I have never seen any of these proofs - I graduated HS in '56 - got a BSEE in '60 - worked in Aerospace 38 years - retired and became a HS math teacher in 2001 - retired in 2014. Never offered any of these proofs to my students. I think they (and I) were shortchanged. Love this stuff.
I really have no idea how US schools still manage to produce some mathematically literate people. Once one of my colleagues brought his 1st or 2nd grade son homework apparently designed to teach subtraction. 3 engineers sat puzzling over (1 UK-educated, 1 Chinese-educated and 1 Polish-educated) until 4th came who's kids were past basic schools and solved it for us. I knew both of Pytagoras proofs before I was taught them. From my POV US math teaching is bizarre.
I have indeed been taught this at school. Not in the math class, but during Latin class. Our Latin teacher also knew ancient Greek and had found the Greek text about this interesting enough to teach us about it. I found it very enlightening.
@@jannikheidemann3805 If you mean the United States, then yes, they teach Latin in many high schools, it's gaining in popularity, many Catholics study it independently to deepen their study of the faith, and many more people study the Latin roots of English in preparation for standardized tests like the SAT and GRE.
My school in Britain taught Latin. I think Latin was an essential requirement if you wanted to be accepted as a student in Oxford and Cambridge Universities. A little indicator of a well educated candidate. Not any more of course!
No, I do not. I expect it to be good, but unless I have watched the entire thing I can not evaluate the quality of the video. There is an important difference between 'expect' and 'know', and that is the same difference as between science and faith, for example.
The first proof is by far the best I’ve ever seen, and it was a big lightbulb moment for me. I’ve seen UA-cam videos of teachers doing it with felt triangles taped on the blackboard, but I never got any lessons like that as a kid. I will be sending my local elementary school a bill as suggested.
The animation in this video is just insane - about the best I've ever seen on UA-cam in fact. I can't imagine all the time it must have required to get all that to flow so smoothly. Well done. I especially liked the part about the bugs, shows you how to solve such a problem in visual steps - a great problem solving technique. And the animation was just beautiful.
Burkard, rather than heap additional praises regarding the dependable brilliance of your channel's content, I thought I might remark on your delivery: Your pacing, pitch, tone and inflections are such that, rather than lulling my mind to sleep in class, you have a special way of always educing and coaxing my brain forward (stubborn mule that it is).
For the bug problem: An equivalent question would be to ask how long it takes for any one ant to reach the ant it is chasing. The chased ant always moves orthogonally to the chaser, so it is neither going towards it nor away from it. Thus the distance between the two ants simply decreases by the speed of the chaser. It thus takes 1 unit of time for the ants to meet.
I guess that kind of depends on, what you mean, by ”moving away from or toward the chasing bug”. If you mean that the Euclidian distance between the chased bug and the stationary chasing bug has changed (increased for ”away from”, and decreased for ”toward”), by the end of 1 segment of the chased bug’s path; then, I think it’s pretty clear that moving orthogonally to the chasing bug, means moving away from it; since the distance between a square’s center and 1 of its corners is greater, than between its center and the midpoint of 1 of its edges. 🤔
@@raphaellfms The video’s been edited since he put it out- there was an incorrect generalization there. If you look closely, you can see- usually, he fades in and out, but a couple of seconds earlier, he cuts out instead.
I mean I cheated or crammed my way through high school I was too bored to try, so this is the first time I’m seeing these design elements explained from a mathematical standpoint which helps me understand them deeper: It’s amazing how intertwined art, beauty, mathematics, numbers, algorithms, patterns etc. Truly all connected. I am all that is, because all that is - is within me. 🙏🏻
Thanks for another fantastic video! I really liked the first proof in the introduction because during school I only learned the second algebraic proof. It was neat how you could prove identities about sin, AM-GM, etc using these diagrams!
Mesmerizing. Reminds me of being a kid in early years of learning math and just drawing and playing with shapes. Wish I had this kind of instruction back then.
i wonder the same thing. those students. in ancient greece, didn't have calculators. don't know if they had compases, either. but they solved a lot, calculated Pi, and figured calculus...
I first became familiar with Pythagoras theorem while surveying field sizes with my father ( a crop farmer in lower Michigan, USA). I was privileged as a seven-year-old boy to be the hold-down boy for the end of the rod-chain. I would look Dad square in the eyes as he circled round me, ending with kneeling with the rod stretched tightly across the ground, and Dad pounding a peg in at my next point. We measured many fields, but Dad had five sons, so the privilege got spread around. I always loved this time, because it was both intimate with my father, and mathematically practical--my teachers were surprised in grade school how well I did with geometry, almost all aspects which I experienced with practical applications on the farm with my father. He's now 96, and we still write to one another often, though now it is his love of philately that drives the mail truck!
This is one of the many things I enjoy about traditional geometric patchwork design quilt patterns , (like Flying Geese, Irish Chain, Feathered Star, Grandmother's Flower Basket, Disappearing 9-patch, Bear's Paw, Log Cabin and all it's variations, Pinwheels, etc.) with 1/2 square or equilateral triangles, rectangles, squares, etc. and shuffling them around to create different patterns. Now I have a name for them, and may design quilt patterns with the names of Pythagoras or Trithagoros, or the long ago Chinese mathematician's name! Great visual explorations of these relationships, thank you. (of course I enjoy even more, the random sewing together of odd shaped and sized pieces, called, "crumb quilting"! lol) (see also, the 'golden ratio' seen so much in nature and used from antiquity in architecture to produce unconsciously, naturally pleasing, building facades.)
You are a brilliant mind. Leaving me smiling from what you teach. I'm literally talking to myself in full conversation when I watch your videos. Thank you for the intelligent video and teaching.
I learned the derivation of the Pythagorean Theorem when I took geometry. We were on the topic of similar triangles and found that a right triangle could be made up of two smaller right triangles. It was very straight forward and consistent with the topic we were on.
I love these visual proofs. The animation of the visual proof at 22:40 is extra fun because two of the triangles are re-scaled. But they are re-scaled instead of having a more complicated animation. Well done!
Yes, and it happens five times from 22:17. @Mathologer has the left-most and right-most triangles swapping sizes instead of swapping places. IMHO it detracts from the quality of this video, because the visual proof depends on the pieces only rotating and translating - not expanding.
The next one at 22:40 really confused me when i tried to work it out because the side length of the blue inside square and the side length of the blue isosceles triangle are not actually the same in general. They do happen to be the same for the 3-4-5 triangle though
The area of the central square in the overlapping twisted square is 1/5 the area of the original square. You can prove this physically as each of the four original triangles can be broken up into five copies of the smaller overlap triangles. As the overlap and original triangles are similar, the ratios of their short to long legs are both 2:1, so the central square is equal to tour of the overlsp triangles. Adding up the total number of triangles, you get 5 overlap triwngles times 4 original triangles, minus the 4 overlapping triangles at the corners, plus the 4 in the center, to get 20. 4/20 = 1/5. Mathematically, we can use Pythagoras and similar squares to determine the area. a² + b² = c² (1/2)² + 1² = c² c² = 1/4 + 1 = 5/4 c = √(5/4) = √5/2 Now we can determine the dimensions of the smaller overlap triangles via similar triangles, since we fan see that c = √5/2 and c' = a = 1/2. c/a = c'/a' (√5/2)/(1/2) = (1/2)/a' (√5/2)a' = (1/2)(1/2) a' = (1/4)(2/√5) = 1/2√5 b'/a' = b/a b'/(1/2√5) = 1/(1/2) b' = 2(1/2√5) = 1/√5 By observation, the central square has side length of b', so: A = b'² = (1/√5)² = 1/5
An alternate solution to the bugs problem: Let each of the bugs have velocity 1 unit/s. Split each velocity vector into a vector that points towards the center of the square and one that points 90 degrees to that one. Using the 45-45-90 triangle created with the original velocity vector and two component vectors, we find that the two new vectors have magnitude sqrt(2)/2. Thus the bugs are going towards the center at a rate of sqrt(2)/2 units/s. The center is sqrt(2)/2 units away so it will take the bugs one second to get to the center. Distance = rate*time so distance = 1 unit/s * 1 s = 1 unit.
I think it's even simpler than that. Each bug always moves perpendicular to the motion of its "target" bug, so the target's motion doesn't affect the rate of closure. Therefore, each bug must travel exactly one unit to reach its target, at the center. Fred
@@gammaknife167 Thanks. But I think I may have simply remembered it from the Mathematical Games column. Not sure. After digesting enough beautiful arguments, they tend to become ingrained in your thinking, without your knowledge of their origin. But whoever came up with it, it is indeed a nice insight.
@@ffggddss I really like that argument. I originally encountered the problem when the bugs were in an equilateral triangle, so I not think that the argument applies to other regular polygons. A fun exercise is to use the method mentioned above to prove that the distance traveled by a bug starting in a regular n-gon configuration with side length 1 is 0.5*sec^2(90(n-2)/n). Another interesting thing to look into is other arbitrary starting configurations of the bugs.
In my retirement I have been working occasionally as a substitute public school teacher. (Math, Science, Spanish, only subjects I know well). I like to show these proofs to the students and jokingly quote your "get your money back" comment. One day I was talking with a math faculty member and told him of your channel and some of the things I like to share with the students to demonstrate that math is fun. He had never seen a proof of the Pythagorean Theorem , and in fact had believed it was an axiom. So thanks for your channel, you do more good than you know.
35:45 Because the motion is smooth, we consider the local movement of one bug (being chased) relative to another (which I will refer to in the first person - I am the bug that chases), and we see that it's linear. Further, in this case we see that the movement is perpendicular (movement being it's change in position all happening relative to me) - this is because I am always facing it, and more importantly, the angle between me, it, and the bug it's chasing (the direction of it's movement) starts off at 90° and never changes - THIS is due to symmetry. It follows that (locally) relative to me, it's moving in a circle around me (and also I'm rotating in place to face it) - never getting either closer or farther from me. I'm the only one that has any affect on the distance between us, so the final distance covered is the initial distance I needed to cover.
The hexagonal theorem is what just brought it all into perspective for me! I could not figure out why we were subtracting the “AB” and then the visual for the hex came with the “-6T” and it all clicked with the extra triangles! And then dividing the hex by six and getting the trithagorean buttoned it all up like a beautiful winter coat…
In art we look at the negative space of a sculpture and in technical drawing the extended construction lines - to me this is what helps to visualise the maths.
I fear most schools prefer not to promote "visual proofs" since it could have pupils do them when the visual is "close enough" yet still false. Meanwhile, the algebraic ones have "rigid" rules, that can be more easily explained and corrected.
That is because every line you draw has a thickness. While in mathematical reality there is no thickness of a line. The mathematical line would become invisible.
@@BartvandenDonk Not only that. Mostly because visual proofs are easy to become deceptive and non generic. The infinite chocolate bar is an example. Things deceptively similar but different, etc. There are books like The Trissectors, A Budget of Paradoxes (by De Morgan) and Mathematical Cranks filled with examples of false visual proofs of wrong theorems or impossible constructions that mathematicians received in letters or in self published books.
i still think that visual proofs are a lot more educational and make math a lot less cryptic, this definitively should be taught at schools as many students run from math teachings only because they cannot see the algebraic connections as anything more than juggling letters and numbers that have nothing to do with reality
I think that good math students can visualize these proofs in their head with perfect lines. A student might become a good math student from a struggling one by seeing one presented and learning the concept and then trying to visualize more concepts himself.
Prior to completing a calculus course taught via uncannily mundane and l laborious methods, I loved math. Watching this video felt like reuniting with a friend I hadn't seen since I was 15. Thank you!!
I tend to approach your videos with an open mind and willingness to learn and I am never disappointed. Your teaching methods are wonderful and refreshing as ever. Thank you for all your hard work and dedication. Your love and passion for math is unmatched. We’re kindred spirits, but your grasp of mathematics is astounding. The final problem I believe the answer is 1 if my math is correct. Though I have been wrong before. Thank you for such a wonderful video!❤❤
Your videos are perfect, when you think you can raise an objection you address the issue immediately and even when I think know where you are going to 2 or 3 steps in advance you frequently surprise me. I like that.
I'm a mandala artist and I find this fascinating. I've been creating eggs before the chicken... :-) Geometry has always been a passion of mine and I thank my high school teacher Mrs. Locke for making geometry interedting and thank you to you too for expanding my understanding !
Very interesting. I wasn't taught either proof in school. Being an amateur math enthusiast, I actually did manage to figure out the algebraic proof for myself a while ago. But I've never seen the first proof before, very elegant and amazing. Thank you.
For the bugs distance problem: Fix the frame of reference to one bug. The bug moving towards it, in that frame of reference, just travels along the shortest distance towards it, meaning the covered distance is exactly 1
You are really the best educator I have ever encountered and make supreme use of this medium. I have an advanced degree in chemical engineering and have worked with r&d at three Fortune 500 companies, so I have encountered some fantastic people until now, yet you are the best.
Golden rule. I love it. I’m a humble carpenter. I hang doors and put shelves up. I use a very sketchy and basic knowledge that Pythagoras exists to position things. When challenged I mention Pythagoras. I urge anybody to learn more about this. It makes the world more understandable.
At 17:37: an A-A-A equilateral triangle clearly has area A²F and nestles into the corner of the larger A-B-? 60 degree triangle. So notably, the two have the same height. Now scale our A-A-A triangle horizontally (i.e., along the base, not the height) by B/A. This new triangle has area A²FB/A = ABF. It's not the same size as our large triangle, but it does have the same base (A*B/A = B) and the height wasn't changed by our horizontal scaling. Any two triangles with the same base and height have the same area, hence the A-B-? 60-degree triangle has area ABF, QED.
For the riddle at the end: instead of rearranging triangles I came up with algebra and the cartesian coordinate system. You get the corner points of the blue area with solving linear equations like -1/2x+1=2x --> x=2/5 y=4/5 and so on. You get the difference by subracting two of these points from another and then you have data to work with Pythagoras. The width of the square is then sqrt(5)/5 and its area 1/5. ^^ Spectacular video by the way. A++ P.S. after that I came to the conclusion that the areas of the smaller triangles are 1/20 and therefore fit 4 times into the blue square. So all the trapezes together make up 3/5.
A very weird rabbit hole I went down related to the four bugs problem: The function that gives the path length in terms of the step size is f(x) = x/(1 - sqrt(x^2 + (1-x)^2)), and we saw that the limit of f(x) as x -> 0 is 1. What I did was look at the Taylor series of f(x) around x = 0 and what I noticed was the terms up to x^15 all had positive coefficients, but from x^16 on the signs of the coefficients are periodic with period 8 (starting from x^19 the pattern is 4 +'s then 4 -'s). I think I understand why it's periodic with period 8; it has to do with the roots of x^2 + (1-x)^2 being (1+i)/2 and its conjugate, where the angle corresponding to (1+i)/2 in the complex plane is 1/8 of a turn. And I also sort of have a formula for the coefficients: if the Taylor coefficients of 1/sqrt(x^2 + (1-x)^2) are a[n] and the Taylor coefficients of f(x) are b[n], then b[n+1] - b[n] = a[n-1] - a[n] + a[n+1]/2. But the a[n] sequence itself is probably not expressible in closed form so I'm not sure how to get more info on b[n].
I'd never seen either of the two proofs you showed in the beginning of the video. I remember a proof that was way more complicated and hard to understand. These are beautifully simple.
Nice one! I have also done the versions of AM-GM and Cauchy Schwarz like you did here - love this diagram for those visual proofs! Used it to animate Priebe’s and Ramos’ sine of a sum too. Was gonna create a mashup of them to show they are all the same but here you’ve done it better. Thanks!
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I consider myself very intellegent. I do excell at certain things but when I come to the Mathologer channel, I find out how much I truly do not know. I like how you explain things yet I have trouble absorbing it. I am in awe of your intellect but also I am in awe of the people who comment on your videos. That in our dumbed down society there are still people that fully understand what is hidden to most of us mere mortals is just amazing to me. I am drawn to these videos but much like a moth is drawn to a flame. I see and then I crash and burn! I am hoping for a breakthrough moment when I shout Eureka! Keep up the good work and I will keep coming back and try to grasp it all!
36:34 Challenge accepted (Spoils the answer) Consider that the small right angle triangle at the top right is similar to the same triangle but including the trapezium below it. The ratio of sides is 0.5:1=1:2 (by considering the ratio of hypotenuses). Therefore, big triangle area = 4×small triangle area. White area = 16×small triangle area then. Now, add the small triangle at the bottom and you get a right triangle with sides 0.5 and 1, so area = ½×0.5×1=0.25. It also happens that's the area of 5 small triangles. 5∆=0.25 ∆=0.05 16∆=0.8 Hence, area of the square is 1-0.8=0.2 Hope my proof is pretty
My proof: let x be the length of the sides of the blue square. The big square has area 1. The two big triangles on the right and left of the blue square have area 2(½½1)=½. The two small triangles at the top say have legs of length y and z then considering the diaganoal of the big triangle at the top we get z+x+y = √5/2. Hence z+y = √5/2 - x The two trapezoids (trapeziums) below and above the blue square have area 2x(z+y)/2 = x(√5/2-x) = (√5/2)x - x^2 putting this together we get 1 = ½ + x^2 + (√5/2)x - x^2 so x =1/√5 and finally x^2 = 1/5
I'm getting a bit confused with this solution. I followed a similar method to @Prometheus. To summarise there are 16 small triangles that make up the white area and four of those same small triangles which make up the area of the smaller square in question. The area of each small triangle is 0.05 so 4*0.05 = 0.2 = 1/5 = the area of the blue square. So far all is good. However if we then reverse engineer this problem to calculate the sides of the smaller triangle I get a bit confused. The hypotenuse of the smaller triangle is 1/2 (the long side of it spans 1/2 the length of the original square). If I refer to the other two sides of this small triangle as y and z where y is the longer side and z is the shorter side (and z = 1/2y), then the inner square that we were finding the area of should be y*y? So y^2 should be the area of the inside square = 1/5 = 0.2. That means that the shorter side of the smaller triangle z should be 1/2y and its square z^2 should be 1/4y^2 which is I think 1/20 (0.05). But then if we take Pythagoras Theorem that the square of the hypotenuse = sum of the square of the other two sides then y^2 +z^2 should be = 1/2 (because the hypotenuse is half the length of the original square). But 1/5 +1/20 = 1/4 not 1/2. I'm obviously getting something wrong here. I'd appreciate any help in resolving this.
I have never seen these proofs before. I am not the brightest bulb but I admire and appreciate mathematics very much. It is always so satisfying to take a concept and make it easy to understand or to amplify the idea in some way. There is a creativity and sophistication into seeing the relationships between the logic of the pure math and its applications.
@@davidmeijer1645 Very glad to here that you do. How about your colleagues? Would you say that showing these proofs to kids is something that is commonly done in Canada?
@@Mathologer i would say that the reality is..no. Most teachers are concerned with covering the curriculum, which doesn’t direct teachers to explore these expansive variations on a theme, say this Pythagorean Theorem. That’s why your videos are great…for us teachers to recommend (to the interested student). I went through 3B1B lighthouse explanation of the Basel Problem yesterday with a class of 6 Gr. 10s. I really think it was quite a bit above them, but I’ll see when we next meet. And I think that’s why most teachers are reluctant to delve….concerned that it’s a bit above the students. Have you been amongst high school students lately? I mean, most are at the stage of struggling with something like, how to get the angle, when a trig ratio is known….I.E. learning to use the inverse trig function on a calculator. It’s quite amazing actually how developmental stages are quite uniform across geography and time.
At 55 seconds what the diagram is the same way the ancient mathematician of India named bodhayan described the pythagorus theorem many centuries before pythagorus. I am now convinced that pythagorus definitely visited India and learned here.
I don't remember doing anything in primary school involving indices including ^2 for 'squared', we did areas of shapes by counting squares on squared paper, nothing about algebra either. It was all new in secondary school. The first year there (ages 11-12) was basically doing catch up for what we should have/could have all learnt in primary school, but without a 'national curriculum', each junior school had probably taught different things and missed some things out. I think things have probably improved since then with maths teaching, but some (10-12 year olds) still seem to struggle with the basics - because of not learning times tables, or because new maths concepts are not introduced in a simple way, avoiding using complicated language.
I was always terrible at math and great at science and now since I'm 40 years old I'm trying to learn all sorts of things I love your channel you explain things nicely and calmly 😊
Learned both when i was 13 by my teacher. I loved math more than i can remember! Thank you for your content once again! Hope to see an ecology "ecosystem stability" math video from you, they are quite intriguing!
I learned the first one of those proofs first, and the second one before I got out of high school. I had no idea that they aren't taught anymore! 😮 Geometry has always been, as you called it, pretty to me. Thanks for this!
I had seen the twisted squares in a textbook before, but I didn't connect them to the proof of the theorem until I received a book specifically on the Pythagorean Theorem called Hidden Harmonies. It had a full chapter of different proofs, half of which led back to the twisted squares. I didn't quite understand the rest of the book at the time, but it really is quite fun to see the history of humans proving the same thing dozens of different ways.
for the first part of the video. I am not familiar with either proof. I am 67 years old, BUT I am pretty sure none of my teachers even hinted at such a visual proof. WOW so nice to see that the ancients could have actually figured this out without knowing any trig function values. I've often wondered how the ancients figured PI to even significant values (3.1416) before having calculus and only having simple geometry. Thanks very much! The demonstration that a person knows his field is that they can dumb it down or explain it to a high school or even middle school person.
Well, I'd say to show one of the proofs that I show at the beginning should not take any longer than in the video, i.e. 30 seconds. And as far as I am concerned there is absolutely no excuse not to do this in school :)
Love all your videos and this is a first ever post; as a small token of appreciation for all your stuff, here’s a link to a beautiful proof for Pythagoras by Jacob Bronowski - first aired in 1973 on UK television. Since you tube apparently doesn’t allow links, it can be found on DailyMotion with a search for “the ascent of man. Episode 05. Music of the Spheres” - from 7:00. The question is does this mean maths is an alien language?
5:18 is the first proof I ever saw (or that I remember). For some reason I think it was featured in the “Cosmos” book but I have it in storage and I can’t check right now; I’m probably misremembering.
@@Mathologer proof #1 at 2:00 is perhaps the nicest of the Pythagoras proof presented. But overall the sin(a+b) proof at 30:30 is the one that really made me applaud with giddiness, what a delightful approach!
My favourite Pythagorean triple is 20, 21, 29 because it so close to an isosceles triangle and thus if you actually make it physically out of wood or meccano with inherent measurement errors, it will be more accurately a right angle than say 5,12,13 would be, subject to the same measurement errors.
@@colinpountney333 (5, 5, 7) would appear to be ever so slightly better engineering-wise than (7, 7, 10), but (12, 12, 17), (17, 17, 24), and 29, 29, 41) are seemingly progressively ever better!
17:31 First stretch one side of the F triangle to A. The altitude perpendicular to A stays the same so the area is now AF. The same argument applies when stretching another side to length B. So the area is ABF. 36:17 The area of the small square is big square - 4 * a different right triangle, one with a hypotenuse flat against the big square. You can get to that triangle by cutting the altitude of the main right triangle with its hypotenuse flat on the bottom. Because the angles remain the same within he cut triangles, the ratio between their sides is still 2:1, and so it's easy to see the two bases that wind up making the hypotenuse of the big triangle are in a ratio 4:1. The hypotenuse of the original triangle is sqrt(5)/2. The base of the new triangle is 4/5 * sqrt(5)/2, and the height of the new triangle is 2/5 * sqrt(5)/2. So the area of the new triangle is 1/5. Now we are ready to go back to our original question: what is the area of the small square? It's the area of the big square minus 4 * the area of the new triangle. This is 1 - 4/5 = 1/5. So the area of the small square is 1/5.
@@Mathologer American schools are focused on the "utility." They teach you what works but not how and it's really sad. I didn't understand how integrals literally functioned until I was in college despite taking 3 years of calculus in high school
Professors Polster and Ross, I express my sincere appreciation for what you provide the world of mathematic enthusiasts at large. Your content and delivery never disappoint, thank you. While not pertinent to your specific topic, the subject matter in general is of particular interest. I would be remiss not pointing out one of my favorite aspects of the Pythagorean (et al.) Theorem. By careful geometric construction, three dissimilar non-right triangles can become the lateral faces of tetrahedra with right triangle base. Each of the three areas are correspondingly Pythagorean compliant. Thus, an oblique triangular pyramid or harmonious Pythagorean tetrahedron. Unfolded along the edges of the right triangle base, the “net” is quite counter intuitive to that of the conventional similar shape example you depict. Perhaps this of passing interest to some. Again, thanks for your amazing contributions and my enjoyment.
That's interesting. Rings a bell but a very faint one. Anyway, just read up on it and also stumbled across this write-up by you www.reddit.com/r/Geometry/comments/sio7tt/help_with_the_harmonious_pythagorean_tetrahedra/
The second one with stating how the big square area is simply equal to its components (small square and four triangles) is my favourite proof, it’s just stating the obvious, doing some simplification and you’re left with a²+b² = c²
Very beautiful and simple proofs of Pythagoras. Thank you. Now three easy challenges for you, dear mathologer: a) prove the horizontal line of sight of an observer looking across the sea from a beach makes a right angle with the earth's radius. b) prove that a 2,000 foot mountain 60 miles across the sea cannot be seen by the observer if earth is 24,901 miles in circumference. c) find a large body of water with nearby mountains and see them from a beach on the other side and prove by contradiction earth is not a ball 24,901 miles round. Good luck!
Wow... It's very, very amazing. Haven't been learn this since in the high school or university. There would be so many things can be covered with the rectangle and triangle. It's a unique things.
3:41 1046-256 BCE Look into Anatoly Fomenko's Empirico-Statistical Analyses regarding Isaac Newton's and other chronologists' timelines. It's more plausible that Chinese Imperialists added fifty-nine years to Li Qingyun's life, thus extending their chronology and thereby the longevity of the empire's mandate. He claimed to be 197 when he died. They said he was 256 and that he lied about his ago pretending to be younger. He was a mountain herbalist. The imperialists were autocrats, totalitarians. Fomenko's New Chronology holds that this type of invented history is more common than not, that Pharaoh ruled Egypt through the 1700s, that 2022 CE (7530 Slavic Aryan Vedic Calendar) is 869 AD, and that The Great Wall of China is still being built. Or maybe you had a typo, and it was 146-256 BCE? I didn't look it up yet, I was just reminded of all the above. Thank you for excellent reference!
1046-256 BCE People just kept adding to it over a long period of time. I had heard of Fomenko's New Chronology. Interesting but in the end a crackpot theory I'd say :( Having said that I am a fan of his mathematics and his maths illustrations :)
I absolutely agree with you, it is a crime that schools don't introduce proofs. Even the most simplistic proofs, like the ones you showed. The only time they are put into my curriculum is if you take the hardest math course, which only 5% of the cohort takes. 95% of students aren't even introduced to proofs.
@@Mathologer If pupils just learn ways to shove numbers around there is nothing that elevates thier mathematical abilities above those of a computer, which will always be faster than them. What they miss out on is learning to strategically and creatively think about numbers and the world those might describe.
@@jannikheidemann3805 Exactly. Following a given algorithm is calculation, not mathematics, and reducing maths to caulculation defies the very purpose of mathematics, which is essentially an ever-expanding toolkit for solving problems.
Based on your T-shirt the following proof of the Pythagorean theorem came to my mind: Consider a right angle triangle with an hypotenuse 1, and call its area m. Scale it by a factor of c. Then the area of the new triangle is E = m*c². Split it into two triangles by its altitude (which is not any of its sides). These triangles will be right angle as well, and their hypotenuse-which length I will call a, and b-are the short sides of the triangle with area E. If you check their angles you can see that these two new triangle are also similar to the original triangle, so we can get them by scaling it by a factor of a and b, respectively, and therefore their areas are F = m*a, and G = m*b². Finally, consider the fact that E = F + G, substitute: m*c² = m*a² + m*b², and divide by m (what is strictly positive, because it was an area of a triangle). This proof was told me by one of my maths teachers as a handy proof for matura exam, and he referred it as "Einstein's proof". As I typed it in, I recalled that probably you had a short animation on this very same idea, maybe with some other notation. (Indeed: ua-cam.com/video/r4gOlttnJ_E/v-deo.html But the choosing of the letters like this was my idea :) Even though I know, probably someone else has done it long before me…) About the question at 3:00, both proofs were shown to me in high school, and I don't really remember which came first. But I remember that I was told to always emphasize that the quadrilateral in the middle is indeed a square because of the sum of the angles of a triangle is π.
I don't remember which proof I saw first but a few weeks ago I wanted to show a proof to one of my kids and realized I forgot all the proofs I had learned in school. The second (algebraic rearrangement) was the one I worked out first.
36:20 we can start by recognizing that a right triangle with legs of 1 and 1/2 is a 30-60-90 triangle. We can then see that the overlaps are similar triangles with the triangles that overlap to form them. They share one angle (30 degrees) and they have a right angle, as given by the problem stating the middle, shaded figure is a square. Truncation by a line parallel to one of the legs produces a similar triangle. So all three sizes of triangles are demonstrably similar, ad 30-60-90 cases at that. Since we know that the side length of the large triangles is 1 and 1/2, we can say that each of those hypotenuses bisects the hypotenuse of the medium triangles. From this we can conclude that a line that bisects hypotenuse of the mid-sized right triangle must do so perpendicularly to the long leg of the mid-sized triangle. From this we can conclude that since the small triangle's long leg bisects the long leg of the mid-sized triangle, the long leg of the small triangle must be equal to the side length of the shaded square. Knowing that the small triangles are 30-60-90 and that the smaller leg of the small triangle is one half the length of the long leg of the small triangle, we can prove that the small triangle fits into the space left by the right-angled trapezium by constructing a perpendicular bisector of the trapezium's longer parallel base. This bisector will prove to be parallel to the right leg of the trapezium and intersects the endpoint of the other base of the trapezium. This point is also the midpoint of the side of the larger square as demonstrated before. We can then demonstrate by corresponding angles and similar side lengths that the newly constructed small triangle is identical to the existing small triangle. The rectangle resulting from the perpendicular bisector can be bisected with a line parallel to the large square, that intersects the perpendicular bisector of the trapezium base and also intersects the corner of the shaded square. Thus we can demonstrate that the trapezium is composed of three triangles which are identical to the smaller triangle provided by the puzzle. We can also prove that the small triangle completes the trapezium into a square by demonstrating that it has complementary angles and both side lengths that would complete the square's sides. By knowing that this hypothetical square is indeed a square and by showing that it shares a side with the shaded square, we can prove that the shaded square has the same area as the hypothetical square. If we subtract from the overall figure the area of one smaller triangle and add to it that same area to complete the trapezium into a square, and do this on all four squares, we will have not changed the overall area, but will have demonstrated that the overall shape has the same area as the sum of five similar squares, one of which is the shaded square shown. Thus, without calculating the side length of the shaded square, we can prove that the shaded square's area is equal to one-fifth of the overall figure. And we can also rigorously prove that flipping the small triangles over is a valid way to demonstrate the shaded figure's proportional area.
Incidentally, we can conveniently compute that the side length of the shaded square is the square root of one-fifth. Or 1/sqrt(5). Or sqrt(5)/5. This also means that the leg lengths of the small triangle are 1/2sqrt(5) and 1/sqrt(5). Which also demonstrates that the hypotenuse of the large triangle is 1/sqrt(5)+1/sqrt(5)+1/2sqrt(5) or (sqrt(5))/2.... I think I did something wrong. Luckily, what I did wrong is not unfixable by changing the 30-60-90 to just the triangle initially given, which I misidentified as a 30-60-90. The exact angle of the triangle never really comes into play in the proof, only the demonstrably similar angles, which can stand on the side-angle-side identification rule and the ratios present.
So, according to Einstein, e = mc^2, so that e/m = c^2. But according to Pythagoras, a^2 + b^2 = c^2. Therefore, e/m = a^2 + b^2, so that e = m(a^2 + b^2). 😜🤪😝😛
I mean even if their areas are incalculable you can just do the Archimedes thing and bound the area between a polygon inside the circle and another outside? (Admittedly Archimedes was trying to work out the exact length of the circumference but I'd say my point still stands)
In my school (in Poland) the teacher made the difference stand out between a circle (pol. okrąg) and a disc (pol. koło). Circle is not "filled", and the disc is (holds up in English as well) because of the following definitions: 1. an open ball is a set of all points which distance from a centre P is smaller than the radius r, or in other words {x : d(P, x) < r} 2. a closed ball is a set of all points which distance from a centre P is smaller or equal to the radius r, or in other words {x : d(P, x) ≤ r} 3. a sphere is a set of all points which distance from a centre P is equal to the radius r, or in other words {x : d(P, x) = r} where d is a metric (simply: the distance between two points, doesn't have to be the euclidean one). Circle is, strictly speaking, a 2-dimensional sphere, and a disc (strictly: closed disc) is a 2-dimensional closed ball. What this means is that while a circle technically only has the circumference (because it only consists of points on the circumference), the disk also has a surface. For circumference we can calculate it's length (e.g. the length of the circumference of a circle with a radius r is 2πr) and for the surface we can calculate it's area (e.g. the surface area of a disc with radius r is πr²). Technically, neither a circle nor a disc have an area. A disc has a surface, which has an area, so a disc has a surface area. In less strict language we usually just say "area" instead of "surface area" and also often say "circle" when we mean a "disc". When someone says "area of the circle" we usually assume they mean "surface area of the region bounded by the circle". The important thing is that everything stays unambiguous, so that everybody understands what we mean. Hope I helped ¯\_(ツ)_/¯
Final Puzzle at 36:01, Area of the inside square is a²(1-(ab/(a²+b²))) , where "a" is side of the bigger square, & "b" is the segment cut. You can solve it using similarity of triangles, and ratios of sides.
For the last puzzle, the small square's area is 1/5. The big square is made of 4 triangles, 4 trapeziums, and the small square. If you fit the triangles to the trapeziums, you get 4 squares sharing a side with the small square, therefore they all have the same area. The large square with an area of 1 divides into 5 equally sized small squares, so each small square's area is 1/5 I can see a couple other ways to solve it, and there's probably more that I haven't thought of. I noticed the small triangles are similar to the large triangles, that could be useful.
That is a clever way to solve it (and probably one Burkard would be thrilled to animate)! I can confirm though that the small triangles and the original triangles are similar (and that the fact is useful) as that is how I reached the same result of 1/5 for the area of the small square.
I grew frustrated quickly with trying to calculate each line segments length and just did that but, as noted earlier, the only way to check that is to precisely cut out a trapezium and smal triangle and see if they perfectly overlap the small square, whereas the proof is fool "proof." This is why I and so many other people have trouble with plane geometry; such is logic. Prosecutor: I can't believe they acquitted someone so OBVIOUSLY guilty! Judge: Of course he's guilty and of course that's obvious--but you failed to PROVE it, counselor.
@@herbpowell343Sorry, I don't understand what you're trying to say? There's no need to calculate any numbers or do any perfect cutting, just the given values are easily enough to prove it. It's given that the edges of the large square are each length 1, and the outside edge of each triangle is 1/2. This means the outside edge of each trapezium is also 1/2. If you match each triangle to each trapezium along that edge, forming a quadrilateral, it must be a rectangle since all the corners are right angles (two given in the trapezium, a third from the angles that form each corner of the large square, and if three are equal then the last must be as well), and the quadrilateral's length and width are both equal to the height of the trapezium, which is equal to the length of each side of the central square. Due to having all right angles and equal sides, each quadrilateral must be a square, and since they all share sides with the central square, all of the squares' areas are equal. Since the total area is 1 and is evenly split into five, the area of each little square must be 1/5. I do like that quote at the end. Is it an analogy you came up with, or a specific reference to something? Either way, it certainly applies to maths in general quite well.
I learned the theorem at school but not the proof. At uni it was assumed that everyone knew the proof so it wasn’t proven to me there either. I commented to a fellow student that I’d never seen the proof, and he explained your second proof to me. Today at 56 is the first time I saw the first proof!
I've been watching your videos for years, THIS ONE! was the proof I was asked to solve in my grade 12 academic maths course. You brought me to a new realization that I hadn't seen before. My teacher was kind enough to give me a passing grade but I missed the mark. Now I want to make a learning tool that will physically describe the geometric relationship you've described. Thank you for your continued insight.
36:17 1) In a 3x3 square grid, where one single grid square has side A, draw a twisted square with vertices at 1/3 of the 3A side length square 2) take a moment to realize that the twisted square drawing corresponds to the puzzle presented 3) the blue area is therefore A², but given the problem data (2A)² + A² = 1. So A² = 1/5.
or by arranging the four smallest triangles next to the trapezoids to complete the squares, you will end up with a cross made of 5 squares with side of length A. Thus 5A^2=1.
For the third square: Since a is 1/2 and b is 1, the base of the parallelogram that's created inside each triangle by the overlap must be 1 - 1/2 = 1/2. Since the hypothenuses of the triangles are parallel this means, the sides of the inner square must also be of length 1/2, so the square has an area of 1/4. PS: great videos! Lot's of things I didn't know before in this one as well. Always fun to learn, especially when the person teaching is as excited about the information as you are!
Oooh, i see my error now. Wow my head made a lot of sense of this... After reading other comments and looking at the diagram again, I now also think that the area of the small square is 1/5. (For anyone whose head had the same idea: the "base" of the parallelograms isn't parallel to the sides of the inner square, so they can't be of equal length)
learned the second proof in school, the first in an earlier Mathologer video. Since then I watch all your videos and bought several of your books. I would recommend your videos to everyone who has the slightest interest in maths.
First I would like to thank you for an excellent presentation - I vaguely recall being taught the geometric proof 55 years ago at age 11. Second, I became a great fan of Martin Gardiner's puzzle books - very entertaining Third, the problem you set to find the area of the square - my solution is as follows: Area of the four large triangles equals the area of the large square, so the small square equals the ovelap - i.e. the area of the four small triangles. Hypotenuse of large triangle equals (sqrt 5)/2 Since small and large triangles are similar and hypotenuse of small triangle is 1/2, the remaining sides will be 1/sqrt5 and 1/(2*sqrt5) So, area of four small triangles = 4*(1/sqrt5 * 1/(2*sqrt5) * 1/2) = 1/5 As a check, side of small square = Hypotenuse of large triangle minus smaller sides of small triangle = (sqrt 5)/2 - 1/sqrt5 - 1/(2*sqrt5) = 5/(2*sqrt5) - 2/(2*sqrt5) - 1/(2*sqrt5) = 1/sqrt5, so area of small square = (1/sqrt5)^2 = 1/5
The solution to the final puzzle is 1/5. Perhaps the most visually pleasing way is to rearrange the 4 small triangles so they each fit onto one of the trapeziums to form 4 identical squares. We know that these are also identical to the middle square with unknown area because the ratio of original lengths create similar triangles. The result is the large square of area 1 is separated into 5 smaller squares of equal area, one of which is the middle square.
@0:27 your explanations helped me to better understand what the Uni teachers were trying to say. Some times I have the “Oh that’s what they were talking about and I couldn’t get it”…
I'm a math teacher in Germany - I teach both of the proofs shown in the intro in my classes. The first proof is usually discovered by the students themselves - i bring cardboard triangles to class and let the students re-arrange those themselves until they discover the proof. Can be done even by students who are usually not that strong regarding mathematics.
Y equals r cubed over three. And if you determined the rate of change in this curve correctly, I think you will be pleasantly surprised. Don't you get it Bart? RDRR Kidding. That was awesome, and I strongly wish this channel existed when I was learning Math. I love looking at things I know well in ways I never considered before, continually making me wonder if I actually knew much of anything in that class.
Graduated HS in 1977. We had pretty good education system. Grew up in a small town on Texas coast with one HS Went up to Trig. We had a math teacher who could teach in any college. Great man But, Never seen this. But I understannd.
35:30 this is not formal by any means, but intuitively I would say it is because the bugs are travelling perpendicularily at every instant. I will try to formalize it using calculus and vectors: let b1=(x1,y1) be the first bug, while b2 be the second bug. Let R be a clockwise rotation of 45 deg about the center of the square. WLOG we say that point is the origin, meaning that R can be expressed as a matrix. It is clear that b2=RR*b1=(y1,-x1)=(x2,y2) let d(t) de the difference between b2 and b1: d(t):=b2(t)-b1(t)= (RR-I)b1(t) Computing the matrix of RR-I=[[-1 1],[-1 -1]] we can see that this is a cw rotation by 135 degs, and an enlargment by sqrt(2) => RR-I=sqrt(2)*RRR let D(t) be the size of d(t): D(t):=|d(t)|=|sqrt(2) RRR b'|=sqrt(2)* |b1| because rotations don't affect length size. We also have D(t)=sqrt(d.d). We will prove D(t)'=1: D(t)'=sqrt(d.d)'=1/2*1/sqrt(d.d)* (d.d)'=(d.d')/2*D(t) let dT be the transpose of d (d.d)'=(dT * d)'= dT'*d+dT*d'=(d'.d)+(d.d')=2(d.d')=2 * Transpose(sqrt(2)*RRRb1') * (sqrt(2)*RRRb1) = 4 * Transpose(RRRb1') * (RRRb1) = 4 *transpose(b1') * R^(-3)*R^3*b1= 4 (b1.b1') Since b1 moves in the direction of d, then the angle(b1,b1')=angle(b1,d)=135 degrees because d=sqrt(2)*RRR*b1. Since the speed is 1 we have |b1'|=1 Solving for (d.d)' we get (d.d)'=2*d.d'=4 (b1.b1')=4 |b1| cos(135) Pugging back in we have D'(t) = 4*|b1| cos(135)/2*D(t) =4*|b1| cos(135)/2/sqrt(2)/ |b1| = 1 D(0) = sqrt(2)*|b1| => T = D(0)/v=sqrt(2)*|b1| which is equal to the time it takes to reach on point of the square. This was much harder than I expected. Probably you can get better proofs using drawings and geometry. Not only is my proof long but it requires a lor of familiarity with linear algebra and some non trivial properties
I showed my maths teacher the diagram of a square and another square with width being the diagonal of the first. Therefore, half the area of the first square was a quarter of the other. This meant the diagonal is sqrt(2)xside of the square. I was ten and it was in France, and because it was my fourth country, I was always learning the language of my new country. However I spent my younger years loving maths and walking round the playground (clockwise). And am proud of that proof with a stick in the Southern French dust.| However I was told I would be taught it next year. So glad to watch this video where no-one delays the learning
I have never seen any of these proofs - I graduated HS in '56 - got a BSEE in '60 - worked in Aerospace 38 years - retired and became a HS math teacher in 2001 - retired in 2014. Never offered any of these proofs to my students. I think they (and I) were shortchanged.
Love this stuff.
I really have no idea how US schools still manage to produce some mathematically literate people.
Once one of my colleagues brought his 1st or 2nd grade son homework apparently designed to teach subtraction. 3 engineers sat puzzling over (1 UK-educated, 1 Chinese-educated and 1 Polish-educated) until 4th came who's kids were past basic schools and solved it for us.
I knew both of Pytagoras proofs before I was taught them.
From my POV US math teaching is bizarre.
genuine questions, in 14 years of teaching you never thought of going beyond the curriculum you were teaching? like, no curiosity at all?🤔🥺
I have indeed been taught this at school. Not in the math class, but during Latin class. Our Latin teacher also knew ancient Greek and had found the Greek text about this interesting enough to teach us about it. I found it very enlightening.
Do they even have Latin class in the country most viewers here come from?
@@jannikheidemann3805 I don't know. I'm from Germany. I grew up in the 90s an 00s.
@@jannikheidemann3805 There are Latin language classes available in some universities in the United States. But Latin Math class is not in the states!
@@jannikheidemann3805 If you mean the United States, then yes, they teach Latin in many high schools, it's gaining in popularity, many Catholics study it independently to deepen their study of the faith, and many more people study the Latin roots of English in preparation for standardized tests like the SAT and GRE.
My school in Britain taught Latin. I think Latin was an essential requirement if you wanted to be accepted as a student in Oxford and Cambridge Universities. A little indicator of a well educated candidate. Not any more of course!
Every month when Mathologer comes out of his hiding, you know it's going to be a great video
Absolutely! Masterpiece, one after the other!!!
and a cool T-Shirt :)
No, I do not. I expect it to be good, but unless I have watched the entire thing I can not evaluate the quality of the video. There is an important difference between 'expect' and 'know', and that is the same difference as between science and faith, for example.
If it is every month I think i missed some videos, going back to watch them
@@SuperJuiceman11 yes true I noticed that too lol
The first proof is by far the best I’ve ever seen, and it was a big lightbulb moment for me. I’ve seen UA-cam videos of teachers doing it with felt triangles taped on the blackboard, but I never got any lessons like that as a kid.
I will be sending my local elementary school a bill as suggested.
The animation in this video is just insane - about the best I've ever seen on UA-cam in fact. I can't imagine all the time it must have required to get all that to flow so smoothly. Well done.
I especially liked the part about the bugs, shows you how to solve such a problem in visual steps - a great problem solving technique. And the animation was just beautiful.
I too enjoyed the animations. The 3Blue1Brown youtube channel animations are also quite impressive. It’s math-oriented, in case you don’t know of it.
Burkard, rather than heap additional praises regarding the dependable brilliance of your channel's content, I thought I might remark on your delivery: Your pacing, pitch, tone and inflections are such that, rather than lulling my mind to sleep in class, you have a special way of always educing and coaxing my brain forward (stubborn mule that it is).
Wow, thanks!
For the bug problem: An equivalent question would be to ask how long it takes for any one ant to reach the ant it is chasing. The chased ant always moves orthogonally to the chaser, so it is neither going towards it nor away from it. Thus the distance between the two ants simply decreases by the speed of the chaser. It thus takes 1 unit of time for the ants to meet.
Wow. Such a simple way of looking at it, that the first reaction is to see if something is missing. But it is not a paradox... thanks.
For added “fun” this shows that the extra path length for the finite cases is the sum of the non-orthogonal movement introduced
other way to look: change the reference system to one of the ants. Now the world is rotating, and the ant goes straight ahead 1 unit.
I guess that kind of depends on, what you mean, by ”moving away from or toward the chasing bug”. If you mean that the Euclidian distance between the chased bug and the stationary chasing bug has changed (increased for ”away from”, and decreased for ”toward”), by the end of 1 segment of the chased bug’s path; then, I think it’s pretty clear that moving orthogonally to the chasing bug, means moving away from it; since the distance between a square’s center and 1 of its corners is greater, than between its center and the midpoint of 1 of its edges. 🤔
At 28:42, your generalization is wrong- it should be that cbrt(ABC)
Ah that bug explanation is lovely! Wish I’d given it more thought myself first haha.
_"your generalization is wrong"_ - he does not give a generalisation, so it cannot be wrong either.
Where have you seen that C in the video? The generalization shown is sqrt (AB)
@@raphaellfms The video’s been edited since he put it out- there was an incorrect generalization there. If you look closely, you can see- usually, he fades in and out, but a couple of seconds earlier, he cuts out instead.
@@renedekker9806 See above.
I mean I cheated or crammed my way through high school I was too bored to try, so this is the first time I’m seeing these design elements explained from a mathematical standpoint which helps me understand them deeper:
It’s amazing how intertwined art, beauty, mathematics, numbers, algorithms, patterns etc.
Truly all connected. I am all that is, because all that is - is within me. 🙏🏻
Thanks for another fantastic video! I really liked the first proof in the introduction because during school I only learned the second algebraic proof. It was neat how you could prove identities about sin, AM-GM, etc using these diagrams!
Thank you very much :)
Mesmerizing. Reminds me of being a kid in early years of learning math and just drawing and playing with shapes. Wish I had this kind of instruction back then.
Same!
i wonder the same thing. those students. in ancient greece, didn't have calculators. don't know if they had compases, either. but they solved a lot, calculated Pi, and figured calculus...
I first became familiar with Pythagoras theorem while surveying field sizes with my father ( a crop farmer in lower Michigan, USA). I was privileged as a seven-year-old boy to be the hold-down boy for the end of the rod-chain. I would look Dad square in the eyes as he circled round me, ending with kneeling with the rod stretched tightly across the ground, and Dad pounding a peg in at my next point. We measured many fields, but Dad had five sons, so the privilege got spread around. I always loved this time, because it was both intimate with my father, and mathematically practical--my teachers were surprised in grade school how well I did with geometry, almost all aspects which I experienced with practical applications on the farm with my father. He's now 96, and we still write to one another often, though now it is his love of philately that drives the mail truck!
This is one of the many things I enjoy about traditional geometric patchwork design quilt patterns , (like Flying Geese, Irish Chain, Feathered Star, Grandmother's Flower Basket, Disappearing 9-patch,
Bear's Paw, Log Cabin and all it's variations, Pinwheels, etc.) with 1/2 square or equilateral triangles, rectangles, squares, etc. and shuffling them around to create different patterns. Now I have a name for them, and may design quilt patterns with the names of Pythagoras or Trithagoros, or the long ago Chinese mathematician's name! Great visual explorations of these relationships, thank you. (of course I enjoy even more, the random sewing together of odd shaped and sized pieces, called, "crumb quilting"! lol)
(see also, the 'golden ratio' seen so much in nature and used from antiquity in architecture to produce unconsciously, naturally pleasing, building facades.)
You are a brilliant mind. Leaving me smiling from what you teach. I'm literally talking to myself in full conversation when I watch your videos. Thank you for the intelligent video and teaching.
I learned the derivation of the Pythagorean Theorem when I took geometry. We were on the topic of similar triangles and found that a right triangle could be made up of two smaller right triangles. It was very straight forward and consistent with the topic we were on.
Your laugh absolutely makes my day! The sound of the joy of learning and knowledge.
I love these visual proofs. The animation of the visual proof at 22:40 is extra fun because two of the triangles are re-scaled. But they are re-scaled instead of having a more complicated animation. Well done!
Yes, and it happens five times from 22:17. @Mathologer has the left-most and right-most triangles swapping sizes instead of swapping places. IMHO it detracts from the quality of this video, because the visual proof depends on the pieces only rotating and translating - not expanding.
The next one at 22:40 really confused me when i tried to work it out because the side length of the blue inside square and the side length of the blue isosceles triangle are not actually the same in general.
They do happen to be the same for the 3-4-5 triangle though
@@EricSeverson Did you mean 23:40?
The area of the central square in the overlapping twisted square is 1/5 the area of the original square. You can prove this physically as each of the four original triangles can be broken up into five copies of the smaller overlap triangles. As the overlap and original triangles are similar, the ratios of their short to long legs are both 2:1, so the central square is equal to tour of the overlsp triangles. Adding up the total number of triangles, you get 5 overlap triwngles times 4 original triangles, minus the 4 overlapping triangles at the corners, plus the 4 in the center, to get 20. 4/20 = 1/5.
Mathematically, we can use Pythagoras and similar squares to determine the area.
a² + b² = c²
(1/2)² + 1² = c²
c² = 1/4 + 1 = 5/4
c = √(5/4) = √5/2
Now we can determine the dimensions of the smaller overlap triangles via similar triangles, since we fan see that c = √5/2 and c' = a = 1/2.
c/a = c'/a'
(√5/2)/(1/2) = (1/2)/a'
(√5/2)a' = (1/2)(1/2)
a' = (1/4)(2/√5) = 1/2√5
b'/a' = b/a
b'/(1/2√5) = 1/(1/2)
b' = 2(1/2√5) = 1/√5
By observation, the central square has side length of b', so:
A = b'² = (1/√5)² = 1/5
An alternate solution to the bugs problem: Let each of the bugs have velocity 1 unit/s. Split each velocity vector into a vector that points towards the center of the square and one that points 90 degrees to that one. Using the 45-45-90 triangle created with the original velocity vector and two component vectors, we find that the two new vectors have magnitude sqrt(2)/2. Thus the bugs are going towards the center at a rate of sqrt(2)/2 units/s. The center is sqrt(2)/2 units away so it will take the bugs one second to get to the center. Distance = rate*time so distance = 1 unit/s * 1 s = 1 unit.
I think it's even simpler than that. Each bug always moves perpendicular to the motion of its "target" bug, so the target's motion doesn't affect the rate of closure.
Therefore, each bug must travel exactly one unit to reach its target, at the center.
Fred
@@ffggddss Wow nice argument!
@@gammaknife167 Thanks. But I think I may have simply remembered it from the Mathematical Games column. Not sure. After digesting enough beautiful arguments, they tend to become ingrained in your thinking, without your knowledge of their origin.
But whoever came up with it, it is indeed a nice insight.
@@ffggddss I really like that argument. I originally encountered the problem when the bugs were in an equilateral triangle, so I not think that the argument applies to other regular polygons. A fun exercise is to use the method mentioned above to prove that the distance traveled by a bug starting in a regular n-gon configuration with side length 1 is 0.5*sec^2(90(n-2)/n). Another interesting thing to look into is other arbitrary starting configurations of the bugs.
@@manicmaths2861 I put some links to interesting write-ups in the description of this video :)
In my retirement I have been working occasionally as a substitute public school teacher. (Math, Science, Spanish, only subjects I know well). I like to show these proofs to the students and jokingly quote your "get your money back" comment.
One day I was talking with a math faculty member and told him of your channel and some of the things I like to share with the students to demonstrate that math is fun. He had never seen a proof of the Pythagorean Theorem , and in fact had believed it was an axiom.
So thanks for your channel, you do more good than you know.
35:45 Because the motion is smooth, we consider the local movement of one bug (being chased) relative to another (which I will refer to in the first person - I am the bug that chases), and we see that it's linear. Further, in this case we see that the movement is perpendicular (movement being it's change in position all happening relative to me) - this is because I am always facing it, and more importantly, the angle between me, it, and the bug it's chasing (the direction of it's movement) starts off at 90° and never changes - THIS is due to symmetry.
It follows that (locally) relative to me, it's moving in a circle around me (and also I'm rotating in place to face it) - never getting either closer or farther from me. I'm the only one that has any affect on the distance between us, so the final distance covered is the initial distance I needed to cover.
That's it :)
The hexagonal theorem is what just brought it all into perspective for me! I could not figure out why we were subtracting the “AB” and then the visual for the hex came with the “-6T” and it all clicked with the extra triangles! And then dividing the hex by six and getting the trithagorean buttoned it all up like a beautiful winter coat…
Yes, that one just clicked beautifully. Also one of the pleasures of what I do when I see something like this for the first time :)
Like a beautiful winter coat.
In art we look at the negative space of a sculpture and in technical drawing the extended construction lines - to me this is what helps to visualise the maths.
I fear most schools prefer not to promote "visual proofs" since it could have pupils do them when the visual is "close enough" yet still false. Meanwhile, the algebraic ones have "rigid" rules, that can be more easily explained and corrected.
That is because every line you draw has a thickness. While in mathematical reality there is no thickness of a line. The mathematical line would become invisible.
@@BartvandenDonk Not only that. Mostly because visual proofs are easy to become deceptive and non generic. The infinite chocolate bar is an example. Things deceptively similar but different, etc. There are books like The Trissectors, A Budget of Paradoxes (by De Morgan) and Mathematical Cranks filled with examples of false visual proofs of wrong theorems or impossible constructions that mathematicians received in letters or in self published books.
i still think that visual proofs are a lot more educational and make math a lot less cryptic, this definitively should be taught at schools as many students run from math teachings only because they cannot see the algebraic connections as anything more than juggling letters and numbers that have nothing to do with reality
I think that good math students can visualize these proofs in their head with perfect lines. A student might become a good math student from a struggling one by seeing one presented and learning the concept and then trying to visualize more concepts himself.
Prior to completing a calculus course taught via uncannily mundane and l laborious methods, I loved math. Watching this video felt like reuniting with a friend I hadn't seen since I was 15. Thank you!!
That's great! Mission accomplished :)
I tend to approach your videos with an open mind and willingness to learn and I am never disappointed. Your teaching methods are wonderful and refreshing as ever. Thank you for all your hard work and dedication. Your love and passion for math is unmatched. We’re kindred spirits, but your grasp of mathematics is astounding. The final problem I believe the answer is 1 if my math is correct. Though I have been wrong before. Thank you for such a wonderful video!❤❤
That's great. Also have a look at this tinyurl.com/3eebxn2k tinyurl.com/
Your videos are perfect, when you think you can raise an objection you address the issue immediately and even when I think know where you are going to 2 or 3 steps in advance you frequently surprise me. I like that.
Glad that this video worked so well for you :)
I'm a mandala artist and I find this fascinating. I've been creating eggs before the chicken... :-) Geometry has always been a passion of mine and I thank my high school teacher Mrs. Locke for making geometry interedting and thank you to you too for expanding my understanding !
That's great :)
From here we could go into the Mandelbrot set via the golden ratio... Damn I wish I had a working laptop.
Very interesting. I wasn't taught either proof in school. Being an amateur math enthusiast, I actually did manage to figure out the algebraic proof for myself a while ago. But I've never seen the first proof before, very elegant and amazing. Thank you.
For the bugs distance problem: Fix the frame of reference to one bug. The bug moving towards it, in that frame of reference, just travels along the shortest distance towards it, meaning the covered distance is exactly 1
To complete the proof you need to show that the second bug is moving orthogonal to the first (which follows from the symmetry of the setup)
You are really the best educator I have ever encountered and make supreme use of this medium. I have an advanced degree in chemical engineering and have worked with r&d at three Fortune 500 companies, so I have encountered some fantastic people until now, yet you are the best.
Glad you think so :)
It’s been years since I’ve seen these proofs but I was taught the second one first. I love the visual explanations. Great video.
Golden rule. I love it. I’m a humble carpenter. I hang doors and put shelves up. I use a very sketchy and basic knowledge that Pythagoras exists to position things.
When challenged I mention Pythagoras. I urge anybody to learn more about this. It makes the world more understandable.
At 17:37: an A-A-A equilateral triangle clearly has area A²F and nestles into the corner of the larger A-B-? 60 degree triangle. So notably, the two have the same height. Now scale our A-A-A triangle horizontally (i.e., along the base, not the height) by B/A. This new triangle has area A²FB/A = ABF. It's not the same size as our large triangle, but it does have the same base (A*B/A = B) and the height wasn't changed by our horizontal scaling. Any two triangles with the same base and height have the same area, hence the A-B-? 60-degree triangle has area ABF, QED.
For the riddle at the end: instead of rearranging triangles I came up with algebra and the cartesian coordinate system. You get the corner points of the blue area with solving linear equations like -1/2x+1=2x --> x=2/5 y=4/5 and so on. You get the difference by subracting two of these points from another and then you have data to work with Pythagoras. The width of the square is then sqrt(5)/5 and its area 1/5. ^^ Spectacular video by the way. A++
P.S. after that I came to the conclusion that the areas of the smaller triangles are 1/20 and therefore fit 4 times into the blue square. So all the trapezes together make up 3/5.
A very weird rabbit hole I went down related to the four bugs problem: The function that gives the path length in terms of the step size is f(x) = x/(1 - sqrt(x^2 + (1-x)^2)), and we saw that the limit of f(x) as x -> 0 is 1. What I did was look at the Taylor series of f(x) around x = 0 and what I noticed was the terms up to x^15 all had positive coefficients, but from x^16 on the signs of the coefficients are periodic with period 8 (starting from x^19 the pattern is 4 +'s then 4 -'s).
I think I understand why it's periodic with period 8; it has to do with the roots of x^2 + (1-x)^2 being (1+i)/2 and its conjugate, where the angle corresponding to (1+i)/2 in the complex plane is 1/8 of a turn. And I also sort of have a formula for the coefficients: if the Taylor coefficients of 1/sqrt(x^2 + (1-x)^2) are a[n] and the Taylor coefficients of f(x) are b[n], then b[n+1] - b[n] = a[n-1] - a[n] + a[n+1]/2. But the a[n] sequence itself is probably not expressible in closed form so I'm not sure how to get more info on b[n].
That sounds like fun. Have to have a close look myself :)
I'd never seen either of the two proofs you showed in the beginning of the video. I remember a proof that was way more complicated and hard to understand. These are beautifully simple.
Nice one! I have also done the versions of AM-GM and Cauchy Schwarz like you did here - love this diagram for those visual proofs! Used it to animate Priebe’s and Ramos’ sine of a sum too. Was gonna create a mashup of them to show they are all the same but here you’ve done it better. Thanks!
Cool!
I'd be interested to read your take on Zog from Betelgeuse animated satirical math lampoons.
Hi~!
@@SuperYoonHo 😀
@@MathVisualProofs 🖐🖐how's ur day goin' sir
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I consider myself very intellegent. I do excell at certain things but when I come to the Mathologer channel, I find out how much I truly do not know. I like how you explain things yet I have trouble absorbing it. I am in awe of your intellect but also I am in awe of the people who comment on your videos. That in our dumbed down society there are still people that fully understand what is hidden to most of us mere mortals is just amazing to me. I am drawn to these videos but much like a moth is drawn to a flame. I see and then I crash and burn! I am hoping for a breakthrough moment when I shout Eureka! Keep up the good work and I will keep coming back and try to grasp it all!
36:34 Challenge accepted (Spoils the answer)
Consider that the small right angle triangle at the top right is similar to the same triangle but including the trapezium below it. The ratio of sides is 0.5:1=1:2 (by considering the ratio of hypotenuses). Therefore, big triangle area = 4×small triangle area. White area = 16×small triangle area then.
Now, add the small triangle at the bottom and you get a right triangle with sides 0.5 and 1, so area = ½×0.5×1=0.25. It also happens that's the area of 5 small triangles.
5∆=0.25
∆=0.05
16∆=0.8
Hence, area of the square is 1-0.8=0.2
Hope my proof is pretty
My proof:
let x be the length of the sides of the blue square.
The big square has area 1.
The two big triangles on the right and left of the blue square have area 2(½½1)=½.
The two small triangles at the top say have legs of length y and z then considering the diaganoal of the big triangle at the top we get z+x+y = √5/2. Hence z+y = √5/2 - x
The two trapezoids (trapeziums) below and above the blue square have area 2x(z+y)/2 = x(√5/2-x) = (√5/2)x - x^2
putting this together we get 1 = ½ + x^2 + (√5/2)x - x^2 so x =1/√5 and finally x^2 = 1/5
This is very well constructed as far as an answer with the proof goes! Than you for the hard work.
I'm getting a bit confused with this solution. I followed a similar method to @Prometheus. To summarise there are 16 small triangles that make up the white area and four of those same small triangles which make up the area of the smaller square in question. The area of each small triangle is 0.05 so 4*0.05 = 0.2 = 1/5 = the area of the blue square. So far all is good.
However if we then reverse engineer this problem to calculate the sides of the smaller triangle I get a bit confused. The hypotenuse of the smaller triangle is 1/2 (the long side of it spans 1/2 the length of the original square). If I refer to the other two sides of this small triangle as y and z where y is the longer side and z is the shorter side (and z = 1/2y), then the inner square that we were finding the area of should be y*y? So y^2 should be the area of the inside square = 1/5 = 0.2.
That means that the shorter side of the smaller triangle z should be 1/2y and its square z^2 should be 1/4y^2 which is I think 1/20 (0.05).
But then if we take Pythagoras Theorem that the square of the hypotenuse = sum of the square of the other two sides then y^2 +z^2 should be = 1/2 (because the hypotenuse is half the length of the original square).
But 1/5 +1/20 = 1/4 not 1/2. I'm obviously getting something wrong here. I'd appreciate any help in resolving this.
I have never seen these proofs before. I am not the brightest bulb but I admire and appreciate mathematics very much. It is always so satisfying to take a concept and make it easy to understand or to amplify the idea in some way. There is a creativity and sophistication into seeing the relationships between the logic of the pure math and its applications.
I don‘t recall seeing any of these proofs in school. Most relevant formulas were simply introduced to us without ever being proven before college.
... which is a crime :(
@@Mathologer from NWT Canada. I assure you, I prove all of these to Gr. 7 and up students.
Same, and even then I never saw it till my elective History of Math class my junior year.
@@davidmeijer1645 Very glad to here that you do. How about your colleagues? Would you say that showing these proofs to kids is something that is commonly done in Canada?
@@Mathologer i would say that the reality is..no. Most teachers are concerned with covering the curriculum, which doesn’t direct teachers to explore these expansive variations on a theme, say this Pythagorean Theorem. That’s why your videos are great…for us teachers to recommend (to the interested student). I went through 3B1B lighthouse explanation of the Basel Problem yesterday with a class of 6 Gr. 10s. I really think it was quite a bit above them, but I’ll see when we next meet. And I think that’s why most teachers are reluctant to delve….concerned that it’s a bit above the students. Have you been amongst high school students lately? I mean, most are at the stage of struggling with something like, how to get the angle, when a trig ratio is known….I.E. learning to use the inverse trig function on a calculator. It’s quite amazing actually how developmental stages are quite uniform across geography and time.
At 55 seconds what the diagram is the same way the ancient mathematician of India named bodhayan described the pythagorus theorem many centuries before pythagorus. I am now convinced that pythagorus definitely visited India and learned here.
Your video on the many proofs of Pythagoras is one of my favorites, so I’m sure I’m gonna enjoy this one!
If only we had teachers like you. Maths would be a subject way more people can enjoy.
I don't remember doing anything in primary school involving indices including ^2 for 'squared', we did areas of shapes by counting squares on squared paper, nothing about algebra either. It was all new in secondary school. The first year there (ages 11-12) was basically doing catch up for what we should have/could have all learnt in primary school, but without a 'national curriculum', each junior school had probably taught different things and missed some things out. I think things have probably improved since then with maths teaching, but some (10-12 year olds) still seem to struggle with the basics - because of not learning times tables, or because new maths concepts are not introduced in a simple way, avoiding using complicated language.
Same in the Netherlands, sadly. No equations, no Pythagoras or pi, not even long division. Arithmetic education got wrecked here since the 80's.
I was always terrible at math and great at science and now since I'm 40 years old I'm trying to learn all sorts of things I love your channel you explain things nicely and calmly 😊
Learned both when i was 13 by my teacher. I loved math more than i can remember! Thank you for your content once again! Hope to see an ecology "ecosystem stability" math video from you, they are quite intriguing!
I learned the first one of those proofs first, and the second one before I got out of high school. I had no idea that they aren't taught anymore! 😮 Geometry has always been, as you called it, pretty to me. Thanks for this!
I had seen the twisted squares in a textbook before, but I didn't connect them to the proof of the theorem until I received a book specifically on the Pythagorean Theorem called Hidden Harmonies. It had a full chapter of different proofs, half of which led back to the twisted squares. I didn't quite understand the rest of the book at the time, but it really is quite fun to see the history of humans proving the same thing dozens of different ways.
Art as the inspiration for math, now that is something!
17:30
Just remember that the triangle can be thought of as a half of a parallelogram of sidelengths A and B.
Thank you
how does that help :(
for the first part of the video. I am not familiar with either proof. I am 67 years old, BUT I am pretty sure none of my teachers even hinted at such a visual proof. WOW so nice to see that the ancients could have actually figured this out without knowing any trig function values. I've often wondered how the ancients figured PI to even significant values (3.1416) before having calculus and only having simple geometry. Thanks very much! The demonstration that a person knows his field is that they can dumb it down or explain it to a high school or even middle school person.
Well, I'd say to show one of the proofs that I show at the beginning should not take any longer than in the video, i.e. 30 seconds. And as far as I am concerned there is absolutely no excuse not to do this in school :)
I am a math uni student and I did not know like half of these proofs... This was amazing. The sin(a+b) formula in particular is just so beautiful.
Love all your videos and this is a first ever post; as a small token of appreciation for all your stuff, here’s a link to a beautiful proof for Pythagoras by Jacob Bronowski - first aired in 1973 on UK television. Since you tube apparently doesn’t allow links, it can be found on DailyMotion with a search for “the ascent of man. Episode 05. Music of the Spheres” - from 7:00. The question is does this mean maths is an alien language?
5:18 is the first proof I ever saw (or that I remember). For some reason I think it was featured in the “Cosmos” book but I have it in storage and I can’t check right now; I’m probably misremembering.
Interesting, so which one of all the proofs in this video do you like best?
@@Mathologer proof #1 at 2:00 is perhaps the nicest of the Pythagoras proof presented. But overall the sin(a+b) proof at 30:30 is the one that really made me applaud with giddiness, what a delightful approach!
I was never that great at Math. But I love learning what ever I can about it. A constant fascination for me.
That's great!
Last puzzle: The area of the middle square is 0.2, since you can divide the big square of area 1 in five equal squares of the desired area. 1/5=0.2
Don't you mean 0.2 of the big square. According to the given scale that's area 1 unit^2
I got ((1/2)^2+1^2)(2/5)^2... the first part is the hypotenuse (C) of the original triangle. The second from its divisions 0.2, 0.4, 0.4... (0.4 C)^2.
in the last couple of years watching people like yourself - I was at school hundreds of years ago - and yes, beautiful
My favourite Pythagorean triple is 20, 21, 29 because it so close to an isosceles triangle and thus if you actually make it physically out of wood or meccano with inherent measurement errors, it will be more accurately a right angle than say 5,12,13 would be, subject to the same measurement errors.
How about 696, 697, 985?
@@colinpountney333 Nice but I haven't a hope of memorising that.
@@donaldasayers The 96 97 98 bit is memorable (I think). But you could just go for 7 7 10 to an engineering tolerance.
@@colinpountney333 (5, 5, 7) would appear to be ever so slightly better engineering-wise than (7, 7, 10), but (12, 12, 17), (17, 17, 24), and 29, 29, 41) are seemingly progressively ever better!
17:31 First stretch one side of the F triangle to A. The altitude perpendicular to A stays the same so the area is now AF. The same argument applies when stretching another side to length B. So the area is ABF.
36:17 The area of the small square is big square - 4 * a different right triangle, one with a hypotenuse flat against the big square.
You can get to that triangle by cutting the altitude of the main right triangle with its hypotenuse flat on the bottom. Because the angles remain the same within he cut triangles, the ratio between their sides is still 2:1, and so it's easy to see the two bases that wind up making the hypotenuse of the big triangle are in a ratio 4:1.
The hypotenuse of the original triangle is sqrt(5)/2. The base of the new triangle is 4/5 * sqrt(5)/2, and the height of the new triangle is 2/5 * sqrt(5)/2. So the area of the new triangle is 1/5.
Now we are ready to go back to our original question: what is the area of the small square? It's the area of the big square minus 4 * the area of the new triangle. This is 1 - 4/5 = 1/5.
So the area of the small square is 1/5.
American math student, even though we got up to differential equations I never saw either proof of Pythagoras until much later on UA-cam lol
Don't you also find this very strange? How can they not show this proof in school?
@@Mathologer American schools are focused on the "utility." They teach you what works but not how and it's really sad. I didn't understand how integrals literally functioned until I was in college despite taking 3 years of calculus in high school
@@HonkeyKongLive Yes, very sad :(
Professors Polster and Ross, I express my sincere appreciation for what you provide the world of mathematic enthusiasts at large. Your content and delivery never disappoint, thank you. While not pertinent to your specific topic, the subject matter in general is of particular interest. I would be remiss not pointing out one of my favorite aspects of the Pythagorean (et al.) Theorem. By careful geometric construction, three dissimilar non-right triangles can become the lateral faces of tetrahedra with right triangle base. Each of the three areas are correspondingly Pythagorean compliant. Thus, an oblique triangular pyramid or harmonious Pythagorean tetrahedron. Unfolded along the edges of the right triangle base, the “net” is quite counter intuitive to that of the conventional similar shape example you depict. Perhaps this of passing interest to some. Again, thanks for your amazing contributions and my enjoyment.
That's interesting. Rings a bell but a very faint one. Anyway, just read up on it and also stumbled across this write-up by you www.reddit.com/r/Geometry/comments/sio7tt/help_with_the_harmonious_pythagorean_tetrahedra/
Great video, as always!
I think that, at 28:43, the AM-GM inequality for 3 numbers should really be
(ABC)^1/3 =< (A+B+C)/3
Correct :)
The second one with stating how the big square area is simply equal to its components (small square and four triangles) is my favourite proof, it’s just stating the obvious, doing some simplification and you’re left with a²+b² = c²
personally, never seen either twisted square proof. but I think I am one of these children you speak of :)
So, are you going to ask for your money back ? :)
That selection of music packed some powerful emotion at the end. Amazing stuff, as usual!
Great video!
Very beautiful and simple proofs of Pythagoras. Thank you. Now three easy challenges for you, dear mathologer: a) prove the horizontal line of sight of an observer looking across the sea from a beach makes a right angle with the earth's radius. b) prove that a 2,000 foot mountain 60 miles across the sea cannot be seen by the observer if earth is 24,901 miles in circumference. c) find a large body of water with nearby mountains and see them from a beach on the other side and prove by contradiction earth is not a ball 24,901 miles round. Good luck!
Have to wear my "Show me the curvature" t-shirt in one of these videos.
Thank you so much what a great video
Wow... It's very, very amazing. Haven't been learn this since in the high school or university. There would be so many things can be covered with the rectangle and triangle. It's a unique things.
3:41 1046-256 BCE
Look into Anatoly Fomenko's Empirico-Statistical Analyses regarding Isaac Newton's and other chronologists' timelines.
It's more plausible that Chinese Imperialists added fifty-nine years to Li Qingyun's life, thus extending their chronology and thereby the longevity of the empire's mandate.
He claimed to be 197 when he died. They said he was 256 and that he lied about his ago pretending to be younger. He was a mountain herbalist. The imperialists were autocrats, totalitarians. Fomenko's New Chronology holds that this type of invented history is more common than not, that Pharaoh ruled Egypt through the 1700s, that 2022 CE (7530 Slavic Aryan Vedic Calendar) is 869 AD, and that The Great Wall of China is still being built.
Or maybe you had a typo, and it was 146-256 BCE? I didn't look it up yet, I was just reminded of all the above.
Thank you for excellent reference!
1046-256 BCE People just kept adding to it over a long period of time. I had heard of Fomenko's New Chronology. Interesting but in the end a crackpot theory I'd say :( Having said that I am a fan of his mathematics and his maths illustrations :)
I absolutely agree with you, it is a crime that schools don't introduce proofs. Even the most simplistic proofs, like the ones you showed.
The only time they are put into my curriculum is if you take the hardest math course, which only 5% of the cohort takes. 95% of students aren't even introduced to proofs.
As far as I am concerned there is no justification for calling a subject :mathematics" if you don't show any of this
@@Mathologer If pupils just learn ways to shove numbers around there is nothing that elevates thier mathematical abilities above those of a computer, which will always be faster than them.
What they miss out on is learning to strategically and creatively think about numbers and the world those might describe.
@@jannikheidemann3805 Exactly. Following a given algorithm is calculation, not mathematics, and reducing maths to caulculation defies the very purpose of mathematics, which is essentially an ever-expanding toolkit for solving problems.
Geometry was all proofs.
Proving the visually obvious. Seemed to be a grand waste of time and effort.
Based on your T-shirt the following proof of the Pythagorean theorem came to my mind:
Consider a right angle triangle with an hypotenuse 1, and call its area m. Scale it by a factor of c. Then the area of the new triangle is
E = m*c².
Split it into two triangles by its altitude (which is not any of its sides). These triangles will be right angle as well, and their hypotenuse-which length I will call a, and b-are the short sides of the triangle with area E. If you check their angles you can see that these two new triangle are also similar to the original triangle, so we can get them by scaling it by a factor of a and b, respectively, and therefore their areas are
F = m*a, and
G = m*b².
Finally, consider the fact that E = F + G,
substitute: m*c² = m*a² + m*b²,
and divide by m (what is strictly positive, because it was an area of a triangle).
This proof was told me by one of my maths teachers as a handy proof for matura exam, and he referred it as "Einstein's proof". As I typed it in, I recalled that probably you had a short animation on this very same idea, maybe with some other notation. (Indeed: ua-cam.com/video/r4gOlttnJ_E/v-deo.html But the choosing of the letters like this was my idea :) Even though I know, probably someone else has done it long before me…)
About the question at 3:00, both proofs were shown to me in high school, and I don't really remember which came first. But I remember that I was told to always emphasize that the quadrilateral in the middle is indeed a square because of the sum of the angles of a triangle is π.
I had not seen this one before. Thank you very much for sharing :)
I don't remember which proof I saw first but a few weeks ago I wanted to show a proof to one of my kids and realized I forgot all the proofs I had learned in school. The second (algebraic rearrangement) was the one I worked out first.
@@Mike_Greene yes
36:20 we can start by recognizing that a right triangle with legs of 1 and 1/2 is a 30-60-90 triangle. We can then see that the overlaps are similar triangles with the triangles that overlap to form them. They share one angle (30 degrees) and they have a right angle, as given by the problem stating the middle, shaded figure is a square. Truncation by a line parallel to one of the legs produces a similar triangle. So all three sizes of triangles are demonstrably similar, ad 30-60-90 cases at that. Since we know that the side length of the large triangles is 1 and 1/2, we can say that each of those hypotenuses bisects the hypotenuse of the medium triangles. From this we can conclude that a line that bisects hypotenuse of the mid-sized right triangle must do so perpendicularly to the long leg of the mid-sized triangle. From this we can conclude that since the small triangle's long leg bisects the long leg of the mid-sized triangle, the long leg of the small triangle must be equal to the side length of the shaded square. Knowing that the small triangles are 30-60-90 and that the smaller leg of the small triangle is one half the length of the long leg of the small triangle, we can prove that the small triangle fits into the space left by the right-angled trapezium by constructing a perpendicular bisector of the trapezium's longer parallel base. This bisector will prove to be parallel to the right leg of the trapezium and intersects the endpoint of the other base of the trapezium. This point is also the midpoint of the side of the larger square as demonstrated before. We can then demonstrate by corresponding angles and similar side lengths that the newly constructed small triangle is identical to the existing small triangle. The rectangle resulting from the perpendicular bisector can be bisected with a line parallel to the large square, that intersects the perpendicular bisector of the trapezium base and also intersects the corner of the shaded square. Thus we can demonstrate that the trapezium is composed of three triangles which are identical to the smaller triangle provided by the puzzle. We can also prove that the small triangle completes the trapezium into a square by demonstrating that it has complementary angles and both side lengths that would complete the square's sides. By knowing that this hypothetical square is indeed a square and by showing that it shares a side with the shaded square, we can prove that the shaded square has the same area as the hypothetical square. If we subtract from the overall figure the area of one smaller triangle and add to it that same area to complete the trapezium into a square, and do this on all four squares, we will have not changed the overall area, but will have demonstrated that the overall shape has the same area as the sum of five similar squares, one of which is the shaded square shown.
Thus, without calculating the side length of the shaded square, we can prove that the shaded square's area is equal to one-fifth of the overall figure. And we can also rigorously prove that flipping the small triangles over is a valid way to demonstrate the shaded figure's proportional area.
Incidentally, we can conveniently compute that the side length of the shaded square is the square root of one-fifth. Or 1/sqrt(5). Or sqrt(5)/5.
This also means that the leg lengths of the small triangle are 1/2sqrt(5) and 1/sqrt(5). Which also demonstrates that the hypotenuse of the large triangle is 1/sqrt(5)+1/sqrt(5)+1/2sqrt(5) or (sqrt(5))/2.... I think I did something wrong. Luckily, what I did wrong is not unfixable by changing the 30-60-90 to just the triangle initially given, which I misidentified as a 30-60-90. The exact angle of the triangle never really comes into play in the proof, only the demonstrably similar angles, which can stand on the side-angle-side identification rule and the ratios present.
Look at the 👕 Pythagoras and Einstein fighting for C^2 😂😂
So, according to Einstein, e = mc^2, so that e/m = c^2.
But according to Pythagoras, a^2 + b^2 = c^2.
Therefore, e/m = a^2 + b^2, so that e = m(a^2 + b^2).
😜🤪😝😛
This a glorious episode! Great job to you and the helpers listed in the credits.
In elementary my math teacher told the class circles had no area 😭😭😭
I mean even if their areas are incalculable you can just do the Archimedes thing and bound the area between a polygon inside the circle and another outside? (Admittedly Archimedes was trying to work out the exact length of the circumference but I'd say my point still stands)
Was she trying to say that a circle was just the bounds? That's a mean trick anyway if that's what she meant.
In my school (in Poland) the teacher made the difference stand out between a circle (pol. okrąg) and a disc (pol. koło). Circle is not "filled", and the disc is (holds up in English as well) because of the following definitions:
1. an open ball is a set of all points which distance from a centre P is smaller than the radius r, or in other words {x : d(P, x) < r}
2. a closed ball is a set of all points which distance from a centre P is smaller or equal to the radius r, or in other words {x : d(P, x) ≤ r}
3. a sphere is a set of all points which distance from a centre P is equal to the radius r, or in other words {x : d(P, x) = r}
where d is a metric (simply: the distance between two points, doesn't have to be the euclidean one).
Circle is, strictly speaking, a 2-dimensional sphere, and a disc (strictly: closed disc) is a 2-dimensional closed ball.
What this means is that while a circle technically only has the circumference (because it only consists of points on the circumference), the disk also has a surface. For circumference we can calculate it's length (e.g. the length of the circumference of a circle with a radius r is 2πr) and for the surface we can calculate it's area (e.g. the surface area of a disc with radius r is πr²).
Technically, neither a circle nor a disc have an area. A disc has a surface, which has an area, so a disc has a surface area.
In less strict language we usually just say "area" instead of "surface area" and also often say "circle" when we mean a "disc".
When someone says "area of the circle" we usually assume they mean "surface area of the region bounded by the circle".
The important thing is that everything stays unambiguous, so that everybody understands what we mean.
Hope I helped ¯\_(ツ)_/¯
Mine told the class that, in the binary number system, 0 represents 0, and 1 represents all the other numbers.
😭😭😭
Final Puzzle at 36:01,
Area of the inside square is a²(1-(ab/(a²+b²))) ,
where "a" is side of the bigger square, & "b" is the segment cut.
You can solve it using similarity of triangles, and ratios of sides.
For the last puzzle, the small square's area is 1/5.
The big square is made of 4 triangles, 4 trapeziums, and the small square.
If you fit the triangles to the trapeziums, you get 4 squares sharing a side with the small square, therefore they all have the same area.
The large square with an area of 1 divides into 5 equally sized small squares, so each small square's area is 1/5
I can see a couple other ways to solve it, and there's probably more that I haven't thought of. I noticed the small triangles are similar to the large triangles, that could be useful.
That is a clever way to solve it (and probably one Burkard would be thrilled to animate)! I can confirm though that the small triangles and the original triangles are similar (and that the fact is useful) as that is how I reached the same result of 1/5 for the area of the small square.
I grew frustrated quickly with trying to calculate each line segments length and just did that but, as noted earlier, the only way to check that is to precisely cut out a trapezium and smal triangle and see if they perfectly overlap the small square, whereas the proof is fool "proof." This is why I and so many other people have trouble with plane geometry; such is logic.
Prosecutor: I can't believe they acquitted someone so OBVIOUSLY guilty!
Judge: Of course he's guilty and of course that's obvious--but you failed to PROVE it, counselor.
@@herbpowell343Sorry, I don't understand what you're trying to say? There's no need to calculate any numbers or do any perfect cutting, just the given values are easily enough to prove it.
It's given that the edges of the large square are each length 1, and the outside edge of each triangle is 1/2. This means the outside edge of each trapezium is also 1/2. If you match each triangle to each trapezium along that edge, forming a quadrilateral, it must be a rectangle since all the corners are right angles (two given in the trapezium, a third from the angles that form each corner of the large square, and if three are equal then the last must be as well), and the quadrilateral's length and width are both equal to the height of the trapezium, which is equal to the length of each side of the central square. Due to having all right angles and equal sides, each quadrilateral must be a square, and since they all share sides with the central square, all of the squares' areas are equal. Since the total area is 1 and is evenly split into five, the area of each little square must be 1/5.
I do like that quote at the end. Is it an analogy you came up with, or a specific reference to something? Either way, it certainly applies to maths in general quite well.
I learned the theorem at school but not the proof. At uni it was assumed that everyone knew the proof so it wasn’t proven to me there either. I commented to a fellow student that I’d never seen the proof, and he explained your second proof to me. Today at 56 is the first time I saw the first proof!
Very cool.
I've been watching your videos for years, THIS ONE! was the proof I was asked to solve in my grade 12 academic maths course. You brought me to a new realization that I hadn't seen before. My teacher was kind enough to give me a passing grade but I missed the mark. Now I want to make a learning tool that will physically describe the geometric relationship you've described. Thank you for your continued insight.
I applied at my school to get my money back, and now they charged a processing fee.
:)
36:17
1) In a 3x3 square grid, where one single grid square has side A, draw a twisted square with vertices at 1/3 of the 3A side length square
2) take a moment to realize that the twisted square drawing corresponds to the puzzle presented
3) the blue area is therefore A², but given the problem data (2A)² + A² = 1. So A² = 1/5.
or by arranging the four smallest triangles next to the trapezoids to complete the squares, you will end up with a cross made of 5 squares with side of length A. Thus 5A^2=1.
@@antonioaguilera8710 Right, but that solution was already in some other comment with a lot of votes, so I had to come up with another one.
I saw the first twisted square proof first, and it was from either you or numberphile lol. Am in America for context
For the third square:
Since a is 1/2 and b is 1, the base of the parallelogram that's created inside each triangle by the overlap must be 1 - 1/2 = 1/2. Since the hypothenuses of the triangles are parallel this means, the sides of the inner square must also be of length 1/2, so the square has an area of 1/4.
PS: great videos! Lot's of things I didn't know before in this one as well. Always fun to learn, especially when the person teaching is as excited about the information as you are!
Oooh, i see my error now. Wow my head made a lot of sense of this...
After reading other comments and looking at the diagram again, I now also think that the area of the small square is 1/5.
(For anyone whose head had the same idea: the "base" of the parallelograms isn't parallel to the sides of the inner square, so they can't be of equal length)
That's it :)
Ty
learned the second proof in school, the first in an earlier Mathologer video. Since then I watch all your videos and bought several of your books. I would recommend your videos to everyone who has the slightest interest in maths.
I was never taught the proof.
:(
First I would like to thank you for an excellent presentation - I vaguely recall being taught the geometric proof 55 years ago at age 11.
Second, I became a great fan of Martin Gardiner's puzzle books - very entertaining
Third, the problem you set to find the area of the square - my solution is as follows:
Area of the four large triangles equals the area of the large square, so the small square equals the ovelap - i.e. the area of the four small triangles.
Hypotenuse of large triangle equals (sqrt 5)/2
Since small and large triangles are similar and hypotenuse of small triangle is 1/2, the remaining sides will be 1/sqrt5 and 1/(2*sqrt5)
So, area of four small triangles = 4*(1/sqrt5 * 1/(2*sqrt5) * 1/2) = 1/5
As a check, side of small square = Hypotenuse of large triangle minus smaller sides of small triangle = (sqrt 5)/2 - 1/sqrt5 - 1/(2*sqrt5) = 5/(2*sqrt5) - 2/(2*sqrt5) - 1/(2*sqrt5) = 1/sqrt5, so area of small square = (1/sqrt5)^2 = 1/5
Very good :) Also, huge Martin Gardner fan here.
The solution to the final puzzle is 1/5.
Perhaps the most visually pleasing way is to rearrange the 4 small triangles so they each fit onto one of the trapeziums to form 4 identical squares.
We know that these are also identical to the middle square with unknown area because the ratio of original lengths create similar triangles.
The result is the large square of area 1 is separated into 5 smaller squares of equal area, one of which is the middle square.
You are truly a great science communicator - a wonderful teacher. I humbly thank you
its funny how u should say science communicator instead of math communicator, but Pythagoras tends to do that
@0:27 your explanations helped me to better understand what the Uni teachers were trying to say. Some times I have the “Oh that’s what they were talking about and I couldn’t get it”…
I'm a math teacher in Germany - I teach both of the proofs shown in the intro in my classes.
The first proof is usually discovered by the students themselves - i bring cardboard triangles to class and let the students re-arrange those themselves until they discover the proof.
Can be done even by students who are usually not that strong regarding mathematics.
Y equals r cubed over three. And if you determined the rate of change in this curve correctly, I think you will be pleasantly surprised. Don't you get it Bart? RDRR Kidding. That was awesome, and I strongly wish this channel existed when I was learning Math. I love looking at things I know well in ways I never considered before, continually making me wonder if I actually knew much of anything in that class.
This was great! The bugs things was especially beautiful!
Graduated HS in 1977.
We had pretty good education system.
Grew up in a small town on Texas coast with one HS
Went up to Trig. We had a math teacher who could teach in any college. Great man
But, Never seen this.
But I understannd.
35:30 this is not formal by any means, but intuitively I would say it is because the bugs are travelling perpendicularily at every instant.
I will try to formalize it using calculus and vectors:
let b1=(x1,y1) be the first bug, while b2 be the second bug. Let R be a clockwise rotation of 45 deg about the center of the square. WLOG we say that point is the origin, meaning that R can be expressed as a matrix.
It is clear that b2=RR*b1=(y1,-x1)=(x2,y2)
let d(t) de the difference between b2 and b1: d(t):=b2(t)-b1(t)= (RR-I)b1(t)
Computing the matrix of RR-I=[[-1 1],[-1 -1]] we can see that this is a cw rotation by 135 degs, and an enlargment by sqrt(2)
=> RR-I=sqrt(2)*RRR
let D(t) be the size of d(t): D(t):=|d(t)|=|sqrt(2) RRR b'|=sqrt(2)* |b1| because rotations don't affect length size.
We also have D(t)=sqrt(d.d).
We will prove D(t)'=1:
D(t)'=sqrt(d.d)'=1/2*1/sqrt(d.d)* (d.d)'=(d.d')/2*D(t)
let dT be the transpose of d
(d.d)'=(dT * d)'= dT'*d+dT*d'=(d'.d)+(d.d')=2(d.d')=2 * Transpose(sqrt(2)*RRRb1') * (sqrt(2)*RRRb1) = 4 * Transpose(RRRb1') * (RRRb1)
= 4 *transpose(b1') * R^(-3)*R^3*b1= 4 (b1.b1')
Since b1 moves in the direction of d, then the angle(b1,b1')=angle(b1,d)=135 degrees because d=sqrt(2)*RRR*b1.
Since the speed is 1 we have |b1'|=1
Solving for (d.d)' we get (d.d)'=2*d.d'=4 (b1.b1')=4 |b1| cos(135)
Pugging back in we have D'(t) = 4*|b1| cos(135)/2*D(t) =4*|b1| cos(135)/2/sqrt(2)/ |b1| = 1
D(0) = sqrt(2)*|b1| => T = D(0)/v=sqrt(2)*|b1| which is equal to the time it takes to reach on point of the square.
This was much harder than I expected. Probably you can get better proofs using drawings and geometry. Not only is my proof long but it requires a lor of familiarity with linear algebra and some non trivial properties
I showed my maths teacher the diagram of a square and another square with width being the diagonal of the first. Therefore, half the area of the first square was a quarter of the other.
This meant the diagonal is sqrt(2)xside of the square.
I was ten and it was in France, and because it was my fourth country, I was always learning the language of my new country. However I spent my younger years loving maths and walking round the playground (clockwise). And am proud of that proof with a stick in the Southern French dust.|
However I was told I would be taught it next year. So glad to watch this video where no-one delays the learning