UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle. Take two pythagorean triples: (u, v, w) and (x, y, z); Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz). It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5). I don't know if there are other ways to do it. Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring.
I also noticed that all of the lengths in your example had a common factor of two, so you could have scaled the whole thing down by a half and had a more simplified version.
Got curious, made a piece of code that brute forces it, get four integers and check if they make a triple in a 2x2 grid("manually" checking the result of the function below). With only (those) two triplets i can't form 4 unique rectangles (the cross product seems to not generate a new whole amount hypothenuse triangle either, only the scaled up version does.) (Not sure if that is what you meant, and if did the exactly that.) If you have "square root of x= integer y or zero else" function you can do it in less steps and in reverse. (At first i tried searching for a general formula for generating P-triples but got to a formula with 3 unidentified numbers and the three is integer restrictions at step 0 (the most famous math formula after 2+2=4), so i just used it as a restriction on code and it works as expected) Around 3 days averagely spent.
I've been watching you in videos for over a decade now. I'm convinced that your close proximity to a proper enigma machine has some kind of supernatural fountain of youth effect on you. You've barely aged a day. Maybe I need to see it in person some day!
@@vimmiduggal6658 Disproval of alleged disproval by appeal to definition: Nepal's flag doesn't have an aspect ratio, therefore its aspect ratio is neither rational nor irrational.
@@JivanPal Disproval of alleged Disproval of original disproval: The definition states, " The aspect ratio of a geometric shape is the ratio of its sizes in different dimensions". It is nowhere mentioned that the shape need be a rectangle (there is precedent in the case of aspect ratios of ellipses). Hence, it does have an aspect ratio, which according to Nepal's constitution, is irrational.
Professor Grimes, I understand fully your comment regarding 'feeling uncomfortable reading positive comments'. I feel the same way at the conclusion of each semester when reading student evaluations. I tend to just ignore the ones in which disgruntled students are making nasty comments and focus on the positive ones. If I had a maths professor like you, I might actually be a mathematician and not a lowly physiologist. ;)
You came to my school (Trinity High School in Manchester) a few years ago and I met you and you were very friendly. You handed pieces of paper with different problems to solve on each one and it was very fun to see you. Just wanted to let you know a few years later that I still remember you and you deserve a thank you from me. 👍
Love the videos both on Numberphile and your own channel, been watching since I entered high school and I'm out of college now. Thanks for the videos James!
I'm such a stupid and lazy person these days that I watch these videos exactly for others to prove them and wake up my brain or at least stimulate some thought or memory. Also I enjoy that there's been a couple more videos on this channel this year. It was also nice to hear you shouted out on Stronger by Science podcast too, I like these crossovers of interests.
The difference could be interpreted as 2xy cos α Where x,y are the side lengths of the parallelogram and α is an angle of the parallelogram. I like this interpretation because it uncoincidentaly connects to correction factor in the law of cosines.
It wasn't till you mentioned applying BFT to other countries' flags that I realised this, but British Flag Theorem would be better named Scottish Flag Theorem.
Games Jrime… I love that I have found this channel about 3 months ago and have mostly watched your videos from a decade ago and now I see a newer one show up in my feed. I really am wondering where the channel got its name. :D
For the integer solutions ones you can think about solving a²+b²=c²+d² as solving (a+d)(a-d)=(c+b)(c-b) and now a simple way to find an infinite number of solutions is to take d=a+1 and choose a such that 2a+1 is not prime, then you can take some non-trivial factors and use them to give you some integer b and c. E.g a=10, d=11 so 2a+1=21=3×7=(5+2)(5-2) so b=5, c=2 so 10²+5²=11²+2²
I really liked the format on this--it felt like sitting in the pub and talking through an interesting problem. Thanks also for putting up the proofs. I always appreciate being able to check my working out.
I really, really liked this video. It's really nice to hear you just talk about interesting stuff that wasn't in the scripted video. You're still excellent, off-the-cuff, talking about all this stuff. I just recently burned my brain's energy figuring out a group theory question for one of my students (so I have no brain power today), but the Pythagorean triples question intrigues me, so I might look into it tomorrow.
Update: I have made no progress :P I have thought quite and tried working some stuff out, but I haven't made pretty much any headway at all. Tough problem! (I'll keep thinking. If I ever discover anything, I will let you know.)
@@singingbanana I did see your update! I had mainly been looking at Richard Holmes's example and messing around with it, seeing if I can tease out any information, but I haven't been able to get anything interesting with it so far. I spent a lot of today working on it again... and I ended up replicating your method. I wouldn't be surprised if I thought of it the same way you did, since the tricky thing was getting the third and fourth to line up. I did come up with a relationship that the legs of the four Pythagorean triples should satisfy, but it's hard to construct rectangles from that relationship.
I think an even better way to express the parallelogram rule would be 2xycos(theta), where x and y are the two sides and theta is the angle between them. That actually reminds me a lot of the generalization of Pythagoras, the "Law of Cosines" for a triangle: a^2 + b^2 - 2 a b cos(gamma) = c^2
It seems the British Flag Theorem works even if you choose a point completely outside the rectangle. I'm inclined to call that the Australia or New Zealand Flag Theorem - because they have the British Flag up in one corner, and you could choose any of the stars away from that corner as the point to draw the diagonals from. (Even though I am neither Australian nor New Zealandian.)
There is an alternative way of expressing the difference in the parallelogram case, which I personally think is cuter, namely this: (a² + b²) - (c² + d²) = ½ (A² - B²) where A and B are the two diagonals, A being the longer one. Proof, building on the proof @13:50 In the figure, let h be the height of the parallelogram, A the longer diagonal, and B the shorter diagonal. Then we get by Pythagoras’ theorem: A² = (s + t)² + h² = s² + 2st + t² + h² B² = (s - t)² + h² = s² - 2st + t² + h² If we subtract the second equation from the first one, we get A² - B² = 4st ⇔ ½ (A² - B²) = 2st So we can substitute ½ (A² - B²) for 2st in your formula. And that's it.
I suspect some of the more difficult flags to have theorems for would be ones where it's not simple, basic shapes like circles and rectangles - such as the flags of Saudi Arabia and Canada. I wonder how many of THOSE we can actually find mathematical or scientific theorems to fit? And.. another video next week? Does this mean maybe another the week after as well? I guess.. if you have been.. thank you for recording!
In the Cuboid theorem, would the sums of the squares of all diagonals equal each other? In other words, would it suffice to say that a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = g^2 + h^2? Or is the abcd relationship different from the efgh relationship? P.S. I’m not a mathematician at all. I just enjoy watching math(s) videos.
Yep! If you can prove it for any 2 sets of diagonals, you prove it for _all_ pairs of sets of diagonals. Consider the relationship between two pairs of opposite corners, A-B and C-D: · either A is adjacent to C (and B is adjacent to D), · or A is adjacent to D (and B is adjacent to C). So, both of those cases are the same. Proving one is the same as proving the other. And thus, showing that a²+b² = c²+d² is exactly equivalent to showing that a²+b² = e²+f² and a²+b² = g²+h².
The British flag theorem is just 1 special case in the parallelogram version where t=0, hence (a^2 + b^2) - (c^2 + d^2) = 2st =0 Or (a^2 + b^2) = (c^2 + d^2)
Did anyone mention the Nepalese flag whose design is completely defined by equations? And thank you and I just wake up, so the rest of my day is in front of me.
A question for Professor Grimes or well anyone else as well really: James showed us that The British Flag Theorem works in a 2-D space and in a 3-D space as well. My question is: Is there a way to prove (or disprove) that The British Flag Theorem always works in any n-D space 2-D or higher (essentially the dimensional space n is any integer 2 or greater)? If this extended theorem is false, is it true for only certain values for n instead of all values and again can this be proven/disproven? I sort of have a visualization of it in my brain but no actual proofs/disproofs yet.
Smallest integer rectangle I can find uses Pythagorean triples (15, 20, 25), (20, 48, 52), (48, 36, 60), and (36, 15, 39). More or less brute force search with a Python script, but turns out it's the method you mention applied to (3, 4, 5) and (5, 12, 13). Does that method generate ALL integer rectangles? Added: NO it doesn't. Counterexample: (25, 60, 65), (25, 312, 313), (91, 312, 325), (60, 91, 109)
For the American flag there is some interesting thing that can be done with the way the stars are arranged. eg How many ways are there to arrange n stars into rows such that the number of stars in each row differ no more than 1.
- 13:28 You _do_ have a sign-off catchphrase: _"and if you have been…"_ - So no compliments, huh? Okay, then would you be more comfortable with insults? That would be awkward for us to have to insult you, you're too endearing and popular with viewers, but we'd make that sacrifice if it makes you feel better. 😛
You can fit together 8 copies of a pythagorean triangle into a rectangle; this already leads to infinitely many fully integer solutions, since (even up to scaling) there are infinitely many Pythagorean triples.
@@singingbanana Cool, that's beautiful! There should certainly still be more... Indeed, if we label the non-hypothenuse sides of the Pythagorean triangles with a,b,c,d (where a,b are in the horizontal direction and c,d are in the vertical direction; these are the w,x,y,z from the original video), then your solutions all satisfy ab = cd. But now consider a = 5, b = 9, c = d = 12 which corresponds to the Pythagorean triples (5,12,13) and (9,12,15). Then clearly ab ≠ cd...
So basically, if the two pairs of points are the same distance apart this will work. So if the opposite vertices are the pair A and B and the pair C and D, the new point is O, and AB = CD, then this works.
9:20 so 3 triples fix the dimensions of the fourth but don't guarantee that it's got a rational hypotenuse? interesting. seems so: first 3 Pythagorean triples (3, 4, 5),(5,12,13),(8,15,17). scale them up (27,36,45),(15,36,39),(8,15,17) the fourth has got to fit the edges 27 and 8. sqrt(27^2 + 8^2) = sqrt(793). it's irrational.
James on your comment #3 on Pythagorean triples you have 280, 210, 250 as a triple. That's a typo as its 280, 210, 350 (the standard 3, 4, 5 multiplied by 70)
How long has the British "Union Jack" flag been in its current configuration (the English red t-cross over while field married with the Scottish white x-cross over blue field and Northern Irish red x-cross over white field)?
So whats the maximum number of stripes you can do with integers? And what about different number of columns? Given n columns, whats the maximum m rows such that all the diagonals in the stripes are integers?
When thinking about trapezoids, consider the degenerate case of a triangle. That is, where points A and D are the same because side AD has collapsed to zero.
I do not understand how infinity can be a property of our real world. If we as people are not able to make use of continuous stuff (like we need to aproximate using descreet things, and can we actually meaningfully describe continuity without aproximations? ), how could the universe (laws of physics) know how to make sense of continuity? It just don't make sense to me.
So what's the general solution for all quadrilaterals, with special cases for trapezoids, parallelograms, and rectangles? Py-flag-orean theorem? You do know that math puns are the first sine of madness. richard -- i -𝜋/2 i = e
I think Matt Parker is the ideal candidate to find other integer solutions, especially because it involves the sums of squares.
parker sum of squares
Of Parker, Some squares.
UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle.
Take two pythagorean triples: (u, v, w) and (x, y, z);
Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz).
It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5).
I don't know if there are other ways to do it.
Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring.
Very nice that a generative triple was able to be made! Although I do want to point out, I'm not mad about anything.
I also noticed that all of the lengths in your example had a common factor of two, so you could have scaled the whole thing down by a half and had a more simplified version.
Whoops! Sorry!
I could have. I saw the example (as presented) first. Then I saw the common factor, but thought, 350m sounds like a path in a park.
Got curious, made a piece of code that brute forces it, get four integers and check if they make a triple in a 2x2 grid("manually" checking the result of the function below). With only (those) two triplets i can't form 4 unique rectangles (the cross product seems to not generate a new whole amount hypothenuse triangle either, only the scaled up version does.) (Not sure if that is what you meant, and if did the exactly that.)
If you have "square root of x= integer y or zero else" function you can do it in less steps and in reverse. (At first i tried searching for a general formula for generating P-triples but got to a formula with 3 unidentified numbers and the three is integer restrictions at step 0 (the most famous math formula after 2+2=4), so i just used it as a restriction on code and it works as expected)
Around 3 days averagely spent.
A follow-up video AND new scheduled video?! It feels like early Christmas! Thank you, watching your videos always make my day brighter 🌞
I've been watching you in videos for over a decade now. I'm convinced that your close proximity to a proper enigma machine has some kind of supernatural fountain of youth effect on you. You've barely aged a day. Maybe I need to see it in person some day!
I agree that we need more positivity on youtube, and your upbeat and wholesome videos are a help with that! Thanks for brightening my day.
Love ur videos Prof. James Grime. Brings so much info to me. Hope the next video on numberphile is with you
I was able to workout all of them. Thanks for this. Wide smiley on your face is very contagious. 😄
Excellent! Well done.
Togolese Flag Theorem : There is exactly one national flag which has an irrational aspect ratio.
Oh cool. I've looked it up, and it's a golden rectangle.
Disproven by counter example: flag of Nepal
@@vimmiduggal6658 Disproval of alleged disproval by appeal to definition: Nepal's flag doesn't have an aspect ratio, therefore its aspect ratio is neither rational nor irrational.
@@JivanPal Disproval of alleged Disproval of original disproval: The definition states, " The aspect ratio of a geometric shape is the ratio of its sizes in different dimensions". It is nowhere mentioned that the shape need be a rectangle (there is precedent in the case of aspect ratios of ellipses). Hence, it does have an aspect ratio, which according to Nepal's constitution, is irrational.
I know you said "comment of the week, no prizes" but being declared comment of the week by you its in itself an achievement
Lovely to see you making videos again, in whatever format!
Professor Grimes,
I understand fully your comment regarding 'feeling uncomfortable reading positive comments'. I feel the same way at the conclusion of each semester when reading student evaluations. I tend to just ignore the ones in which disgruntled students are making nasty comments and focus on the positive ones. If I had a maths professor like you, I might actually be a mathematician and not a lowly physiologist. ;)
The cuboid version actually follows from the 3d rectangle version, since the four corners in use connects to a rectangle.
Fascinating Dr. Grime, you and Numberphile have awaken in me the "Love of Math" at my advanced age
Thank you for your positive content! ❤️
You came to my school (Trinity High School in Manchester) a few years ago and I met you and you were very friendly. You handed pieces of paper with different problems to solve on each one and it was very fun to see you. Just wanted to let you know a few years later that I still remember you and you deserve a thank you from me. 👍
I love this! Thank you.
Love the videos both on Numberphile and your own channel, been watching since I entered high school and I'm out of college now. Thanks for the videos James!
I'm such a stupid and lazy person these days that I watch these videos exactly for others to prove them and wake up my brain or at least stimulate some thought or memory. Also I enjoy that there's been a couple more videos on this channel this year. It was also nice to hear you shouted out on Stronger by Science podcast too, I like these crossovers of interests.
Oh? Why was I mentioned on the podcast?
Been quite few years since i last saw you and you dont look a day older!
The difference could be interpreted as
2xy cos α
Where x,y are the side lengths of the parallelogram and α is an angle of the parallelogram.
I like this interpretation because it uncoincidentaly connects to correction factor in the law of cosines.
It wasn't till you mentioned applying BFT to other countries' flags that I realised this, but British Flag Theorem would be better named Scottish Flag Theorem.
Games Jrime… I love that I have found this channel about 3 months ago and have mostly watched your videos from a decade ago and now I see a newer one show up in my feed. I really am wondering where the channel got its name. :D
Keep posting more.🙏🇿🇦
Great, fun content. Thank you for your work on this channel and on Numberphile!
I know barely anything about maths but I still find the channel entertaining
@@zoewells3160 and that's the beauty of it all! At the end of it you'll know a lot of things about math
For the integer solutions ones you can think about solving a²+b²=c²+d² as solving (a+d)(a-d)=(c+b)(c-b) and now a simple way to find an infinite number of solutions is to take d=a+1 and choose a such that 2a+1 is not prime, then you can take some non-trivial factors and use them to give you some integer b and c. E.g a=10, d=11 so 2a+1=21=3×7=(5+2)(5-2) so b=5, c=2 so 10²+5²=11²+2²
I really liked the format on this--it felt like sitting in the pub and talking through an interesting problem. Thanks also for putting up the proofs. I always appreciate being able to check my working out.
Thanks, I don't like long videos. But maybe this is a way to do asides and things.
@@singingbanana Do you dislike long videos for editing purposes, or for attention span purposes?
I think it's better to make your point, and then stop. Long videos seem to dilute the point.
I really, really liked this video. It's really nice to hear you just talk about interesting stuff that wasn't in the scripted video. You're still excellent, off-the-cuff, talking about all this stuff.
I just recently burned my brain's energy figuring out a group theory question for one of my students (so I have no brain power today), but the Pythagorean triples question intrigues me, so I might look into it tomorrow.
Update: I have made no progress :P I have thought quite and tried working some stuff out, but I haven't made pretty much any headway at all. Tough problem! (I'll keep thinking. If I ever discover anything, I will let you know.)
I'm glad you're having a go. I hope you saw my update in the description, with one way to do it.
@@singingbanana I did see your update! I had mainly been looking at Richard Holmes's example and messing around with it, seeing if I can tease out any information, but I haven't been able to get anything interesting with it so far.
I spent a lot of today working on it again... and I ended up replicating your method. I wouldn't be surprised if I thought of it the same way you did, since the tricky thing was getting the third and fourth to line up.
I did come up with a relationship that the legs of the four Pythagorean triples should satisfy, but it's hard to construct rectangles from that relationship.
We probably did do the same way. I played with Richard's Holmes's example too, and got nowhere.
The cuboid proof is a direct result of the 3D result you described before, because those four corners of the cuboid form a rectangle.
Let people have their fun.
James, i love your videos. Thank you.
I think an even better way to express the parallelogram rule would be 2xycos(theta), where x and y are the two sides and theta is the angle between them. That actually reminds me a lot of the generalization of Pythagoras, the "Law of Cosines" for a triangle:
a^2 + b^2 - 2 a b cos(gamma) = c^2
That would be a good way to express it.
I'd laugh if next week's video has something to do with the differences between sums. "sum differences in it" instead of "some differences in it."
I noticed you used all integers in the example! I just didn't think to comment about it!
I had wondered about the paralellagram case, I wish I'd said something now. I'd assumed sine and cosine would have showed up.
"there will be sum differences in that"
I get it! I get it!
I really appreciate this follow-up video. Have a great day. :)
It seems the British Flag Theorem works even if you choose a point completely outside the rectangle. I'm inclined to call that the Australia or New Zealand Flag Theorem - because they have the British Flag up in one corner, and you could choose any of the stars away from that corner as the point to draw the diagonals from.
(Even though I am neither Australian nor New Zealandian.)
I like your thinking.
There is an alternative way of expressing the difference in the parallelogram case, which I personally think is cuter, namely this:
(a² + b²) - (c² + d²) = ½ (A² - B²)
where A and B are the two diagonals, A being the longer one.
Proof, building on the proof @13:50
In the figure, let h be the height of the parallelogram, A the longer diagonal, and B the shorter diagonal. Then we get by Pythagoras’ theorem:
A² = (s + t)² + h² = s² + 2st + t² + h²
B² = (s - t)² + h² = s² - 2st + t² + h²
If we subtract the second equation from the first one, we get
A² - B² = 4st ⇔ ½ (A² - B²) = 2st
So we can substitute ½ (A² - B²) for 2st in your formula. And that's it.
This is a nice way to express the answer. If I did it again and made it a puzzle, that would definitely be a more fun way to write it
What a great and positive guy
i think videos like this compliment your normal ones quite nicely. if you were to do more like this i think it would be well received :)
I suspect some of the more difficult flags to have theorems for would be ones where it's not simple, basic shapes like circles and rectangles - such as the flags of Saudi Arabia and Canada. I wonder how many of THOSE we can actually find mathematical or scientific theorems to fit?
And.. another video next week? Does this mean maybe another the week after as well?
I guess.. if you have been.. thank you for recording!
Dutch people and their blue balls. Interesting.
I'm so chuffed that James Grime is back in town 😊
How about this sign off line? "And if you have been, thanks for watching."
Now I got curious 8f this works for other regular polygons besides squares/rectangles
Weren’t you the guy who flipped 10 heads in a row haha ? Love the vids keep it up :)
In the Cuboid theorem, would the sums of the squares of all diagonals equal each other? In other words, would it suffice to say that a^2 + b^2 = c^2 + d^2 = e^2 + f^2 = g^2 + h^2? Or is the abcd relationship different from the efgh relationship?
P.S. I’m not a mathematician at all. I just enjoy watching math(s) videos.
Yep! If you can prove it for any 2 sets of diagonals, you prove it for _all_ pairs of sets of diagonals.
Consider the relationship between two pairs of opposite corners, A-B and C-D:
· either A is adjacent to C (and B is adjacent to D),
· or A is adjacent to D (and B is adjacent to C).
So, both of those cases are the same. Proving one is the same as proving the other.
And thus, showing that a²+b² = c²+d² is exactly equivalent to showing that a²+b² = e²+f² and a²+b² = g²+h².
The British flag theorem is just 1 special case in the parallelogram version where t=0, hence (a^2 + b^2) - (c^2 + d^2) = 2st =0
Or (a^2 + b^2) = (c^2 + d^2)
It is. Which is nice, isn't it.
Did anyone mention the Nepalese flag whose design is completely defined by equations?
And thank you and I just wake up, so the rest of my day is in front of me.
Good Job Dr. James, now being a Doctor and a Mathematician, can you throw in a bit of medical advice too? (badum tss)
Such a lovely guy.
A question for Professor Grimes or well anyone else as well really: James showed us that The British Flag Theorem works in a 2-D space and in a 3-D space as well. My question is: Is there a way to prove (or disprove) that The British Flag Theorem always works in any n-D space 2-D or higher (essentially the dimensional space n is any integer 2 or greater)? If this extended theorem is false, is it true for only certain values for n instead of all values and again can this be proven/disproven? I sort of have a visualization of it in my brain but no actual proofs/disproofs yet.
Smallest integer rectangle I can find uses Pythagorean triples (15, 20, 25), (20, 48, 52), (48, 36, 60), and (36, 15, 39). More or less brute force search with a Python script, but turns out it's the method you mention applied to (3, 4, 5) and (5, 12, 13). Does that method generate ALL integer rectangles?
Added: NO it doesn't. Counterexample: (25, 60, 65), (25, 312, 313), (91, 312, 325), (60, 91, 109)
That counterexample is great. Those four pythag triples are actually different. I can't see how to construct other examples like that yet.
For the American flag there is some interesting thing that can be done with the way the stars are arranged. eg How many ways are there to arrange n stars into rows such that the number of stars in each row differ no more than 1.
That's a fun one.
maybe for the non-SB video sign off, "Thank you for watching,,, if you have been."
10:46 A mathematician who doesn't compute??!!
3 dimensional British flag theorem should be called Egyptian pyramid theorem.
Love the video!
- 13:28 You _do_ have a sign-off catchphrase: _"and if you have been…"_
- So no compliments, huh? Okay, then would you be more comfortable with insults? That would be awkward for us to have to insult you, you're too endearing and popular with viewers, but we'd make that sacrifice if it makes you feel better. 😛
You can fit together 8 copies of a pythagorean triangle into a rectangle; this already leads to infinitely many fully integer solutions, since (even up to scaling) there are infinitely many Pythagorean triples.
Ah, true, I was going for non-trivial solutions. If you are interested, have a look in the description for a more interesting solution.
@@singingbanana Cool, that's beautiful! There should certainly still be more... Indeed, if we label the non-hypothenuse sides of the Pythagorean triangles with a,b,c,d (where a,b are in the horizontal direction and c,d are in the vertical direction; these are the w,x,y,z from the original video), then your solutions all satisfy ab = cd. But now consider a = 5, b = 9, c = d = 12 which corresponds to the Pythagorean triples (5,12,13) and (9,12,15). Then clearly ab ≠ cd...
Regarding the next video... "some differences" or "sum differences"? 🤔🤔🤔
So basically, if the two pairs of points are the same distance apart this will work. So if the opposite vertices are the pair A and B and the pair C and D, the new point is O, and AB = CD, then this works.
9:20 so 3 triples fix the dimensions of the fourth but don't guarantee that it's got a rational hypotenuse? interesting.
seems so:
first 3 Pythagorean triples (3, 4, 5),(5,12,13),(8,15,17).
scale them up (27,36,45),(15,36,39),(8,15,17)
the fourth has got to fit the edges 27 and 8.
sqrt(27^2 + 8^2) = sqrt(793). it's irrational.
I've found one way to construct examples, I've put it in the description.
Hi James! 👋😀
unscripted video = scripted video - "if you have been"
Looks like it.
This time he read through the screen so he knew who have been watching.
By the way, I nearly mentioned your comment too Adheesh. But then, it becomes another list.
Make a series on Non Eucledian Geometry Please.
James on your comment #3 on Pythagorean triples you have 280, 210, 250 as a triple. That's a typo as its 280, 210, 350 (the standard 3, 4, 5 multiplied by 70)
Thanks. I've corrected that. Also, I found how to construct other examples like that.
hello? aren't you james grime on numberphille?
Do you have a party every time the queen's been queen for ages??
Constant parties.
Wow please make a video about binomial theorem
How long has the British "Union Jack" flag been in its current configuration (the English red t-cross over while field married with the Scottish white x-cross over blue field and Northern Irish red x-cross over white field)?
missed you
So whats the maximum number of stripes you can do with integers? And what about different number of columns? Given n columns, whats the maximum m rows such that all the diagonals in the stripes are integers?
Sum differences? As in 1+2 != 1+3?
Good on you! Too many people are normalizing narcissisms anyway.
Awesome!
Truth theorem. Thounds like I'm lithping, doethn't it.
If it doesn't work for parallelograms, does it work (or not) for trapezoids?
When thinking about trapezoids, consider the degenerate case of a triangle. That is, where points A and D are the same because side AD has collapsed to zero.
I do not understand how infinity can be a property of our real world. If we as people are not able to make use of continuous stuff (like we need to aproximate using descreet things, and can we actually meaningfully describe continuity without aproximations? ), how could the universe (laws of physics) know how to make sense of continuity? It just don't make sense to me.
No waistcoat in this one? Disappointing!
more skin crawling incoming :)
Here another positive comment. But not for you, for maths! Maths is beautiful!
No views, what is this!! :)
So what's the general solution for all quadrilaterals, with special cases for trapezoids, parallelograms, and rectangles?
Py-flag-orean theorem? You do know that math puns are the first sine of madness.
richard
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i -𝜋/2
i = e