How to lie using visual proofs

[@QDWhite]

My Calc II prof would always do this. He'd start going through a proof like it was critical to the course and we'd all be furiously copying it off the board. Then he'd get to the 1=2 conclusion at the end and laugh at how none of us saw it coming. Then we'd spend time trying to see where the error was (usually a very well hidden "divide by 0" mistake) I'll always remember him. He made calculus actually fun and interesting. The world needs more educators like him

[@pik33100]

Let a+b=c Then 4a+b=3a+c (added 3a) 4a+4b=3a+3b+c (added 3b) 4a+4b+3c=3a+3b+4c (added 3c) 4a+4b-4c=3a+3b-3c (now substracted 7c from both sides) 4(a+b-c)=3(a+b-c) 4=3

[@user-tf2oh3cu7l]

@@pik33100 I understand now! Because a+b=c, a+b-c is actually 0, and because every number times 0 equals 0, so 4(a+b-c)=3(a+b-c) is valid. So it's 4.0 = 3.0, which is true. How is it 198?!?!?! Idk but this is a lot for me. Thanks and also, ⛽.

[@jonathanhagerty9807]

@@user-tf2oh3cu7l exactly right, so you could divide a+b-c from each other because 0/0 is undefined. As a result, the proof given is correct until you try to say 4=3, which makes sense

[@user-tf2oh3cu7l]

Tanks (i mean literal tanks and thanks)

[@Noelciaaa]

Damn, I wish I had one like this. Would've been super engaging!

[@Mutual_Information]

You know you’re a good math communicator when you can lie with proofs.

[@MouseGoat]

yeah, guy could easily have convinced me with these false proofs. but knowing why they wrong gives a great insight into weird aspects of our world and how things can be folded up into small spaces

[@Mutual_Information]

@@MouseGoat 100%, pointing out how proofs can be false highlights how precise these tools really are.

[@bohanxu6125]

your statement lacks rigor. I accidentally lied with proofs before, yet I'm not a good math communicator :] or... your statement is true, I just don't know I'm a good math communicator yet~

[@Mutual_Information]

@@bohanxu6125 yea it’s not intended to be precise. Maybe I should have said can “lie convincingly”.. but even with that, it’s not serious. He’s not actually lying.

[@l.3ok]

Yeah Grant Sanderson is indeed an exceptional math communicator; the thing here is that all these false proofs were not invented by him, they've been very well known for a long while.

[@mCoding]

This is the video mathematicians want to make after being told convergence proofs are "unnecessary details, it's obvious" for the 1000th time. Great work!

[@artey6671]

Who are the people that tell mathematicians things like that?

[@aguyontheinternet8436]

@@artey6671 engineers?

[@Etern1tyOne]

@@artey6671 I will admit: some physicists 🙈 I try not to, but... sometimes it's hard to resist. Not with convergence, but with some other "formal" proofs. 🙈

[@artey6671]

@@Etern1tyOne Well, some mathematicians try to avoid conflict. If they are ok with your proof, they might secretly be like "well, he tried".

[@TomJones-tx7pb]

yup I used to hate physicists who did this. Then I got really good at functional analysis and realized that if you sat inside a good function space you can do this at will. I realized that the physicists were doing correct math, but they were probably oblivious as to why.

[@HunterJE]

The pi=4 "method" can also be used to "prove " √ 2 = 2 by approximating a diagonal with finer and finer scale taxicab paths

[@Loony-cz3xb]

hello. I agree but what do you think about my opinion: in pure mathematics this kind of proove is obviously wrong, because when you shrink and double the starecases to infinity you get 3,1415... (or √ 2 in the other example) BUT in our reality, in OUR universe with a minimum physical length (planck) you can't go to infinity, you cannot iterate to infinity; -> so the starecases have the planck length in horizontally and vertically direction; -> so the aproximation stops at this point -> and this means pi is still 4 (√ 2 = 2) in this universe...

[@ThomasFackrell]

@@Loony-cz3xbthis is fascinating. However a problem is that for the 45-45-90 triangle, under this Planck taxicab metric the hypotenuse length would be 2, not ~root 2. Any practical doodling with a ruler shows that the Euclidean length cannot be 2 though. So how do we define the length of a curve or line that requires 2+ dimensions to be constructed? Root 2 is incommensurable (not rational) which is another way of saying it cannot be measured in one dimension. So does it exist in one dimension? Is root 2 actually on the number line? No one has found it, so I say no.

[@Jayantea]

Math people are dumb and smart at the same time

[@zern7617]

@@Loony-cz3xb square root of 2 cannot be 2. 2 * 2 = 4, simple maths can completely dislodge the entire point of pi = 4. lmfao

[@Loony-cz3xb]

@Jaaaco i agree, there is no proove.. But...still you can''t build a perfect circle with inifinitive small triangles; in my opinion, when you reach with the short side of the triangel (the arc length) the planck length, nobody could say in which way the line goes between these two edges. You cannot double again the triangles to iterate the formula of PI. So the real path of the arc lengh is unnown. Of course this is only my opinion. :-) Have a nice day.

[@Tutorial7a]

As a non-math person who does a lot of CGI work, I’m rather proud of myself for getting the first one right. That sort of warping is something we have to watch out for all the time in UV Unwrapping. Yay!

[@StormBurnX]

As a 3D printing need that deals with a lot of low-poly work I knew from the thumbnail exactly what the problem was! Yay!

[@johnson42069]

yeah the moment he unwrapped it i thought "wait those are triangles when they're supposed to be curved on the sides"

[@kyrawr83]

I'm studying geography and I caught that too, unsurprisingly.

[@tilenkos2065]

@@kyrawr83 CG artists 3D printers 🫱🏻‍🫲🏼 Geographers

[@ajaiyp8679]

its strange how math connects all the professions and fields together.

[@HelixSnake]

I got the third one. I'm usually not smart enough to figure out anything that this channel says is "tricky" by myself so I was really satisfied with that

[@MasterHigure]

You should be. That is indeed a known and famously tricky false proof. P is actually outside the triangle (as long as the triangle isn't isosceles), as is exactly ONE of E and F. (Oh, right, I see now that the second half of the video is going through the errors.)

[@bloom945]

Same!

[@ekisacik]

I figured it out by going, "Okay, what does that point P look like in a real isoceles triangle?", before realizing that the perpendicular bisector of the bottom line and the angle bisector of the top angle would be *the same line*. Then, I knew something had to be up with where point P actually lied... and once I understood that it was always outside the triangle, then I finally understood where the proof was wrong.

[@petrsmital7340]

I'm proud of you, bro

[@squeakybunny2776]

Me too. Point p was the one thing that "came out of nowhere" and I didn't like how it was assumed to be inside the triangle. Especially because, is the triangle actually would be equilateral there ouldnt exist a point p because the perpendivual bisector and angle bisector lines would overlap. Really happy with myself😊 only thing is I didn't do the math to then conclude it was the last step where it all went down... Missed oppertunity for full points.. I'll take 2 of those gold stars then.. Still happy

[@williamtomlinson85]

One of my favorite ways to determine if a statement is plausible is to draw an extreme example. For the triangle proof, I drew a scalene triangle with one very short side. As soon as I tried to connect P, it was obvious it wasn't going to work because the intersection points existed outside of the original triangle. TL;DR I got three gold stars from Grant and I'm very proud of myself.

[@bangscutter]

That is actually a good approach to test by setting extreme limits. Like the equations for special relativity in physics, should reduce to classical newtonian mechanics in the limit of speeds much smaller than c.

[@Metal_Master_YT]

I did the _exact same thing_ wow, and I came to the same conclusion, its a good strat.

[@XxZeldaxXXxLinkxX]

@@bangscutter Good ol' binomial approximation

[@stevencraeynest7729]

Exactly! I did the same and it was obvious. Must admit, if I hadn't done that the fake proof does sound convincing

[@waroftheworlds2008]

Loads of theories fail at 0, 1, or approaching infinity or approaching 0.

[@gabiliorcoolkid2651]

I am a seamstress, an knew from my experience taking 2D shapes to make 3D objects that the "triangles" of the sphere should have a curve. It was very exciting learning the maths behind my empirical knowledge, thank you so much!

[@haveatyou1]

Are you the one that ghosted me on Tinder?

[@pommeverte531]

​@@haveatyou1I don't use tinder, so no 😅

[@haveatyou1]

@@pommeverte531 Why flirt then ghost me though 😒?

[@pommeverte531]

​@@haveatyou1why write when you can't read tho?

[@haveatyou1]

@@pommeverte531 I thought we got on pretty well.

[@usualunusualkid7149]

A note on the sphere proof explanation: We can also prove that those "triangles" don't have flat sides by looking at their angles. The dividing lines on the sphere are all perpendicular to the "equator" of the sphere, so each of those triangles would have two right angles.

[@chadchampion7985]

what

[@4...3]

@@chadchampion7985 all straight lines from the equator to the poles are perpendicular to the equator

[@anonymous_4276]

Nah that argument would apply to the 2D case of a circle too when the normal is perpendicular to the tangents. In the 2D case you get a sector which becomes closer and closer to a triangle with even the summation of error of the areas of all such sectors (difference between the areas of the thin sectors and the corresponding triangles) adding up to 0. This doesn't happen in the 3D case for a sphere because the shape we get is not even a sector and the error is relatively greater in the 3D case such that the summation of errors (difference between the areas of those weird thin slices and the corresponding triangles) doesn't converge to 0. Basically I'm saying that even sectors satisfy the property (sectors got flat sides) you're talking about but the argument still works in the 2D case where we get sectors as the thin slices of the circle.

[@4...3]

@@anonymous_4276 I'm not quite sure what you're saying but the same logic doesn't apply in the 2D case because the base of each "triangle" is an arc rather than a straight line.

[@martinepstein9826]

@@anonymous_4276 Interesting point. So in either case you need to shave off some piece(s) of a sector to get an approximating triangle, but in the circle case you shave off the short side of the sector while in the sphere case you shave off the long sides of the sector. As you say, when the part you're shaving off is comparable in size to the sector itself then the error won't go to 0 in the limit.

[@benjaminsmith3151]

I had a SAT prep book way back in the 80s that said whenever the test says "Not drawn to scale", you should immediately redraw it to scale. That rule of thumb has served me well my entire life. Unfortunately, most people will throw out everything they know in the face of a confident deception.

[@andrewharrison8436]

Yes, nice

[@altrocks]

Like the infamous telephone wire question. A quick sketch shows the answer while the math just frustrates.

[@coloradowestaerialarts1316]

I agree. Very good video.

[@a_commenter]

@@altrocks What question is that?

[@reidflemingworldstoughestm1394]

@@altrocks Yes and no. It depends on what answer you are asked to provide. If the problem states the wire hangs in a parabolic arc and asks for values of that arc, then the correct answer is independent of the fact that real wires don't hang in parabolic arcs. If on the other hand the solution asked for is to use the real world model despite the parabolic example given, then it's appropriated to go all pedantic on they ass.

[@tylerbeaumont]

The fact that any circle drawn on a TV will have the same edge length as a square or rectangle which perfectly houses that pixel-based circle is honestly mind boggling. It makes perfect mathematical sense, but at the same time I just can’t wrap my head around it! Thank you for this enlightening piece of information - I’ll be sure to spread it to random people next time I’m drunk!

[@ZachAttack6089]

Note that this only works if the circle is drawn with crisp, pixelated borders. Most of the time, anti-aliasing will be used to "smudge" the border and make it look more round. If you look at any picture of a circle on a screen and zoom in a lot, you can see the effect. So unfortunately the edge length is inexact in most cases. :(

[@ffs4141]

In 5:57, where 3B1B is drawing, the text in the book says next: "The point of rigour is not to destroy all intuition; instead it should be used to destroy bad intuition while clarifying and elevating good intuition. It is only with a combination of both rigorous formalism and good intuition that one can tackle complex mathematical problems" -Terence Tao "When you yourself are responsible for some new application in mathematics in your chosen field, then your reputation, possibly millions of dollars and long delays in the work, and possibly even human lives, may depend on the results you predict. It is then the need for mathematical rigor will become painfully obvious to you." -Richard Hamming

[@kaszagryczana2260]

You can read

[@realperson9951]

@@kaszagryczana2260 i cant

[@PacoCotero1221]

@@kaszagryczana2260 no joke, this may be helpful to blind people

[@PacoCotero1221]

@ThatOneRookie 🚫🧢

[@aureliontroll2341]

@@PacoCotero1221no joke blind people probably will never like that video at all because is a video with VISUALS proofs in KZbin .

[@Stop.Arguing]

as a former geometry teacher, i was very excited when I saw "SSA" used as a reason. if i ever teach again, I will most certainly used this video to engage my students!

[@hopechr]

the ASS congruence.

[@anandixitin]

I think that can be said to be RHS congruence.

[@doggo6517]

I was a sailor on the HMS Congruence

[@mnek742]

Actually the SSA "congruence" is a false congruence unless a right angle is involved

[@balrighty3523]

@Murtaza Nek Yep, I was thinking the same thing. The second S in the ASS "congruence" can potentially reflect and create two valid triangles that aren't the same. The only times this is guaranteed to not be the case is when the A in the ASS is obtuse (because the reflection ends up being outside the triangle entirely) or if, as you say, there's a right angle involved (meaning the second S can still reflect, it just doesn't become anything new).

[@mute1085]

The fun thing about the second "proof", you don't even need to do it with curves. Apply it to a diagonal of a square, and you can "prove" that the square root of 2 equals 2.

[@karunk7050]

Could you please elaborate?

[@FaranAiki]

@@karunk7050 Just watch this video: LWPOlZBXtD8. (put it on the KZbin link)

[@buttonasas]

@@karunk7050 Imagine a diamond instead of a circle. And you get perimeter 4×2=4×sqrt(2), which is false.

[@slightlybluish3587]

@@karunk7050 Take a unit square, ABCD, and draw the diagonal AC. The perimeter is 4. Now, fold the two edges of the square B and D inwards toward the center of the diagonal. The perimeter is still 4. You have now created 4 new "corners." Fold those in as well. The perimeter is still 4. Keep on folding the corners in, infinitely. The perimeter approaches two times the length of the diagonal (because the square is "getting closer" on both sides). The perimeter is still 4, so therefore the diagonal of the circle is 4/2=2. idk why it's wrong

[@stephaneduhamel7706]

@@karunk7050 You first draw an isosceles right triangle of sidelength 1, by definition its hypothenuse is of length sqrt(2). Then you can do the same manipulation as with the circle, by floding in the corner with the right angle to make it go closer to the hypothenuse. This way, you can construct a sequences of jagged lines of length 2, and the limit of this sequence is the hypothenuse, which would make you think sqrt(2)=2.

[@evilgirlypop]

This was the last video I watched before leaving my aunt's house. This was around 3 months ago, the last conversation I had with her was about my college classes and this math video I watched on KZbin (being you). She passed away roughly a month later and every time I watch this video I get reminded of her. I know you'll never read this comment, but your channel does mean a lot to me and you've helped me learn and nurture my curiosity about mathematics. I don't have any sort of deep statement, I just love these videos and they remind me a lot about days gone by. Thanks for making these, keep up the good work.

[@Phoenix-nh9kt]

Sorry for your loss man.

[@JR-White]

Wishing you well. My condolences.

[@evilgirlypop]

Thank you both.

[@somepunkinthecomments471]

Sorry to hear about your loss. I hope you are doing well.

[@3blue1brown]

I'm very sorry for your loss, and thank you for such kind words.

[@moocowpong1]

A good way to catch yourself on the triangle proof at the end: you’re arguing that a triangle is isosceles, and the argument constructs a perpendicular bisector and an angle bisector and makes use of their intersection. But if you do that argument with an isosceles triangle, those two lines are the same and they don’t have just one intersection. The argument makes itself incoherent. That doesn’t fully explain how the argument is wrong, but it should be a red flag, and that kind of self-consistency check will often help to flag this kind of faulty assumption in practice.

[@gregoryfenn1462]

Good point! A retrospective sanity check on something you think you've proven goes as long way :D

[@fares8005]

Good point, but it's also worth noting that many euclidean geometry theorems have a "generic proof" that only works for non-degenerate cases and those special cases (e.g isoceles triangle) are verified separately.

[@thomasjunker5415]

This was the same conclusion I arrived at as well. As P isn’t well defined if the triangle is in fact isosceles, the entire proof goes out the window by contradiction

[@mishaerementchouk]

The proof starts to look suspicious from the very beginning, when it suggests that P is inside the triangle. It could be inside or outside. The suggestion is turned into an assumption at the end, when it is concluded that from AF + FB = AE + EC follows AB = AC. This is the only mistake in the proof, since the conclusion holds only if both E and F are inside the triangle, which doesn’t follow from anything. Such transitions from «geometry» to «algebra» are always a weak point because some effort is required to establish signs geometrically: one needs to keep track of the mutual arrangements of the points.

[@mishaerementchouk]

Oh, it turns out that this is explained at the end of the video.

[@syllight9053]

These fake proofs have been boggling in the back my mind for months(specially the p=4 proof) and I'm *very* thankful to finally get porper answers and finally let my mind rest in peace. I'm *also* very thankful that you gave tips on being careful when looking at proofs. You're truely the best Math teacher I've ever seen!

[@jansenart0]

To me it was more about calculus caring about the area under the curve instead of the length of the curve.

[@martinepstein9826]

@@jansenart0 Calculus does care about the lengths of curves. Arclength is defined as an integral.

[@tychobrahe6379]

I have also had that pi=4 proof on my mind lately, and I’ve seen some other people attempt to explain what it gets wrong, but to very unsatisfactory results (like saying the curve is always going to have jagged edges, ignoring that the limit will indeed be the circle). I’m glad 3b1b can come and show how it’s done.

[@derekdreery]

Another interesting angle for this problem is that functions can converge in different ways. One type of convergence is where every point on the curve converges to the corresponding point on the limit curve (where 'corresponding' means 'the same t for a parametrisation'). Another is where you integrate the area under a curve to get a number, and then those numbers converge to the area under the limit function. Neither of these implies the other and it's fun to think of counterexamples. You don't even need curves, you can find counterexamples for functions that are straight lines joined together.

[@chlli]

porper

[@AlphaPhoenixChannel]

Oh man I stared at the TV for a solid few minutes of pain on that 3rd one before realizing that the intersection point wasn't inside the triangle. Unfortunately I had convinced myself that it lived at point D, so I can only take partial credit for finding the error and immediately replacing it with my own lol. awesome video

[@MasterHigure]

The point being inside or outside is not really relevant. The relevant part is that exactly one of E and F is outside the triangle, which makes one leg a sum and the other a difference, instead of both being a sum (which allows the two legs to be unequal in length). Of course, the fact that one of them is outside is easier to hide if P is drawn inside.

[@hens0w]

@@MasterHigure you can show that they amount to the same thing

[@MasterHigure]

@@hens0w My point is, they could've drawn P outside the triangle and still E and F both inside (or both outside), and the faulty conclusion would still be that the triangle is isosceles. Whether P is drawn inside or outside the triangle is rather irrelevant to the actual crux of this mistaken proof. If you want to use this proof to fool someone, where you draw P is of rather minimal importance. It is the placements of E and F that are crucial.

[@hens0w]

@@MasterHigure can you give an example, I don't think you can I think if p is inside then so are E and F

[@megamaser]

Initially I made the same guess but then I realized the same conclusions would follow so I thought for a bit longer and I realized an isosceles triangle from the same angle would require the left segment to shrink or the right segment to stretch, which made it clear that the midpoint would have to fall on the right side of the current midpoint.

[@kipchickensout]

when he drew the triangle and split the base in "two" i was like "wait that's not even close to the middle"

[@kyx5631]

Right? Even when eyeballing it, that was too sloppy to happen by accident.

[@saurabhkumarsingh3986]

And that's exactly where the trick lied

[@kipchickensout]

@@saurabhkumarsingh3986 yes

[@rosepinkskyblue]

Angle bisector wasn’t right either, but I thought maybe I was wrong and maybe he did measure it 😭

[@joshc5613]

I think the bisector for the side was fine but the angle bisector was kinda freehanded with extreme generosity

[@michalnemecek3575]

The triangle trick reminds me of the infinite chocolate trick. If you cut a chocolate bar on the diagonal of one of the rows, cut off the longest column from the top piece and switch the top pieces, you seemingly get an extra piece, but in reality, the pieces along the diagonal cut get smaller and the missing chocolate in that row accounts for the extra piece.

[@atanvardecunambiel8917]

To quote the gr8 jan Misali, math prank’d!!!!!!!!

[@abhinavdp7376]

Isn't it that pizza trick

[@Android480]

That’s funny. I knew the sphere example was wrong, I know you can’t flatten sections of a sphere without the pieces looking deformed. But you flattened it out and I said “huh, I must be wrong”. Blind trust I guess.

[@mrpedrobraga]

I was like that, too, except it was when he said "this is innacurate, but as we take the limit" and there I was with "uh... i guess..."

[@ed_iz_ed]

he does mention they arent flat, but the same goes for the circle

[@freshrockpapa-e7799]

@@ed_iz_ed a circle is flat man

[@wiggles7976]

That's true, a sphere is famously not flat, because there's no isometry from the surface of a sphere to the plane. Though a sphere is a manifold, so locally, it approaches being flat. But, if you are taking a bunch of pieces and laying them next to each other, even though they get smaller and flatter, the error due to the non-flatness could accumulate. That would be my guess as to why the proof is wrong, before watching the rest.

[@Felipe-sw8wp]

@@ed_iz_ed but the circle IS flat (this is the flat-circle theory and I`m a flatcircler) No, jokes aside, the circle is flat as a surface, it isn't flat as a curve.

[@JakubS]

When you showed the first "proof", my immediate thought was: you can't just convert from spherical geometry into euclidean geometry like that, otherwise there would be no need for the Mercator projection

[@squirrelcarla]

i know next to nothing about math but im a 3d artist so when i saw him do that i immediately went "boy, IF ONLY it was that easy" 🤣

[@feynstein1004]

I had the same thought. However, that does make me wonder. Why does it work for the 2D proof though? Like, you can take the curved circumference of a circle and bend it into a straight line without problems.

[@SioxerNikita]

@@feynstein1004 The thing is, you don't do that. He explained that, that is the error, but with smaller and smaller slices it approaches a straight line, and if we take an infitesimally small slice, it is essentially a straight line.

[@SgtSupaman]

@@feynstein1004 , for the 2-dimensional circle, slicing smaller and smaller vertically is all you need to approach the limit where a curve and a straight line become the same thing, because only one side of those slices is even curved in the first place, and that is what you are making smaller with each successive division. For a 3-dimensional sphere, however, the vertical slices are still just shortening the one side, but now the other sides also have curvature. That's why the video points out that you need to slice both vertically and horizontally to approach something that would be equivalent to flat.

[@feynstein1004]

@@SgtSupaman Ah okay. That helps a lot. Thanks 😀

[@little_wintry3098]

I've never been great at math. I barely passed geometry and algebra, and I've never even touched a calculus textbook. But something about your math videos make me feel like I understand it all, it brings back a sort of childlike curiosity where I don't have to feel bad about making mistakes or being wrong, I just get to sit here and learn things. It's oddly soothing for me and I love your videos :)

[@extrapathos]

Same, that's why I love math youtube. No pressure, like when school was easy and everything clicked

[@tejing2001]

I'm really glad you took the time to talk about this, since your channel spends most of its time presenting visually intuitive proofs of mathematical things. It's important to acknowledge that while visual intuition is really useful for aiding understanding, it is NOT a substitute for carefully checking every aspect. The devil is in the details often enough, even in the beautifully idealized world of mathematics. I got the triangle one, but probably only because I've seen a similar fake proof before, which purported to prove a 90 degree angle was equal to a 100 degree angle. It really is very sneaky. Actually the most confounding one for me was the sphere one. I couldn't quite put my finger on what the issue was. Thanks for the clear explanation there. I was very glad to see how you explained the pi=4 one. I've seen a lot of bad explanations out there, which make the error seem complicated, when in fact, it's exactly what you said: the limit of the property is not the same thing as the property of the limit. It's probably the most broadly applicable cautionary tale among these, though "make sure your diagrams are at least topologically correct" is pretty broadly applicable too.

[@purplegill10]

Ironically, it was the other two that got me, but the sphere one was easy for me purely because I had a map projection/sewing phase for most of my late-teen life. The second the lines turned straight sent off alarm bells in my head since, both in sewing and map projection, straight lines on a curved object almost never translate to straight lines when dealing with 2d things. It took many confused pillows and plushies before I finally got around to that in my head.

[@RocketsNRovers]

speech op

[@int0x80]

Another thing to note about the isosceles triangle proof: In an actual isosceles triangle, the perpendicular bisector of the base *is* the angle bisector for the angle opposite the base. (This is because of SSS congruency between the two triangles you get when you draw this line.) When we assume the triangle ABC is isosceles, point P, the intersection between the perpendicular bisector of the base and the vertex angle bisector, is not well defined because they are the *same* line. They have infinite intersection points. Therefore, to speak of the distance between a point on the triangle and point P is meaningless if the triangle is isosceles (which is what the proof tries to prove)

[@arcguardian]

That's why I concluded his first step is where the lie went wrong. There's no sense in me over thinking it, a yellow flag in math is usually a red one.

[@XyntXII]

​@@arcguardian I think you are right, that this is the point. If as he said, the hidden assumption, that would be needed for the proof, was that the point P would have to be inside the triangle, then this would prove, that all triangles where P is inside would be equilateral triangles.

[@sihaskumarasingha8148]

*breaking bad noises intensify*

[@linkhyrule5800]

@@XyntXII Indeed. It therefore follows that for all scalene triangles (triangles where no two sides are equal), P must be outside the triangle. As @ Tomáš Slavík discovered while trying to follow along. There's no contradiction there.

[@ca1498]

The fact that for an isosceles triangle we get one and the same line in no way breaks the proof; it only makes it incomplete. It just means that with those you are encountering a special case, which you have to acknowledge and possibly have to do something different. In this case, you can take any point P on the line inside or outside of the triangle and repeat the rest of the arguments. You will find that the triangle is... drumroll... isosceles! So, the problem is not that it breaks down for a triangle that you already know is isosceles; the problem is that if the triangle is not isosceles (I mean actually AB != AC), then it turns out to be isosceles, because we incorrectly assume that E and F are on the same side of the the BC line.

[@toebel]

when you started drawing the triangle I immediately noticed "hey that isn't the halfway point" and wondered why you didn't just redo the drawing. Tried drawing it myself and got a point outside the triangle

[@Benoit-Pierre]

Ah yes, AF=FB clearly does not seem right ...

[@redstonulo]

yea the middle point was about 1/22 off (1/22 of the distance between B and C)

[@totoshampoin]

Also expalins why he didn't use manim this time

[@redstonulo]

@@totoshampoin yea pretty much

[@toebel]

@Jul W kinda, because the last step of the "proof" is only convincing if you mess up enough steps while drawing the diagram that points E and F both end up looking like they're on the border of the triangle

[@bastienfelix4605]

There’s something so frustrating about knowing you’re right, but not being able to prove it… good job, magic numbers man, that last example really got me.

[@MK73DS]

I remember, when I was a "child" (14 years old), I asked my math teacher why the length of the diagonal of a square isn't twice the length of one side, with the same reasoning as you did with the circle, bending the corner over and over until, at the limit, there's just a straight line which is the diagonal. This really haunted me for days, I couldn't figure out why that was wrong (I obviously knew about Pythagoras' theorem). He tried to explain to me the way you did, how it is not always possible to assume that f(lim x_n) = lim f(x_n), but he couldn't really give me a formal answer to **why** this doesn't work. By that I mean, yes, we can't say the length of the limit is the limit of the lengths because our example shows it's not true, but it doesn't explain why it's not true. It's only years later, when I learnt about function sequences and their convergences, and also about what's a curve and what's the length of a curve (at least of class C^1 piecewise), that I could fully understand what's going wrong. For the ones who don't know, the length of a curve is given by the integral of it's derivative (in modulus, or norm if you consider it over a normed vector space). If you have a sequence of functions (f_n) that converges uniformly to a limit f, even if they all are of class C^1, the sequence of their derivatives doesn't not necessarily converges to f'. This means the sequence of their lengths doesn't necessarily converges to the length of f. In other words, the length function isn't continuous. Knowing that, it is also quite easy to create a sequence of curves that converges to a circle (or a segment), but with their sequence of length diverging towards infinity (think of the sequence of function (1/n sin(n^2 x)) )

[@mystic839]

i struggled with this well into adulthood, and still find my heuristics leading me towards these continuous functions that prove things incorrectly. i definitely have to remember that sometimes, they aren't.

[@ragnkja]

The zig-zag approximation works for area, but not for distance.

[@saschabaer3327]

There's a generalization of the definition of length for not necessarily differentiable curves that works in a pretty general setting (any continuous function from a compact interval into a metric space I believe): Let I = [a,b] be the interval on which the curve c is defined, and let d be the metric on the target space. Say a subdivision of I is a finite collection of points x0 = a < x1 < ... < xn = b. Pick a sequence of subdivisions of I with the property that the n-th subdivision satisfies d(c(xi), c(x{i+1})) < 1/n for all n, this exists by continuity of c and compactness of I. To the n-th subdivision assign the value L_n as the sum of all d(c(xi), c(x{i+1})) and the length of c as the limit of the L_n, which may be infinite (e.g. in the case of certain fractal curves). One can prove that if c is piecewise differentiable, this definition coincides with the one you gave, but it gives a well-defined answer for some curves that aren't differentiable. It's also incidentally very similar to the procedure that doesn't work, as you can imagine it as approximating the curve with straight line segments - the big differences being that (a) both endpoints of each segment must be on the curve and (b) the segments must get shorter uniformly.

[@_4y4m3_ch4n_]

so baiscally.... in the fake proof, the curve is continuous, but the approximation isn't, right?

[@MK73DS]

@@_4y4m3_ch4n_ Not really, what is not continuous is the length function for a curve. This means that the limit of the lengths of a sequence of curves isn't necessarily equal to the length of the limit of this sequence of curves. And this is exactly what's happening there, the limit of the lengths of the curves is 8, but the length of the limit of these curves is 2pi (because the limit of these curves is a circle). In a more extreme case, if you give me any length you want, as large as you want, I can find a curve arbitrarily close to a circle that has a length greater than the one you chose. This means I can find a sequence of curves, which converges to the circle, but their lengths diverges to infinity! (you can do that by oscillating very quickly around the circle) If the length of a curve was a continuous function, this proof would have worked. For example, the area delimited by a curve is continuous, so this means that the area delimited by these modified squares converges to exactly pi (if there was a way to easily calculate the area delimited by each of these curves, this would give us a way to approximate pi). But the length is not a continuous function, and therefore it is incorrect to say that because the curves converges to a circle, their lengths converges to the length of the circle too. Proving that one function is continuous isn't always easy, but proving that one function isn't continuous is often done by finding a sequence that doesn't satisfy the "limit of f(sequence) = f(limit of sequence)". For example, how can I prove the step function isn't continuous (the step function is the function defined by f(x) = 0 if x < 0 and 1 if x >= 0), I just look at the sequence (-1/n). This sequence converges to 0, but the sequence (f(-1/n)) is the sequence (0,0,0,...) and does not converge to 1 = f(0). Therefore this function is not continuous. What I wanted to do in my first comment, is to give another proof why the length of a curve is not continuous. Proofs by counter-example are great, they are quick ways to prove the non-continuity, but they don't always show why, what was the issue, what's special about that function that makes it not continuous. I wanted (at least, I wanted to try) to give an explanation of why the length function is not continuous, not just a proof that it isn't. Sometimes, proofs and understanding aren't the same things.

[@DylanNelsonSA]

There's a sort of extension to the isosceles triangle proof. Once someone correctly mentions that P is outside the triangle, you draw another inaccurate diagram where the points E and F are also both outside of the triangle. (i.e. On extensions of the sides that they correspond to.) Then you still get the conclusion that the triangle is isosceles, but this time by subtracting at the end instead of adding. In reality, one of E and F lies on a side of the triangle strictly between the two vertices, and the other one lies on an extension of the side that it is on.

[@gubblfisch350]

Not only is P outside the triangle, it's actually on the circumscribed circle of the triangle. In Germany this fact is called "Südpolsatz" (South pole theorem).

[@michamiskiewicz4036]

Nice observation! Moreover, the two projections E, F are colinear with the middlepoint D (in particular, one of them has to be inside and the other outside the triangle). That's a special case of Simson's line theorem.

[@jay_13875]

Also, the intersection point P doesn't exist at all for an isosceles triangle because the angle bisector and the opposing line segment bisector coincide. But you can pick any point P on the coinciding bisectors that is within the triangle and make the construction valid. In all other cases, P, as well as either E or F lie outside the boundary of the triangle, making the construction invalid.

[@caspermadlener4191]

The Dutch mathematical olympiad came up with "BOM-lemma", but "zuidpoolstelling" sounds way better. I also just heared that New Zealand also doesn't have a name for this. Thank you, random German Math person!

[@ericzhu6620]

I think it can be proved using Incenter/Excenter Lemma

[@lc1777]

@@ericzhu6620 yes

[@AirPodzol]

Found the flaw. Obviously it was because he assumed that it was a triangle. It was actually just a shape with three angles and three sides.

[@adventurousclash6323]

But it is close to a triangle. As we make finer and finer slices, it will get closer and closer to a triangle.

[@zerokun2655]

@@adventurousclash6323 if you're talking about the sphere, no, it isn't. The sides will be bulged, non straight, creating an overlap that doesn't go away. That's the only thing I was able to figure out on my own haha

[@networkofneurons]

an eighth sized spherical segment is just an open box |_|

[@francescovultaggio2540]

I'm an engineer and while getting my degree I had my fair share of math. This is the perfect elucidation of why I could have never been a mathematician even tho I can work with math with no problem. I tried really hard to spot the mistakes and, except in the first case, I wasn't able to spot the error in the other two before they were explained. Amazing video!

[@pannekook2000]

euclidation 💀💀💀

[@ionymous6733]

Though to any inspiring mathematicians, just because Francesco Vultaggio believes he "could have never been a mathematician" because he didn't get it before the explanations, that doesn't prove you can't, ironically perhaps.

[@JM-us3fr]

I’m a mathematics grad student, and I only spotted the mistake in the second one. Proofs are very unnatural for our minds, which is why rigor is required. My friends often ask me “How are you so naturally good at proofs?” The reality is no one is “naturally” good at proofs. I had to practice a lot before gaining important intuitions.

[@stefanperko]

That doesn't mean at all that you couldn't be a good mathematician. The issue with these "proofs" is that they aren't proofs in the first place. By that I mean the logic is so subtle and opaque that it's really hard to find faults with them. They are mostly based on intuition But our initial intuition is often times just wrong. That's why in the end you need to write down the argument in symbols to really convince yourself it works. And then it will usually become very obvious that it simply doesn't or at least that there is a huge logical leap somewhere.

[@francescovultaggio2540]

I admit I wrote the comment in a way that could be misconstrued quite easily. Not being able to spot the mistakes does not mean that you are not cut to be a mathematician, just like not being able to spot a bug in a code at first reading does not mean that you are not cut for CS. Growing up I noticed that what really makes the difference is not your natural raw talent or intuition but your dedication and passion. The true indication that one can be a mathematician is not necessarily if you found the mistake but how hard you tried, did it stayed in your mind afterwards, will you read more about this kind of puzzles/problems? Honestly I love the channel and the clarity in which he explains difficult concepts but I won't do any of that, mainly because math to me is a mean to an end and not an end in itself. That's the difference btw a mathematician and an engineer/CS/etc. I hope I didn't discouraged the next field medal recipient from getting into math with my previous comment, lol

[@rosiefay7283]

Thank you for stating that triangle fallacy, and for explaining its flaw. It's one of the few lasting contributions to mathematics by an obscure nineteenth-century mathematician called Charles Dodgson, better known to us as Lewis Carroll.

[@vigilantcosmicpenguin8721]

"Beware the Jabberwock, my son! The fallacies that seem consistent!"

[@MichaelRothwell1]

Nice! The origin of the alias Lewis Carroll is rather fun. From Charles Lutwidge Dodgson, Charles translated to Latin is Carolus and Lutwidge translated to Latin is Ludovicus, so (after swapping) we get Ludovicus Carolus, anglicised to Lewis Carroll. I think the Dodo character comes from Dodgson.

[@rosiefay7283]

@@MichaelRothwell1 You are correct. Mr. Do-Do-Dodgson had a stammer.

[@BrooksMoses]

Excellent choice of which pages to use for drawing the diagram for the third "proof"; the quotes are nicely apropos indeed.

[@ahsan4306]

Wow. I missed that little detail. Or maybe it was just coincidental. Nah! Can't be. GS is a perfectionist.

[@james64ibm]

I actually got that last one, but even then it took me 2 minutes or so, despite me spending essentially all of my life until the age of 19 at math competitions. What I'm saying is: If you construct a false proof cleverly enough, you can fool anyone. Including yourself, sometimes. And not just in mathematics.

[@alinpopescu4147]

the intersection of the angle bisector and perpendicular bisector is on the circumcircle (since they both pass through the midpoint of arc bc) , which is not even close on the drawing.

[@penguin_reader_yt9510]

Ima try it on my dad later. Hes an engineer. It should be interesting. Ill report how it goes. Maybe.

[@ca1498]

@@alinpopescu4147 Nice - I was wondering why P seems to always have to be on the outside. Now I know.

[@ca1498]

It is surprising how many known mathematicians have made known mistakes. Including Euler. Probably not Gauss, though.

[@sleepntsheep1169]

​@@penguin_reader_yt9510so, how did it go?

[@tomkerruish2982]

As many have noted, in the third 'proof', P actually lies outside the triangle. However, that's not enough; you can rework it so that it still seems valid. The actual sticking point is that, of points E and F, one of them lies on a side of the triangle while the other one only lies on an extension of a side of the triangle. For example, F is actually on the segment AB, while E is beyond C, relative to A. (I must confess I saw this explained in a book some time ago. It's a really good one.)

[@Kerostasis]

That explains why I was having trouble. I quickly realized that P could sometimes be outside the triangle, but I couldn't figure out how that actually changed the solution.

[@solanaceous]

What book?

[@tomkerruish2982]

@@solanaceous I'm sorry, it's been at least 40 years. (Also, I was referring to the puzzle as being a really good one. I wish I could remember the book.)

[@ahsan4306]

@@solanaceous Probably Mathematics and the Imagination. At least that's where I saw it first

[@TheEulerID]

It is simply not true that we have to go through the whole process to invalidate a proof. Finding a single flawed step is sufficient. It's then up to the original mathematician to submit a corrected version. The goes for any mathematical proof; if there is a single flawed step, then the proof is invalid and the onus is always on the one making the claim or, "onus probandi", as it is in Latin. In this case we know the "proof" must be flawed from the outset, as we can start with a counter-example. However, it's not always so clear. For example, Andrew Wiles' original published proof of Fermat's last theorem was found to be flawed, even though the conjecture turned out to be true. Andrew Wiles was able to publish a valid proof of Fermat's last theorem a couple of years later when that flaw was corrected.

[@ericdculver]

When reading The Elements, I often find multiple propositions proving what looks like the same thing, but different cases where some point is on the outside instead of the inside. This shows that Euclid really needed to include those different cases as separate proofs.

[@allank8497]

When he said: "they get increasingly subtle", i took that to mean I should get the first one; maybe the second, and that the third one was out of reach. Missed the first, missed the second, felt terrible about myself, and then got the third. Nice.

[@SgtSupaman]

It just goes to show how considering things more or less difficult is often just one person's perspective, because we can all have different thought processes and approaches (though, of course, a consensus can be reached). For instance, I agree with you that the third one was easier to see through than the second one, but I agree with the video that the first one was the easiest of the three to find the issue.

[@wiener_process]

I've been through too much courses on mathematical analysis to get fooled by the first two, but the third one gave me more trouble. Until I tried drawing it myself, and then it became pretty clear.

[@TlalocTemporal]

And still more conversely, I completely missed the first one, saw the second one as soon as the first fold happened, but only got halfway on the third one (saw that p was outside the triangle, but didn't realize what that meant).

[@vaish_diarys]

This is why maths is the most interesting subject

[@rakinhasan1360]

Agreed 💯

[@Rócherz]

*17:51** “The point in all of this is that while visual intuition is great and visual proofs often give you a nice way of elucidating what’s going on with otherwise opaque rigor, visual arguments and snazzy diagrams will never obviate the need for critical thinking. In Math you cannot escape the need to look out for hidden assumptions and edge cases.”* *-3Blue1Brown*

[@black_platypus]

For the sphere "triangles", an intuitive way of phrasing the problem is that, proportionally, the overlap stays the same as we approach the limit. It may shrink as we go finer, but so does the main area of our triangles (both their widths are scaled by the same "interval" of approximation while their height remains the same). Thus, the overlap may shrink out of view, but it doesn't shrink away from our problem :P

[@JamesChurchill]

I struggled with this for many years until I realized it was a failing in my understanding of how a limit works. Any approximation is the true value plus some error. For a limit to work (ie reveal the true value) the successive approximation has to *reduce* the error. But in these examples the error is maintained, so the limit still ends up with the original wrong value.

[@black_platypus]

@@JamesChurchill Exactly. 3b1b phrases it pretty much the same when he says the error must approach zero in the limit. I just felt like looking at this example, realizing that the width of both the triangles and their overlap are scaled by the same factor, this might serve as an intuitive way of coming to that realization

[@WanderTheNomad]

I feel like you could probably apply that to life somehow. The errors disappear from view over time, but are still there.

[@black_platypus]

@@WanderTheNomad whoa... 🤯😁

[@Oturan20]

@@WanderTheNomad and come back to haunt you in your nightmares!😈

[@The_Jacobian_Hotspot]

It's interesting how the Pi listeners in the animation react when something is noteworthy or thought provoking. I noticed myself understanding better when I saw the pi people understanding together. Very kewl

[@asdf7219]

Beautiful. A set of correct axioms plus a hidden incorrect assumption leads to an incorrect conclusion. Even if all the pieces fit perfectly together. This just goes further to show that something that is plausible isnt necessarily true.

[@TheGamer583]

If one wants to know more rigorously what happens in the second "proof", it essentially means that the uniform limit of curves, that is, the limit in the C^0 topology, does not necessarily preserve lengths of curves. This is indicative of the fact that "length" is not an innate property of curves that are only continuous, as many continuous curves do not necessarily have a well defined length (fractal curves for example). You would need what is defined as a rectifiable curve in order to talk about it's length. If one considers, on the other hand, curves that are piecewise smooth and they have a well defined limit in the C^1 topology, that is, their derivatives (ergo their velocities) also have a limit, then length is in fact preserved. The jagged curves approximating the circle do not have their velocities converging uniformly to the "velocity" (or the tangent velocities) of a circular path, hence it does not converge in the C^1 topology. Even more thoroughly, one could imagine the most general space on which a C^1 topology makes sense, and one would have to start dealing with Sobolev spaces from functional analysis. I find this remarkable in that it shows that even though our intuition fails, mathematics finds a way to rescue a significant and meaninguful portion of it on a maybe more subtle or advanced context, but nethertheless a correct and rigorous one.

[@WarDaft]

I've seen this one before and to be frank it's always baffled me that it baffles anyone. Behaviour at the limit is not behaviour approaching the limit unless you explicitly force it to be so. Consider lim n->2 of n. At no point is n equal to 2, but the limit is. Why should any given derived property be conserved if you haven't proven that it is? Proving what is conserved is why we use them in the first place and should be the first thing anyone learns about them!

[@londonl.5892]

@Joji Joestar The 1, 1/2, 1/3 one makes more sense because they do seem to approach something non-positive because each one is getting smaller. Here, each perimeter in the sequence is 8. And it seems really bizarre that a sequence of 8s doesn't have a limit of 8. Like, if you do lim(len(curve)), that's 8s all the way down and at the limit. But if you do len(lim(curve)), for some reason, each curve individually has a len of 8, but the limit doesn't? That's very, very odd. I still don't understand. The explanation above helps a little bit, but it's still not quite clear.

[@cr10001]

@@londonl.5892 Not a mathematician! But it seems to me that, with the second example, as you zoom in and add more and smaller right-angle 'kinks' to the rectilinear line, the ratio of [length of line] to [length of circle perimeter] doesn't get any better. Therefore the approximation isn't tending towards the correct value. (Whereas, comparing it with integration, the finer you subdivide it, the closer the area under the rectangles gets to the curve.) Similarly in the first example - in the case of the sphere the 'triangles' all bulge in the middle, and as you narrow them down the degree of 'bulginess' stays the same. Whereas with the circle, the triangles all have straight sides, with a tiny curve at the top, and as the triangles reduce in width that degree of 'bulginess' reduces (as a proportion of the triangles' area).

[@amarbapat8599]

@@cr10001 congratulations Your intuition has led you to discover the very primitive form of Convergence criterias. I am not being sarcastic! Not being from a mathematical background but still having that kind of intuition is crazy amazing.

[@fredgoodyer4907]

@@londonl.5892 So I think the problem is that you can’t swap lim and len like that. In the 1 ½ ¹/₃ case, notice that for every term in the sequence sign(1/n)=1. But lim(sign(1/n))=1 and sign(lim(1/n))=0. It’s actually a remarkably similar example! I think the takeaway is that limits are not always intuitive 😅

[@onemightsay248]

This video was honestly genius. You communicated everything so well, and it really helped my understanding in the basis of proofs. Thanks!

[@Victor-sw4ne]

This reminds me of a ‘problem’ we have when teaching calculus: How to explain why you can calculate a volume by integrating the area of the cross-section but you can’t calculate a surface area by integrating the perimeter of the cross-section.

[@shre6619]

I never thought about this before, can u explain more

[@matheusalmeidadamata]

I would also like to understand why it is not possible to do this. Thanks!

[@johnsmith-gq5jw]

@@matheusalmeidadamata ​ @Shre You can see the problem if you try to find the length of a the diagonal of a triangle by integrating the width of cross sections. For example, the length from (0,0) to (1,1) would be \int_0^1 1dx = 1. It might also make sense to compare it do what is should be. You might recall that ds=\sqrt{1+[f'(x)]^2} dx.

[@ladyravendale1]

If I remember correctly from my calc 3 course, this happens because of missing information. Integrating in multiple dimensions is (I'm fairly sure, at least) generizable to a series of nested integrals of the value 1. When integrating just a perimeter of a shape with the value 1 you are given no information about the shape inside the perimeter that you are trying to get the surface area of. If I remember correctly, you need to integrate sqrt(1+ fx^2 + fy^2) or ||ru x rv|| depending on parametrization, which gives you the information in the form of partial derivatives to actually know what's happening inside the perimeter.

[@Victor-sw4ne]

Consider a sphere, for simplicity, and a cross-section which does not pass through it’s center. While the volume of a cylinder whose base is this cross-section is a good approximation for the volume of a thin ‘slice’ of the sphere around this section, the lateral area of the cylinder is not a good approximation for the surface area of the slice, since the surface of the sphere meets the cross-section plane at an angle. That’s what creates this problem. Basically, you would have to consider this ‘angle’ due to the change in the perimeter, and, in the case of the sphere (or actually any surface of revolution), think of a sum of areas of cone trunks rather than cylinders. In the general case, this leads to you having to considering an integral involving partial derivatives, which carry information about the direction of the tangent planes.

[@bsmanu112]

I'm in 8th grade and you videos are the only reason I am able to do calculus. My favorite part was when I caught the mistake you made in your all triangle are isoceles proof. "SSA" is not a suffiect congruence statement. This reminds me off my math 2 teacher who would make sure we never forget that SSA is never a sufficient proof. This video just reminded me of him.

[@AKWLMath]

I thought that was the error at first as well, but as it turns out, it does work for right triangles. I did figure out the actual error before he revealed it, though I wasn′t 100% sure at first.

[@Drails]

Excellent video as always, you managed to verbalise/visualise the whole "limiting process needs to reduce the error" part so succinctly when I've failed to convince a few students of mine about that for a long time now - definitely showing this video to them. Cheers!

[@maxthexpfarmer3957]

That's why I think finding the area in a way analogous to Darboux integration as Archimedes did is a better way to do it.

[@UCXEO5L8xnaMJhtUsuNXhlmQ]

You can't forget the delta from your epsilon delta proofs

[@mvmlego1212]

I wish that he spent an extra minute or so deriving the limit of the error, because otherwise he's basically asking us to trust us that it doesn't approach 0, but it was a great video overall.

[@singularity3724]

I found all the flaws quite quickly, and it only took an entire 3 years of undergraduate mathematics to prepare me for this moment. Great vid!

[@johnchessant3012]

9:06 just wanted to point out, the step that turned out to be flawed was obfuscated by that favorite word among math professors, "clearly". three gold stars to you Grant if that was intentional!

[@ok1989fish]

As an Engineer and lifelong lover of math (but working with Finance&Accounting), I am astonished with your skills of explaining so many complex concepts on a simpler way - which makes it so much fun to watch (back in school I never thought I’d be watching a math video in YT just for fun, but you changed that and I’m grateful 😊) Congrats for all you do and wish you all success!

[@bleepblop]

I love math as well and am wondering about how you like finance/accounting/engineering in reference to math. I am close to college and wondering about all of these fields

[@svensorensen7693]

I was practically shouting at my screen "You can't just make a sphere flat like that!" it doesn't translate to straight-lined triangles on a flat surface. Wonderful explanations, wonderful video.

[@smlckz]

I'd like to remind you of your series on Differential Equations and request for its completion. It is such a great pleasure to be able to comment on a video which was just released a few minutes ago.

[@iankelley9302]

I second this!

[@iOSMinecraft120]

agreed, but perfection requires patience!

[@guilhermeantonini1777]

Curiously, the one that got me was the second one, because since I knew they were fake visual proofs I was very careful with the assumptions of shapes and sizes. But the second one also preys on our hidden assumption that 2D limits are about converging areas, when in this case it's about lengths, and the length of the curve doesn't approach the length of the circle even if the area between them goes to zero. Very cool

[@slightlybluish3587]

How can that be? I’m a bit lost, because if the area between them approaches 0 then you have the same line

[@andrewnazario2253]

im confused also by this :/ can you explain?

[@slightlybluish3587]

​@@andrewnazario2253 i've discovered that it can be true. Imagine you have a straight line going from point A to point B. Now imagine another line that starts from point A, goes midway to B, then up 10 units (arbitrary), and then down 10 units before heading towards B. There is no area because it didn't have any horizontal movement, but you end up with a tiny spike that is longer than the original straight line. I find it a little complicated to apply it to this problem, but oh well :P

[@Herio7]

@@slightlybluish3587 I think its better to visualize 2D things with 2D things. I mean your example isn't wrong but operating on 0 area is a bit unintuitive. It would be easier to visualize and compare simple and similar shapes say square 10x10 (area = 10, circumference=40) and long pole like rectangle 1x100, area is the same but circumference is 202.

[@slightlybluish3587]

@@Herio7 but in this specific problem the part that is troubling is the fact that the area *between* these two different lines (the perimeter of the circle and the zig-zagged perimeter) approaches zero, even as their perimeter is different; rephrased, the shapes are getting more alike and approaching the same area, while still having different perimeters. In your example the shapes are obviously different aren't they? I mean it's obvious that they have the same area but different perimeter, but in this problem what's confusing is how the two different shapes seemingly get closer to being identical while still having different perimeters but the same area.

[@raymondwang3110]

Hey, just wanted to let you know that my school now has this in the official syllabus for discrete math. Congrats and hope you become even more widespread in popularity!!

[@aleph-null3820]

I think I'll use the triangle "proof" as a collaborative exercise for my discussion in geometry on Monday, it ties in nicely with our exercises in thinking about the different ways to present diagrams and the importance of construction.

[@PowerhouseCell]

This is a really important topic! As a science animator, I always do my best to depict everything accurately and honestly, and it's great to see this in action amongst the community. Much love 💛

[@everythingisalllies2141]

you must have a problem being honest when trying to make the lies of Einstein's Relativity look convincing.

[@thecanmanification]

The sphere problem boggled me into right before you explained it and I remembered that you can’t protect a sphere onto a rectangle perfectly. Then I saw your proof, and as an electronic engineering student who’s literally studying different coordinate systems I felt a little disappointed that I didn’t realize the problem sooner lol

[@kyx5631]

At first I was about to make a joke about the first two proofs were very well animated while the third was drawn on paper. Turns out, that was actually required for it to fool us. That said, I'm pretty proud that I got the third one. The longer the drawing went on, the more I was convinced that those lines just wouldn't intercept like that. Fantastic video, love it.

[@IqbalHamid]

Oh man, he gave the answers to all the easy questions but left out discussion of the most challenging and intriguing curiosities @16:55. I hope you do please cover these in the future because I have ALWAYS been curious about these questions ever since I learnt calculus. And I've never found a calculus books that has discussed these very questions either, ever!

[@MarioFanGamer659]

If I'm honest, the second proof was for me much harer than the third one. The first one is quite easy (curvature) while the third one can be disproven by drawing the "proof" correctly. Second one really requires you to understand how limits work, instead, and even with your explanation, I still don't understand the issue that much.

[@SimonBuchanNz]

Basically the problem is that it assumes that the perimeter is the same between the two curves (the true circle and the limit of folding the square), but that's not a guarantee unless you have the third part of the proof showing that the limit of the error between them is 0. For the classic example of the area under a curve, you can prove that with the upper bound as well as the lower bound forming a rectangle with area greater than the error, and showing that the limit of the area of the error bounding rectangle is 0, and thus the limit of the error must also be 0 (handwaving detail about discontinuities and turning points)

[@rdaysky]

You could try to apply the same faulty reasoning to claim that the length of a straight line from (0, 0) to (1, 1) is 2. The problem is that the error doesn’t go to zero as you add more segments. For each individual right triangle the error does go to zero but only because the triangles become smaller; the error would need to go to zero faster than 1/n because you’ll be adding n of those error terms, where n is the number of the triangles. It’s similar to why 1+½+⅓+¼+⅕+… doesn’t add up to a finite number, the addends do get closer and closer to zero but not fast enough.

[@n0ame1u1]

Calculating the arc length of a function (which is how we would find the lengths of these curves) is not so simple. It involves an infinite sum (which in some cases can be simplified into an integral). And the limit of an infinite sum is not in general the same as the infinite sum of a limit. There are some conditions that need to be true for that to be the case, which must not be true here.

[@Michael-sq5ju]

I think the difference in difficulties for most people is that almost all pure math students will have taken a real analysis course, where arguments like the second one are discusssed in great detail, whereas fewer pure math students know about curvature (real analysis is often a first year course, whereas courses covering geometry for pure math majors are typically later) and fewer pure math students will remember Euclidean geometry from their high school days.

[@jeffsamuelson7221]

It is actually even possible to find a sequence of curves which hug the circle whose lengths blow up to infinity. What is not possible is that a sequence of approximating curves exists whose lengths converge to a smaller value than the true length (in technical terms, arc length is lower semicontinuous)

[@dvoid4968]

6:05 Given that the triangle is not already isosceles, the point where the angle bisector and the perpendicular bisector meets always lies outside the triangle (on the circumcircle of the triangle). In this case either the point E or the point F will lie outside of lines AB or AC. That means that one of the lengths AE or AF is already longer than the side of the triangle; by subtracting the other length you will get the length of the side of the triangle (I might try to formally prove this later on my own just for fun)

[@sinocchi3321]

I guess the argument in the video already proves it

[@dvoid4968]

@@sinocchi3321 I didn’t actually get that far lol

[@mcclellanj424]

I was so close to nailing the mistake in the isosceles triangle proof -- my first thought when he intersected the angle bisector and the perpendicular bisector was "how do you know those two lines intersect? They could have been parallel, for all we know!" And I thought about how, in an actual isosceles triangle, the angle bisector of the vertex angle ks also the perpendicular bisector of the base. So once again, I figured "that point can't exist." But I was still thrilled to see the subtlety of the fact that the point COULD exist but it's location in relation to the triangle is everything. Love it!

[@anjnagupta5985]

SSA doesn't exist

[@AKWLMath]

​@@anjnagupta5985It does for right triangles.

[@ai_serf]

This is so good. As someone self studying rigorous math, this man speaks to me.

[@conure512]

I'm quite proud of myself for actually figuring out that isoceles triangle proof. And I had a few things to add about it: - If the triangle truly was isoceles, then the angle bisector of A and the perpendicular bisector of B and C are actually THE SAME LINE, meaning their intersection point (P) is undefined. So if anything, the existence of point P is enough to prove that the triangle is not isoceles (or rather, that those two particular sides are not equal to each other). - AF really does equal AE, and FB really does equal EC. However, since either E or F must be outside the triangle, that means that one of the sides must equal AE+EC and the other must equal AE-EC. So essentially what we've done with this "proof" is, we found a new way to construct a line equalling the average of AB and AC, plus another line equalling half the difference between AB and AC. (Of course, if that was your original goal, there are far easier ways to do this - but I think it's interesting nonetheless.)

[@Dragon30ficationXD]

Such an excellent video! The clarity in your explanations as well as the production quality never cease to amaze me! Bravo!

[@Haluna11]

I'm honestly just impressed at the graphics/animations in this video, they either took a lot of work or were done using an algorithm

[@EebstertheGreat]

The fun thing about this is that Euclid must have known about this problem, as did the geometers whose theorems he wrote about. Euclid uses several visual proofs in his _Elements_ which are fallacious by modern standards, but whenever his axioms were sufficient for a more rigorous proof, he gave the better one. It's as if he recognized the deficiencies in some of his proofs but just knew of no better proof. Some of these are sort of trivial, like his very first proposition for the construction of an equilateral triangle (which assumes without proof that two circles, each passing through the other's center, must intersect), but others are absolutely essential to the axiomatic system he is setting up. His fallacious proof of SAS congruence might be the best example, because it is pretty much impossible to interpret in the modern day. He simply does not invoke his own axioms at all, instead using some sort of argument by translation, presumably based on what he felt was obvious. And I like the example of arclengths too, because ancient geometers suspected that it was simply impossible to define the length of an arc that wasn't piecewise straight. Archimedes disagreed and introduced some clever elementary axioms that allowed him to prove things about the circumference of a circle and surface areas of a cylinder, cone, and sphere. He was mostly interested in convex curves, because proving anything about general curves was not feasible using the tools of the time. This example of nonconvex polygons approaching a circle pointwise might not have been known to Archimedes, but a similar example of them approaching a right triangle probably was. These apparently paradoxical cases (where the limit of the arclengths of a sequence of curves does not equal the arclength of their pointwise limit, even though both are defined) might have dissuaded him from tackling nonconvex curves altogether. This is an example where modern calculus is genuinely superior to ancient reasoning, and where we couldn't make meaningful progress without the set of real numbers.

[@theodorealenas3171]

I loved reading this. Do you happen to know, when was it that mathematicians decided it's worth defining points in space using numbers? I don't mean which year, I mean what triggered them. It sounds obvious that if you turn points in space into number triplets, you should be able to prove anything, such as these intersecting circles or the theorem that perpendicular on perpendicular makes parallel. But apparently they didn't always think it's worth the effort? Or they imagined it would make things too complicated? Or they didn't have negative numbers to make things elegant? Or they had bad experiences from coordinate systems? Is religion and tradition in the mix? It's really strange to me. I assume that's what you mean by modern calculus. I don't know enough but it looks like points in space are defined using numbers, so I guess that's what you're referring to. Thanks!

[@EebstertheGreat]

@@theodorealenas3171 I don't know too much about the early history of analytic geometry. The person usually credited with first assigning coordinates to points on a plane is René Descartes, which is why we call it the Cartesian plane. However, the modern notion of real numbers is much more recent and is based on the work of Cantor, Cauchy, and Dedekind. The set of real numbers is a much more complicated concept than some people realize. Issues of things like uniform continuity of sequences of curves or rectifiability of curves started to be addressed in the early 18th century, I think, while real numbers were formalized in the 19th century. The Greek philosophers did not understand numbers in the same way we do at all. They did in fact draw a correspondence between numbers and geometry, but their numbers were all natural numbers greater than 1, i.e. {2,3,4,...}. Numbers showed up in geometry when "measuring" one line segment with another; that is, using a compass to mark off lengths of one line segment onto another. That's the sense in which you could say that segment AB was, say, five times as long as segment CD. And it was this sort of construction with which Euclid did number theory. For instance, 3 is a prime number because any line segment that can measure a segment 3 times the unit length is either of unit length or 3 times the unit length. Greeks considered ratios and proportions to be objects in their own right but not numbers. So there were no rational numbers, and of course no negative numbers and no zero (and the unit was usually also not considered a number). They were also aware that some magnitudes were incommensurable; for instance, you could not put the side of a square and its diagonal into a proportion with whole numbers. But that doesn't mean they had a concept of irrational numbers. Moreover, in modern terms the Euclidean axioms are not actually sufficient to construct all real numbers. They can only be used to construct the "constructible numbers," which turn out to be the numbers that you can define using just the integers and the operations +, −, ×, ÷, and √. Numbers like ³√2 and π cannot be constructed, which is why it is impossible to double the cube or square the circle.

[@theodorealenas3171]

@@EebstertheGreat ouch. So what I'm getting out of this is, they really didn't have the tools to use coordinates in geometry. They'd need a replacement for negative numbers such as naming a direction, they'd need to sum a natural number and a fraction... Then they'd need to fill in the gaps... Their system would get more edge cases if they used their numbers instead of their geometry. It seems like that's why it's a big deal to have a unified system for everything. And why web developers thrive as time goes on. Don't worry about that one. Thank you for the reply!

[@EebstertheGreat]

@@theodorealenas3171 Yeah, the number of special cases they had to treat separately is pretty heinous. This problem wasn't solved for a LONG time. Even if you look at the way medieval Europeans did algebra, they at least comprehended the continuum a little bit, but they still had no negative numbers or zero. So for instance, the equations x² + bx = c and x² = bx + c and x² + c = bx had to all be treated separately. Writing down the quadratic formula in a single equation would have been unimaginable.

[@vigilantcosmicpenguin8721]

It's a truth about all of science, really, that the first person to figure something out never gets it completely right. After all, the entire field is traced back to Thales, who thought everything is made of water.

[@vladmunteanu5864]

That video would have been so useful to me a few weeks ago! But hey, at least it gave me the opportunity to think by myself instead of referring to your videos at the slightest obstacle. But thanks to you now I know why my approximation of a sphere based on cylindres was giving me π²R² and not something else! I really like how your videos make you understand the topic but also the quirks that come with it so as to leave the viewer prepared for anything it has to throw at it. In a nutshell, great video as always!

[@pavelkalugin4537]

I made that mistake too sometime ago :) I calculated ∫ 2π√(1 - x²)dx for x ∈ [-1; 1] and got the same π² answer. It was only after I had looked at the extreme case of cylinder being close to the upper sphere that I realized the error wasn't limited there. I guess, as always, special intuition is needed here.

[@Joephnarro]

The third proof sounded fishy when I saw the intersection of the perpendicular bisector and the angle bisector and thought "Where should that intersection point really be?" The answer I thought was that on a true Isosceles triangle, that intersection lies directly on the unequal side. Then after seeing the explanation, it clicked further that the moment one side elongates or shortens bit to make a scalene, that intersection point begins to lie outside the triangle.

[@diabl2master]

In fact, for an isosceles triangle, the perpendicular bisector and the angle bisector are the same line. So they don't have a single point of intersection.

[@Theimtheimtheim]

Inrerestingly, these two lines always intersect on the circumference of the triangle, thus if they are not equal, P will always lie outside of the triangle.

[@TheZombersLMAO]

"You eat my food?" "No." "Prove it" "Alr let me show you"

[@kerofrog]

7:54 I think I got it. I decided to attempt following along with the construction in desmos geometry tool, and quickly found that point P always lies outside the triangle, except in the case of an isosceles triangle, where the lines are equal and there is no single way to define P. I'll have to watch the rest of the proof to know *exactly* where this throws a wrench in things, though

[@Xeridanus]

That's how I got there, about 19 secs later than you :P. I used proof by absurdum. If AB and AC are the same, P cannot by defined so the whole thing falls apart. Notably, you can't state that BP and CP are the same length. Or FP and EP for that matter.

[@YLLPal]

"You can't always trust the conclusions of SCEM communicators. Sometimes there is some hidden assumption which was missed. For example, you thought this was an educational video, when in fact it was an ad for my new book." 😉 10/10, would trust 3B1B to deliver insights AND make me notice the cool book hes using.

[@FBI-zb6do]

This channel is absolutely fantastic. Absolutely LOVE you guys! You explain everything very well, and I especially appreciate explanations for the simple terms that I, being a non-native english speaker, don't know from school! Absolutely fantastic stuff!

[@Abstract_zx]

1:29 when i saw this it seems like some part of "unraveling" the slices into flat shapes is an approximation for the actual curvature, and the actual difference between the 2 areas boils down to approximating 4 as pi (or vice versa) in some capacity

[@abbashashimov4395]

When I was learning geometry in school i hated the rigor and proofs because I was studying with INCREDIBLY rigor soviet textbook. For example, there were exercises where you needed to prove that line cannot intersect circle at 3 point or if the first line is parallel to the second one and second is parallel to the third one then first is parallel to third. I really hated all of this because it looked very stupid to prove such obvious things. But throughout your math journey you learn to appreciate rigorous proves of the simplest things. And this video is a brilliant explanation example why rigour is so important!

[@divisix024]

It’s often the most intuitively obvious fact that’s the trickiest. Similarly, the most intuitive falsehood. Also let me try one of your proofs. The first one: Let A,B be two distinct points on the circumference of the circle and choose P another point on the circumference so that PAB is a triangle inscribed in the circle. Then by definition the sides of the triangle except the vertices lies inside the circle, hence AB does not intersect the circle at any point other than A and B.

[@francesco1777]

@@divisix024 This is a bit circular: you are assuming that there is a point P such that ΔPAB is a triangle inscribed in the circle and this kinda begs the question. I would proceed as follows. Let's suppose that there were 3 distinct points A,B,C in the intersection between the line and the circle. This means that there are 3 distinct aligned points in a circle. Let O be the center of the circle (clearly O does not coincide with A,B or C because the center doesn't belong to the circle, except in the degenerate case in which the radius is 0, but this case is trivial). Without loss of generality I can assume that B is in the segment AC. Now I consider the two triangles ΔOAB and ΔOBC. They are both isosceles because OA,OB,OC are radiuses. Moreover the angles ∠OBA+∠OBC=∠ABC=180°. So at least one between OBA and OBC has to be obtuse or rect. Without loss of generality let's assume OBA is obtuse or rect. Since ΔOAB is isosceles and OBA is obtuse and rect also OAB has to be obtuse or rect. Since the sum of the angles of a triangle is 180°, ∠OAB=∠OBA=90° and ∠AOB=0°. Since ∠AOB=0° then A, O, B must be aligned. Moreover since OA=OB, there are two possibilities B=A or B is the symmetric of A with respect to O. In the second case ∠AOB=180°, so we must have B=A. This contraddicts the fact that A,B,C were distinct points. I'm not saying that this is a rigorous proof, because Euclid axioms are not rigorous. A rigorous proof in this style should be done through Hilbert axioms or affine spaces axioms. i just think that this is a more convincing "proof".

[@divisix024]

@@francesco1777 Tbh the part you pointed out was on my concern, but I wasn’t able to resolve it. I just decided to throw it out anyway and see if someone else would be able to supplement a proof without invoking the result.

[@mujtabatahir7834]

07:29 Side Angle Angel prove is valid only if one side, and angles except opposite angle are congruent. In other words, if the common line is congruent, for triangles to be equal angle and both ends, alpha and angle at centre of triangle must be same

[@dwightpoop3369]

SAA is always valid. SSA means two angles and one side that is not between the said two angles. The 'one side and angles except opposite angle' you mentioned is ASA, which is also valid. The SSA in the video is actually invalid though.

[@mujtabatahir7834]

I think i messed up the nomenclature, but i think that i have made my point ☝️

[@Owen_loves_Butters]

@@dwightpoop3369SSA is always valid for right triangles.

[@trba7996]

The only reason the sphere “proof” set off an alarm immediately was because I knew about map projections. Otherwise that logic is so deceptively sound

[@vigilantcosmicpenguin8721]

Yeah, I felt so smart being able to immediately point at the screen and shout, "THOSE AREN'T TRIANGLES!"

[@NotKartikeySingh]

* 12:48 Simply u r taking that triangle from a sphere which is not a plane so it will not follow the angle sum property and thus it is also not a triangle. So if it is not a triangle then it will not form a rectangle when combined. U can also see 2 parallel lines meet at the poles of a sphere, draw a line perpendicular those 2 lines like the equaters and now u see that the triangle formed by those three lines has 2 right angles because u draw the third line perpendicular to those 2 lines when they were parallel and ur still left with the third angle wich has a considerable angle. So clearly 90+90+the third angle >180 and thus its not a triangle.

[@Andrew-tk9lh]

The second “proof” is very similar to one constructed in one of the first videos by Looking Glass Universe :). I don’t think they ever made an update video addressing the missing mathematical rigor of accounting for successive errors, so I’m really glad you do that here.

[@SantoLucasST]

I never get tired of watching this stuff, even though I understand about 3% of it

[@theseusswore]

In your proof, you gave it away with the "I made no assumption" lines. You actually did assume that the point P, the point of intersection of Perp. Bisector and Angular Bisector of angle A was inside of the triangle. It could very well be outside the triangle too.

[@theseusswore]

More accurately, for this reason you cannot assume that |AC| = |AE| + |EC| as E may not even be on line segment AC

[@pantoffelkrieger8418]

It is also very easy to show that exactly one of E and F lie on the side and one is outside unless they are equal to B and C (apart from using the contradiction that ABC would be isosceles): It is well known that P is on the circumcircle, specifically, the midpoint of arc BC. Then D,E,F are the Simson line of P with respect to ABC, so they are collinear

[@ThePharphis]

I thought the giveaway was that he clearly didn't measure the angle before bisecting it

[@person8064]

The perpendicular bisector of BC doesnt look like its bisecting.

[@kerstinhoffmann2343]

this sort of thing is precisely why I hate doing euclidean proofs

[@jonahansen]

I think the real underlying moral is that you have to be a smart cookie to do things right.

[@tunafllsh]

Haha! I got it! 9:53 Point P must not necessarily be inside the triangle ABC. It can be outside, in which case either AB=AF-FB or AC=AE-EC

[@Winnubooiii]

It will be on BC side of the triangle

[@n0ame1u1]

The issue with the first proof is that the sections aren't triangles, even in the limit. The issue with the second proof is that limit(len(c_n)) = x does not imply len(limit(c_n)) = x. The length of c infinity is not the same as the limit of the the lengths of c_n. I'm not sure about the third proof. My first instinct would be one of the SSA or SAS equivalences isn't valid, but that's mainly because I don't remember pre-trig geometry that well. EDIT: Wow, that last one really was subtle

[@Kuvina]

14:26 Fun Fact: I actually have an entire video about this phenomenon, where there are so many potential ways to flatten a globe , but it's inevitable to lose some detail, and depending on the map projection, that can be in orientation, area, angles, or some combination of the 3.

[@kkim5000]

this is a message to all those people on "can you solve this math problem?" videos who claim that their approximations are good enough to answer any math problem with great accuracy.

[@jan_kulawa]

While running the third proof, I thought about formalizing it on a formal axiomatic system such as Hilbert's or Tarski's, because that always helps me show the holes in an intuitive mathematical proof, whether they can be ultimately filled or point to a flaw in it. I didn't actually do it, because the video is kind of gripping, and I wanted to see through it to the end first, but now I wish I did! That funny step would be immediately visible as flawed if I were to run the proof myself on this setting. That's a good moral to learn, I suppose. Also, and it goes without saying, but great video! I always learn so much from your work, even when I believe I already understand the subject matter fairly well. Not only is the surprise entretaining, but also engaging, and it makes me return to your channel as soon as a new video comes out, probably in response to the serotonin rush I always get from them because of this. I hope to one day teach and communicate mathematics, and in particular my subjects of interest within it (mathematical logic and philosophy of mathematics), as well as you have taught me. Cheers from Brazil!

[@theodorealenas3171]

This reminds me of a programming language, "Rust", which can predict when the program you are writing is going to have certain flaws, just like that methodology you wish you used. The cost of using Rust is that coding in it is so tangled and restrictive that a lot of people give up on it immediately. I didn't expect to see a mathematical Rust! I wonder if I'll see a Rust in other places in life too. Funny!

[@HelixSnake]

***SPOILERS FOR PROBLEM 3*** If you want a proof that the angle bisector never crosses the perpendicular bisector of the opposite side inside the triangle: 1. imagine a non-isoceles triangle like the one in the proof. Picture the top point leaning towards the right. 2. Bisect the bottom side of the triangle and draw a line from the top angle to the bisection point. 3. Let's call the top two angles A1 (on the left) and A2 (on the right) and the lengths of the corresponding opposing sides (that together make up the bisected line) 'a' (we can use the same for both because they're equal). Let's call the line from the top to the bisection point L, and the remaining angles of the original triangle B (on the left) and C (on the right). 4. The law of sines says that the proportion of the sine of an angle to the length of the opposite side is equal for all angles of the triangle. So for the left triangle section, Sin(A1) / a = Sin(B) / L and for the right triangle section, Sin(A2) / a = Sin(C) / L. Divide one equality by the other and you get Sin(A1)/Sin(A2) = Sin(B)/Sin(C) Remember that our top point is biased towards the right side of the perpendicular bisector, which if you look at the triangle, you can tell that angle B is greater than angle C. This leaves two possibilities: 1. Both B and C are acute which means, since Sin between 0 and 90 degrees only goes up as the angle increases (look at a graph of a Sine wave if you don't believe me, the top of the first "hump" going right is 90 degrees), and angle C is greater than angle B, so Sin(C) > Sin(B) 2. B is potentially obtuse, which means you have to do a bit more work The sine of an obtuse angle equals the sine of its supplement (180 degrees - that angle) and since there are only 180 degrees in the triangle in the first place, A + B + C = 180, so 180 - C has to be B + A, which means the supplement of C is more than B and therefore Sin(C) > Sin(B) Now that we've proven Sin(C) > Sin(B), that means Sin(B) / Sin(C) is less than 1 (larger denominator than numerator). Which since Sin(A1)/Sin(A2) = Sin(B)/Sin(C), Sin(A1)/Sin(A2) is also less than 1. Again, the larger denominator than numerator required to get a quotient less than 1, means Sin(A2) > Sin(A1). I will now prove A1 is acute. The angle between L and the left side of "a" is obtuse, because the top point is to the right of the perpendicular bisector; since A1 is in the same triangle as that angle, A1 must be acute (a triangle can't have two obtuse angles). This leaves two possibilities: 1. A2 is obtuse, which since A1 isn't, means A2 > A1, which means the angle bisector line is to the right of the line we've currently drawn 2. A2 is acute, which for the reason stated earlier, Sin(A2) > Sin(A1) when both are acute means that A2 > A1, which means the angle bisector line is to the right of the line we've currently drawn. Boy that wound up being a lot longer than I intended it to be. I'm sure there are more elegant proofs but this is the best I could come up with.

[@sahilsohani2759]

Goddamn bro

[@f5xs757]

I didn't expect a Skate 3 youtuber to be here and give us some math proofs. Keep being awesome!

[@TechToppers]

The point of intersection of perpendicular bisector and angle bisector lies on circumcircle of ABC which is a well-known fact in Olympiad Geometry.

[@Aeropig_]

Well there actually is a proof of this in the video! Suppose that the angle bisector crosses the perpendicular bisector of the opposite side inside the triangle. Then, as shown in the video, all triangles are isosceles. But this is a contradiction. QED

[@billymanilli]

Oh yeah. How about that!

[@racheline_nya]

finding flaws in proofs seems fun, and i'm glad to find out i enjoy it right before i'm actually required to do it a lot (if everything goes well and i don't get scared of the necessary social interactions), so thanks for making this video! :D

[@blokin5039]

Social interaction?

[@racheline_nya]

@@blokin5039 i want to help organize a math competition, which involves checking proofs and some other fun stuff, but also interacting with other organizers and with the competitors and that's scary

[@lax4695]

I showed this to my math teacher and he started screaming like a maniac and saying "PLEASE LET ME LEAVE, I WANT TO SEE MY KIDS " crazy

[@jacefairis1289]

for the second one - I bet you could do this correctly if, instead of doing something that leaves the perimeter constant, you did something that reduces it as you get closer to the circle, so the error in the length actually goes down. my immediate thought is: instead of folding, cut off the corners of the square with a line tangent to the circle, making it into a regular octagon; then you turn it into a 16-gon, and so on. according to Wolfram Alpha, the perimeter of a 16-gon with apothem 1 is about 6.365, which seems pretty close to 2pi given its only n=2, so I'd say it seems promising

[@ajjdgj6tmgedvnmtmek]

The limit of a regular N-gon is a circle, and some of the earliest records attempts at calculating digits of pi are handmade tables of the areas of N-gons with apothem 1. You can try it yourself with some small values of N if you want. As N approaches infinity, the apothem becomes the radius and the area of a circle of R=1 is pi. You could do same approximation with the perimeter as the sum of side lengths of a regular N-gon with apothem 1, which would converge on 2*pi.

[@IsaacC20]

This is what is missing from school curriculum: an explanation of mathmatical rigor and what it means to think critically (thinking about hidden assumptions). I think kids crave this stuff (I certaintly did anyway in highschool and math teacher just brushed it off) but teachers simply don't have these kinds of insight.

[@hetox1994]

My best guess for the last one was that the point P may or may not be inside the triangle. While preparing for IIT-JEE, they teach us a chapter called solutions of triangles and anyone who has put extensive hard work on that chapter might come to the same guess as me.

[@thelittlestkovo]

During the triangle proof, I was having to try not to scream the minute you labelled FP == EP. I was at work and muttering under my breath. Thank you for that.

[@T3sl4]

What's interesting about the 3rd "proof", is -- geometry being what it is, it's quite simple, it doesn't mind signs, or even complex numbers -- *as long as you are consistent with them*. Which is of course the problem as shown: a lack of consistency. Handled properly, geometry doesn't much care if you have negative or complex lengths. You can get some interesting (if maybe not particularly useful) insights when considering "out of range" values, when applied to a lot of theorems/problems. And, I would say, building an intuition about this expanded problem space (including negative lengths, areas, etc.) helps hone ones' intuition about what equations can arise from that geometry. Even entering "wrong" numbers, say to the Pythagorean Theorem, gives a certain amount of rightness: an imaginary length can be thought of as, say, an effective width/thickness of a line, i.e. a dimension perpendicular to its apparent direction; or as a "height" in a new dimension perpendicular entirely to the plane. Consider a dumb-3-4-5 right triangle: let c = 4, b = 5, and a = sqrt(c^2 - b^2) = ±3i. A hypotenuse of 4, with a base length of 5, does not, a right triangle, make! There is no angle we can draw the hypotenuse at, which connects with the altitude (length a), while still drawing it perpendicular to the base (length b). At best, we could draw the hypotenuse parallel to the base, coming up short by 1 unit from the base length. But we could also claim that the altitude is 3i long, effectively extended widthwise to reach the "hypotenuse". (That the minimum length difference is 1, while the complex magnitude is 3, is less easy to imagine, I'll admit!) And this kind of I think fits nicely with other applications of complex numbers, that they show up so often in rotational symmetry, or in this (fully imaginary) case, right angles.

[@Xeridanus]

I'm trying to imagine what this would look like and my mind is spitting out an MC Escher sketch.

[@DoktorTaiko]

In my third semester I was taking a shower, looking at the tiles and thinking: if I follow the edges from one corner to the corner on the diagonally other side of the tile, the length is 2*1=2. If I where to put the corner of the diagonal line in between I'd get a new path with length 4*0.5=2. If I were to do that an infinite amount of times I'd get the diagonal line but also never change the length and therefore get 2=√2. It was at that point that I realized: math has gotten into my head and became part of me. It was the weekend, I still had a hangover from last night and I was just taking a shower. But math was what my brain wanted to think about anyway.

[@diogeneslaertius3365]

7:26 the triangles AFP and AEP are not the same or congruent, because the angles alpha and the right angles are not angles between the shared side and the other sides leaving the possibility for the angles APE and APF to be different. So the proof collapses. It's much harder to see the error in the second proof though.

[@fiyanshuNahar]

U r right

[@fiyanshuNahar]

Even I an exam in India called isi bstat they actually asked these combinations in multiple choice questions, likes shows how much people actually care they say they know geometry but I feel geometry is quite different bqz so much algebra that comes with it not direct as it's all in shapes etc ,in my opinion. I'm learning geometry and still thinking about very small things bqz its quite different from other braches

[@Jivvi]

The first thing I noticed wrong with the triangle proof was trying to find the intersection point (P) between the perpendicular bisector of BC and the bisector of ∠A. If the triangle was really isosceles, they would be the same line, so APD would be collinear and P could be anywhere along AD, which is the triangle's axis of symmetry. I don't think this is the key to the proof being wrong, since the intersection point actually does exist for non-isosceles triangles (or even for isosceles triangles if you pick one of the equal sides instead of the odd one out), but I just thought it was interesting that you're using a point in the proof that would not exist if what you were trying to prove was true.

[@sandhuekam14]

So true even i think the same and also the perpendicular bisector is not correct its a bit off centre the point must be somewhere else

[@MK-13337]

Yes, the point P must exist, although for all non isosceles triangles it is outside of it, and the proof falls apart.

[@christophercripps7639]

Fantastic graphics. The isosceles/equilateral triangle "proof" is akin to one where the result is 2 = 1 (through the deceptive concealment of a step involving division by zero).