Talk:Heron's formula

message: 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)
replies: text: 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)

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

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

message: 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)
thread_title: Errors

message: 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)
replies: text: 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  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  is almost universal and that is what we should do here. --Bill Cherowitzo  (talk) 20:09, 5 December 2018 (UTC)
thread_title: "A"

message: 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)
replies: text: 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)
replies: text: 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)/2Your 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)
thread_title:  First formula has an oddity and so does the 2nd one.

message: 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)
thread_title: Intrinsic nature of the formula

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

message: 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} a h = \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: 
Lfahlberg (talk) 06:31, 8 August 2013 (UTC)
replies: text: 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)
thread_title: examples 

message: 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)
replies: text: 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)
replies: text: 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. besiegedtalk</font> 04:22, 9 September 2013 (UTC)
thread_title:  Sum of squares 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title:  LATEX error 

message: 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)
replies: text: 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)
thread_title:  Confusing variable names 

message: The section "History" mentions this Chinese formula for computing the area of a triangle:
:A=\frac1{2}\sqrt{a^2c^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 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)
replies: text: 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)
thread_title:  Chinese formula and Hero's formula 

message: 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)
thread_title:  Heron's own proof 

message: The formula for the area of a triangle (with sides lengths ,  and ):
* must be invariant under permutation of sides lengths ,  and ;
* assuming the area formula is a polynomial, it would have to be a homogeneous polynomial of degree 2 (since multiplying all side lengths by , the area must be multiplied by );
* for degenerate triangles, i.e. when a side length equals the sum of the other 2 sides lengths, the area must be , so the polynomial would have to be divisible by the product of the three trinomials : 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.  for some positive constant ;
* using the Pythagorean theorem for a right triangle with 2 orthogonal sides of length , and thus hypothenuse  and area , we have  which yields .
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: .
Now, how can we turn this heuristic strategy into a proof?
A homogeneous trivariate (in ,  and ) homogeneous polynomial of degree 3 is of the form (with 10 coefficients)
: 
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  
: 
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  to be determined?
::  — TentaclesTalk or ✉ mailto:Tentacles 18:21, 21 January 2023 (UTC)
thread_title:  Heuristic strategy to get Heron's formula 

message: 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)
thread_title: Other Proofs

message: 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)
thread_title:  In Heron's ''Metrica''