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!
Hi. I'm brazilian too, I'm 58 years old and I remember I've learned the Heron's formula when I was a student in a military school, back in the 1970 decade. I have a masters degree in Math and I've worked with bicentric quadrilaterals, it is the theme of my master thesis (sorry for my bad english). One of the funny facts about them is that their area is simply the square root of the product of their sides. This is the best Mathologer video I've already watched. Thank you all!
@@patrickwebster3152 Mesmo olhando o conteúdo deste canal, fica claro que não caberia em currículo algum, nem de um curso de faculdade. Fica para o estudante, ao menos os de matemática, se aprofundar em seus conhecimentos. Mas mesmo assim, ainda irá surgir muita coisa que não foi descoberta ou que está em estudo.
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!
I wish my Biochemistry degree had such great teachers. I might have actually finished it. Instead; most of my teachers and professors were lackluster nobodies, who didn’t really know, how to make stuff accessible. For example, my Physical Chemistry professor just gave us, like, 50 different formulas for rote memorization, without really motivating any of them (like, why do you need integration, for temperature-/pressure-systems 🤯?). 😔
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
It’s actually good that actual math proofs aren’t very geometrical. The geometry behind a lot math is definitely extremely beautiful but it is very easy to get things wrong if you rely on pictures and drawings. Most geometric proofs of complicated results are derived a posteriori since we can back up the geometrical proof with a more precise proof based on other techniques.
It's how we used to do it before! But once we started using more complex stuff like powers of 4 ie. x⁴ We would have needed to use 4d brains to comprehend And this is why we still thank the arabic mathematician Al-Khawarizmi who created algebra and the hindu-arabic numbers (1, 2...)
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. 🙂🙂
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.
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!
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
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!
On the contrary, the formula had enough symmetry to easily memorize, for me personally at least. But I agree, the derivation is so simple, I'd never forget or have to memorize it! And never learned about the quadratic form.
In 1967, I took an Engineering Measurements course at U.C. Davis. One of the measuring tools was a polar planimeter which allows you to find the area of any figure by tracing its perimeter. If you have a plot of land drawn on a map. you can easily find the enclosed area. This is a very practical application of Heron's Formula.
Planimeters don’t use Heron’s formula. A polar planimeter adds up the changes in angle multiplied at each point by the square of the radius. A linear planimeter adds up x dy - y dx, a continuous version of the "shoelace formula" about which Mathologer made another nice video a few years ago th-cam.com/video/0KjG8Pg6LGk/w-d-xo.html
Planimeters are connected to the divergence theorem, which green's theorem is derived from, and which the shoelace algorithm can be derived from. And you can derive Heron's formula from green's theorem (divergence theorem). Heron's formula relates area to perimeter of a closed boundary. With the more general divergence's theorem, you can calculate area from the closed boundary of its 2D shape. And going deeper you can also calculate volume from the closed surface boundary of a 3D shape. This is what's amazing about the geometric relationship of a shape's boundary. You can also calculate metrics like center of gravity, volume, inertia tensor of 3D shapes (which are typically expressed with triple integrals) by a simple surface integral. (For example Volume = 1/3 Surface integral of dot product of field [x, y, z] and surface normal * dS. Most 3D models are Boundary Reps, so this is a handy relation in computational geometry. Heron's formula is mapping a 1D metric to a 2D metric just like (perimeter to area) just like this example of divergence theorem is mapping a 2D metric to a 3D metric. The trick is to recognize the special field. In this video, the trick is recognize the geometric relation of the triangle and inscribed unit circle. But there is also a trick in recognizing the field to apply with the divergence's theorem.)
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
I was taught Heron's formula in high school, in Greece. I found it rather interesting, but considered it a curiosity, mostly, as it wasn't connected to anything else I was taught.
Actually, here in Russia we are in fact taught Heron's formula at school, but I'm not really used to it since it was a lockdown year when we were taught this. So if a problem requiring Heron's formula to solve it occurs, I'm always a bit perplexed as it doesn't come to mind at first.
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"
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.
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.
A truly beautiful visual and mathematical feast. The complicated symmetry is an extra detail that (for those of us obsessed with symmetry) elevates this to astronomical heights. Thank you, Burkhard for reminding me how much I love Mathematics. A special round of applause to the crystal clear animations.
I'm homeschooling my children, and interestingly I just taught them Heron's formula yesterday as an example of where square roots can be very practical in real life problems. Specifically, we talked about how convenient it can be if you are trying to determine the size of a plot of land you want to buy when all you can measure are distances and no idea if the land is square. Just walk the perimeter, walk the diagonal, apply Heron's formula to the 2 triangles, and you know immediately how big the plot is. For 7th graders who don't yet know trigonometry, the example worked beautifully. It should absolutely be taught in schools. It is a crime that it is not considered essential these days.
You need the square root for computing accelerations from position or distances from a plan or much much more. And when surveying land, you need square roots all over the place. But you will never once use Hero's formula. In particular, accurately measuring angles is easier than accurately measuring distances.
Yes, in profesional landsurveying you use angles, because you had equipment (theodolite, total station) that measure angles and distances. Surveying in the core, is to translate points between carthesian coordinate (2,2) to polar coordinate (45°, 2root(2)) and vice versa. And that translation is all about solution for right triangles. Heron's formula is niche one, and good if you only have ruler to measure sides. I would say its Identity should to teach in schools. (And cut some amout of mindles computing) (And why not throw Ptolemy's theorem in to play and calculate solution to last unknown measurement. Actually, it would be fun to first calculate missing border, blocked by huge tree, with Ptolemys. And then using Heron's to calc. the area. )
@@EebstertheGreat Well, you'll have to measure at least on side of a triangle even for the standard 1/2 base times height formula :) Anyway, speaking for myself, Heron's formula has definitely proved a very useful addition to my mathematical toolbox on many occasions :)
@@EebstertheGreat I use Pythagoras _all the time_ in structural engineering, as every problem has to be turned into, first, their vertical and lateral components (think diagonal braces, roof members, trusses, anything not perfectly vertical or horizontal), and then every member needs to be checked for axial stresses and flexural stresses (column buckling and failing vs beam bending and sagging).
Opposing angles subtend a partition of the whole circle. The measure of an angle is half the measure of the arc it subtends. Therefore, the sum of the opposing angles equals half the sum of opposing arcs, which is half of 360 degrees, which is 180 degrees.
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 👍
I learned Heron's formula when I was in middle school in the state of California (4 years ago) and I'm pretty sure it's still being taught, but I've never used it outside of math competitions. Lots of great information in the video that I didn't know before though. Thanks!
Whilst in Romania I sat down with a man once who looked JUST EXACTLY like you. He was a Romanian math teacher... we sat there and wow'd each other for about 4 hours. LOL the restaurant was trying to close and we had this deep mathematical conversation going on... You remind me of him so much. (This was about a year ago... seems like a million years ago.)
@@Mathologer High school maths (Algebra and Geometry) teacher here. The reason that we don't do classical geometry anymore is because school is compulsory. Back when classical geometry was taught at this level, school was an elective or privileged activity. The people who were taking classes did so because they wanted to and/or understood the value of it. Most people do not value math. Take a look at your subscriber count compared to mindless entertainment like Markiplier or something. I have good reason to believe that most of my students are not only uninterested in math (in fact, they hate it -- even the cool stuff), but are incapable of comprehending it. So how did the school curriculum respond? By boiling all of the math down so that the students with the lowest math performance could still pass the classes -- that is to say, teaching kids how to push buttons in a calculator, but not why the formulas work or what they mean. It's a waste of everyone's time, but that's what you get when teachers become glorified babysitters for families with working parents.
I keep coming back to watch the animation for finding Brahmagupta’s formula from Heron’s formula. It’s simply incredible, both the animation and the actual proof.
There's an extended formula for three dimensions which shows that meshes consisting entirely of triangles have a fixed volume even if they aren't rigid. Would love to see a video on that.
How could a deformed geodesic dome have the same volume? The same surface area for sure, but if I squish the top of a dome down there's definitely less area.
Apparently the great Alexander Grothendieck independently discovered this formula as a schoolchild while being hidden from the Gestapo at an orphanage in Southern France during WWII....it would be nice to be able to give him a mention!
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.
8:54. I never applied myself in school at all but watching your videos the past year or so has been enlightening! I was a bit intimidated I don’t remember the first topic I saw but I noticed you are a great teacher! … Anyway the time stamp is because I knew it was going to have something to do with a circle lol.
I'm from Brazil, and I was taught Heron's formula very early on, around 3rd or 4th grade. It always puzzled me, because it seemed to have been dropped on my lap out of nowhere (as far as my education was concerned, it really was). Still, I'm personally thankful it happened, because my first memory of doing maths by myself, out of sheer curiosity and without any obligation to do so, was trying to verify it from the usual formula for the area of a triangle, equating the two and working through the algebra. I wasn't successful, of course, but it got me here eventually, so that's nice.
I'm from India and study in a particular syllabus called ICSE. Heron's formula is a compulsory part of the mathematics syllabus starting from Year 8. However, we only learn and use it in the √s(s-a)(s-b)(s-c) form.As for my opinion on whether it is useful, I think it is very handy to have learnt because you wont always have the height of the triangle given as data. Thanks for the amazing clarification on how the formula was derived.
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 .
I was taught this formula and found it to be a hidden gem for many problems. Unfortunately, I cannot remember whether we proved it or not, but we did prove most things we have learned.
I was taught about Heron's formula. My high school teachers said that there is never a need to use it. I think it's always better to use some other tricks to get the area unless all sides are known
Yeah it really depends on which parameters of triangles are most often known or measured. I almost never saw anyone measure or know all three sides of a scalene triangle in practice, since it's way easier to just find two sides and one angle.
Thanks a lot. This is the first time I have seen someone prove the Heron's formula in this manner. The proof of Brahmagupta's formula is also very interesting and unique.
Big fan of Heron's formula here, so I may go on a bit. Here in the US, New York State, under the "Regents" math curriculum, in the late 1960's, we did learn Heron's formula. We were not required to memorize the derivation, but it was in the textbook. Decades later, I decided to test my algebraic chops, and try to derive it with the only clue I remember from the book, that it relied on factoring the difference of two squares. If you drop an altitude on one of the sides, you can solve for the altitude and one of the unknown segments on the base simultaneously, using the Pythagorean Theorem. Mathologer takes a shortcut by using the law of cosines, but my method is how you get the law of cosines as well (and you can get rid of the fraction in the Mathologer's version with some convenient cancellations). But once I was there, and I got the formula knowing what I was looking for, I thought of four justifications for "discovering" Heron's formula instead of calling it a day having a formula for the area in terms of the sides: 1) it lacks symmetry. There is nothing special about any side, other than that you chose one to drop an altitude on; 2) it's pretty nasty to calculate from. Many of us have probably calculated nastier ones, but we can do better; 3) it is badly scaled. You end up raising numbers to the fourth power, which usually results in something large, then subtracting them, leading to truncation or round-off errors if you don't keep a lot of decimal places (I realize Heron wasn't thinking about floating point calculations. Heck, he didn't even have a decimal system) 4) it's not obvious from the original formula, at least to me, that your area isn't going to turn out to be the square root of a negative number. If you expand the trinomial and collect terms, you do get something symmetrical, but all the other objections remain. That might tell you that, since expanding didn't work out very well, maybe the opposite--factoring--is the solution. In addition to factoring the difference of two squares, twice, at one point you have to collect some terms and recognize it as the square of a binomial. It's all quite pretty. But it's 100%, algebra, none of the geometric insight of this video. With the final formula, in addition to being much prettier and easier to calculate from (if the need arises which, truthfully, it rarely does) you can see at once that for a legitimate triangle, no one side being longer than the sum of the other two (or equivalently, no side being longer than the semi-perimeter) you'll never get a negative number and moreover, if a "triangle"s one side is exactly equal to the sum of the other two it has zero area, as expected.
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.
That makes perfect sense. As usual, I was looking for a profound reason and making it more complicated than it needed to be. You'd think I'd know better by now...
@@Mathologer Why did you chose purple then ? You had so many nice color available. Saphire or Sky. Ocean, Turquoise or Teal, lavender , Iris even Mauve, Lillac and Violet is better as a blue than "purple", If you must use P why not persian blue, prussian blue or pacific blue. I'm am _SO MUCH_ falsely offended right now...
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 TH-cam.
Hero's formula for the area of a triangle is one of those things introduced as a curiosity in a math textbook way back in middle school that I never really got a handle on. To a sixth grader, that is an impressively complex formula for an ancient to have discovered, and no proof was forthcoming. Through high school I saw it a couple more times, always in passing, a sort of neat oddity that seems compact but rarely gets used in practice. It was really neat to see an intuitive proof and motivation after all these years. That said, it doesn't exactly seem useful. Even if you somehow do know the lengths of a triangle but not its angles, this formula is still not the fastest way to find the area. Typically, if you're doing this by hand, you will either have a table of square roots (for Hero's method) or of logs and logs of sines (for the law of sines method). That method is still faster, because you skip all the multiplication steps. If you want to compute the area of the triangle with a computer, you can use Newton's method to get the square root, and I assume Heron's formula really is faster. But the thing is, you basically never know all the side lengths of a triangle (and nothing else) before trying to find its area. Rather, you probably have coordinates, in which case the shoelace formula is by far the fastest. So like, what is this formula actually good for? Is it just a novelty like the quartic formula? If it's never used, then no, I don't think it should be taught as part of a standard curriculum. The brief mentions in books for interested students are probably enough. There is _so much_ I want to add to the math curriculum, and the curriculum is already packed as it is. It's hard to justify cramming in more random formulas to teach, prove, and memorize. (BTW, although the phrase "Heron's formula" is seen pretty often in mathematical texts, in pretty much all other contexts in English, "Hero" is far more common than "Heron." Similarly, we say "Plato" rather than "Platon." The practice of Latinizing ancient Greek names is pretty standard in English. In classical Latin, the nominative singular would be "HERO," and the genitive singular would be "HERONIS." Since the Latin stem is still Heron-, the English adjective would be "Heronic" rather than "Heroic." Again, that's like the adjective "Platonic" rather than "Platoic. Other examples include "Pluto/Plutonic" and "Apollo/Apollonic." Admittedly, there are some exceptions, like the word "gnomon.")
One minute into the video, and I was already surprised: I had never learned/noticed that the 3-4-5 triangle has inradius 1! The fact that the sum/product identity generalizes to any triangle with inradius 1 is amazing.
To answer your question about other countries (although you might already know this): Suggesting to include something like Heron's formula into the mathematics curriculum in Germany would make the other commission members look at you as if you had told them to leave the building because a marsian space ship has landed on the roof.
When I was at the Gymnasium in Germany, in 9th grade, approx. 1990, the formula was included in the problem section of the book we used (Lambacher Schweizer). If I remember correctly, the problem even consisted in proving the formula (given many hints on how to proceed).
@@bjornfeuerbacher5514 Yeah, that was long ago. Lambacher Schweizer is still there, but anything beyond "draw a rectangle around the triangle and compute half the area of it" is nowadays beyond expected student abilities regarding "proofs". (Actually, most students don't even (need to) get the difference between "proving" and "using" the formula.) Fun fact: Just last week I was asked to comment on a complaint of several principals (obviously urged by students and parents) about the recent "Abitur" in the state of Niedersachsen. One of the exam tasks could not be solved by the students, because they had to calculate the area of a triangle without being allowed to look up the formula (one part of the exam is to be solved with pen and paper only).
I teach math and introduced Heron's formula as part of an exam question. There the students had to use the formula to calculate the area of a triangle with sides 2, 7 and 11 cm. The trick about this question is that this is an impossible triangle (A+B < C), which quite beautifully does not work in Heron's formula as you get the square root of a negative number. Sadly nobody figured out that the question itself was incorrect and probably assumed that the fault was either in the formula itself, or in their calculations. Once the solution became known I got some complaints about it (lol), but I think they learnt a good lesson.
I didn't learn Heron's formula in school and I was taught in Portugal. Very good video as always, just found a typo at 25:03 (the third term is A+B+D-B)
"There should be an equation in there; let's go and find it!" For a moment, I felt that spark of adventure I used to feel when doing math in my youth. I missed that feeling.
Even some articles written in English use p (but usually by authors who don't speak English natively). In English-language textbooks, it's usually a lowercase s. Rarely, you might see it with no s at all, in the fully expanded form in terms of a, b, and c.
Calculating the area of a triangle is part of the Mathematics Olympiad program (IMO training), because counting things in multiple ways can reveal complicated identities. The proof used the cosine rule. Ew. I actually discovered (and proved) the formula myself, when I was about ~13, just by using the right-angle theorem (Pythagoras). More recently, I found out (thanks to Wikipedia) that the maximum area of a quadrilateral happens with a cyclic quadrilateral. This is beautiful, because we can extend this for any polygon. I used some ugly math to prove this, but I did found out about some nice things. I also found out about the correction term. I think that many people discovered this, even before 1800. Your video's encourage me to seek for simple proves myself. The method for the proof of p=a²+b², can be extended for all natural numbers. You can prove that numbers that shouldn't be written as sum of two squares, have an even number of representations. Next, you parameterise n=a²+b²=c²+d², to show that n is the product of two numbers (greater than one), as the sum of two squares. Next, you can say that we had the lowest number that works (with some extra things), to finish the prove.
We were taught this in school (7th/8th standard Maths) in Maharashtra, India. No proof was derived sadly but I did one by myself later when learning about incircles. I was talking about it to an Aussie kid years later and he very proudly said they didn't learn any of that. Like I was the idiot for learning something extra.
I was taught Heron's formula in Belgian high school. 12th grade, the most intensive math option they had (8 hours/week). And only as a side tangent, without proof.
I do NOT remember it being taught in my school in England in the 1960s. I do remember coming across it in my mothers school textbooks - but not with Heron's name attached. Always thought it was fabulous - particularly like the way it gives a zero if S is any one of A, B or C which corresponds to the triangle collapsing into a line.
No, Heron's formula has never been part of the UK maths curriculum. Like a lot of interesting things actually, because we expect students *not* to do any maths at university and so try to rush as quickly as possible to calculus.
The remarkable fact that a 3-4-5 triangle has an incircle of radius 1 was new to me. It just goes to show that Truth can be sitting there, just 1 step away, and you may not see it-- for decades. I am both humbled and joyous whenever Mathologer opens my eyes in this way.
In Italy, heron's formula is taught in the middle school, when pupils are about 12 years old. Many among them will forget it forever, some of them see it again in high school.
Teaching mathematics of that level to children of that age is a waste of time. Until ~ age 13-14, most children haven't developed the abstract reasoning enough for anything beyond numeric arithmetic. It would be better to start teaching logical reasoning sooner, so kids don't freak out when asked to write proofs.
Took me way too long to realize that I've already learned some of this (nearly all to the cyvlic quads, including proofs for them) in highschool (special, math-heavy class, it is not in the base material)
@@Mathologer How long did you live in Germany, and how long have you been in Australia? (I promise to avoid my natural inclination to add any numbers I see.)
@@godfreypigott I grew up in Germany, did military service there, a bit of uni and then spent a couple of years studying and postdocing abroad. I've been in Australia since 1995 :)
@@Mathologer But you chose Melbourne - WHHHYYYY? Sydney is quintessential Australia. Oh wait ... then you might have had to work with Norman Wildberger ... I think I understand.
One great use of Heron's formula, which I used not long ago, was to calculate the distance from a point to the line joining two other points, all in 3d space. I'm sure there is a formula just for that, but I couldn't easily find one. But I did remember Heron's formula, so worked out that I could calculate the area of the triangle using Herons formula and then the perpendicular distance using the ordinary formula for the area. I had to calculate that for several million points, so all done very fast on a computer.
Excellent video, the mathematical proof at the end impressed me. I struggled a lot to get the challenge at minute 11:36. First, I drew the diagonals e and f, with an angle x between them. I added the area of the 4 small triangles that are formed (I divided the diagonals into e1,e2 and f1,f2) with the formula A1=(1/2)*e1*f1*sinx, etc. and got A=(1/2)*ef*sinx. Then, I worked the rule of cosines with the same little triangles to get cosx and replace: (sinx)^2=1-(cosx)^2, then: A^2=(1/4)*((ef)^2) *(1-(cosx)^2). In the end I got: A=(1/4)*sqrt(4(ef)^2-(a^2-b^2+c^2-d^2)^2). In this formula there are no trigonometric elements and the formula that appears in the video can be derived algebraically. Maybe there are many easier ways to do it.
I teach Art of Problem Solving Geometry and I've taught the proof of Herons formula and even the Area=(perimeter • inradius)/2 for years. I never even thought that perimeter/2 is S from Herons formula. I had no idea there was a connection and my mind was blown at every step along this beautiful journey.
I learnt about heron's formula in 9th grade but our textbooks didn't give a proof of it. Luckily our math teacher didn't like that and he showed us a proof of it by deriving it with just algebra. The geometric proof in the video was something I've never seen before, it is really nice.
I earned a PhD in mathematics in the U.S. but was unaware of Heron's formula until reading William Dunham's great "Journey through Genius." And then you say there's a formula. for. the. area. of. ANY. quadrilateral. (convex or not) Thank you, Mathologer!!! 💥💥💥 ❤❤❤
There is a beauty when it is explained with visuals which you have well explained. You have seemed have done a beautiful job at that for which i can't thank you enough.Great job.
How wonderful! At 1:44 I even realized (with small simple prove) that this triangle is also divided in areas of 1, 2 and 3! One ssssimple SSSSS - WOW! And there is a general term for a triangle with a right angle and inlying unit circle. The value of the area AREA (using small prove) = G*1 + P*1 + R*1 = (A-1) + (B - 1) + 1 = A + B - 1. Ready! And: There is hidden something more universal... ;-) Once again - just wonderful.
After you drew the incircle for heron's formula my mind immediately went to the Tan identity i.e, Tan x +Tan y + Tan z= (Tan x)(Tan y)(Tan z) if x+y+z=nπ Using it, it was pretty straightforward to see how heron's formula worked.The visual proof was really elegant to see though.
10:30 i guess another thing to mention is that it doesn't matter which pair of angles, since a+b+c+d = 360 in a square, meaning a+b = 360-(c+d), and so cos((a+b)/2) = cos(180 - (c+d)/2) = cos((c+d)/2)
Mind blown. I've always liked all your videos, but for this one, with each new step, I kept exclaiming to myself: "Oh my goodness", "Oh my goodness", "Oh my goodness".
Wurde mir noch gelehrt:-) Doch wie bereits erwähnt ohne skizziertem Beweis, jener folgte dann auf der Uni. Danke für jene Videos! Sie sind eine Bereicherung! Ein großes Komplement, und Danke für die Jahre.
I introduce my highschool students to this, but without the proof, as an alternative way to find the area of a triangle. This video is going to my "maths plus ultra" students, who like things like this. It's super accessible. Thank you.
In Brazil we learn about Heron's formula in high school, but it's extremely rare for it to be on college admission tests (Vestibular/ENEM). And I had never heard of Brahmagupta's formula.
1:21 Can we all just appreciate the fact that the 3-4-5 -triangle features both, a unit circle (as its incircle/”heart”), and a unit square (in its right-angled corner) 😮?
”1:21 Can we all just appreciate the fact that the 3-4-5 -triangle features both, a unit circle (as its incircle/”heart”), and a unit square (in its right-angled corner) 😮?” Of course; the area of the unit circle: A(C) = π*1² = π, and the area of the unit square: A(S) = 1² = 1. 🙂
Learned it in USA; knew it for the test, haven't stumbled on many applications. Heron's formula is numerically unstable in floating point. I feel like there should be a straightforward algebraic proof involving a system of two quadratics, but it's more interesting to know how the ancient Greeks convinced themselves.
For those dazzled by the geometric argument that shows PQR = P+Q+R, a quicker way is to use complex numbers: From the observation that the angle sum of the (P,1), (Q,1) and (R,1) right triangles is 90°, we know the sum of the arguments of the three complex numbers P+i, Q+i and R+i is also 90°. The argument of their product is therefore 90°, which implies that the real part of this product is zero. Since Re[(P+i)(Q+i)(R+i)] = PQR -P -Q -R, we get PQR = P+Q+R.
If you liked that you'll probably also like this proof of Pythagoras's theorem th-cam.com/video/r4gOlttnJ_E/w-d-xo.html I also featured this earlier in a video on this main channel th-cam.com/video/r4gOlttnJ_E/w-d-xo.html
@@Mathologer Thanks! It immediately reminded me of this one, of course. The one in this Heron’s formula video was just perfectly timed for some reason. And might I say, just that extra tad more magical.
2:32 Because phi is defined as the number that satisfies this equation: x=(1/x)+1., therefore 2*x*[(1/x)+1]=2*(x+1)=2x+2. So this means that the left side of the equation is the same as the right side of the equation.
It took me a long time to get to what the Mathologer pulls out of his hat after the first minute (following "whoa! What just happened?") Part of the problem was I steadfastly refused to accept that the three angle bisectors of the triangle meet at the center of the inscribed circle, without proving it according to Euclid--perhaps because of it's similarity to the flawed proposition that leads to the "theorem" that all triangles are isosceles. Getting past that hurdle, though, it's pretty straightforward to derive a formula for any triangle with a unit circle inscribed in it. Each colored segment (BTW of course you wouldn't use "blue" because it's name would be the same as side B) is simply equal to the cotangent of the half angle of each vertex. For the 3-4-5 you can apply the half-angle formulas to get the cotangents in terms of the cosines (sqrt(1+cosine/1-cosine) ), which for the 3-4-5 are 4/5/ 3/5 and 0, and you miraculously get whole numbers for each side. Time to unpause the video. It follows that the sum and product of the cotangents of any 3 angles that add up to 90 degrees must be the same. I guess the only hope of proving that is expressing them in terms of imaginary exponentials, and grinding through a lot of algebra. No geometric insight in that though. For another time. BTW while Heron's formula in terms of the sides instead of the semi-perimeters might be preferable in some ways, if you're actually calculating it you annoyingly have to form 4 sum-difference combinations of the sides, whereas in terms of the semi, you can just stick it in a memory on your pocket calculator and re-use it. Did Heron have a pocket calculator? He was an engineer, wasn't he? In all seriousness, Heron being an engineer clears up a question I've had for years: were this Heron of Alexandria and the one credited with inventing the first steam engine--which consisted of a steam boiler with jets on the side that set it spinning-- one and the same? Apparently so. It's inefficiency, and other logistical problems problems, probably explains why the chariots of the day weren't powered by Heron engines.
I love Mathologer's videos. His videos are very informative and great for anyone who has an affinity for math. That being said, I also LOVE his shirts. I like to collect witty t-shirts, and his make me jealous that I don't have them!
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!
Hi. I'm brazilian too, I'm 58 years old and I remember I've learned the Heron's formula when I was a student in a military school, back in the 1970 decade.
I have a masters degree in Math and I've worked with bicentric quadrilaterals, it is the theme of my master thesis (sorry for my bad english). One of the funny facts about them is that their area is simply the square root of the product of their sides.
This is the best Mathologer video I've already watched. Thank you all!
Já eu estudei em ensino médio normal mesmo e nunca ouvi falar dessa fórmula na vida kkk
@@aureapureza8324 E quanto a Proporção, Áurea ? :P
Não sei pq não aprendemos essa fórmula aqui na Austrália, ela é bonita né? (desculpa por meu português ruim)
@@patrickwebster3152 Mesmo olhando o conteúdo deste canal, fica claro que não caberia em currículo algum, nem de um curso de faculdade. Fica para o estudante, ao menos os de matemática, se aprofundar em seus conhecimentos. Mas mesmo assim, ainda irá surgir muita coisa que não foi descoberta ou que está em estudo.
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!
Oh you make me jealous now.
@@aradhya_purohit Indeed....
@@aradhya_purohit Yep 😌. Too bad I’m not Aussie 😢🇦🇺.
I wish my Biochemistry degree had such great teachers. I might have actually finished it. Instead; most of my teachers and professors were lackluster nobodies, who didn’t really know, how to make stuff accessible. For example, my Physical Chemistry professor just gave us, like, 50 different formulas for rote memorization, without really motivating any of them (like, why do you need integration, for temperature-/pressure-systems 🤯?). 😔
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
It’s actually good that actual math proofs aren’t very geometrical. The geometry behind a lot math is definitely extremely beautiful but it is very easy to get things wrong if you rely on pictures and drawings. Most geometric proofs of complicated results are derived a posteriori since we can back up the geometrical proof with a more precise proof based on other techniques.
It's how we used to do it before!
But once we started using more complex stuff like powers of 4 ie. x⁴
We would have needed to use 4d brains to comprehend
And this is why we still thank the arabic mathematician Al-Khawarizmi who created algebra and the hindu-arabic numbers (1, 2...)
Then explain 4:25
@@motherisape Area = Sr
And Sr² = RGP is correct
When r was 1
Sr wad S
And Sr² was S since 1² = 1
@@praharmitra also the fact that algebra and calculus works for any number of dimensions, while geometrical intuition stops at 3.
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. 🙂🙂
Nice 👍🏻. Love to India 🇫🇮❤🇮🇳.
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.
Blackpenredpen has got an amazing video with the derivation of Heron's formula. It's a bit old, so you might have to scroll down a bit.
@@chessematics too bad that TH-cam punishes people who post useful links in comments :(
@@chessematics a workaround might be to post only the portion after the dot Belgium
No wonder why all our tables are freaking Squares. We never learned how to make a rectangle.
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!
I was about to point out the same but decided to scroll the comments to see if someone had already done so. Lickily I did.
@@j100j Same thing, here.
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
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!
Mission accomplished :)
On the contrary, the formula had enough symmetry to easily memorize, for me personally at least. But I agree, the derivation is so simple, I'd never forget or have to memorize it! And never learned about the quadratic form.
🤯
In 1967, I took an Engineering Measurements course at U.C. Davis. One of the measuring tools was a polar planimeter which allows you to find the area of any figure by tracing its perimeter. If you have a plot of land drawn on a map. you can easily find the enclosed area. This is a very practical application of Heron's Formula.
Planimeters don’t use Heron’s formula. A polar planimeter adds up the changes in angle multiplied at each point by the square of the radius. A linear planimeter adds up x dy - y dx, a continuous version of the "shoelace formula" about which Mathologer made another nice video a few years ago th-cam.com/video/0KjG8Pg6LGk/w-d-xo.html
Planimeters are connected to the divergence theorem, which green's theorem is derived from, and which the shoelace algorithm can be derived from. And you can derive Heron's formula from green's theorem (divergence theorem). Heron's formula relates area to perimeter of a closed boundary. With the more general divergence's theorem, you can calculate area from the closed boundary of its 2D shape. And going deeper you can also calculate volume from the closed surface boundary of a 3D shape. This is what's amazing about the geometric relationship of a shape's boundary. You can also calculate metrics like center of gravity, volume, inertia tensor of 3D shapes (which are typically expressed with triple integrals) by a simple surface integral. (For example Volume = 1/3 Surface integral of dot product of field [x, y, z] and surface normal * dS. Most 3D models are Boundary Reps, so this is a handy relation in computational geometry. Heron's formula is mapping a 1D metric to a 2D metric just like (perimeter to area) just like this example of divergence theorem is mapping a 2D metric to a 3D metric. The trick is to recognize the special field. In this video, the trick is recognize the geometric relation of the triangle and inscribed unit circle. But there is also a trick in recognizing the field to apply with the divergence's theorem.)
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
I was taught Heron's formula in high school, in Greece. I found it rather interesting, but considered it a curiosity, mostly, as it wasn't connected to anything else I was taught.
Actually, here in Russia we are in fact taught Heron's formula at school, but I'm not really used to it since it was a lockdown year when we were taught this. So if a problem requiring Heron's formula to solve it occurs, I'm always a bit perplexed as it doesn't come to mind at first.
not only is it taught but 90% of other Greek formulas are also taught. I knew most greek scholars from all the formulas!
but i know this with the semi-perimeter denoted by p
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"
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.
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.
A truly beautiful visual and mathematical feast. The complicated symmetry is an extra detail that (for those of us obsessed with symmetry) elevates this to astronomical heights.
Thank you, Burkhard for reminding me how much I love Mathematics. A special round of applause to the crystal clear animations.
I agree 100%. This was a spectacular video.
2 minutes in and you've already blown my mind Mathologer! Thanks! Love your videos
I'm homeschooling my children, and interestingly I just taught them Heron's formula yesterday as an example of where square roots can be very practical in real life problems.
Specifically, we talked about how convenient it can be if you are trying to determine the size of a plot of land you want to buy when all you can measure are distances and no idea if the land is square. Just walk the perimeter, walk the diagonal, apply Heron's formula to the 2 triangles, and you know immediately how big the plot is. For 7th graders who don't yet know trigonometry, the example worked beautifully. It should absolutely be taught in schools. It is a crime that it is not considered essential these days.
It's taught very carefully in India, but i doubt how many students actually care about it.
You need the square root for computing accelerations from position or distances from a plan or much much more. And when surveying land, you need square roots all over the place. But you will never once use Hero's formula. In particular, accurately measuring angles is easier than accurately measuring distances.
Yes, in profesional landsurveying you use angles, because you had equipment (theodolite, total station) that measure angles and distances.
Surveying in the core, is to translate points between carthesian coordinate (2,2) to polar coordinate (45°, 2root(2)) and vice versa.
And that translation is all about solution for right triangles.
Heron's formula is niche one, and good if you only have ruler to measure sides. I would say its Identity should to teach in schools. (And cut some amout of mindles computing)
(And why not throw Ptolemy's theorem in to play and calculate solution to last unknown measurement.
Actually, it would be fun to first calculate missing border, blocked by huge tree, with Ptolemys. And then using Heron's to calc. the area. )
@@EebstertheGreat Well, you'll have to measure at least on side of a triangle even for the standard 1/2 base times height formula :) Anyway, speaking for myself, Heron's formula has definitely proved a very useful addition to my mathematical toolbox on many occasions :)
@@EebstertheGreat
I use Pythagoras _all the time_ in structural engineering, as every problem has to be turned into, first, their vertical and lateral components (think diagonal braces, roof members, trusses, anything not perfectly vertical or horizontal), and then every member needs to be checked for axial stresses and flexural stresses (column buckling and failing vs beam bending and sagging).
Opposing angles subtend a partition of the whole circle. The measure of an angle is half the measure of the arc it subtends. Therefore, the sum of the opposing angles equals half the sum of opposing arcs, which is half of 360 degrees, which is 180 degrees.
Excellent point, and an easy way to remember this characteristic of cyclic quadrilaterals!
That's it :)
@@Mathologer Hello sir ,i am from India and i am one of the biggest fan of your. ...and i have wish that i want to talk to you...
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 👍
1+2+3 = the amount of seconds between the release of the video and me clicking on it
For me, it’s: (1+2+3)-(1*2*3).
Starke Herleitungen!
Die Schlussmusik wird immer besser!
I learned Heron's formula when I was in middle school in the state of California (4 years ago) and I'm pretty sure it's still being taught, but I've never used it outside of math competitions. Lots of great information in the video that I didn't know before though. Thanks!
Whilst in Romania I sat down with a man once who looked JUST EXACTLY like you. He was a Romanian math teacher... we sat there and wow'd each other for about 4 hours. LOL the restaurant was trying to close and we had this deep mathematical conversation going on... You remind me of him so much. (This was about a year ago... seems like a million years ago.)
If I'd lived in Classical Europe, I'd totally have worshiped geometry as sacred! Great video!
Not much classical geometry left in high school curricula. These days a lot there is about learning how to push buttons on a calculator :(
@@Mathologer High school maths (Algebra and Geometry) teacher here. The reason that we don't do classical geometry anymore is because school is compulsory. Back when classical geometry was taught at this level, school was an elective or privileged activity. The people who were taking classes did so because they wanted to and/or understood the value of it. Most people do not value math. Take a look at your subscriber count compared to mindless entertainment like Markiplier or something.
I have good reason to believe that most of my students are not only uninterested in math (in fact, they hate it -- even the cool stuff), but are incapable of comprehending it. So how did the school curriculum respond? By boiling all of the math down so that the students with the lowest math performance could still pass the classes -- that is to say, teaching kids how to push buttons in a calculator, but not why the formulas work or what they mean. It's a waste of everyone's time, but that's what you get when teachers become glorified babysitters for families with working parents.
@@mr.johnson3844 Totally agree. High school teacher from USA.
@@mr.johnson3844 I totally agree and want to add: there is also a shift from "learn how to think" to "do as instructed“.
@@mr.johnson3844 true ma'am
I keep coming back to watch the animation for finding Brahmagupta’s formula from Heron’s formula. It’s simply incredible, both the animation and the actual proof.
There's an extended formula for three dimensions which shows that meshes consisting entirely of triangles have a fixed volume even if they aren't rigid. Would love to see a video on that.
How could a deformed geodesic dome have the same volume? The same surface area for sure, but if I squish the top of a dome down there's definitely less area.
Apparently the great Alexander Grothendieck independently discovered this formula as a schoolchild while being hidden from the Gestapo at an orphanage in Southern France during WWII....it would be nice to be able to give him a mention!
Were you going to answer brian's question?
@@mikeoffthebox What is the formula?
@@briancooke4259 There are multiple possible volumes but you can't continuously go from one to the other by flexing, even when it isn't rigid
After appreciating geometry and history, I must commend the script. The coverage and sequencing is well planned and executed. Thanks for everything.
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.
8:54. I never applied myself in school at all but watching your videos the past year or so has been enlightening! I was a bit intimidated I don’t remember the first topic I saw but I noticed you are a great teacher! …
Anyway the time stamp is because I knew it was going to have something to do with a circle lol.
*explain this **4:25*
I'm from Brazil, and I was taught Heron's formula very early on, around 3rd or 4th grade. It always puzzled me, because it seemed to have been dropped on my lap out of nowhere (as far as my education was concerned, it really was). Still, I'm personally thankful it happened, because my first memory of doing maths by myself, out of sheer curiosity and without any obligation to do so, was trying to verify it from the usual formula for the area of a triangle, equating the two and working through the algebra. I wasn't successful, of course, but it got me here eventually, so that's nice.
I'm from India and study in a particular syllabus called ICSE. Heron's formula is a compulsory part of the mathematics syllabus starting from Year 8. However, we only learn and use it in the √s(s-a)(s-b)(s-c) form.As for my opinion on whether it is useful, I think it is very handy to have learnt because you wont always have the height of the triangle given as data. Thanks for the amazing clarification on how the formula was derived.
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 .
*In the second paragraph I mean to say that Pythagoras's theorem gives the result above _for the height._
I was taught this formula and found it to be a hidden gem for many problems. Unfortunately, I cannot remember whether we proved it or not, but we did prove most things we have learned.
I was taught about Heron's formula. My high school teachers said that there is never a need to use it. I think it's always better to use some other tricks to get the area unless all sides are known
Yeah it really depends on which parameters of triangles are most often known or measured. I almost never saw anyone measure or know all three sides of a scalene triangle in practice, since it's way easier to just find two sides and one angle.
Thanks a lot. This is the first time I have seen someone prove the Heron's formula in this manner. The proof of Brahmagupta's formula is also very interesting and unique.
Big fan of Heron's formula here, so I may go on a bit. Here in the US, New York State, under the "Regents" math curriculum, in the late 1960's, we did learn Heron's formula. We were not required to memorize the derivation, but it was in the textbook. Decades later, I decided to test my algebraic chops, and try to derive it with the only clue I remember from the book, that it relied on factoring the difference of two squares. If you drop an altitude on one of the sides, you can solve for the altitude and one of the unknown segments on the base simultaneously, using the Pythagorean Theorem. Mathologer takes a shortcut by using the law of cosines, but my method is how you get the law of cosines as well (and you can get rid of the fraction in the Mathologer's version with some convenient cancellations). But once I was there, and I got the formula knowing what I was looking for, I thought of four justifications for "discovering" Heron's formula instead of calling it a day having a formula for the area in terms of the sides: 1) it lacks symmetry. There is nothing special about any side, other than that you chose one to drop an altitude on; 2) it's pretty nasty to calculate from. Many of us have probably calculated nastier ones, but we can do better; 3) it is badly scaled. You end up raising numbers to the fourth power, which usually results in something large, then subtracting them, leading to truncation or round-off errors if you don't keep a lot of decimal places (I realize Heron wasn't thinking about floating point calculations. Heck, he didn't even have a decimal system) 4) it's not obvious from the original formula, at least to me, that your area isn't going to turn out to be the square root of a negative number.
If you expand the trinomial and collect terms, you do get something symmetrical, but all the other objections remain. That might tell you that, since expanding didn't work out very well, maybe the opposite--factoring--is the solution. In addition to factoring the difference of two squares, twice, at one point you have to collect some terms and recognize it as the square of a binomial. It's all quite pretty. But it's 100%, algebra, none of the geometric insight of this video.
With the final formula, in addition to being much prettier and easier to calculate from (if the need arises which, truthfully, it rarely does) you can see at once that for a legitimate triangle, no one side being longer than the sum of the other two (or equivalently, no side being longer than the semi-perimeter) you'll never get a negative number and moreover, if a "triangle"s one side is exactly equal to the sum of the other two it has zero area, as expected.
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.
as a teacher, I'd avoid using Blue because of how easy it is to confuse with side B of the triangle.
And that's exactly the reason why I used purple :)
That makes perfect sense. As usual, I was looking for a profound reason and making it more complicated than it needed to be. You'd think I'd know better by now...
@@johnopalko5223 I was thinking, "Is there some weird historical reason?" Nope, of course not.
@@Mathologer Why did you chose purple then ? You had so many nice color available. Saphire or Sky. Ocean, Turquoise or Teal, lavender , Iris even Mauve, Lillac and Violet is better as a blue than "purple",
If you must use P why not persian blue, prussian blue or pacific blue.
I'm am _SO MUCH_ falsely offended right now...
@@hurktang because purple is a common color name,
the other one is Orange with looks like 0, and martin did campaign for Yellow
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 TH-cam.
Hero's formula for the area of a triangle is one of those things introduced as a curiosity in a math textbook way back in middle school that I never really got a handle on. To a sixth grader, that is an impressively complex formula for an ancient to have discovered, and no proof was forthcoming. Through high school I saw it a couple more times, always in passing, a sort of neat oddity that seems compact but rarely gets used in practice. It was really neat to see an intuitive proof and motivation after all these years.
That said, it doesn't exactly seem useful. Even if you somehow do know the lengths of a triangle but not its angles, this formula is still not the fastest way to find the area. Typically, if you're doing this by hand, you will either have a table of square roots (for Hero's method) or of logs and logs of sines (for the law of sines method). That method is still faster, because you skip all the multiplication steps. If you want to compute the area of the triangle with a computer, you can use Newton's method to get the square root, and I assume Heron's formula really is faster. But the thing is, you basically never know all the side lengths of a triangle (and nothing else) before trying to find its area. Rather, you probably have coordinates, in which case the shoelace formula is by far the fastest.
So like, what is this formula actually good for? Is it just a novelty like the quartic formula? If it's never used, then no, I don't think it should be taught as part of a standard curriculum. The brief mentions in books for interested students are probably enough. There is _so much_ I want to add to the math curriculum, and the curriculum is already packed as it is. It's hard to justify cramming in more random formulas to teach, prove, and memorize.
(BTW, although the phrase "Heron's formula" is seen pretty often in mathematical texts, in pretty much all other contexts in English, "Hero" is far more common than "Heron." Similarly, we say "Plato" rather than "Platon." The practice of Latinizing ancient Greek names is pretty standard in English. In classical Latin, the nominative singular would be "HERO," and the genitive singular would be "HERONIS." Since the Latin stem is still Heron-, the English adjective would be "Heronic" rather than "Heroic." Again, that's like the adjective "Platonic" rather than "Platoic. Other examples include "Pluto/Plutonic" and "Apollo/Apollonic." Admittedly, there are some exceptions, like the word "gnomon.")
Heron's formula was one of my favorite from trig class! Good stuff! Thank you for sharing!
I remember coming across this formula at the corner of one page in my textbook in high school, and we just glossed over it.
One minute into the video, and I was already surprised: I had never learned/noticed that the 3-4-5 triangle has inradius 1!
The fact that the sum/product identity generalizes to any triangle with inradius 1 is amazing.
To answer your question about other countries (although you might already know this): Suggesting to include something like Heron's formula into the mathematics curriculum in Germany would make the other commission members look at you as if you had told them to leave the building because a marsian space ship has landed on the roof.
When I was at the Gymnasium in Germany, in 9th grade, approx. 1990, the formula was included in the problem section of the book we used (Lambacher Schweizer). If I remember correctly, the problem even consisted in proving the formula (given many hints on how to proceed).
@@bjornfeuerbacher5514 Yeah, that was long ago. Lambacher Schweizer is still there, but anything beyond "draw a rectangle around the triangle and compute half the area of it" is nowadays beyond expected student abilities regarding "proofs". (Actually, most students don't even (need to) get the difference between "proving" and "using" the formula.)
Fun fact: Just last week I was asked to comment on a complaint of several principals (obviously urged by students and parents) about the recent "Abitur" in the state of Niedersachsen. One of the exam tasks could not be solved by the students, because they had to calculate the area of a triangle without being allowed to look up the formula (one part of the exam is to be solved with pen and paper only).
@@hugo3222 Did you know that I actually grew up in Germany? That last fun fact is actually pretty depressing.
@@Mathologer What's going on here? My answer disappeared twice.
I was looking for an Intuitiy proof of this some time ago. Excited to see you explain it. Thank you :)
I teach math and introduced Heron's formula as part of an exam question. There the students had to use the formula to calculate the area of a triangle with sides 2, 7 and 11 cm. The trick about this question is that this is an impossible triangle (A+B < C), which quite beautifully does not work in Heron's formula as you get the square root of a negative number. Sadly nobody figured out that the question itself was incorrect and probably assumed that the fault was either in the formula itself, or in their calculations. Once the solution became known I got some complaints about it (lol), but I think they learnt a good lesson.
I didn't learn Heron's formula in school and I was taught in Portugal. Very good video as always, just found a typo at 25:03 (the third term is A+B+D-B)
Really loved this video.
The most practical way is the geometric way, that's you have the mastery.
Hats off to you mr. Mathologer.
"There should be an equation in there; let's go and find it!"
For a moment, I felt that spark of adventure I used to feel when doing math in my youth. I missed that feeling.
Visual animated proof at the end + good music = piece of art !
We were taught Heron's formula in Bulgaria. And half the perimeter is denoted as p, not S. That's the area.
Its just notation, the letters truely do not matter in the slightest.
We were given Heron's formula in Poland as a side note, but our teacher still went over it. The sum (a + b + c) / 2 was also denoted as p.
Here in Denmark, the half perimeter was denoted with small s, and the area is A.
It most probably depends on your language.
in Italia we learn to denote the perimeter as "2p" in school long before we encounter the semiperimeter in formulas (and it of course is p).
Even some articles written in English use p (but usually by authors who don't speak English natively). In English-language textbooks, it's usually a lowercase s. Rarely, you might see it with no s at all, in the fully expanded form in terms of a, b, and c.
Calculating the area of a triangle is part of the Mathematics Olympiad program (IMO training), because counting things in multiple ways can reveal complicated identities.
The proof used the cosine rule. Ew.
I actually discovered (and proved) the formula myself, when I was about ~13, just by using the right-angle theorem (Pythagoras).
More recently, I found out (thanks to Wikipedia) that the maximum area of a quadrilateral happens with a cyclic quadrilateral. This is beautiful, because we can extend this for any polygon.
I used some ugly math to prove this, but I did found out about some nice things. I also found out about the correction term. I think that many people discovered this, even before 1800.
Your video's encourage me to seek for simple proves myself.
The method for the proof of p=a²+b², can be extended for all natural numbers.
You can prove that numbers that shouldn't be written as sum of two squares, have an even number of representations.
Next, you parameterise n=a²+b²=c²+d², to show that n is the product of two numbers (greater than one), as the sum of two squares.
Next, you can say that we had the lowest number that works (with some extra things), to finish the prove.
We were taught this in school (7th/8th standard Maths) in Maharashtra, India. No proof was derived sadly but I did one by myself later when learning about incircles.
I was talking about it to an Aussie kid years later and he very proudly said they didn't learn any of that. Like I was the idiot for learning something extra.
At a rural high school in PA a few years ago, I was taught Heron's Formula in a Trigonometry class.
I was taught Heron's formula in Belgian high school. 12th grade, the most intensive math option they had (8 hours/week). And only as a side tangent, without proof.
I knew there was a gem in herons formula ... just couldn't see it until your video.... thank you for your amazing insight!!...
I do NOT remember it being taught in my school in England in the 1960s.
I do remember coming across it in my mothers school textbooks - but not with Heron's name attached.
Always thought it was fabulous - particularly like the way it gives a zero if S is any one of A, B or C which corresponds to the triangle collapsing into a line.
No, Heron's formula has never been part of the UK maths curriculum. Like a lot of interesting things actually, because we expect students *not* to do any maths at university and so try to rush as quickly as possible to calculus.
The remarkable fact that a 3-4-5 triangle has an incircle of radius 1 was new to me.
It just goes to show that Truth can be sitting there, just 1 step away, and you may not see it-- for decades.
I am both humbled and joyous whenever Mathologer opens my eyes in this way.
In Italy, heron's formula is taught in the middle school, when pupils are about 12 years old.
Many among them will forget it forever, some of them see it again in high school.
I actually wasn't taught the formula in high school but when I stumbled across it myself I still remember it making a big impression on me :)
In India about age 11-12
Teaching mathematics of that level to children of that age is a waste of time.
Until ~ age 13-14, most children haven't developed the abstract reasoning enough for anything beyond numeric arithmetic.
It would be better to start teaching logical reasoning sooner, so kids don't freak out when asked to write proofs.
Its india bro
Most finest mathematicians like ramanujan,aryabhatta were born here
We have their blood running bro..
I learned Heron's formula in high school and it is so great to see a proof of it! Thank you for such a great video! :)
Took me way too long to realize that I've already learned some of this (nearly all to the cyvlic quads, including proofs for them) in highschool (special, math-heavy class, it is not in the base material)
Actually Herons method for approximating square roots is still taught here in Germany from time to time, depending on the teacher.
Interesting. I actually grew up in Germany and was never taught anything due to Heron :(
@@Mathologer How long did you live in Germany, and how long have you been in Australia? (I promise to avoid my natural inclination to add any numbers I see.)
@@godfreypigott I grew up in Germany, did military service there, a bit of uni and then spent a couple of years studying and postdocing abroad. I've been in Australia since 1995 :)
@@Mathologer But you chose Melbourne - WHHHYYYY? Sydney is quintessential Australia. Oh wait ... then you might have had to work with Norman Wildberger ... I think I understand.
One great use of Heron's formula, which I used not long ago, was to calculate the distance from a point to the line joining two other points, all in 3d space. I'm sure there is a formula just for that, but I couldn't easily find one. But I did remember Heron's formula, so worked out that I could calculate the area of the triangle using Herons formula and then the perpendicular distance using the ordinary formula for the area. I had to calculate that for several million points, so all done very fast on a computer.
Ohh I'm early! Yo mathologer sir ! Big fan I'm of yours! It's my physics exam in couple of days so I'll watch the video later! Excited ! ❤❤❤
I am late and early 1:26 am here in Melbourne. Won't last much longer. Good luck with the physics exam :)
@@Mathologer THAT MEANS A LOT TO ME SIR!! THANK YOUUUU!
ALL THE BEST IN EVERYTHING YOU DO SIR! 👍👍👍
@@black_jack_meghav good luck!!
@@nugboy420 thanks a lot 🥺 😇😇😊
Excellent video, the mathematical proof at the end impressed me.
I struggled a lot to get the challenge at minute 11:36. First, I drew the diagonals e and f, with an angle x between them. I added the area of the 4 small triangles that are formed (I divided the diagonals into e1,e2 and f1,f2) with the formula A1=(1/2)*e1*f1*sinx, etc. and got A=(1/2)*ef*sinx. Then, I worked the rule of cosines with the same little triangles to get cosx and replace: (sinx)^2=1-(cosx)^2, then: A^2=(1/4)*((ef)^2) *(1-(cosx)^2). In the end I got: A=(1/4)*sqrt(4(ef)^2-(a^2-b^2+c^2-d^2)^2). In this formula there are no trigonometric elements and the formula that appears in the video can be derived algebraically. Maybe there are many easier ways to do it.
25:02 small typing mistake here inside the third pair of parethsesis, you wrote "A+B+D-B" instead of "A+C+D-B"
I teach Art of Problem Solving Geometry and I've taught the proof of Herons formula and even the Area=(perimeter • inradius)/2 for years. I never even thought that perimeter/2 is S from Herons formula. I had no idea there was a connection and my mind was blown at every step along this beautiful journey.
I'm Ukrainian. We did learn Heron's formula and used it a couple of times. Just enough to still remember it.
I learnt about heron's formula in 9th grade but our textbooks didn't give a proof of it. Luckily our math teacher didn't like that and he showed us a proof of it by deriving it with just algebra.
The geometric proof in the video was something I've never seen before, it is really nice.
This is truly some mind boggling math
Also Heron's formula isn't really taught in school here in Singapore, except in Math Olympiads.
I'm from India and we have Heron's Formula in our schools! I'd be sure to share this video onwards!
I want that shirt man lol
I earned a PhD in mathematics in the U.S. but was unaware of Heron's formula until reading William Dunham's great "Journey through Genius."
And then you say there's a formula. for. the. area. of. ANY. quadrilateral. (convex or not) Thank you, Mathologer!!! 💥💥💥 ❤❤❤
There is a beauty when it is explained with visuals which you have well explained. You have seemed have done a beautiful job at that for which i can't thank you enough.Great job.
How wonderful! At 1:44 I even realized (with small simple prove) that this triangle is also divided in areas of 1, 2 and 3! One ssssimple SSSSS - WOW! And there is a general term for a triangle with a right angle and inlying unit circle. The value of the area AREA (using small prove) = G*1 + P*1 + R*1 = (A-1) + (B - 1) + 1 = A + B - 1. Ready! And: There is hidden something more universal... ;-) Once again - just wonderful.
Seeing the numbers all click perfectly into place over and over is giving me the warm fuzzies.
After you drew the incircle for heron's formula my mind immediately went to the Tan identity i.e,
Tan x +Tan y + Tan z= (Tan x)(Tan y)(Tan z)
if x+y+z=nπ
Using it, it was pretty straightforward to see how heron's formula worked.The visual proof was really elegant to see though.
Yes, the cotangent version of this also features at the bottom of the cut-the-knot page that I show in the video :)
10:30 i guess another thing to mention is that it doesn't matter which pair of angles, since a+b+c+d = 360 in a square, meaning a+b = 360-(c+d), and so cos((a+b)/2) = cos(180 - (c+d)/2) = cos((c+d)/2)
Mind blown. I've always liked all your videos, but for this one, with each new step, I kept exclaiming to myself: "Oh my goodness", "Oh my goodness", "Oh my goodness".
I actually cried when watching the last segment. What a beautiful experience.
Wow. You're constantly outdoing yourself with the animations. Heron himself would be mind blown!!!
Wurde mir noch gelehrt:-) Doch wie bereits erwähnt ohne skizziertem Beweis, jener folgte dann auf der Uni. Danke für jene Videos! Sie sind eine Bereicherung! Ein großes Komplement, und Danke für die Jahre.
I wish my math lectures at university were like this. The animations are insanely good.
I introduce my highschool students to this, but without the proof, as an alternative way to find the area of a triangle. This video is going to my "maths plus ultra" students, who like things like this. It's super accessible. Thank you.
maths plus ultra, I like it. Where are you from? Maths so it's probably U.K., Australia, New Zealand?
@@Mathologer UK and Germany, but teaching in Cambodia at the moment :)
In Brazil we learn about Heron's formula in high school, but it's extremely rare for it to be on college admission tests (Vestibular/ENEM). And I had never heard of Brahmagupta's formula.
The ending izz soo elegant, the music, the animation, and the proof……
1:21 Can we all just appreciate the fact that the 3-4-5 -triangle features both, a unit circle (as its incircle/”heart”), and a unit square (in its right-angled corner) 😮?
”1:21 Can we all just appreciate the fact that the 3-4-5 -triangle features both, a unit circle (as its incircle/”heart”), and a unit square (in its right-angled corner) 😮?”
Of course; the area of the unit circle: A(C) = π*1² = π, and the area of the unit square: A(S) = 1² = 1. 🙂
I live in America and learned this in my second or third year in highschool. Very cool seeing some more info about all these years later.
Learned it in USA; knew it for the test, haven't stumbled on many applications. Heron's formula is numerically unstable in floating point. I feel like there should be a straightforward algebraic proof involving a system of two quadratics, but it's more interesting to know how the ancient Greeks convinced themselves.
I wish we all could be taught math like this. I love this geometrical explanations!
The intuition this video provides is incredibly beautiful.
Thank you very much!!
The final section, the proof of bhrama gupta formula is so beautiful a soothing. Thank you.
Nice! I could have used this recently when working out hole coordinates in a metal place knowing nothing but the distances between them
For those dazzled by the geometric argument that shows PQR = P+Q+R, a quicker way is to use complex numbers:
From the observation that the angle sum of the (P,1), (Q,1) and (R,1) right triangles is 90°, we know the sum of the arguments of the three complex numbers P+i, Q+i and R+i is also 90°. The argument of their product is therefore 90°, which implies that the real part of this product is zero. Since Re[(P+i)(Q+i)(R+i)] = PQR -P -Q -R, we get PQR = P+Q+R.
The 20eth minute gave me the most satisfied feeling since a long time. ♥
If you liked that you'll probably also like this proof of Pythagoras's theorem
th-cam.com/video/r4gOlttnJ_E/w-d-xo.html
I also featured this earlier in a video on this main channel
th-cam.com/video/r4gOlttnJ_E/w-d-xo.html
@@Mathologer Thanks! It immediately reminded me of this one, of course. The one in this Heron’s formula video was just perfectly timed for some reason. And might I say, just that extra tad more magical.
@@jasperiscool Yes, an absolute beauty :)
2:32 Because phi is defined as the number that satisfies this equation: x=(1/x)+1., therefore 2*x*[(1/x)+1]=2*(x+1)=2x+2.
So this means that the left side of the equation is the same as the right side of the equation.
It took me a long time to get to what the Mathologer pulls out of his hat after the first minute (following "whoa! What just happened?") Part of the problem was I steadfastly refused to accept that the three angle bisectors of the triangle meet at the center of the inscribed circle, without proving it according to Euclid--perhaps because of it's similarity to the flawed proposition that leads to the "theorem" that all triangles are isosceles. Getting past that hurdle, though, it's pretty straightforward to derive a formula for any triangle with a unit circle inscribed in it. Each colored segment (BTW of course you wouldn't use "blue" because it's name would be the same as side B) is simply equal to the cotangent of the half angle of each vertex. For the 3-4-5 you can apply the half-angle formulas to get the cotangents in terms of the cosines (sqrt(1+cosine/1-cosine) ), which for the 3-4-5 are 4/5/ 3/5 and 0, and you miraculously get whole numbers for each side. Time to unpause the video. It follows that the sum and product of the cotangents of any 3 angles that add up to 90 degrees must be the same. I guess the only hope of proving that is expressing them in terms of imaginary exponentials, and grinding through a lot of algebra. No geometric insight in that though. For another time.
BTW while Heron's formula in terms of the sides instead of the semi-perimeters might be preferable in some ways, if you're actually calculating it you annoyingly have to form 4 sum-difference combinations of the sides, whereas in terms of the semi, you can just stick it in a memory on your pocket calculator and re-use it. Did Heron have a pocket calculator? He was an engineer, wasn't he?
In all seriousness, Heron being an engineer clears up a question I've had for years: were this Heron of Alexandria and the one credited with inventing the first steam engine--which consisted of a steam boiler with jets on the side that set it spinning-- one and the same? Apparently so. It's inefficiency, and other logistical problems problems, probably explains why the chariots of the day weren't powered by Heron engines.
I learned Heron's formula in a math formula book without knowing the name Heron or proof. Glad to see this simple proof! thank you a lot.
I love Mathologer's videos. His videos are very informative and great for anyone who has an affinity for math. That being said, I also LOVE his shirts. I like to collect witty t-shirts, and his make me jealous that I don't have them!