Heron’s formula: What is the hidden meaning of 1 + 2 + 3 = 1 x 2 x 3 ?

Today's video is about Heron's famous formula and Brahmagupta's and Bretschneider's extensions of this formula and what these formulas have to do with that curious identity 1+2+3=1x2x3.

00:00 Intro
01:01 1+2+3=1x2x3 in action
02:11 Equilateral triangle
02:30 Golden triangle
03:09 Chapter 1: Heron
06:18 Heron's formula
08:50 Brahmagupta's formula
10:20 Bretschneider's formula
11:52 Chapter 2: How? The proof
12:57 Heron via trig
20:09 Cut-the-knot
21:16 Albrecht Hess
21:46 Heron to Brahmagupta proof animation
25:10 Thank you!

Heron's formula on the Cut-the-knot site:

Simple derivation of Heron's formula just using Pythagoras's theorem:

Job Bouwman's maths posts on Quora (you'll have to scroll a bit to get to Heron's formula)

A very comprehensive book about quadrilaterals:
Claudia Alsina, Roger B. Nelsen - A Cornucopia of Quadrilaterals (Dolciani Mathematical Expositions) (2020, American Mathematical Society)

Albrecht Hess's paper "A Highway from Heron to Brahmagupta"

If you liked the rectangle proof of the sum = product identity you'll probably also like this proof of Pythagoras's theorem:
I also mentioned this one earlier in a video on this main channel

Two more interesting notes on the cut-the-knot page:
2. A supercute way to derive Pythagoras from Heron with one line of calculus

For a cyclic quadrilateral that also has an incircle we have a+b=c+d and it follows that the area is just square root of the product of all of the sides.

A 3d counterpart to Heron's formula:
A different 3d connection (de Gua's theorem)

A couple of links to get you started on generalisations involving cyclic n-gons:

Another interesting observation extending the fact that the 3-4-5 right-angled triangle has incircle radius 1: In general, the incircle radius of any right-angled triangle with integer sides is an integer.

Have a look at this for a related proof that arctan 1 + arctan 2 + arctan 3 = pi:

Not many integer solutions for x+y+z=xyz:
0+0+0=0x0x0
1+2+3=1x2x3
(-1)+(-2)+(-3)=(-1)x(-2)x(-3)

Other interesting little curiosities (some mentioned in the comments):
2+2=2x2=2^2 (of course)
3^3+4^3+5^3=6^3 = 6*6*6=216 illuminati confirmed
6+9+6*9 = 69
a+9+a*9 = 10a+9 (sub any digit)
log(1+2+3)=log(1)+log(2)+log(3) follow from 1+2+3=1x2x3

Grégoire Locqueville 2:32 "Maybe one of you can check in the comments" is the new "left as an exercise to the reader" :)

length, area and "volume" start out the same with radius 1: length=area=volume.
When you scale by r, these values scale in this way Length = length * r, Area = area * r^2 and Volume = "volume" r^3. Therefore, Length = length * r = area *r and so (multiply through with r) Length* r = area *r ^2 = Area, etc.

Typo spotted: At the very end, in Brahmagupta's Formula the third bracket should be (A+C+D-B) not (A+B+D-B).

Nice variation of the t-shirt joke by one of you: M - I - I = V :)

The Chrome extension I mentioned in this video is called CheerpJ Applet runner.

Music used in this video: Aftershocks by Ardie Son and Zoom out by Muted

Enjoy!

Burkard

[patrickwebster]

A few years ago I used to binge watch this channel. You might imagine my surprise when I started my science degree (undergraduate biochemistry where maths is thankfully a necessary subject to take) and walked into a maths lecture and was confronted by this wonderful person in the flesh. That was about 4 or 5 years ago. Keep up the saintly work you do!

[murillonetoo]

I'm Brazilian and I've studied the high school in a technical school. I recall learning about Heron's formula as a secondary topic, but no proof nor any thorough explanation, as the one presented, was given. Thanks for the lecture!

[chessematics]

7:00 yes actually. In 9th grade (or, as we call it here, Class IX), there's an entire chapter in the book called Area of Triangle and it's simply filled with good old Heron.
Respect from India

[PaceTheAce]

There's a mistake at the end. In Brahmagupta's Formula the third bracket should be (A+C+D-B) not (A+B+D-B).
But great video. That are some beautiful equations and you explained all of it really nicely with all these brilliant animations. Love it!

[grumpyparsnip]

I learned Heron's formula in school in the US state of New Jersey in the 1990s. Unfortunately, it was not proven, though I did derive my own proof using Law of Cosines similar to the one in the video. I am always on the lookout for better proofs, and the one in the video is definitely beautiful. I love the way the exposition on the 345 triangle gently leads the viewer into the appropriate conceptual space. Well done.

[Jouzou]

Finnish engineer student here. Heron's formula is in our textbook and might have been mentioned like once during a lecture. But we mainly used laws of sine and cosine to solve triangles.

[yaskynemma]

For me is impressive how almost everything can be explained geometrically but commonly isn't thaught like that, even when I find it more beautiful and comprehensible

[gonshi]

1+2+3 = the amount of seconds between the release of the video and me clicking on it

[DavidBeddard]

If I'd lived in Classical Europe, I'd totally have worshiped geometry as sacred! Great video!

[professorpoke]

In India we have a whole chapter by the name *Herons Formula* . After that I decided to derive my very own formula for finding area of a triangle, but I accidentally deriverd Herons Formula. I was so Happy at that time. Those were the days. 🙂🙂

[gregwochlik]

I remember using this "unusual" formula around 1995 in a math olympiad (16 years old at the time). I picked it up in a textbook, and managed to memorise it to use it "blindly"

[sreedharperupally]

I learnt Heron's method of finding area of any triangle by myself from a mathematical formulas Book, when I was in 8th class in the year 1988.
Even today I find it useful in my field works 👍

[ibrt]

I learned of Heron's formula when I was about 8 years old from a table of mathematical formulae in the back of a dictionary, but I was never taught it in school, and I spent most of my life being curious about how to derive it in an elegant and geometrical manner. Your video is fantastic, it has unlocked some remaining pieces of the puzzle I had not figured out.

[dr.kraemer]

as a teacher, I'd avoid using Blue because of how easy it is to confuse with side B of the triangle.

[xCorvusx]

2:34, check of the calculations up to that point:
For the equilateral triangle, the height is the length of one of its legs times the cosine of π/6, so the height equals 2*sqrt(3) times sqrt(3)/2 .
Thus, the area is half its base, i. e. sqrt(3), times 2 times sqrt(3) times sqrt(3) times 1/2 which amounts to 3*sqrt(3).

For the isosceles triangle with legs of the length 2+φ with a base of 2*φ, Pythagoras's theorem gives sqrt( (2+φ)^2 - φ^2 ) = sqrt(4 + 4φ + φ^2 - φ^2) = sqrt( 4(1+φ) ) .
The golden ratio is a solution to the the equation x^2 = x+1, so sqrt( 4(1+φ) ) = sqrt( 4φ^2 ) = 2φ .
(Since we're talking about lengths, we can ignore negative results.)
Finally, the area of this isosceles triangle is then half its base times its height: φ times 2φ = 2φ^2 .

[mohammadazad]

I already know Heron's formula and Brahmagupta's formulas but I haven't seen a proof for them until now, also that general formula for the area of a quadrilateral is beautiful, thanks

[KSignalEingang]

This has to be a record - two minutes in and I'm already pausing the video to pick exploded bits of my mind out of the carpet. How did I never learn this before? In school, Heron's formula was presented as a side note, and I never really understood the derivation (meaning I'd have to look it up again any time I wanted to use it). Given a little bit of reflection it all seems crystal clear, now. Nice!

[MusicalRaichu]

no, in australia we didn't learn the formula, but luckily i came across it somewhere and was so enthralled by it, i used it in class to my teacher's dismay.

[JatinSanghvi]

I had requested long back as comment in one of the earlier videos, an intuitive graphical explanation for why the Heron's formula for area of the triangle works, without needing to use trigonometric formulae. Thanks a ton for releasing a video that exactly answers to my request. I haven't found such a great shortcut to reach to the Heron's formula anywhere else in internet.

[mlmimichaellucasmontereyin]

WOW!!! Bravo again! This is another indisputably perfect example that supports my theory, metatheory, proofs & metaproofs that show how and why nature's astrophysical geometry, geometry, numbers, maths, and logic are enabled & sustained by the natural metalogical principles of being (i.e., the "cosmos"). I will definitely cite (& link) this episode in my next draft of "Astronomy, Geometry, and Logic" (and formally request permission to use a pic or 2 from the video). Dear Burkard & Marty, thanks again for doing the best, most useful maths series on Youtube.