Ptolemy’s Theorem and the Almagest: we just found the best visual proof in 2000 years

[@Claudius_Ptolemy]

Thanks for making a video about my theorem!

[@Mathologer]

Pleased to make your acquaintance :)

[@seventhtenth]

when worlds collide you can't deny when worlds collide

[@obinator9065]

nice

[@nitayg1326]

Congrats 👏

[@johanclaes8546]

Numberphile also a Ptolemy video 4 years ago, featuring Zvezdelina Stankova. Also enthusiastic, but with very different twists.

[@benpaz9548]

A day with a new Mathologer video is a better day

[@Mathologer]

For me a day with a new Mathologer video is the day I start working on the next Mathologer video :)

[@klausolekristiansen2960]

@@Mathologer Please keep it that way.

[@jotch_7627]

im already a big fan of long-form content in general, but this channel has such a distinct and enjoyable format that it easily sits in the top of the list. instead of continually building to a large and complex problem/explanation, its like you meander through a selection of related small problems that can be appreciated on their own or as a collective. it makes them very nice to watch and rewatch

[@norskradiofabrikk]

Your comment hit the nail on the head!

[@kajdronm.8887]

What could go wrong at 28:16 ? The corner, where the three tetrahedrons meet, corresponds to a spherical triangle. The side lengths of wich are the angle wich meet there. These side length have to conform the triangle inequality, otherwise the corner can't be form. Lucky as we are, the angles are just the angles of the corner, where the lower case edges meet. So all is fine.

[@Mathologer]

Exactly, you are the first one to figure that out (and leave a comment :)

[@JCOpUntukIndonesia]

I've now realized how easy it is to miss the edge cases when using visual proof. It's a crucial lesson in mathematical proof. I'll keep in mind when I'm dealing with visual proof in the future. Thank you for the beautiful insight professor. Yet another gem you've shared with us 🎉

[@Mathologer]

Glad you picked on this aspect of the video to comment on. It's a point that's really not made very often :)

[@josephyoung6749]

Didn't learn about Ptolemy's theorem until my late 20's, well after having learned calculus and other pretty sophisticated algebraic properties up through grade school and all. Should have been the first thing I learned.

[@Mathologer]

Yes, as I said a real scandal :(

[@christymccullough7306]

Shhhhh.... they don't want us knowing these things, hahaha. Thank goodness you are here!!!! What a time to be alive!!!

[@jacemandt]

The 2nd of the basic geometry theorems (that the inscribed angle is half the measure of the subtended arc) can also be used to easily prove the 1st: two opposite angles of a cyclic quadrilateral subtend arcs that clearly add to the entire circle, and their measures would then add to half of a circle, or 180°.

[@guiorgy]

This is how I remember it

[@williamhutchins8423]

When he flipped that triangle for the difference rule 14:07 I almost jumped out of my seat, Great work as always !

[@Mathologer]

Actually that particular transition between the two formulas occured to me while making the video :)

[@Anmol_Sinha]

This was an amazing video. Ptolemy theorem is one of my fav theorems now!

[@Mathologer]

Glad this video worked so well for you. Now make sure to also tell all your friends (and enemies) about Ptolelmy :)

[@nicolaslj]

Your videos never disappoint, great work. This proof reminded me of that for the Pythagorean theorem that we take 3 copies, scale and bring together.

[@Mathologer]

Rainer told me that he was actually inspired by that proof after I showed it in one of my videos :)

[@Supremebubble]

I am Rainer, and yes the video was exactly what inspired me :D It was a literal heureka moment where something suddenly struck me and all dots connected.

[@andrewkepert923]

@@Supremebubble this connection doesn’t surprise me - it is my favourite Pythagoras theorem proof. … and nice work on nerd-sniping Burkard.

[@Mathologer]

Andrew, mathematically what keeps you awake at night these days ? :)

[@ibbiradar2419]

After preparing for competitive math Olympiads i hated geometry, this video has shown me the true light of how beautiful it is.

[@Mathologer]

Yes, I've seen a lot of Olympiad books that, in the way they treat geometric proofs, really butcher the beauty of the ideas involved. A real pity :(

[@Mathologer]

Here is a nice application of Ptolemy's theorem to a Olympiad problem kzbin.info/www/bejne/hHnNpXuFepafodU

[@ibbiradar2419]

@@Mathologer Thank you so much sir for being so considerate, unfortunately i was unable to clear the Olympiad i was preparing for, but yet thanks to your videos my love for mathematics has not died down and hopefully never will.

[@blackholesun4942]

00:00 intro. 04:30 Geometry 101 08:20 Applications (Very excited to watch the whole thing!) By the way, what are the notes being played in the piano jingle at the start? Thanks in advance 😁

[@Mathologer]

You are probably reminded of Kate Bush's Babooshka?

[@blackholesun4942]

​@@Mathologer yeah there is a similar melody in that song!

[@shmupshmuppewpew5260]

Thanks for this clear explanation. As a non-mathematician I could still follow along and appreciate the elegance of the math here. Also, your little giggle at 5:52 sent me. I love your enthusiasm!

[@Mathologer]

Great to hear!

[@mfaynberg]

Mr. Burkard, first of all thank you so much for your Mathologer channel. It is wonderful. I enjoy each and every video and keep waiting for the next one :) I would like to share one thought I've got after watching the latest episode about the Ptolemy's theorem - suspecting it is quite trivial and seems different to me only - for which I apologize in advance. Here is what I thought: the Ptolemy's imparity can be generalized a little bit. Let say, we have a quadrilateral - any in fact. Add the diagonals, i.e. get all vertices mutually connected. We will have a number of line segments (original shape's sides and diagonals). Now split those in pairs such that the segments in each pair would not have common vertices. Then apply that naming convention you've been using in your video, where one pair would have A and a segment, the next - B and b, and C and c. Then it seems that the rule Aa+Bb>=Cc will hold despite which pair is a''s, b's or c's, that is c-pair should not necessarily be of the diagonals. And it is pretty obvious based on the proof you have presented. Then, the case of equality follows with all vertices located on a circle, and c-pair being the diagonals. Thank you so much for your time - and again sorry for being not enough educated to recognize a trivial result. Am already waiting for your next video! Best regards, Mike Faynberg

[@Alexander-Bunyip]

You are very special. Thank you.

[@PC_Simo]

24:30 For a while, I struggled with figuring out, why the ”>”-sign can’t become a ”/= Cc. Very satisfying, I must say. 😌

[@PC_Simo]

*_*Specification:_* I, of course, mean: ”Aa+Bb ≥ Cc”, with: ”Aa+Bb >/= Cc”. But the symbol: ”≥” wasn’t part of my iPhone’s repertoire, at the time; before I copied it (from Wikipedia, I think); and I’m not gonna edit my original comment and lose my heart (❤). 😅

[@_majortom_]

your videos are becoming more and more magical. thank you for doing this.

[@NoLongerBreathedIn]

Note that a proof of an equality that only works in a special case that's still sufficiently generic (in other words, where there are just as many free variables in the special case as in a generic situation) can often be generalized by analytic continuation. Inequalities need more careful handling.

[@Mathologer]

That's right and to be honest I am not really worried about focussing on sufficiently general specific cases in explanations. Just thought that it would be a good idea to spell out that this is what everybody is doing almost by default :)

[@MichaelRothwell1]

I thought of analytic continuation also. This gives comfort when proving the compound angle formulae for sine and cosine, for example, given that the diagrams used in any proof only apply in very limited cases. Of course, I don't mention this to my students! On the other hand, when proving the sine and cosine rules, I do separate diagrams and proofs for the the cases of acute and obtuse angled triangles.

[@SaturnCanuck]

Thanks again. I love to watch your videos on a Sunday afternoon with my coffee. Oh and cool T shirt

[@harrybarrow6222]

When I learned about imaginary numbers and their exponentials, it became much easier to remember and derive the angle sum formulae.

[@Jack_Callcott_AU]

At 10:30 let us note a corollary: 2*q is the harmonic mean of r and s.

[@a6hiji7]

One of the best channels! Just good quality content that requires focus and thinking. 🙏

[@yinq5384]

Wonderful video as always! The 3D-to-2D continuity argument is brilliant! 10:42 Let θ=pi/7, the identity ⟺ 1/sin3θ+1/sin2θ=1/sinθ ⟺ sinθsin2θ+sinθsin3θ=sin2θsin3θ ⟺ (cosθ-cos3θ)+(cos2θ-cos4θ)=cosθ-cos5θ ⟺ cos2θ+cos5θ=cos3θ+cos4θ ⟺ 0=0. P.S: Not long after I wrote down the proof above, I saw the relation between the trig identities and Ptolemy's thm in the video. :P 13:30 We can move the top vertex to be antipodal to the bottom-left vertex, and the angle remains α+β. Thus the chord's length is sin(α+β). 19:50 We can consider the two smallest equal angles to be negative, then the red angle is still the "sum" of two angles and the argument works as before. 28:12 They fit together because the three angles at the vertex "inside" Aa, Bb and Cc are exactly the same as those at the pink vertex (from 27:52).

[@Mathologer]

Perfect :)

[@gamerpedia1535]

Always such a treat :) You're the best!

[@KettaTabuAaron-n9e]

Thanks a lot mathologer... But I'm badly missing videos on number theory and algebra... You seem to have gone full board to geometry. Still waiting for your promised video on Abel's proof of why there can't be formulae for some higher degree polynomials

[@Mathologer]

Looks like the next one will be pretty algebraic :)

[@Inspirator_AG112]

*@[**05:05**]:* If I had to guess, it involves the omni-directional symmetry of a circle.

[@misfitt2969]

Great video! Get him to 1 million already! :)

[@Mathologer]

Would be nice but things are moving a lot slower than they used to in this respect and so probably not anytime soon :)

[@gaetanomontante5161]

Oh, how I miss the days when such elegant demonstrations would come sooth my brain like butter on hot toast...

[@Mathologer]

Well, these days I am aiming to feed something like this to a brain like yours once every four or five weeks :)

[@gaetanomontante5161]

@@Mathologer ❤

[@diddykong3100]

Kudos to Rainer for a lovely proof. The opposite angles of a cyclic quadrilateral adding up to a half turn is easily obtained, even when the quadrilateral all lies on one side of the centre, by remembering to draw the diagonal connecting the two corners we're not looking at; this is a chord, that subtends at each of the corners not on it half the angle it subtends at the centre, on the relevant side; and they're on opposite sides, so those two angles at the centre add up to a whole turn, so the opposite-corner-angles add up to half that, so half a turn.

[@JaybeePenaflor]

Wow! I’m this early for a Mathologer video!

[@ozzyvocal]

Amazing theorem. It's a funny description, that Ptolemy "beats" Pythagoras... Like ellipse does to circle. The proof video at the end is amazing. Thanks for the super work. I'm wondering if there is a work for finding the sides with - smallest - integer values of the convex ones (incl. diagonals). BTW I liked also "1/s+1/r=1/q".

[@Mathologer]

In terms of smallest just consider a rectangle with sides 3 and 4 (and diagonals 5). Yes, I know what you are going to ask next :) Have a look at my notes on integer sided/diagonaled quadrilateral in the description of this video.

[@ozzyvocal]

@@Mathologer many thanks. I've read and understood it well. In the example in the description is the second diagonal also an integer (56). I took another two triplets (15-8-17 and 12-5-13)as example; the first diagonal is 221 (13x17), and the second diagonal is exact 220. Tadaaa :)

[@ozzyvocal]

@@Mathologer Mr. Polster, I made a small study. I chose 5 different type of Pythagorean triangles. These are: 1) 3-4-5 ; 2) 5-12-13 ; 3) 8-15-17 ; 4) 7-24-25 ; 5) 20-21-29 I did what you explained. There are 10 combinations (couples). I used a CAD software and the 4th diagonals are in all cases integer (for me very surprising). I thank you one more time for this very interesting information. Combinations: 1st Diagonal 2nd Diagonal C1) 1&2 65 56 C2) 2&3 221 220 C3) 1&3 85 84 C4) 1&4 125 117 C5) 2&4 325 323 C6) 3&4 425 416 C7) 1&5 145 144 C8) 2&5 377 352 C9) 3&5 493 475 C10) 4&5 725 644

[@Mathologer]

That's great ! Glad that you are having fun :)

[@batencheetos]

Great. Another thing I've never heard of!

[@miruten4628]

In higher dimensions (> 3), any quadrilateral is contained in a 3-dimensional subspace (the vertices are… co-3-spatial??), so the 3-dimensional version of the theorem applies directly.

[@Mathologer]

Yep, that's true. I think you are the first person to point this out :)

[@paulhansen5053]

Thanks much for another wonderful, excellent video. Equality and inequality are powerful subjects, and can have quantitative (applied math?) as well as qualitative (pure math?) aspects. For an arbitrary set of 4 planar vertexes, choose any subset of 3 which then specify a circle. The 4th vertex is either on the circle, if Ptolemy's formula gives zero, or non-cyclic if non-zero. My conjecture (wish I had time to work on it) is that the non-zero result gives the "degree of non-cyclicness" and maybe inside/outside measure of distance to the circle. This is related to what may be the simplest geometric equality, the standard form of the line equation. I think that early students don't get the more exciting introduction; the focus tends to be on the slope-intercept form and graphing of functions, helpful but rather boring. Strangely, they short-shrift the general form, and even get students to derive it from the slope-intercept form! But, the general form is much more useful, and is derived more simply by a cross-product type formula of the endpoint coordinates. The resulting Ax + By + C = 0 is then a formula -- if you plug in a point it gives zero if it's on the line, but if >0 then the point is on the right side of the line, on the left if

[@Anmol_Sinha]

I wasn't expecting that part on the last lol. I was so captivated with the proof that i forgot about the Easter egg lol

[@Mathologer]

I have to include more more of these Easter eggs in my videos :)

[@lazarbaruch]

Sensational presentation. Just 3 points. 1) The sine notation, which makes such neat formulas, was invented by Indian mathematicians. Prolemey's trigonometry was much more cumbersome as he used the chord instead of the sinus. I am sure you know this, but maybe it is important to emphasize this to clarify what a great mathematician Ptolemy was. 2) I would recommend also the 2 books by Aaboe that were my introduction to the subject: Aaboe - Early episodes in the history of mathematics Aaboe - Early episodes in the history of astronomy 3) What tool do you use for drawing production?

[@Mathologer]

I mentioned that it's really a table of chords at the beginning :) I was tossing up whether or not to go into more details in this respect but decided against it. Definitely all very interesting but when it comes to crafting a story that flows nicely it's important to pick and choose wisely what to explain in detail, what to only hint at and what to not mention at all.

[@johnchessant3012]

16:42 so satisfying! I had to explain to my roommate why I audibly gasped :D

[@AM-ur7uu]

Thanks!

[@morkdel4084]

This is beautiful, thank you. I remember learning this rules and asking my teacher about where they come from / proofs. And she refused to tell me. Probably she did not know. :(

[@ВикторПоплевко-е2т]

19:48 using the inscribed angle theorem the sum of the opposite angles is equal to a half of the length of the arcs, but in out case the length of the arcs is equal to 360 degrees so the sum of the opposite angles is half of that which is 180 degrees

[@Mathologer]

Yep, actually asking viewers for this proof would have been a nice little challenge, too.

[@ВикторПоплевко-е2т]

@@Mathologer I explained this that way cause I might have got this proof from the teacher

[@Mathologer]

Good to know that there are still teachers who know some geometry :)

[@JimmyMatis-h9y]

trig & geometry were my favorite in school. ty for telling us about Ptolemy. 💜

[@jedglickstein]

This was characteristically great, as usual, but if I may be so bold, please at some point bring back the number theory! While the visual geometric proofs are marvelous, I never quite “get” them the way I do with some of your other topics (nothing new, it’s been that way for me since grade school). I’ll admit it’s not like I have any leverage here. Whatever you put out I’ll keep watching anyway!

[@Mathologer]

I've got something very algebraic lined up for the next video :)

[@williamrhopkins]

Very nice. I like the argument pushing the 2d vertex slightly into 3d and then considering how it moves back. I think Archimedes would like that.

[@faiza7740]

Best mathematics channel so far I came across

[@Mathologer]

Glad you think so :)

[@mathmeetsmachines]

I agree.

[@guiorgy]

The sum of the angles of opposite cyclic quadrilaterals is 180°, because the angle of a cyclic quadrilateral is half of the angle between the two radiuses joined to the same points as the quadrilateral, and the sum of the angles of the radiuses of the opposite quadrilaterals are obviously 360°, since the radiuses divide the circle in two parts, and 360° / 2 = 180°

[@Mathologer]

You should probably read over this again and fix a few things you clearly did not want to say :)

[@guiorgy]

@@Mathologer English is not my native tongue, and haven't touched math since school. Cut me some slack 😅

[@Mathologer]

Sure no problem, just thought that you typed this in a hurry and did not read over it again before clicking the submit button. E.g. opposite cyclic quadrilaterals ? That's definitely not what you meant, right? :)

[@kevinportillo9882]

Awesome! I wish these dropped more often

[@Mathologer]

Me too :)

[@hugh081]

Beautiful proof. Well done to Rainer!

[@OlivierGeorg]

French swiss guy here, we had classes of algebra _and_ geometry at school, and had to do a lot of demonstrations, many of which involved the angle at the center, Thales,... I loved that stuff, and it really framed my mind. I enjoyed that video the same as I did then. Nowadays my children do no more see any proof, not even Pythagore... it's a pity! Btw I still have the book, Géométrie plane, by Delessert

[@Mathologer]

Same here me/you & your schools/schools in Australia :(

[@MichaelRothwell1]

Thanks for this great video. I have taught Ptolemy’s Theorem but was unaware of his inequality - nice! Also (to my astonishment) I was introduced to kinds of quadrilateral that I didn't even know existed, particularly the truly 3D kind. Nice work! A couple of notes: I wonder how many generalisations of Pythagoras' theorem there are? For now, I can see two: the cosine rule and Ptolemy’s Theorem. I was slightly disappointed you didn't use the word "continuity" in your explanations of how 2D quadrilateral cases can be thought of as the limit of 3D quadrilateral cases.

[@Mathologer]

Did you already watch the two videos dedicated to Pythagoras theorem? kzbin.info/www/bejne/pl6ThIKNl9-Ij6s kzbin.info/www/bejne/j2bamop5h550rsU Also have you heard of de Gua's theorem? That's a really nice generalisation into higher dimensions.

[@MichaelRothwell1]

@@Mathologer I hadn't heard of de Gua's theorem (en.m.wikipedia.org/wiki/De_Gua%27s_theorem ) - a nice generalisation indeed. I'll check out your two videos on Pythagoras' theorem. Thanks!

[@mathmeetsmachines]

Absolutely stunning! It is a big joy for me to see such a simple proof. Going to 3D is perhaps not too unnatural if you know Desargue. But still ...

[@Mathologer]

Yes, not the first time that taking a 3d scenario into 3d makes things a lot easier. Still definitely a very nice result :)

[@Supremebubble]

7:17 There is another visual way of seeing this. Call the two yellow points below the midpoint A and B, the yellow point at the top P and the midpoint M. Extend PM beyond M. Draw in the parallel lines to PA and PB through the midpoint M. Now the angle AMB is split into 4 equal parts. Through the well know angle theorems for parallel lines and the fact that PM, AM and BM are all the same length, you can quickly see how the angles APM and BPM fit in there twice each.

[@Grateful92]

I never get what you deliver but considering the positive comments, your videos must be very interesting

[@Mathologer]

Keep watching :) Well, since you keep watching the videos what is it that makes you do so ? :)

[@marjon888]

Aussie is a smarter place for having the Mathologer as a resident

[@Supremebubble]

I love the symmetry in your augmented proof :D

[@Mathologer]

Yes, that symmetry is really very very nice :)

[@illogicmath]

Just one word: ASTONISHING

[@jakobr_]

28:19 It is possible that when you construct things in geometry according to general instructions, in some specific cases you might end up being instructed to make something impossible. Where the pieces don’t fit together. This isn’t an issue here though since these tetrahedra are convex and have triangular faces (which are also necessarily convex) so all the necessary edges are always available and you aren’t required to overlap any of the volumes.

[@Mathologer]

Not quite. Potentially even though the tetrahedra are convex and have triangular faces, they may not fit together, just like three segments may not form a triangle. Not a problem here because the three triangular angles that are fitted together here are the same as the ones meeting at the distinguished vertex of the original tetrahedron :)

[@jyotsanabenpanchal7271]

You're Awesome 😎😎

[@Mathologer]

Good to know :)

[@ayushrudra8600]

I think you can extend the proof for the opposite angle sum by using directed angles

[@mienzillaz]

Transition music bit too loud 😅 otherwise excellent as always. No voiceover this time, but there was a cut at some point, so what was left out?😊

[@Mathologer]

You mean in the animated part at the end? There was just a mistake that I needed to fix :)

[@mienzillaz]

​@@Mathologerno, not that one, but now i see this one too. I was still pondering about the rotations, that's why I missed it. The cut I had on mind happens shortly after 22:01.

[@vivekdabholkar5965]

Absolutely Brilliant. Thanks for wising all of us up!

[@stephengraves9370]

Loved the video! I noticed that the last step of the 2d animation appears to draw the edges of a cube in the final step, and a bipyramid in the concave case. Is this pure coincidence, or is there more 3d shenanigans?

[@Mathologer]

A coincidence. What's there 3d wise is really captured in the animation at the end of the video :)

[@saxbend]

I'm wondering if the cosine rule generalisation of Pythagoras can be derived from this too. Or rather the further I get into this video the more confident I am that it can be. Will have a go next time I have a pencil and paper handy.

[@Graham_Rule]

14:03 Wow. At school I remember that I had to learn a proof of this for my exams. I've completely forgotten it over the last 50 years (and have mislaid my school notebook) but this proof is so much clearer.

[@Mathologer]

Yes, that's a really nice way of deriving the sum and difference formulas. I first came across this proof many many years ago and never had to look up a proof again ever since :)

[@jakobr_]

19:58 The same proof still works if you allow for “negative” angles on the one isosceles triangle that overlaps with the others. There’s a sense in which that triangle’s orientation is flipped compared to the others.

[@Mathologer]

Absolutely right but requires a bit of extra thought :)

[@jakobr_]

@@MathologerI like it because the numbers involved continue smoothly from the familiar “all positive” case when you continuously change the quadrilateral.

[@anaslakchouch202]

4:06 Mathologer: In the next two sections I will be lying a tiny little bit here and there try to spot the lies before I come clean Me the whole two sections: this wise man speaks absolute facts!

[@OjasSinghYadav]

Make a video on circle inversion

[@TheMichaelmorad]

8:05 YES! 14:07 YES WE DO!

[@Mathologer]

Glad you think so :)

[@AbhaySingh-sf9op]

thank you Sir!🙏

[@mitjamastnak9206]

Wow - amazing proof. Fortunately I did learn the Ptolomey's Theorem when I was in elementary school. Unfortunately this implies that I was in elementary school a long long time ago.

[@Mathologer]

:)

[@AntonioLasoGonzalez]

I have never seen a geometric proof like this! Very original! If any of you wants a more motivated proof of ptolemy's inequality, use inversion around any vertex of the cuadrilateral.

[@Mathologer]

That's a nice one too. Never seen a 3d version though :)

[@mananself]

I'd never seen the visual proof from this video. It's great! I also noticed that when you pronounce Ptolemy, "p" isn't completely silent as in normal English pronunciation, but a "p" without release. I wonder if this way of pronouncing Ptolemy is from German.

[@Mathologer]

There are actually some scaling based proofs but nothing is as slick as this one (I've included some links in the description of this vide) and I really like the nice 3d extension of this proof which weights the a, b and c the same. In terms of my pronunciation of Ptolemy, who knows. As you say it's got its roots somewhere in having been exposed in a major way to both German and English :)

[@talastra]

Absolutely adorable and charming video about Ptolemy's theorem by Zvezdelina Stankova on Numberphile. I especially watched this video because that one taught me about Ptolemy's theorem. So the teaching is happening :)

[@caspermadlener4191]

I actually calculated the (signed) area of the triangle with side lengths aA, bB, and cC, which I call the Ptolemian, in an attempt to prove the inscribed square theorem and because I thought it would be an efficient way to check if four points lie on a common circles. It turns out to be a constant time the determinant of the 4×4 matrix, with row entries of 1, x, y and x²+y², for every point (x,y) of the original quadrilateral. This is the algebraic way of saying that the points satisfy a common equation of the form a+bx+cy+d(x²+y²), which is obvious in hindsight. It also satifies some interesting identities using area, and given a quadrilateral ABCD, the Ptolemian of this quadrilateral is equal to a constant times the area of ABC times the power of D to ⊙ABC.

[@Mathologer]

I never really followed up on this but what you describe there looks like something that others have used to generalise Ptolemy's theorem to higher-dimensions.

[@Nat-pk3gc]

For the proof of sine addition, substraction, etc. I get why that specific quadrilateral would have 2 right angles given that its two triangles are inscribed in a semicircumference each, but why does it prove it for all cyclic quadrilaterals, since you could have a cyclic quadrilateral that does not have right angles? Thanks for the great video!

[@stevewithaq]

25:00 Okay, but for this "pseudo-cyclic" quadrilateral, if you draw the diagonals Cc, you've just completed a proper cyclic quadrilateral with sides BCbc and diagonals Aa. So for this quadrilateral, Bb + Cc = Aa, right? This should extend to any quadrilateral which is inscribed in a circle (provided you properly label the sides as you noted.) So a more general rule should be to draw all four sides and two diagonals. Then the sums of the products of the lengths of the two opposing pairs of non-intersecting segments should equal the product of the two intersecting segments.

[@syjwg]

21:48 Going from 2D to 3D in this case was quite easy to understand. Are there some similar theories about four (really five) sided pyramid shapes?

[@Mathologer]

Yes, there are Ptolemy like theorems about higher-dimensional simplices :)

[@rtravkin]

A tiny bit less elementary, but to me much more conceptually transparent, proof can be obtained by applying inversion centered at one of the vertices, and expressing distances between the images of three other vertices as A/bc, B/ac, C/ab (for the unit radius of inversion). (Works in any dimension.)

[@Mathologer]

I like elementary, especially for these videos. The more elementary the better :) But, yes, inversion also works nicely and a video dedicated to inversion powered niceness is also on my to-do list :)

[@caspermadlener4191]

11:13 Ptolemy's theorem is actually completely useless for math competitions. In my three years of IMO training, I have never used Ptolemy's theorem. In order to use Ptolemy's theorem, you have to know at least four different lengths in order to use it, and that is just never the case. Most Olympiad geometry problems don't even mention lengths, let alone four.

[@Mathologer]

When I did maths competitions way back in the days Ptolemy was a must-know :)

[@Mathologer]

Here is a nice application of Ptolemy's theorem to an International Olympiad problem kzbin.info/www/bejne/hHnNpXuFepafodU

[@graf_paper]

@3blue1brown is a patreon sponsor of Mathologer!!!

[@Stef-zp5sj]

Thank you very much for this very interesting video and your friendly reminder that quadrilaterals are not only convex and only 2d! This was very helpful!👍

[@Mathologer]

And, looking at generic 3d quadrangles gets rid of all the many cases noise and shows what is really going on here :)

[@YaNykyta]

Where is that beautiful video about game NIM in the French movie? You deleted it?

[@gabest4]

3:15 Small mistake, sin(2°) is in the blue rectangle from the book.

[@Mathologer]

Actually, no. it's the chord for 2 degrees which corresponds to the sin of 1 degree. The numbers in the list are sexagecimal numbers, the diameter of the circle Ptolemy is working with is 120 units and then there is also some halving of angles and doubling of sines at work to translate Ptolemy's chords into into our sines. In the end it all boils down to this: The numbers in the list are 2, 5 and 40 and from this you calculate the value of the sine of 1 degree to be approximately: (2 + 5/60 + 40/3600)/120. Details can be found here demonstrations.wolfram.com/PtolemysTableOfChords/

[@EmilianoGirina]

Is the degenerated case comparable to a quadrilater in a ellipse? Is Ptolemy's theorem valid also for elliptic quadrilateral?

[@Mathologer]

Sorry, but no :)

[@EmilianoGirina]

@@Mathologer Thanks for answering. 🙂

[@michaelhartl]

Perhaps I missed something, but I don’t see why the marked angles at 17:05 are equal. Any help?

[@gordonn4915]

In the 2D proof the end diagram appears as a 3D square. This means there is a projection angle, which is possibly the amount needed to “fix” the inequality.

[@JurijFajnberg]

Vielen Dank, auch an Reiner!

[@pankajraghavofficial]

hello Sir, i watched your video on the topic of Ramanujam ( 8 years old)...i have seen few more videos on same topic few people strongly arguing that how can we substract infinity from infinity. i have a question sir, as people are saying how can we substract ifinity from infinty but 1+2+3+4......INF and 1-2-3-4-5-6......INF in both scenario the value of infinty must be different. so why we can not substract... i mean to say that the value of infinity must differ in different equations...or if its not true ,we should consider the value of infinity is constant. if not constant than it can be substract ,divide or anything you want. i know, i am very late on the video...but if you see this kindly revert. thank you

[@rossholst5315]

What happens if the lines connecting the points are not flat? Does the inequality hold when quadrilaterals are drawn on more exotic surfaces?

[@rossholst5315]

It would also seem the proof requires that the distance of A to B is equal to the distance of B to A?

[@Mathologer]

Yes, this has also been considered. Have a look at this en.wikipedia.org/wiki/Ptolemy%27s_inequality#In_general_metric_spaces

[@hawlitakerful]

@20:58 ... and sorry if i am wrong my understanding of mathematics is rather poor... But isn't a "cross themselves" tje same as a regular one with extra steps? It might not be "convex but i can make it so ... just exchange a for c both upper and lowercase .... now i have a convex again ... did i miss something critical?

[@mrshodz]

Can you please do a video of how Madhava derived the series for trig functions 🙏🙏🙏

[@Mathologer]

Sort of on the to do list but not sure when I'll get around to covering this :)

[@mrshodz]

@@Mathologer So cool. THANKS!!!

[@VideoFusco]

Is there any book that reports the mathematical part of the Almagest in modern language?

[@Mathologer]

This is a very good English translation classicalliberalarts.com/resources/PTOLEMY_ALMAGEST_ENGLISH.pdf

[@Inspirator_AG112]

*@[**06:44**]:* This will also involve the omni-directional symmetry of a circle, won't it... (;

[@Mathologer]

Happy for people to discuss anything reasonable in these comment sections. Having said that maybe consider giving a bit more context to improve your chances of people actually engaging with you :)

[@TheMichaelmorad]

19:49 if your proof uses directional angles, then the proof is not a lie!

[@Mathologer]

Yes. These things usually automatically work out like this when you use directional angles. And to witness this working out in action, e.g., in a geogebra animation is always a bit magical :)

[@ingobojak5666]

The "crossed over" cases from about 23:00 onwards arise only if one names diagonals and sides arbitrarily and thus abandons their original meaning. Take one of the side pairs - let's say b and B - of a convex quadrilateral and call them now c and C, and rename the original diagonal accordingly - thus here b an B. Then you get your "crossed case". The caveat at the end about cyclic quadrilaterals is an artifact of this arbitrariness. Take the proper equality for the cyclic case aA+bB=cC. Now just switch the names of b and B with c and C. The old equality now reads aA+cC=bB (and holds true, we just changed the names). Now write the inequality for the crossed case as aA+bB>=cC (new names), then we can substitute into this the old (renamed) equality as cC=bB-aA and get aA+bB>=bB-aA or aA>=-aA, and with a and A being positive we get 1>=-1. Equality is obviously not possible, but the greater case is true. Therefore, in the crossed case aA+bB>cC. Thus, if one calls one of the side pairs of a cyclic quadrilateral "diagonals", and those diagonals "sides", creating a crossed case out of a convex one, then one gets the result at about 24:39. There may be reasons for doing this, but here it just confuses in my opinion...

[@FF-ms6wq]

Very nice proof, of course, and great video as always. Though I think some crucial (yet “trivial”) observations are missing. Most notably, in my view it should be explicitly mentioned that each of those “scaled quadrilaterals” have the same respective interior angles. This is the essential point used in the argument at around 17:00 to conclude that the angles are indeed the same. One shall say: “Yes, but this is trivial!” And I shall respond: “Yes, of course it is. Yet, when broken down *all* math is nothing but a sequence of trivial observations.” Saying simply “Scaling the quadrilaterals by different scale factors obviously does not change the interior angles” or something of the likes would have sufficed. Similarly at 17:24 I think the triangle inequality needs mentioning. I get it, it is a visual proof, but even then. “The shortest path between two points is a straight line [in Euclidean geometry]” would add to some people’s understanding, I feel. All around awesome video as always. Keep up the good work!

[@Mathologer]

I would not object to anybody emphasising those points you are making in a similar video :) But of course when it comes to “when broken down all math is nothing but a sequence of trivial observations.” it's always super important to get the balance just right in terms of what to say and not to say, the exact choice of words, etc. to make things work for my intended audience. I actually agonise over this to no end. Should I mention the triangle inequality or not? Final decision: No. Should I point out that exchanging a and A does not change the result of any of the calculations? Final decision: No. Etc. To be honest the scaling preserves angles bit did not even make the list of things to agonise over. Pretty sure nobody will stumble over that one. The shortest path between two points bit did make the list of things but I also decided against spelling this out. I think my little roof gesture when I say Aa+Bb and line gesture when I say CC is really all that's needed here. Actually, in retrospect there is one superfluous/distracting thing I say in this video that I regret saying (around the 15:21 mark)

[@PC_Simo]

I learned of Ptolemy’s Theorem, from a Numberphile-video 🙂.

[@6ygfddgghhbvdx]

3:17 Did the book use modern decimal system for table or is that later interpolation in 19th century.

[@6ygfddgghhbvdx]

Ptolemy did not use the decimal number system as we know it today. The sine and cosine tables found in **Ptolemy's Almagest** were constructed using the **sexagesimal (base-60) system**, which was inherited from the ancient Babylonians. This system was used for astronomical calculations and was common in Hellenistic mathematics at the time.

[@yyeeeyyyey8802]

It seems we can "stretch" any of the 3d quadrilaterals and the non convex ones into a convex quadrilateral by just rotating two sides around one of the diagonals. When this is done the only side that changes size is the other diagonal, and it seems to always be "streched" (i.e. yielding a c' for the new quadrilateral that is greater than c). By having a proof for the convex case one could then argue that Aa+Bb>=Cc' (for the new streched quatrilateral) and Cc'>Cc as c'>c

[@Mathologer]

Yep, that's how it is usually proved in textbooks :)

[@Noconstitutionfordemocrats1]

Maybe 2000 years from now, we solve quantum gravity.

[@Mathologer]

Or, I could do a video about it next month :)

[@kajdronm.8887]

In 2000 years quantum theory is old stuff and science historians, will wonder, how people could do physics like that.