Talk:Heron's formula

Order of variables

how does making a>b>c change the formula? i can't see how it influences it.

It changes the stability of the numerical compuation on a floating-point processor. In the limit case of needle triangles, where one side is very small, you want that side to be 'c' in the formula. In perfect real-number computation it doesn't make any difference, of course, but if you're not worried about numerical stability, you wouldn't use this version of the formula anyway; you'd just use the millenia-old original one. --Delirium 05:51, 9 November 2005 (UTC)

It would seem to be the unique parenthesization minimizing the intermediate results while keeping them all nonnegative, subject of course to the triangle inequality. As such it never uses addition when subtraction is possible. However I'd be interested to see an example where (a+b)+c is less accurate than a+(b+c). --Vaughan Pratt (talk) 20:42, 22 February 2008 (UTC)

When did Heron, otherwise known as Hero discover the fomula?

The article says in the 1st century A.D. Do you need a date? --Gesslein 14:22, 25 July 2006 (UTC)

It would be nice but my homework is due in tomorow so you prob. won't tell me in time. I'm sure 1st Century AD is enough (hopefully). :D Jess

I don't have a date, probably sometime near 50 A.D. it was published. --Gesslein 19:13, 28 July 2006 (UTC)

The section "Ch'in Chiu-shao's fomula" really belongs to the history section. Most of it doesn't add to Heron's formula and can be found on Ch'in Chiu-shao’s. The sentence "This fomula was proved recently by Wu Wenjun" is very strange! Anyone can say were did it came from? Anyway I am deleting this section and making the appropriate changes. Ricardo sandoval 05:19, 9 July 2007 (UTC)

Spelling

is another name for this formula Hero's Formula? —Preceding unsigned comment added by 63.215.27.161 (talk) 00:57, 27 November 2007 (UTC)

Yes, Hero's Formula and Heron's Formula are the same. The spellings "Hero of Alexandria" and "Heron of Alexandria" are both used, and neither is correct, although the Greek language purists would probably prefer the version with the n. --♦IanMacM♦ ^{(talk to me)} 08:11, 27 November 2007 (UTC)

Properly he is Hero in English, not Heron. Just as it is Plato and Meno. I have no idea of the history of this in math circles, but the name is very definitely Hero. If this were in Greek, we could worry about those purists, but even classicists say Hero. Eponymous-Archon (talk) 02:27, 24 March 2014 (UTC)

For anyone who is interested, the situation is as follows. In Greek, the name is Ἡρων, which transliterates into the Latin alphabet as Hērōn. Latin, being reasonably closely related to Greek, had a noun declension recognisably related to the Greek declension that included the name Hērōn. However, at an early stage of development of the Latin language, the n at the end of the nominative case was lost. Romans tended to use a Latinised form of Greek names of this type, in which the n was omitted, by analogy with Latin words in the corresponding declension. Cases other than the nominative retained the n, so that for example the genitive was Heronis. The name "Hero" was in this respect treated in exactly the same way as many other Greek names, such as Plato. In English, it has long been common practice to use ancient Greek names in their Latinised form, thus we refer to Plato, and not Platon, but this is not universal, and some ancient Greek names are used in a form which is a direct transcription of the Greek, even when that differs from the Latin spelling. In the case of the name Ἡρων, both the Greek form Heron and the Latin form Hero have been widely used in English.

What about which form is "correct", or, as Eponymous-Archon puts it "proper"? It is possible to take the view that the Greek form is the "correct" one, since it is a Greek name. It is also possible to take the view that what is "correct" in language is defined by what is used and accepted, and since both forms are widely accepted, both are equally correct. I find it difficult to see, however, any justification for the view that "Hero" is in some sense more proper in English. The editor who uses the pseudonym "JamesBWatson" (talk) 12:48, 24 March 2014 (UTC)

Errors

There is an error in the expression that follows "Expressing Heron's formula with a determinant in terms of the squares of the distances between the three given vertices,". It is missing a necessary minus sign in front of the determinant and under the square root radical. See http://mathworld.wolfram.com/HeronsFormula.html —Preceding unsigned comment added by BillFish321 (talk • contribs) 09:20, 6 March 2008 (UTC)

There is an error in the formula that discusses the equivalency of Heron's formula to that of the Chinese Qin Jiushao, published in A.D. 1247. As stated, the Chinese version does not give the correct value for the area of a general triangle in 3-space with edge lengths a,b,c.

"A"

Is it sensible to use A both for the area of the triangle and for the name of a vertex in the diagram. How about chnaging the diagram and all related text from ABCabc to PQRpqr? -- SGBailey (talk) 09:40, 22 June 2008 (UTC)

The reason that this question has remained unanswered for ten years is that it is a non-issue. Why not change the symbol most commonly used for area to a K as many authors have done? The possibility of confusing a vertex label with a quantity associated with a triangle is very, very small. We report on what is in the literature and do not try to "fix" it or make it better, as this suggestion would have it. Labeling the vertices of a triangle ABC is almost universal and that is what we should do here. --Bill Cherowitzo (talk) 20:09, 5 December 2018 (UTC)

Fixed this -Jam Jamgoodman (talk) 20:06, 5 December 2018 (UTC)

I would prefer reverting to the usual custom of denoting the triangle's sides as a, b and c and angles/vertices as A, B and C instead of the angles being α, β and γ. Area would be spelt out in full (with an explanation why, if deemed necessary). There would be one instance of Area² and one of Area⁻¹, or 1/Area (which I would prefer). The article would be much easier to read and understand for those of us familiar with traditional Trigonometry. I have not attempted to edit the article because I have no skills in rendering graphics and formulae. Chris Carter — Preceding unsigned comment added by 2405:6E00:268D:99BA:C5E3:B9B5:EC31:B8BF (talk) 09:22, 21 October 2022 (UTC)

First formula has an oddity and so does the 2nd one.

Ok whoever wrote this didnt really know what they were doing in math, the formula sqrt(s(s-a)(s-b)(s-c)) does not seem equal to (sqrt((a+b+c)(a+b-c)(a-b+c)(-a+b+c))/4[pulled out of the sqrt sign for ease on the eyes]), but seems to work regardless. Can anyone explain this? — Preceding unsigned comment added by Aliotra (talk • contribs) 19:29, 25 September 2009 (UTC)

It's fine. a+b+c = s/2, a+b-c = (s-c)/2, etc. Multiplying them all together puts 16 in the denominator, and then moving it outside the square root reduces it to 4. --Vaughan Pratt (talk) 20:10, 25 April 2010 (UTC)

You were a bit too eager to simplify, above. From the definition, s-c = (a+b+c)/2 - c = (a+b+c)/2 - 2c/2 = (a+b+c-2c)/2 = (a+b-c)/2
Your basic explanation is correct, though - you repeat similar rewrites for the -a and -b factors, then factor out 1/2 from each of the four factors inside the square root, then multiply the four 1/2s together and perform a separate square root operation on that half of the formula.
sqrt((1/2)^4) = (1/2)^2 = 1/4 --85.227.216.133 (talk) 20:26, 25 February 2018 (UTC)

Intrinsic nature of the formula

Is it worth mentioning in the article that Heron's formula is more natural than the standard formula, because it gives the area of the triangular plane region in terms of its boundary (which is intrinsic to the region (i.e. free of choice)), whereas the standard formula gives the area in terms of the height (which depends on the base and hence requires a choice)? It could be helpful (to students, at least) because the FTC, Stoke's theorem, the residue theorem etc., are formulated this way. Just a thought (I know I should be bold, but I'm not; I prefer collegiality to leadership).Aliotra (talk) 19:29, 25 September 2009 (UTC)

references for proofs

The article needs references for the proofs. Bubba73 ^{You talkin' to me?} 05:22, 30 October 2012 (UTC)

examples

Great article - easy to read, easy to follow for all readers and full of interesting and useful information for the advanced reader.

If nobody objects, I would like to add the following examples (right after the start formulas):

Example: Let ΔABC be the triangle with sides a=7, b=4 and c=5.

The semiperimeter is:   $s={\tfrac {1}{2}}(a+b+c)={\tfrac {1}{2}}(7+4+5)=8$ ,
and the area is:  $A={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {8\cdot (8-7)\cdot (8-4)\cdot (8-5)}}={\sqrt {8\cdot 1\cdot 4\cdot 3)}}={\sqrt {96}}=4{\sqrt {6}}\approx 9.8$

Example: Let ΔABC be the triangle with sides a=3, b=4 and c=5.

The semiperimeter is:   $s={\tfrac {1}{2}}(a+b+c)={\tfrac {1}{2}}(3+4+5)=6$ ,
and the area is:  $A={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {6\cdot (6-3)\cdot (6-4)\cdot (6-5)}}={\sqrt {6\cdot 3\cdot 2\cdot 1}}={\sqrt {36}}=6$.
This is a right triangle, so the side b is also the height h to the base a. Using the standard formula for the area of a triangle, we confirm $A={\tfrac {1}{2}}ah={\tfrac {1}{2}}\cdot 3\cdot 4=6$.

Two other notes:

Mathematical:

• Please change T back to A for area. I think this confuses the reader much, much more than the double use of the letter A as area and A as vertex, which is common everywhere.
• Also, since another writer was also confused and these formulas are used in the secondary formulas and in both proofs, I might add a line like: *Here and in the proofs below we also use the alternative formulas for the semiperimeter: $s-a={\tfrac {1}{2}}(-a+b+c),\,\,s-b={\tfrac {1}{2}}(a-b+c),\,\,s-c={\tfrac {1}{2}}(a+b-c)$.

Technical: I personally think that using the smaller notation ${\tfrac {1}{2}}$ is much easier to read. See formulas in: [1]

Lfahlberg (talk) 06:31, 8 August 2013 (UTC)

a	b	c	s	A	notes
3	4	5	6	6	right
5	5	8	9	12	isosceles
5	5	6	8	12	isosceles
6	8	10	12	24	right, 2× 3-4-5
4	13	15	16	24
5	12	13	15	30	right
9	10	17	18	36
3	25	26	27	36
7	15	20	21	42
10	10	16	18	48	isosceles, 2× 5-5-8
10	10	12	16	48	isosceles, 2× 5-5-6
9	12	15	18	54	right, 3× 3-4-5
13	13	24	25	60	isosceles
10	13	13	18	60	isosceles
8	15	17	20	60	right
6	25	29	30	60
11	13	20	22	66
5	29	30	32	72
13	14	15	21	84
10	17	21	24	84
8	29	35	36	84
7	24	25	28	84	right
12	17	25	27	90
4	51	53	54	90
12	16	20	24	96	right, 4× 3-4-5
8	26	30	32	96	2× 4-13-15

I think an example where all numbers are integers is easier to follow. This table lists all such non-degenerate triangles with area under 100 units. A 4-13-15 triangle is the smallest one which is not isosceles, degenerate or right-angled, so I've used it. cmɢʟee⎆τaʟκ 23:04, 28 April 2014 (UTC)

Sum of squares

Decora (talk · contribs) has today been trying to add a link to the disambiguation page sum of squares to the see-also section of this article, on the basis that some of the formulations of Heron's formula look like sums of squares. I disagree, for two reasons: (1) we should not be linking to disambiguation pages from articles, and (2) I think the connection is too indirect to be informative or helpful. Discussion? Other viewpoints? —David Eppstein (talk) 19:58, 8 September 2013 (UTC)

I agree with David Eppstein on both counts. If someone clicks the link to Sums of squares, the only thing of relevance they find there is a link back to here. Duoduoduo (talk) 23:44, 8 September 2013 (UTC)

Agreed with David Eppstein. If there is a valid reference/comparison, this would be something that could actually be sourced, documented, written, and then properly added to the mainspace article itself: a link to a disambiguation page should not be used to escape the necessity of constructively writing text or providing reliable third-party sources. In addition, WP:INTDAB defines quite plainly when and how disambiguation links are to be used, and this is not one of those times. besieged^talk 04:22, 9 September 2013 (UTC)

LATEX error

Does anybody know how to fix the problem with multiline equations rendering the error code Failed to parse(unknown function '\begin')? Circlesareround (talk) 08:49, 9 February 2014 (UTC)

 Done The problem was caused by this edit, which was followed by this, which was apparently intended to be a revert of the previous edit, but which also removed one whit space character. I have no idea why that caused the problem, but the cure was easy: revert to the last version of the article before the problem occurred. JamesBWatson (talk) 10:44, 9 February 2014 (UTC)

One more point. The problem does not show up to anyone with their Wikipedia preferences set to use MathJax, which may be why the editor who caused the problem was not aware of it, and didn't correct it. This is the second time recently that I have come across mangling of mathematical formulas of this kind, and the other time the editor explicitly told me that he/she had been unaware of the problem for that reason. MathJax is described on the preferences page as "experimental", and perhaps someone involved in the experiment should be alerted to the problem. JamesBWatson (talk) 10:50, 9 February 2014 (UTC)

I have seen the same problem with multiline equations on other pages as well. There is in fact a very simple solution. Just choose Edit and then save the page without doing any edits whatsoever! That will fix the problem and the "edit" will not even show up in the log. Circlesareround (talk) 07:17, 10 February 2014 (UTC)

Probably all you really need to do is purge the page. You can do this under Preferences->Gadgets->Appearance: check the purge tab box, and then on top of every page you should see a tab labeled with a star that will purge the page. That gets rid of any cached copy and makes the Wikipedia servers recompute the page's appearance. There was a Wikipedia bug with image alignment that was recently fixed but that causes this; see WT:WPM for more discussion. —David Eppstein (talk) 07:23, 10 February 2014 (UTC)

Confusing variable names

Is there some reason why the figure at top right of page labels the angles alpha, beta, gamma, then the section "Trigonometric proof using the Law of cosine" labels them A, B, C, but the subsequent formulae label angle C with a caret on top (except not in the altitude formula)? Then again the formulae under "Trigonometric proof using the Law of cotangents" show angles as plain A, B, C, with no carets.

Maybe this caret is some notation system I'm not familiar with, but it seems gratuitous, given the other two angle notations previously introduced on the same page.

Also why is T chosen as the variable for Area, when virtually every other source uses A, or spells out Area? Is it because someone previously commented that it's confusing having "A" appear both for Area and for a vertex in the triangle (and also for an angle, if we're abandoning the greek letters introduced in the figure.)?

There actually is not a confusion relative to the vertex labels, since Area A is a variable and hence italicized. However, there is a confusion relative to the angle variables, since they too are italicized.

So, how about discard the vertex labels, since they aren't used, and stick with the greek angle names, and then A can be used for Area? Otherwise, how about spell out Area where needed? Using T for this quantity which, I claim, is almost universally known as A, is needlessly disorienting. Gwideman (talk) 07:34, 13 February 2014 (UTC)

Yes, the caret as an indicator of the size of an angle ("0.7 radians") as opposed to the angle itself ("these two lines meeting at a vertex") as opposed to the vertex itself ("this point, where you can see two lines meeting") is always confusing. (But not as bad as the old "m∠A" meaning the "measure of the angle A" notation that I was subjected to in junior high.) Be that as it may, I had attempted to use the same letter for both the vertex and the angle at that vertex, which required that I use the non-intuitive letter "T" for the area. I've been outvoted, and it's all for the better. Accordingly, I have uploaded a new picture, with the Greek letter for the angle. "A" is both a vertex and an area, but that's OK. "A=5" clearly refers to the area, not the vertex. SamHB (talk) 17:19, 30 March 2014 (UTC)

Proof section

I just reverted the unexplained blanking of the Proof section by an anonymous user. Normally this sort of thing isn't the cause for much comment. But in this case it caused me to wonder: is the proof section really helpful for this article? What useful information does it convey to the reader, beyond "this statement is true"? Where are the references for all these different proofs? Maybe removal was an improvement. Anyone else have an opinion here? —David Eppstein (talk) 17:14, 7 August 2014 (UTC)

The second proof is mine, and amounts to the only-if direction of a proof I published later in The College Journal of Mathematics in 2009, so I added a reference. The published proof is a fair bit shorter so it may be worth shortening the article's version to match. The if direction (Pythagoras from Heron) may also be worth mentioning. Not sure about the appropriateness of an editor referencing his own publications though. Vaughan Pratt (talk) 19:33, 9 August 2014 (UTC)

The idea in the proof you cite is not so similar to the one given in this article. Besides, I doubt the proof here is your proof. More likely, the proof here has been known longer than any living human on this planet have been alive. A changed the reference to a very similar proof that is much older than 2009. Circlesareround (talk) 11:28, 12 August 2014 (UTC)

You're right! I'd failed to notice that the version at https://en.wikipedia.org/w/index.php?title=Heron%27s_formula&oldid=193084332 , which I'd written in February 2008, had gradually morphed over time, starting in September 2012 with Tal physdancer's edit https://en.wikipedia.org/w/index.php?title=Heron%27s_formula&oldid=512210379. Subsequent edits removed other aspects of my proof, with the result being much closer to Raifaizen's original proof, which requires a lot more detailed algebraic manipulation than my proof.

Personally I'd prefer to see the section restored to the version immediately prior to Tal physdancer's 2012 edit, which doubled the amount of algebraic manipulation for no apparent reason. However in light of the evident conflict of interest I'm happy to leave the question of whether a short or long proof is preferable to others. Vaughan Pratt (talk) 04:48, 14 August 2014 (UTC)

On second thoughts Tal physdancer simply expanded what I'd assumed the reader could do on their own, so I guess that's ok. The subsequent edits were what eliminated my trick of factoring Heron as s(s-a) times (s-b)(s-c). Vaughan Pratt (talk) 04:56, 14 August 2014 (UTC)

Chinese formula and Hero's formula

The section "History" mentions this Chinese formula for computing the area of a triangle:

$A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}$

and calls it "equivalent" with Hero's formula. It is not equivalent since it requires that a ≥ b ≥ c, a requirement not present in Hero's formula. That is why it is in some sense "weaker" than Hero's formula. Nxavar (talk) 13:33, 3 October 2017 (UTC)

Why do you think it requires this variable ordering? When I try it on numerical examples it comes out the same regardless of ordering. —David Eppstein (talk) 18:33, 3 January 2019 (UTC)

This formula holds iff b+c ≥a, c+a ≥b and a+b ≥c. In other words it’s valid in all situations where a, b and c form the edges of a triangle, just like the original Heron formula Overlordnat1 (talk) 10:17, 10 April 2021 (UTC)

Heron's own proof

I was curious to know how Heron himself proved this far-from-obvious formula. The present text of the article is rather vague about this, and links to an external discussion, but as far as I can see this doesn't take us any closer to Heron's own proof. After a good deal of searching (including a wrestle with the German translation of Heron's Metrica ) I found the solution where I should probably have looked in the first place: in Sir T Heath's Greek Mathematics (the condensed version of his History of Greek Mathematics ), at pp. 420-22. Heath's account is very full, and seems faithful to Heron's original version as far as I can decipher it from the German translation. (There is also a recent French translation, but I do not have easy access to this.) Heron's proof is ingenious but very long, so I don't suggest including it in the article, but a reference to Heath's account might be included for the benefit of anyone interested. Curiously, Heath's book does not seem to be available at the Internet Archive, but the longer History is, and contains the same proof. 109.149.185.139 (talk) 14:44, 12 October 2018 (UTC)

Heuristic strategy to get Heron's formula

The formula for the area of a triangle (with sides lengths a, b and c):

• must be invariant under permutation of sides lengths a, b and c;

• assuming the area formula is a polynomial, it would have to be a homogeneous polynomial of degree 2 (since multiplying all side lengths by m, the area must be multiplied by m²);

• for degenerate triangles, i.e. when a side length equals the sum of the other 2 sides lengths, the area must be 0, so the polynomial would have to be divisible by the product of the three trinomials (− a + b + c) (a − b + c) (a + b − c): unfortunately this is a homogeneous polynomial of degree 3;

• assuming the formula for "the square of the area of the triangle divided by the perimeter of the triangle" is a polynomial, a homogeneous polynomial of degree 3 is just what we would need, i.e. ⁠A²/(a + b + c)⁠ = k (−a + b + c) (a − b + c) (a + b − c) for some positive constant k;

• using the Pythagorean theorem for a right triangle with 2 orthogonal sides of length 1, and thus hypothenuse √2 and area ⁠1/2⁠, we have (⁠1/2⁠)² = k (1 + 1 + √2) (−1 + 1 + √2) (1 − 1 + √2) (1 + 1 − √2) which yields k = ⁠1/16⁠.

Thus, assuming the formula for "the square of the area of the triangle divided by the perimeter of the triangle" is a polynomial, the heuristic suggests: A = ⁠1/4⁠ √(a + b + c) (−a + b + c) (a − b + c) (a + b − c).

Now, how can we turn this heuristic strategy into a proof?

A homogeneous trivariate (in a, b and c) homogeneous polynomial of degree 3 is of the form (with 10 coefficients)

α_3,0,0 a³ + α_2,1,0 a² b + α_2,0,1 a² c + α_1,2,0 a b² + α_1,1,1 a b c + α_1,0,2 a c² + α_0,3,0 b³ + α_0,2,1 b² c + α_0,1,2 b c² + α_0,0,3 c³.

Assuming the formula for "the square of the area of the triangle divided by the perimeter of the triangle" is a polynomial, with 10 pairwise non-similar triangles for which we know the sides and areas, we can uniquely determine the above coefficients, to get

− a³ + a² b + a² c + a b² − 2 a b c + a c² − b³ + b² c + b c² − c³ = (−a + b + c) (a − b + c) (a + b − c).

Now, how do we prove that

Assuming the formula for "the square of the area of the triangle divided by the perimeter of the triangle" is a polynomial

is a valid assumption?

(Now correctly using the reply box.) In the heuristic, why is the following not tried, for some k to be determined?

A³ = k [(− a + b + c) (a − b + c) (a + b − c)]² — Tentacles^{Talk or ✉ mailto:Tentacles} 18:21, 21 January 2023 (UTC)

Other Proofs

There are other elegant proofs such as the one using coordinate geometry, starting out with the fact that half the cross product of two vectors is the area of the triangle they form, see the following:- https://www.quora.com/How-can-you-prove-Herons-formula-by-using-coordinate-geometry

There’s also several proofs such as the one using similar triangles and an incircle of unit radius and another using an incircle and an excircle and similar triangles and another using the double angle tangent identity (Euler’s) and another using an identity linking tan and cos, and Heron’s original proof and also a link to a proof using complex numbers, in the following link:-https://www.quora.com/Is-there-any-easy-way-to-prove-Herons-Theorem Overlordnat1 (talk) 23:35, 4 May 2021 (UTC)

In Heron's Metrica

Here’s a link to the original Greek (and German translation) https://archive.org/details/heronisalexandri03hero/page/19/mode/2up and here is an English translation https://web.calstatela.edu/faculty/hmendel/Ancient%20Mathematics/HeroAlexandrinus/Metrica.i.1-9/Metrica.I.1-9.html#prop8 –jacobolus (t) 20:41, 2 December 2022 (UTC)

As a geometric mean

Let

α = (1/2)(− a + b + c), β = (1/2) (a − b + c), γ = (1/2) (a + b − c),

then

A = √(α β γ) (α + β + γ)

so that the area is the geometric mean of the product and the sum of same three variables. — Tentacles^{Talk or ✉ mailto:Tentacles} 03:22, 21 January 2023 (UTC)

Welcome back to Wikipedia after a long absence from editing.

Do you have another problem/topic where this point of view was insightful? It’s plausible one could be found, but there are plenty of other related formulas where these variables show up as $\alpha \beta (\alpha +\beta +\gamma ){\big /}\gamma ,$ $\alpha \beta \gamma {\big /}(\alpha +\beta +\gamma )$ etc. etc., where a geometric mean interpretation may not make much sense.

An alternative point of view that I have seen in at least a couple sources before and seems fruitful is to take $s,s_{a}=s-a,s_{b},s_{c}$ as four variables such that $s_{a}+s_{b}+s_{c}-s=0$ (or for symmetry negate $s$ or negate the other three), and then look what triangles and other objects are generated under various permutations and sign flips of those quantities. But I think this rather belongs in Heron's formula or other trigonometry articles rather than here per se. –jacobolus (t) 04:16, 21 January 2023 (UTC)