelogram with sides a and b, so the area of the triangle is half that. And |a × b| = |a||b| sin θ, where θ is the angle between the two vectors. So if a and b are sides of a triangle and the included angle is θ, the area A is ab/2 · sin θ. Now we want θ is in terms of the third side c. The law of cosines generalizes the Pythagorean theorem to non-right triangles: c^{2} = a^{2} + b^{2} - 2ab cos θ, where θ is the included angle between sides a and b. (In a right triangle, θ = 90°, and the cosine term drops out.) So we have cos θ = (a^{2} + b^{2} - c^{2}) / 2ab. But we want sine instead of cosine, so convert cosine to sine using sin θ = √(1-cos^{2}&theta);, which yields:
$$\sin\theta = {1\over 2ab}\sqrt{4a^2b^2 - {(a^2 + b^2 - c^2)}^2}$$ Expanding out the right-hand side, and remembering that what we really want is A = ab/2 sin θ. we get:
$$A = {1\over4}\sqrt{2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4}$$. Now, that might satisfy anyone, but Hero's answer is better. Hero says: Let p be the "semiperimeter" of the triangle; that is, p = (a+b+c)/2. Then the area of the triangle is just √(p(p-a)(p-b)(p-c)). This is "Hero's formula".Hero's formula is simple and easy to remember, which (2a^{2}b^{2} + 2a^{2}c^{2} + 2b^{2}c^{2} - a^{4} - b^{4}- c^{4})/16 is not. If you expand out Hero's formula, you find that that the two formulas are the same, as of course they must be. But if you have only the complicated fourth-degree polynomial, how would you get the idea that putting it in terms of p will simplify it? There is some technique or insight here that I am missing. Even though such technique is the indispensable tool of the working mathematician, mathematical writing customarily scorns such technique,and has numerous phrases for disparaging it or kicking it under the carpet. For example, having gotten the fourth-degree polynomial, we might say "obviously, this is equivalent to Hero's formula", or "it is left to the student to prove that the two formulations are equivalent." Or we could just skip direct from the polynomial to Hero's formula, and say what Wikipedia says in the same situation: "Here the simple algebra in the last step was omitted." Indeed, if you are trying to prove that Hero's formula is true, it's tedious but straightforward to grovel over the algebra and grind out that the two formulations are the same. But all this ignores the real problem, which is that nobody looking at the trigonometric formulation would know to guess Hero's formula if they had not seen it before. A proof serves two purposes. it is supposed to persuade you that the theorem is true. But more importantly, it is supposed to help you understand why it is true, and to give you some insight that may help you solve similar problems. The proof-by-handwaving-away-complex-and-unexpected-algebra technique serves the first purpose, but not the second. And Hero did not discover the formula using this approach anyway, since he did not take a trig class when he was in high school. Another way to proceed is to drop a perpendicular from the vertex opposite side c. If the length of the perpendicular is h, the area of the triangle is hc/2. But if the foot of the perpendicular divides c into x + y, we also have x^{2} + h^{2} = a^{2} and y^{2} + h^{2} = b^{2}. Grovelling over the algebra again yields something equivalent to Hero's formula. I don't know a good proof of Hero's formula. I haven't seen Hero's, about which MathWorld says:
Heron's proof is ingenious but extremely convoluted, bringing together a sequence of apparently unrelated geometric identities and relying on the properties of cyclic quadrilaterals and right triangles.A "cyclic quadrilateral" is a quadrilateral whose vertices all lie on a circle. Any triangle can be considered a degenerate cyclic quadrilateral, which happens to have two of its vertices in the same place. Indeed, there's a formula, called "Bretschneider's formula", just like Hero's, but for the area of a cyclic quadrilateral: A = √((p-a)(p-b)(p-c)(p-d)), where a, b, c, d are the lengths of the sides and p is the semiperimeter. If you consider that a triangle is just a cyclic quadrilateral one of whose sides has length 0, you put d = 0 in Bretschneider's formula and you get out Hero's formula.
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