- #1

- 350

- 81

A = √[(s(s-a)(s-b)(s-c)]

I tried to develop one but could not. Can any body give or give me an hint to proceed.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Let'sthink
- Start date

- #1

- 350

- 81

A = √[(s(s-a)(s-b)(s-c)]

I tried to develop one but could not. Can any body give or give me an hint to proceed.

- #2

- 15,922

- 14,336

The most interesting proof is via the volume of simplices by the Cayley-Menger determinant.

- #3

- 350

- 81

So you mean to say proof without words will not be possible. Should I explain what I mean by proof without words?

The most interesting proof is via the volume of simplices by the Cayley-Menger determinant.

- #4

- 15,922

- 14,336

I don't know what you mean by "without words". Classic geometric proofs consist of a lot of words. I think the one with the diagonals can be used to write a purely geometric proof, i.e. no coordinates, no determinants, no algebra. Whether you call this without words is a strange, but personal decision. A proof by telepathy is certainly impossible.So you mean to say proof without words will not be possible. Should I explain what I mean by proof without word?

- #5

- 350

- 81

Ok I shall explain. It is a proof with minimum words. For example the proof of Pythagoras theorem for any right angled triangle with c as hypotenuse and a and b as sides can be drawing a square ABCD of side (a+b) then taking point E on AB such that AE is a and EB is b; taking point F on BC such that BF is a and FC is b; taking point G on CD such that CG = a and GD = b and lastly taking point H on DA such that DH = a and HA = b. You can easily see after joining EFGH that it is square with side c. Alternatively you can take four congruent triangles HAE, EBF, FCG and GDH all right angled triangles congruent to original right angled triangle ABC say, and make this figure. As you look at this figure you see that[ {Area of ABCD = (a+b)² - 4 times area of each congruent triangle} = the area of the remaining square EFGH = c². As we know that each congruent triangle is (1/2)ab so the left hand side = a² +b². This one does not have to write one just looks at the diagram gets it (a² + b² = c²). I think you may like to call it telepathy! So my question was can we make a diagram which will make the hero formula will look obviously true. In the literature of mathematics education I understand that there may be a large number of such proofs for Pythagoras and other theorems.I don't know what you mean by "without words". Classic geometric proofs consist of a lot of words. I think the one with the diagonals can be used to write a purely geometric proof, i.e. no coordinates, no determinants, no algebra. Whether you call this without words is a strange, but personal decision. A proof by telepathy is certainly impossible.

- #6

- 37

- 5

- #7

- 919

- 651

I don't think that's proof that such a proof cannot be done.

- #8

- 926

- 483

A = √[(s(s-a)(s-b)(s-c)]

I tried to develop one but could not. Can any body give or give me an hint to proceed.

Ok I shall explain. It is a proof with minimum words. For example the proof of Pythagoras theorem for any right angled triangle with c as hypotenuse and a and b as sides can be drawing a square ABCD of side (a+b) then taking point E on AB such that AE is a and EB is b; taking point F on BC such that BF is a and FC is b; taking point G on CD such that CG = a and GD = b and lastly taking point H on DA such that DH = a and HA = b. You can easily see after joining EFGH that it is square with side c. Alternatively you can take four congruent triangles HAE, EBF, FCG and GDH all right angled triangles congruent to original right angled triangle ABC say, and make this figure. As you look at this figure you see that[ {Area of ABCD = (a+b)² - 4 times area of each congruent triangle} = the area of the remaining square EFGH = c². As we know that each congruent triangle is (1/2)ab so the left hand side = a² +b². This one does not have to write one just looks at the diagram gets it (a² + b² = c²). I think you may like to call it telepathy! So my question was can we make a diagram which will make the hero formula will look obviously true. In the literature of mathematics education I understand that there may be a large number of such proofs for Pythagoras and other theorems.

As I understand it you ask for a proof by a

Now, take note that developing a geometric proof for Heron's formula is - at least as far as I know, somewhat more involved than proof for

For instance take a look at this one from University of Georgia.

- #9

- 350

- 81

I followed the link of the proof for Heron's formula from University of Georgia. It is really involved. But following further links I found Brahmgupta's law on area, A of cyclic quadrilateral with sides a, b, c, d and 2s = a+b+c+d, as;

A = √[((s-a)(s-b)(s-c)(s-d)], which will reduce to Heron's formula in the event when d=0 making the quadrilateral a triangle. So my problem now is reduced to finding the geometric or proof without words for the area of a cyclic quadrilateral. Needless to say that unlike quadrilaterals all triangles are cyclic.

A = √[((s-a)(s-b)(s-c)(s-d)], which will reduce to Heron's formula in the event when d=0 making the quadrilateral a triangle. So my problem now is reduced to finding the geometric or proof without words for the area of a cyclic quadrilateral. Needless to say that unlike quadrilaterals all triangles are cyclic.

Last edited:

- #10

coolul007

Gold Member

- 271

- 8

Here is another link: http://galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html

- #11

- 350

- 81

Archimedes' proof is great. Symmetry argument is also good a+b-c, b+c-a and c+a-b should be factor, but how do we come to the factor(a+b+c)?Here is another link: http://galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html

- #12

epenguin

Homework Helper

Gold Member

- 3,878

- 902

[

Damn! At the end of that there is exactly the solution that I had thought of myself about three days ago part of which I think is beautiful.

You see I had two issues. One was all these proofs are pretty ugly. You do not visualise the answer has to be this strange looking one, You do not at first, or for that matter at the end, see the sense of it.

At the end most people would think, well it least there is a part of that I recognise. ##(a+b+c)## that’s the perimeter, almost nice. But the other part ##(P-a)(P-b)(P-c)## what does that rhyme with? Not nice.

Instead I now see the second expression is very nice, and the perimeter is less nice. Calling this quantity M

$$ M = (a+b-c)(a-b+c)(-a+b+c)~~~~~~~~~~~~(1)$$

you see that M ≥ 0. The combined length of two sides of a triangle will never be less than the length of third side. If there is equality, we have triangle of zero area. To say this again, for any triangle ##(a+b-c) ≥ 0## and by symmetry or renaming the quantities in the two other brackets ≥ 0 .

M = 0 then is a condition for the triangle to have zero area. Therefore M must be a factor of the expression for the area of the triangle.

However, that does not mean that it is a rational factor, in fact it cannot be because it has the dimensions of volume not area. So as far as the argument so far tells us, the area could be something like ##nM^\frac{2}{3}## where n is some number. ( I wondered if there could be some other function of M and numbers alone that could satisfy the requirements so far, but then I thought not - that the dimensionality excludes it.)

The alternative to ##nM^\frac{2}{3}## is the product of M and some other symmetric function of a, b, c, everything raised to some appropriate power. And perimeter, P, even if we were not cheating because we know it already, would be shouting at us. It is the very most simple symmetric function of a, b, c. Plus it is an acquaintance of ours! So it is a very strong candidate for being a factor of the area formula, for not quite sufficient reasons; I’d hope to find better.

I thought to argue that if the perimeter length is 0, the area is 0, and hence by the same argument as before therefore P must be a factor of the expression for area. Trouble I was seeing with that is that there is no way for P to be 0 without all of a, b, c, being 0 and thus also M= 0; It seemed to me that makes the argument fall. I wanted to find a situation where P but not M could be zero. This problem, if it truly is one, arises as a result of the fact that lengths are all positive. I was wondering if we can use some concept of signed lengths which could give us a zero P while leaving the factors of M nonzero. I am sure we could live with negative areas. But this is all a bit abstract mathematical so I haven’t done this.

Further problem was that the obvious combination to give us the area (apart from any numerical factor) is (PM)^{½}. I have just got to the point of convincing myself that there is no other form conceivable. (That is I asked why not something like (αM^{2}P^{4}+ βMP^{7})^{1/5} was the right dimensions. Ah but then we can factor P or M out of all such expressions, in such a way that one of them is not a factor of the sum that remains. Then if the vanishing of the remaining factor does not make the area zero, it can’t be valid, or it adds nothing. Think that’s right, hope covers all possibilities.

There remains to get the numerical factor. For this I had hit on exactly the same method in fact the same example as the link. Hwever I still have a logical and aesthetic problem with that! The beauty of Eq (1) at least Is that*it did not depend on already knowing a formula *for the area of a triangle. Another beauty is that depends only on the ‘triangle inequality’ ##(a+b)≥c##. This holds (and is definitional) for many spaces other than Euclidean, called metric spaces.Therefore I would expect this quantity M defined by Eq 1 to appear in some analog of Heron’s formula in these other spaces.

But then having breathed the air at this lofty level, at the very last stage, I have to descend to ground level, calculating an area by an already known formula and method! Ugh! I might almost as well just have done the other calculations that were given. All they are doing is proving in a routine fashion that one formula is equivalent to another formula already known - essentially trivial in the sense that there is no new math in it, even if superficially unobvious it seems guaranteed that one can get to the answer. It involves knowing stuff (traditional triangle area formula) that I had been trying to do without. I tried to minimise my knowledge utilisation by taking the same special case as the link, a right angled isosceles triangle. I say I know the area of that without knowing the general triangle formula, because it is half a square. And I can allow myself to know the area of a square, because that is really part of the definition of area anyway. But then again knowing it is half a square involves congruence and Euclidean rigid sliding, which is enough to give me the general formula for triangle I think. Plus it requires the calculation of the length of the diagonal, which requires Pythagoras’ theorem. Maybe I can get away with that, and Pythagoras can be proved without rigid motions, invoking only similarity? Maybe I am being hypercritical. But you see what I am trying to do? - to be not practical but mathematically minimal, even if trying to be maximally simple turns out to be complicated.

There are still a couple of matters arising, but not to try anyone’s patience more I will put these in another post.

Here is another link: http://galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html

Damn! At the end of that there is exactly the solution that I had thought of myself about three days ago part of which I think is beautiful.

You see I had two issues. One was all these proofs are pretty ugly. You do not visualise the answer has to be this strange looking one, You do not at first, or for that matter at the end, see the sense of it.

At the end most people would think, well it least there is a part of that I recognise. ##(a+b+c)## that’s the perimeter, almost nice. But the other part ##(P-a)(P-b)(P-c)## what does that rhyme with? Not nice.

Instead I now see the second expression is very nice, and the perimeter is less nice. Calling this quantity M

$$ M = (a+b-c)(a-b+c)(-a+b+c)~~~~~~~~~~~~(1)$$

you see that M ≥ 0. The combined length of two sides of a triangle will never be less than the length of third side. If there is equality, we have triangle of zero area. To say this again, for any triangle ##(a+b-c) ≥ 0## and by symmetry or renaming the quantities in the two other brackets ≥ 0 .

M = 0 then is a condition for the triangle to have zero area. Therefore M must be a factor of the expression for the area of the triangle.

However, that does not mean that it is a rational factor, in fact it cannot be because it has the dimensions of volume not area. So as far as the argument so far tells us, the area could be something like ##nM^\frac{2}{3}## where n is some number. ( I wondered if there could be some other function of M and numbers alone that could satisfy the requirements so far, but then I thought not - that the dimensionality excludes it.)

The alternative to ##nM^\frac{2}{3}## is the product of M and some other symmetric function of a, b, c, everything raised to some appropriate power. And perimeter, P, even if we were not cheating because we know it already, would be shouting at us. It is the very most simple symmetric function of a, b, c. Plus it is an acquaintance of ours! So it is a very strong candidate for being a factor of the area formula, for not quite sufficient reasons; I’d hope to find better.

I thought to argue that if the perimeter length is 0, the area is 0, and hence by the same argument as before therefore P must be a factor of the expression for area. Trouble I was seeing with that is that there is no way for P to be 0 without all of a, b, c, being 0 and thus also M= 0; It seemed to me that makes the argument fall. I wanted to find a situation where P but not M could be zero. This problem, if it truly is one, arises as a result of the fact that lengths are all positive. I was wondering if we can use some concept of signed lengths which could give us a zero P while leaving the factors of M nonzero. I am sure we could live with negative areas. But this is all a bit abstract mathematical so I haven’t done this.

Further problem was that the obvious combination to give us the area (apart from any numerical factor) is (PM)

There remains to get the numerical factor. For this I had hit on exactly the same method in fact the same example as the link. Hwever I still have a logical and aesthetic problem with that! The beauty of Eq (1) at least Is that

But then having breathed the air at this lofty level, at the very last stage, I have to descend to ground level, calculating an area by an already known formula and method! Ugh! I might almost as well just have done the other calculations that were given. All they are doing is proving in a routine fashion that one formula is equivalent to another formula already known - essentially trivial in the sense that there is no new math in it, even if superficially unobvious it seems guaranteed that one can get to the answer. It involves knowing stuff (traditional triangle area formula) that I had been trying to do without. I tried to minimise my knowledge utilisation by taking the same special case as the link, a right angled isosceles triangle. I say I know the area of that without knowing the general triangle formula, because it is half a square. And I can allow myself to know the area of a square, because that is really part of the definition of area anyway. But then again knowing it is half a square involves congruence and Euclidean rigid sliding, which is enough to give me the general formula for triangle I think. Plus it requires the calculation of the length of the diagonal, which requires Pythagoras’ theorem. Maybe I can get away with that, and Pythagoras can be proved without rigid motions, invoking only similarity? Maybe I am being hypercritical. But you see what I am trying to do? - to be not practical but mathematically minimal, even if trying to be maximally simple turns out to be complicated.

There are still a couple of matters arising, but not to try anyone’s patience more I will put these in another post.

Last edited:

- #13

epenguin

Homework Helper

Gold Member

- 3,878

- 902

Archimedes' proof is great. Symmetry argument is also good a+b-c, b+c-a and c+a-b should be factor, but how do we come to the factor(a+b+c)?

As you can see I have had some trouble with that. I have shown that it is plausible. More important that it is consistent – it does not give rise to contradiction. Did not manage to show that it has to be.

I am tired now but I think it could be shown along the following lines: Show that nM

- #14

- 350

- 81

As you can see I have had some trouble with that. I have shown that it is plausible. More important that it is consistent – it does not give rise to contradiction. Did not manage to show that it has to be.

I am tired now but I think it could be shown along the following lines: Show that nM^{2/3}gives the wrong answer in some particular example. (As mentioned, I tried to avoid examples we independently know the answer to. But I could be more accepting to negative examples, provisionally.) If there exists such an example in which that last formula is wrong then the formula must be a (fractional) power of SM, where S Is a polynomial symmetrical in a, b, c. I think I can show that if P is consistent (as I think I have) then no other symmetrical polynomial can be, so the answer must be P. However this must all be known, and maybe somebody will be along with an answer.

Sir this was not my home work. May be I posted it in a wrong section. The question occurred to me on my own and I wanted to search an answer to my own curiosity. I am interested in drawing a diagram which will constitute the proof. I am reading every proof sincerely and trying to develop a diagram. Somememebers also said that this may not be possible at all.

After careful examination of the diagram given in the proof and referring to it one can note that:Here is another link: http://galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html

AF= EA = x,say; FB= BD = y, say and DC = CF = z, say

Then we can easily see,

AF+FB+ BD+DC + CE + EA = x+y+y+z+z+x = a+b+c or

2(x+y+z) = p =2s or s = x+y+z ------------- (1)

but we know that y+z = a, z+x = b and x+y = c so we find that

x = s-a, y = s-b and z =s-c.

I think this knowledge and the diagram wherein in-circle, DEF drawn in the triangle with OD = r, say as its radius. and the fact that the area,A of the triangle ABC can be written as

A = sr, should help us in thinking about a suitable diagram to construct the proof of Heron's formula. I am trying to work on it algebraically as well as trying to think of a diagram.

- #15

- 37

- 5

Do you know how to construct the product of two numbers with ruler and compass? By taking a line segment of length a and another of length b, do you know how to construct a line segment of length ab?

s=(a+b+c)/2 is easily constructible with ruler and compass. And so is the formula given by Heron if you know how to construct products and square roots by ruler and compass.

So, what you are left with is a line segment with length √[s(s-a)(s-b)(s-c)] = ab/2 say, if your triangle were to be right-angled. Now take out your compass and verify that the two line-segments in the above equation that you have constructed are indeed equal.

This is as visual and straightforward as it gets. There are many references on how to construct products and square roots of numbers with ruler and compass.

s=(a+b+c)/2 is easily constructible with ruler and compass. And so is the formula given by Heron if you know how to construct products and square roots by ruler and compass.

So, what you are left with is a line segment with length √[s(s-a)(s-b)(s-c)] = ab/2 say, if your triangle were to be right-angled. Now take out your compass and verify that the two line-segments in the above equation that you have constructed are indeed equal.

This is as visual and straightforward as it gets. There are many references on how to construct products and square roots of numbers with ruler and compass.

Last edited:

- #16

- 350

- 81

my problem is not about geometrically drawing the expression but drawing a diagram which will make this expression represent the area of the triangle.Do you know how to construct the product of two numbers with ruler and compass? By taking a line segment of length a and another of length b, do you know how to construct a line segment of length ab?

s=(a+b+c)/2 is easily constructible with ruler and compass. And so is the formula given by Heron if you know how to construct products and square roots by ruler and compass.

So, what you are left with is a line segment with length √[s(s-a)(s-b)(s-c)] = ab/2 say, if your triangle were to be right-angled. Now take out your compass and verify that the two line-segments in the above equation that you have constructed are indeed equal.

This is as visual and straightforward as it gets. There are many references on how to construct products and square roots of numbers with ruler and compass.

- #17

- 37

- 5

Ok, having constructed the line segment representing the area of the triangle, now also construct its square root line segment with ruler and compass. Now draw the square with that side and you have a square with area your triangle's area. The square with area given by the Heron's formula matches the square with area given by ab/2 ! Collect these steps in a nice geometric diagram and write LOOK beneath.my problem is not about geometrically drawing the expression but drawing a diagram which will make this expression represent the area of the triangle.

Last edited:

- #18

epenguin

Homework Helper

Gold Member

- 3,878

- 902

.............

- #19

epenguin

Homework Helper

Gold Member

- 3,878

- 902

Here is another link: http://galileoandeinstein.physics.virginia.edu/more_stuff/Heron.html

coolul , a critical step in that quote in the last paragraph of your link is

It looks to me the author may have some doubts of the same kind as he writes:

(My difficulty arises from trying to say that if C is zero the area is zero, therefore C Is a factor of the area expression. But that this does not work where in when the whole expression is CM and M is coincidentally zero. If I relaxed the idea that the lengths have to be positive, then actually we see that when C = 0 then M = - 8abc, real nonzero number, even positive, the logic seems to work, but difficult to be comfortable with something like that. )

Last edited:

Share: