A Problem with Rectangles - Numberphile

[@AlanKey86]

I'd love to hear how the Oxford applicants responded to this question. What was the neatest solution Tom Crawford saw from a student? What was the weirdest solution? Did students make embarrassing / common mistakes? What sort of fraction of applicants got the full answer?

[@miggle2784]

I hope Numberphile answers.

[@ragnkja]

@@miggle2784 Or Tom Rocks Maths.

[@luigy0648]

Tag him

[@maaikefiefi]

It's mentioned in the extra footage video in the discription!

[@stpirate89]

I assisted with the zoom interviews of students at York this year for physics undergraduates, and many academics ask questions similar to this. They're not after the correct solution, but more to see how a student approaches the problem, and if they ask sensible questions or not to get to the solution.

[@YouennF]

The math and fun content of this is high, but must not hide in our minds the quality of the "off-brown paper" animations presented in this video !

[@thariqiisafika7776]

Theres any knows.,, how number toto Sidney 04-08-2021???

[@custardtart1312]

They are intensely annoying and distracting.

[@goofoffchannel]

@@custardtart1312 cry about it

[@clapdrix72]

They're world class for sure

[@galahad692000]

I love it. It essentially reminds you that even complex mathematics can be accomplished with just a pen and paper.

[@uplink-on-yt]

I got into Oxford using this test, but the second security guard wasn't as impressed as the first and kicked me out.

[@WhiteHatMatt]

As a security guard, I might very well be persuaded to let someone in if they showed me a neat math trick.

[@mauricereeves7642]

The second guard always requires a rousing recitation of a Lord Alfred Tennyson poem for entry. It’s multilayered security.

[@Triantalex]

??

[@LewisDruid]

Quite off-topic, but I think it is neat to see this guy working at Oxford. I would assume such a prestigious uni would be very picky about how it's faculty and staff appear, but they let him be there just fine with his piercings and tattoos which is wonderful :D

[@TomRocksMaths]

[@CrashSable]

​@@topherthe11th23 "it's" is correct - that is a possessive apostrophe

[@montytiger9700]

@@CrashSable Its is always possessive, it needs no apostrophe

[@Giantcrabz]

it's hardly a public facing role

[@TuberTugger]

I paused the video for the 7 solution and when I did it myself, I got a different answer. It still involves 3 different sized rectangles with a single largest one. But the large one is smaller and has one of the smallest rectangles to finish it off. Instead of fractions, I worked with a 10x10 square for easier math. You need: 1x 4 by 8 rectangle 4x 4 by 2 rectangles 2x 3 by 6 rectangles One half of the square is a total of 4 by 10 (4 by 8 + a 4 by 2) The other half is a total of 6 b 10 (all remaining rectangles) => the 3 by 6s form a 6 by 6 square and the 4 by 2s form a 4 by 6 square. Just in case anyone was curious. It is a unique solution because the number of sized rectangles is different than Tom's.

[@CaptainRuff]

That's the same solution I found first.

[@kubtastic]

When he said "the solution" :/ ... This is unique and also unlocks the n+5 property.

[@polarisraven5613]

Yeah, same here. Posted the result to the Numberphile Reddit thinking I'd stumbled on something new, nope, appearantly there's an army of us.

[@oz_jones]

@@polarisraven5613 Dozens of us, dozens!

[@polarisraven5613]

@anomie nous Depends on what you define as a thought, further if you count how one treats it (explores it, acts on it, rejects it, forgets it, etc.). Does Sensory Input count as a thought? In that case, even the slightest change in viewing angle will offer a different thought. Certainly there are patterns as to how we react, but everyone's somewhat deformed or otherwise atypically molded in some way or other, and that influences the way we think. Maybe the vast majority of our thoughts have already been thought up by others, so what? Originality is over rated, it's not the uniqueness of an idea, it's how effective it is that counts. Now, uniqueness has it's own benefits, occasionally toppling entire meta's and traditions, but this is often done either by chance, or through a deeper study of the principles that make it all work, often a combination of the two. Whatever your goal, the likelihood that you're not the first to consider something has very little to do with the viability of the option, and even then, there are fields of study ripe with research to be done, combinations of thought yet undiscovered, original thoughts are still out there to be discovered, though the paths we take for granted that bring us there have been forged by the efforts of previous like minded (in goal) individuals.

[@fomx2753]

Very cool. I like that this is the kind of problem they ask candidates for uni.

[@toniokettner4821]

they don't

[@Ray25689]

@@toniokettner4821 he said they do, at least in oxford

[@ragnkja]

@@toniokettner4821 Dr Crawford literally said he’s used this question when interviewing people who have applied to study maths at the University of Oxford, so how can you say that they don’t? While they don’t use this _specific_ question any more, they still ask questions where the thing the interviewer is interested in is how the potential student reasons their way through the problem.

[@toniokettner4821]

@@Ray25689 well i should have gone to oxford then

[@Ray25689]

@@toniokettner4821 I think the pressure there is enormous

[@Zarunias]

I tried the 7 rectangles on my own and came up with a completely different solution with one 2/5x4/5 rectangle, two 3/10x3/5 rectangles and four 1/5x2/5 rectangles.

[@MrDoctorDen]

I think the number of WAYS you can do it for any given number N is another interesting question worth another video = )

[@lnx0007]

i got the same rectangles as you: in a 10 by 10 square, one 4x8 and one 2x4 can fill one side, leaving a 6 by 10 space. then 3 more 2x4s sandwiched together like books on a shelf cut off the bottom of that area by 4x6, leaving a 6 by 6 area to be filled by two 3x6s

[@TuberTugger]

This is how I did it as well. Instead of fractions, I used a 10 x 10 square to start. Which is more or less what you did. And I got those same rectangles.

[@DaneWeber]

Cool! I'd be curious how we each approached this. My solution was different: two big 3/4x3/8, three medium 1/4x1/2, and two small 1/4x1/8. I started by covering one side with two equal medium rectangles. I then added two large rectangles to cover the other corners. The gap between fit another medium rectangle, leaving me a square to split.

[@wazaN]

I found the same solution as you Dane. My approach was to think "is there a way that we would have a center rectangle surrounded by the 6 others ?"

[@mathoc5273]

Tom is great! I can see why you use him so much. Perfect numberphile guest. He clearly has FUN with math and I think thats part of what numberphile is all about.

[@TomRocksMaths]

[@mathoc5273]

@@TomRocksMaths 😍

[@shikhanshu]

I didn’t expect much but the video kept me glued all the way to the end!

[@ASOUE]

What I love about numberphile. This is an excellent display of leaping induction without mentioning leaping induction.

[@TomRocksMaths]

:)

[@Verlisify]

I like to imagine Mathematicians from centuries ago being comparably as exciting and quirky as this guy is to us today

[@ragnkja]

I’d be surprised if none of them were.

[@NumbToons]

dude, i keep finding your commetns in science videos and checking your channel out, thinking you also make science videos, and everytime its that pokemon intro and sike deja vu hits

[@Ian.Murray]

I don't

[@lucasng4712]

@@Ian.Murray k

[@vigilantcosmicpenguin8721]

I wonder what kind of arm tattoo Pascal would have gotten.

[@LeviATallaksen]

The first solution I found for 7 actually uses 6 different sizes! It goes like this, always starting with horizontal length: -16x8 on the bottom. -4x8 in the top left. -Two horizontal 6x3s next to each other, on top of the 16x8. -10x5 on top of that. -2x4 and 2x1 to cover the rest. This makes a 16x16 square.

[@HarrisD214]

This was surprisingly interesting. The solution was so clever.

[@TomRocksMaths]

Glad you enjoyed it!

[@haikumagician4363]

@@TomRocksMaths some people in the comments have solution for 7 rectangles. Are they right?

[@ragnkja]

@@haikumagician4363 There’s at least one other solution that works. If the one given in the video is on a 6x6 grid, there’s another on a 10x10 grid, and that one does not have the largest rectangle full half the square.

[@General12th]

It's a numberphile video. You shouldn't be surprised it's interesting.

[@invisibledave]

@@General12th I didn't find this one interesting at all and finally gave up on it. I don't like "proofs". There's way too much going over my head and so I lose interest after about 1 minute.

[@Keepturbo]

I always found it curious how school managed to make me dislike math in the way it was presented/forced upon me in school. Only to re-discover a passion for math/physics in my own free time later on in life (albeit after completing a completely different educational path), partly due to all these great youtube channels. And i guess the point that i'm trying to make is that i am pretty thankful for that.

[@leif1075]

What path did you take?

[@antonio97b]

Because a lot of stuff numberphile shows is really surface level that is meant to inspire the pursuit of learning. But deep level learning is not so fun all the time. There is a reason that historical documentaries are easier to digest than sifting than through loads of the actual historical texts/analysis.

[@Typical.Anomaly]

@joseph crosby mecham -147?

[@rrrajlive]

You can pick up a graduate level physics or math book, in sure you'll start to dislike then again 😃 (Pl. Take it for Fun)

[@Triantalex]

??

[@mihir_sheth]

Thanks!

[@MatSmithLondon]

Tom is an absolutely wonderful teacher. As a 41 year old who didn’t go on to study maths at university (although mg degree was slightly related) I find these videos excellent.

[@randy7894]

I like this guy's enthusiasm for math. All of the Numberphile people actually. Keep 'em up.

[@TomRocksMaths]

[@PanduPoluan]

Dr. Crawford seriously is such a charismatic person, and his explanation is so brilliantly clear & enthusiastic. I wish I had someone like him as my professor back in my univ days...

[@dylanwinestone4625]

Me feeling vaguely smart for seeing the n+3 method and n+4 method from the start, but then realising I hadn't come up with the building up method.......

[@harveyrice8504]

Then what was your n+4 method?

[@simaomarto6140]

​@@harveyrice8504 One alternative way is to do: Have two 1/3-by-1/6 forming a 1/3 side square on the top left corner, then add a 2/3-by-1/3 to complete the top 1/3 of the square. then put a 1/3-by-2/3 on the bottom right, and you are left with an empty square of 2/3-by-2/3 at the bottom left. So you can shrink your original solution by 2/3 into the bottom left and put those 4 rectangles around.

[@dylanwinestone4625]

@@harveyrice8504 that was my point, I didn't have an n+4 method. Hence I wouldn't have come up with a complete solution without the video

[@jackozeehakkjuz]

@@dylanwinestone4625 I bet you would. You just with a bit more time. The tricky part was the n=7 I think.

[@EebstertheGreat]

@@dylanwinestone4625 You said you came up with the n+4 method from the start, then realized you hadn't come up with the n+4 method. I don't understand.

[@jameslima9817]

I love Tom. I like his teaching style and how excited he is about what he’s teaching.

[@TomRocksMaths]

[@szabolcsmate5254]

That's the magic of maths. The excitement this guy has!

[@bidish2224]

Watching numberphile videos is the best part of my day

[@TomRocksMaths]

You and me both :)

[@dannybodros5180]

Cloud Strife retired as a mathematician after defeating Sephiroth.

[@hyknerf]

I LOL’d so hard

[@sasha-2574]

best video I've watched today! very enlightening indeed. thank you very much!

[@DerNesor]

Our didactics professor did a similar thing with us. It was an equilateral triangle and we could basically draw all the lines necessary for a Zelda-esque triforce. quickly finding out that you can turn one triangle into 4 little ones and each of those into smaller ones and so on, constructing every n except a few ones if I remember correctly.

[@millylitre]

Cool. I have been exploring triangle based versions of the original rectangle puzzle. The interesting variant is indeed the question of fitting N equilateral triangles into one larger equilateral triangle. My proofs, so far, show this is impossible for N=2 and for N=3, not yet determined for N=5 or for N=8, and proved to be possible for N=all other positive integers.

[@evanhoffman7995]

@@millylitre 8 is possible. Take one large triangle with a side length of 3/4, put it in the corner, and fill in the remaining strip with 7 triangles of side length 1/4. This method should work for any even number, where the small triangles with side length 1/n will give 2n triangles. And then of course you can subdivide one triangle to add 3 more and get any odd number (greater than 5). I suspect 5 is impossible but I'm not certain.

[@highpath4776]

@@evanhoffman7995 The impossibility (for Square and triangle) is probably due to there being a minimum size that is needed for the sub-size or supersize internal or external number of shapes. One might consider for why filling a circle with circles is both impossible (adjacent whole circles leave little gaps where they join) and (only) infinty ? - filling a circle with another immediately inside- I think the ratio remains the same, then another inside that and so on ?

[@OrangeC7]

It's really neat seeing how many people actually went ahead and tried 7 themselves. I also came up with an answer, and I got it by starting with what Tom was doing when he was disproving 3 and 4, but I used a 4x4 instead of a 1x1. I put in the 2 x 4 and then the 2 x 1, and then tried solving it from there. In the remaining 2 x 3 space I have two 3/4 x 1+1/2s standing upright on each other, and to the right of them a 1+1/4 x 2+1/2. In the final space there's a 1 x 1/2 and a 1/4 x 1/2. Very fun problem to work out! (But maybe explaining it in words alone is somewhat confusing, hehe...)

[@Hunne2303]

nah, I figured 3+4=7 and therefore it should work...out of the depth of my guts...

[@jrjihu]

I got a different solution for the 7 rectangle problem. I started off by marking an 8 by 8 grid. I started with two 6 by 3 rectangles next to each other creating a 6 by 6 square in one corner of the large square. Then on one side add two 2 by 4 rectangles, leaving only a 2 by 6 area empty. Then I added another 4 by 2 rectangle leaving a 2 by 2 area empty. Finish it off by splitting that final square into 2. So in the end I have two 3 by 6 rectangles, three 2 by 4 rectangles and two 1 by 2 rectangles. I wonder how many solutions there are for each size.

[@mapr1049]

I got the same solution.

[@doctajohn07]

I did this one too! I paused the video before they gave the solution and was surprised when they did it differently!

[@donaldhobson8873]

@@doctajohn07 Me too.

[@pataplan]

Same solution here. Two 18 unit rectangles, three 8 unit rectangles, two 2 unit rectangles, add them all up and you get 64 units.

[@Pulsar77]

Yep, I had the exact same solution.

[@jacobstratford1754]

Great video guys! I found a visually appealing rotationally symmetric solution to 7, which was pretty satisfying. Take a 34x34 grid, and the following rectangles 8x16 (1) 10x20 (2) 9x18 (2) 5x10 (2) 8x16 goes horizontally in the center of the grid, 10x20 are both horizontal in the upper left and lower right corners, 9x18 go vertically in the lower left and upper right corners, and the small two fit in the gaps. Considering using it for a new patio stone layout

[@Zwiezwerg92]

This doesn't add up. Your rectangels would form a 34*28 rectangle.

[@jacobstratford1754]

@@Zwiezwerg92Whelp you're right, those numbers didn't make any sense. I plead the sleep-deprived college student at the time. I redid my algebra and got the same sort of pattern I remember seeing originally, but with different numbers. 1x2 (2) 2x4 (3) 3x6 (2) 3x6 goes vertically in lower left and upper right corners, 2x4 goes horizontally in the upper left and lower right corners, other 2x4 goes vertically in the middle, and 1x2 fills the gaps. I just noticed your comment and felt I had to dive in and figure it out once and for all. I remember being so determined to find a solution with a rectangle centered on the board. Thanks for double checking me!

[@GregoMorgan]

In a 4x4 grid (A=16), you have to end up with 2x1 (A=2) or 4x2 (A=8) blocks. You can do 2x8, 1x8+4x2 or 8x2, none of which is 7 blocks. For 5x5 grid you know you'll never make to 25 by summing blocks of 2 and 8. 6x6 now gives you new 3x6 blocks and an even area. Then just find a way of summing 7 blocks of 18, 8 and 2 to get to 36. 18+8+5x2 = seven blocks = bingo.

[@txikitofandango]

Took me 3 days to figure out the n=7 case but it was worth it. Kind of cool that my way was totally different; I ended up with an 8x8 square instead of Tom's 6x6. I had two 1x2, three 2x4, and two 3x6 rectangles.

[@Hackfresse92]

This is a really cool puzzle. Now I'm wondering what the solution is for other ratios.

[@KiLLJoYYouTube]

For all ratio X:1, You can increase the number of rectangles by 4 each time, and go on forever. By "Going up" and "Going down", you could do modular arithmetic each time until you get to your sequence of a length you could go on forever. A solution to this problem will require looking at unique solutions for squares. edit: Wrote this comment like 3 times tripping over myself, Lol.

[@ragnkja]

@@KiLLJoYKZbin So proving that you can make all numbers over a certain size shouldn’t be too hard. The challenge, then, is to find out which small numbers are impossible for each ratio, and if there is a pattern to it.

[@TheBasikShow]

@@KiLLJoYKZbin How can you increase by 4? If you use the going down method you increase by X²-1 (for a ratio X:1), because you can choose any one rectangle, split it into X squares, and split each square into X rectangles. The going up method increases by 4(X-1), because you need four quarter-edges of dimensions X:(X-1) each to make a larger square. Now, there is one other thing we can do. If we split our initial square into four smaller squares, we can solve each sub-square to get a bigger solution. This is, if n, m, j, and k are solvable then n+m+j+k is solvable. In particular, if n is solvable then 4n, 7n, 10n, 13n, 16n, and so on are also solvable. Then, if n is solvable for some X > 2, we have that n+4k(X-1), n+k(X²-1), and n+3kn are solutions for all positive k. Interestingly, we are no longer guaranteed that every sufficiently large n is solvable: for X = 4 our derived solutions are n+12k, n+15k, and n+3kn, which can only cover differences which are multiples of 3. Of course, having a guaranteed n+4k would solve this problem for X = 4, and I think for all values of X.

[@TheBasikShow]

Ah! You can also make a different going-up by building a 2:1 rectangle out of your X:1 rectangles, and then putting four of these 2x1 on the edges. You need at most 2X rectangles to build each of these, for 8X total, which means that: For an X:1 ratio, if there is a solution for a value of n then there is a solution for n+4k(X-1), n+k(X²-1), n+3kn, and n+8kX for any positive whole number k. For even values of X, the greatest common divisor of X²-1 and 8X is 1, so all sufficiently large n have solutions.

[@KiLLJoYYouTube]

@@TheBasikShow you can split a rectangle into 4 smaller rectangles of the same ratio. Giving you +3 new ones. The +4 is from 4 new ones. Not entirely sure why you would quadruple everything

[@LiborTinka]

Reminds me of HV partitioning used in fractal image compression scheme based on PIFS (Partitioned Iterated Function Systems). The HV partitioning better represents horizontal and vertical edges than than the quadtree partitioning.

[@jeremykunath954]

i did 7 in another way (my starting square has sides of 10) i used a 4*8 and a 2*4 then placed them on top of each other that filled 80% of one side then i used 2 3*6 and put them on top of each other and placed them in a corner and then i filled the rest with 3 2*4

[@pantomime5919]

A similar problem was given in the danish Georg Mohr competition in 2002: Show that a square can be partitioned into n > 5 squares. The solution for the square problem can be modified to show why n > 4 works (without treating n = 7 as a special case): Let m = n - 4. Partition the square into (m+1) x (m+1) small squares (all equally large: 1 x 1), and merge m^2 of these squares so that you have a "large" m x m square and (2m + 1) small squares. Merge two small squares to make a 2:1 rectangle. You can systematically do this m times to make m 2:1 rectangles. Then divide the last small square and the m x m square into two 2:1 rectangles. This makes in total m + 4 = n 2:1 rectangles.

[@freerkderuiter8822]

For some reason I’m thinking of tatami mats right now.

[@RGC_animation]

XD Same!

[@fintux]

I did my own version for the 7. It also has three different sizes, but it is rotationally symmetrical. The sizes are (W x H, row by row): 2x4, 6x3 2x1, 4x2, 2x1 6x3, 2x4

[@kumquatlich]

If there's a new one every year I'd live to hear other similar puzzles/questions. This was a great video

[@TomRocksMaths]

:)

[@lk5355]

An equivalent 7-rectangle solution may be directly derived by using an n+5 transformation of an existing 2:1 rectangle into six 2:1 subrectangles. A surprisingly straightforward set of transformations from n total rectangles to n + [any odd number > 1] total rectangles includes this n+5 case as well as the video's n+3 case. (Apologies if this approach is already buried in the 1000+ existing comments.)

[@shapiroyaacov]

So we know that 1,3 & 4 are not possible for the ratio of 1:2. What happens if you change the ratio (e.g., 1:3). What is the general rule for ratio 1:X (if there even is one...)

[@yaeldillies]

Consider a ratio `a:b`. Breaking down and building up still work and give you n + 3 and n + 4 if you have n. So you know that if you can do n, you can do everything strictly greater than n + 5 (Frobenius applied to 3 and 4 gives 3 * 4 - 3 - 4 = 5). So the situation is like this: There's a bunch of stuff you can't do. Then you can do one of them. Call it n_0. Then maybe you can do some of `n_0 + 1`, `n_0 + 2`, `n_0 + 5`, maybe not. You can definitely do `n_0 + 3` and n_0 + 4. Then you can do anything >= n_0 + 6. The only uncertainty is about what is `n_0` and whether you can do `n_0 + 1`, `n_0 + 2`, `n_0 + 5` To determine `n_0`, let's first consider the simpler case of ratio `1:b`. You can definitely do it with `b` rectangles of size 1 x 1/b. And you can't do less because a rectangle has sides c x bc (or bc x c) for some real c and so bc

[@shapiroyaacov]

@@yaeldillies I'm not sure you can say you can do n+3 or n+4 when the ratio is different. Those builds are using the ratio of 1:2...

[@highpath4776]

@@shapiroyaacov Using a ratio of 1:2 , min is 2, 1:3 surely min is 3 1:4 is min 4 and so on.

[@shapiroyaacov]

@@highpath4776 I completely agree with that. But what about the rest of the numbers?

[@zanti4132]

@@shapiroyaacov No, the n+3 and n+4 techniques are still valid. Any a×b rectangle can be broken into four rectangles, each with side lengths a/2 and b/2. That takes care of the n+3 case. For the n+4 case, the "build up" technique still works. It's just a matter of finding the ratio needed to wrap four rectangles around a square. Looking at 3:1 as an example, every n can be ruled out except 1, 2, 4, 5, and 8. 1 and 2 are clearly impossible, leaving only 4, 5, and 8 to investigate.

[@tumbleddry2887]

I really appreciate Tom's explanations of mathematical concepts and proofs. He has a 'way' that I very much understand. Still won't get me into Oxford, but I'll definitely understand much more about the world than I did. I wonder if this is what Professor (or should it be Sir?) Roger Penrose went through when working on tiling the plane......

[@TomRocksMaths]

[@OrangeC7]

This guy has an anime hairdo, and now I want to see an anime where the main character is a mathematician

[@official-obama]

Dr. Stone?

[@Antoine893]

In Moriarty the Patriot the MC is a maths teacher, but we never see him teach

[@TomRocksMaths]

I'd watch

[@aminulhussain2277]

@@dandre8019 Follow your own advice.

[@dandre8019]

@@aminulhussain2277 i deleted my childish comment. thanks for reminding me!

[@gusmichel7035]

I came up with the +3 by splitting by 2, then with a +8 with splitting by 3 each direction. This allowed me to similarly reduce to solving all but a few small case, not nearly as small as the video got with the adding step, but finite.. After checking back in, I was given the +4 I absolutely should have seen but missed. At the idea 7 was possible, I paused. I floundered a bit, but realized that starting with the obvious 2, adding 8 by subdividing one part by thirds to get ten and then subtracting three using the +3 rule in reverse would get me there, and this worked because since I had added 8 by splitting 3 by 3, I could merge 2 by 2, reducing by 3.

[@NoNameNoLastName]

This was awesome, I'll use it in my CS interviews.

[@TomRocksMaths]

:)

[@highpath4776]

Indeed how can you python code or test this ?

[@NoNameNoLastName]

@@highpath4776 Problem solving is also part of an interview.

[@thegreaterconundrum]

@@NoNameNoLastName Pretty much everyone in CS watches numberphile 😂

[@MrRyanroberson1]

15:07 i guess the intuitive way to make sense of it is to start with a 3x3 grid of 2x1 rectangles which makes one large rectangle. you need one more to finish the square, making 10 total, and you can merge four of the rectangles from the 3x3 to make it 7

[@halfplushalfsqrt5]

This was my approach, too, though it took a moment to realize that I could divide by numbers other than 2.

[@MrRyanroberson1]

@@halfplushalfsqrt5 makes me wonder what the limitations are given N different rectangle sizes. for N=1, all the squares must have even area, and therefore even side length, and therefore the number of rectangles must be an even number; other numbers get impossibly complicated very fast

[@carlocatalano9662]

It was very enjoyable,like playing with tangrams when I was a child.

[@TomRocksMaths]

:)

[@drawapretzel6003]

Ive paused the video at "what numbers can we do it for" all of them, all you need to prove is that you can tile a 2:1 rectangle with *other* 2:1 rectangles in a pattern, and then you can infinitely subdivide the square into repeating patterns. Job done.

[@drawapretzel6003]

a few minutes in, i see he did exactly what i reccomended. Well, since it only gets you 3x numbers, you find another tiling, use a covering pattern, and then you can break it down into any addition of numbered tilings above a certain number.

[@safepethaven]

Quilt-makers for generations have used the "building-up" technique to create quilt squares to be pieced together, so high-end academics need not be the only persons to use "theoretical AND practical mathematics. ;-))

[@TheKingofkrypton]

For my 7, I used a 4x4 square with two (1x2)s stacked vertically on one side, two (1 1/2 x 3)s stacked horizontally in the top of the remaining space, another (1x2) below them in the corner, and two (1/2x1)s stacked horizontally in the remaining bottom space. These puzzles are great.

[@Martin-y8v4k]

for information, how many time had the applicants for answering that question ?

[@RobOwenKing]

Interview questions at Oxford aren't really timed. The interview itself will last 20-30 minutes and you'll explore two or three problems in that time. Also, it's not just that they say "do this problem" and you do it and that's it. You will talk through your thinking and what you try with your interviewers. That's what they're interested in seeing: how you tackle new problems and things you've never seen before. Not whether you get the right answers to some specific list of questions.

[@johnbrazier2272]

Robert is right - this is an interview question. But for the formal exam, a science exam will typically be 3 hours, have 10 questions on the sheet, and you'll be asked to answer 6 or 7 of them. And you have to show how you get your answer. And note: you'll be taking more than one exam for each subject (so one classic division for maths is a pure maths paper, and an applied maths one). Lastly, Oxford and Cambridge entrance exams tend to look for "something more" than just regurgitation of facts or repetition of a standard proof: they're looking for evidence that you can, in fact, think - not just remember.

[@nobugsnohugs6040]

There are a lot of solutions to N = 7, I got one looking like this: 4/5 by 2/5 at a corner (for example right bottom), 1/5 by 2/5 above it: we filled right 2/5 of the square completely. Than we place 3 of 1/5 by 2/5 vertically in the left bottom corner to get a smaller square 3/5 by 3/5 at the top left. Than we just devide it into 2 parts as we have done yet and there it is! Beautyful math problem, thx :)

[@sebastianelytron8450]

The Bermuda Triangle used to be known as the Bermuda Rectangle, until one of the sides mysteriously vanished.

[@shubhamraj25]

Then it'll have a hidden conjugate triangle people think is safe but was part of risky rectangle earlier

[@vigilantcosmicpenguin8721]

Oh no, we lost a vertex!

[@killianobrien2007]

I found the hidden 180° angle the government doesn't won't you to see!

[@jj_...]

Ever consider that there might be a complex reason, for which i might be involved?

[@Triantalex]

false.

[@RabbitsRi]

Tom Crawford's Oxford question is also happens to be the 'Brain Buster' puzzle from an ancient IQ test booklet. Allegedly only the top minds could answer it, without looking in the back of the book of course. It was worded differently, it gave out the first bit in as a given, all numbers but 3, 4 and 7. And it asked which of them was possible. Timed response. I didn't answer it correctly, as I was young and not mathmatically inclined.... Or interested in taking an IQ test, but damn. There it is.

[@jeanterre1134]

Is there any sort of relation between the fact that it doesn't work for 3 and 4, and the fact that, if it works for n, then it works for n+3 and n+4?

[@Nemelis0]

The only relation I can come up with is that 3 and 4 are the 'magic' numbers which you need to solve it for all other numbers except 7. But if that is the true relation I don't know.

[@hadrienlart]

I was gonna say the same. The n+x works only with x=3 and x=4. But if n+x=3 works it would mean n=0 or n=-1 which is impossible. Same goes for n+x=4 I means that n=0 or n=1 who. Again is impossible.

[@aldobernaltvbernal8745]

@@hadrienlart but what if n+x=2

[@SgtSupaman]

@@hadrienlart , proving n+3 doesn't imply n-3, because n has to work for n+3 to work. Meaning this doesn't apply to 3 or 4, because neither of them work. By your logic, 2 and 7 also shouldn't work, because neither of them are 3 or 4 more than a possible n. You have to have a base case, but, as we see here, there can be more than one base case.

[@YOM2_UB]

You could take the 7-rectangle solution at 15:06 and replace the rectangles labeled 4 and 5 with any square, creating an n+5 rule, so it's probably coincidence.

[@peterhassack]

Thank you for an intriguing and fascinating video. Looking back I just wish my maths teachers had had your exuberance and energy when I was at school - your presentation is very engaging.

[@maxwellsequation4887]

Oh yeah, Tom is back!

[@TomRocksMaths]

[@drsuper8180]

Great talk, Another solution for the 7 is to divide to square into a 3/4 and a 1/4 piece and make 2 1/2x1/4 pieces on the 1/4 side and two 3x4by 3/8 pieces on the 3/4 side and divide the remaining 1/4 into a 1/2x1/4 piece and 2 1/4x1/8 pieces!

[@alveolate]

what's going on with the stop motion, it's awesome!

[@myrenmusic1611]

there will be a smol fee and a smol donation :)

[@javgarces]

In 12:50, he sort of skips explaining why 4 rectangles doesn't work if each one touches a single corner ("you can play around with it..."). So I attempted to explain it: With each of the 4 rectangles touching one corner each, then there must be a line crossing the whole square (otherwise there would be a gap in the middle). So, the square gets divided into rectangle A and rectangle B, and we need two 2:1 rectangles to cover both A and B. We can prove the ratio of rectangle A, to be coverable by two 2:1 rectangles, must be either 4:1, 5:2 or 1:1. Discarding 1:1, the ratio of rectangle B must be then either 4:3 or 5:3 which is not coverable by two 2:1 rectangles. Therefore, 4 rectangles doesn't work.

[@tonisassano8409]

i'm trying to follow along with the explanation but tom's tattoos are so cool I keep getting distracted

[@TomRocksMaths]

[@WtbgoldBlogspot]

Paused the video and figured out the 2+3x, and came back to the vid. It was funny though, something in the back of my mind just...wasn't satisfied. Love the puzzle. :) Thanks

[@NikolajKuntner]

The academic mathematician shines through when you say "For which numbers is this possible? I can go on with this forever, but I'm gonna stop now at 11 because I'm running out of space." That is to say, the answer is given right then and there, but mathematicians like to think in the formalist box they confined themselves into. Pun intended.

[@Dziaji]

That only gives 1/3 of numbers.

[@SgtSupaman]

Well, once you've discovered the means by which all answers can be reached, there is no point in enumerating them, since, you know, they are infinite.

[@ccityplanner1217]

2n² n²+1 n²+n² Each 1 can be interchanged with any square number, so all numbers that can be made by summing square numbers (except since 1 is a square number that doesn't mean anything). For a grey guess, the sum of a quantity of square numbers that is a power of 2 (defeated by 7).

[@rayhanjasin1589]

Oh my, it literally showed up in my rec tab just 48 seconds after its release

[@jursamaj]

"Rec tab"?

[@duncanhw]

@@jursamaj recent

[@dayawalker]

@@jursamaj Recommendations?

[@jursamaj]

@@dayawalker That makes more sense than 'recent', altho I don't see anything for either that I'd call a 'tab'.

[@haikumagician4363]

@@jursamaj record

[@PiercingSight]

I found another method of doing 7. You create a spiral of rectangles along the outside, however you expand a pair of opposite rectangles (while shrinking the other two) until the gap in the middle has a ratio of 2/3, at which point you can put three small rectangles stacked side by side. I would have to pull out a pen and paper to work out the exact sizes though.

[@TrimutiusToo]

I found 7 by actually coming up with n+5 build up strategy instead somehow... I have a square split in 2 rectangles... I attached to it another square 3 times smaller which is another two, and then build around 2 edges takes exactly 3 more ractangles that are twice as big as the small ones

[@Connorses]

I got to the 7 rectangles, but I did it by breaking up the square into different size grids and experimenting. The solutions tend to line up on a grid within any given square, and it's an easier way to find new arrangements than trying to think in fractions. Also, if you're left with any uncovered grid squares they can obviously be broken down into their own grid.

[@Gunstick]

Liked your stop motion animations!

[@janmelantu7490]

For the 7 rectangle, I used a 10x10 grid (really it’s a 20x20 but I don’t speak French so 20 isn’t helpful). In two of the corners (diagonally opposite) there are 3x6 rectangles, and in the other two there are 3.5x7 rectangles. The central rectangle is 4x3, which can be split up into a 4x2 rectangle and the remaining 1x4 can be split in half. 4 outside rectangles, 3 inside rectangles. I like how in Creating 7 rectangles, some of them have 7 unit sides.

[@caspermadlener4191]

This could be in IMO training! Nice question!

[@TomRocksMaths]

:)

[@johnchessant3012]

I solved it!! (except I initially thought n=7 wasn't possible until Tom said it was) My solution was the same except: 1) Instead of "building up", I got n=10 by making a border of eight (1/5 x 2/5) rectangles around a (4/5 x 4/5) square. 2) For n=7 I used four (1/5 x 2/5) rectangles, two (3/10 x 3/5) rectangles, and one (2/5 x 4/5) rectangle.

[@guyedwards22]

I feel like it needs some appreciation just how beautiful this guy is 😤

[@BlaqueT]

Seconded

[@Riuyilmistico]

Nah

[@axlrose5082]

Focus on the maths, Edwards!

[@archenema6792]

Eewwwww

[@TomRocksMaths]

[@feliomichaels]

Ironically I ended up solving this for squares, as a mathematical faitdivers to myself, when I was younger. The same fundamental principles (of n+3, and modulo 3 solutions being derived from there) apply, even down to the n=7 case (which has an analog in n=6 for squares), so when I saw this, it gave me a smile that someone went for this problem like I would.

[@Iroh72]

I'd take that admission test, knowing that I have no chance passing, just to meet that guy!🥰

[@TomRocksMaths]

if you're ever in Oxford, come say hi :)

[@Iroh72]

@@TomRocksMaths Aaah omg🤩 I take your word for it!🛫

[@CraigGidney]

It was interesting that he described the 7 case as tricky. I thought the building-up and the proving-3-4-impossible was the tricky part! I got the 7 pretty soon after seeing the "split into 2x2" because I realized you could generalize to splitting into NxN and then also merging an NxN into 1, letting you go up and down relatively freely. So e.g. you start with 2 rectangles -> split 1 into 3x3 giving a total of 10 rectangles -> merge a 2x2 within the 3x3 into 1 giving a total of 7 rectangles.

[@poorman-trending]

What about other proportions? 1:3 , 2:3, etc....

[@richardweiss5217]

For m:n with positive natural m and n, you can always add 4 by building up and add m*(n^2) - 1 by breaking down (first fill 1:n by n^2 of its copies, then copy the result m times). If the breaking down number is not coprime with 4, we won't be able to reach all numbers this way. So almost all natural numbers for even:odd ratios and at least almost all numbers in some mod 4 modulo classes for other cases.

[@GynxShinx]

This is an excellent example for how proofs feel as well as being a fun problem.

[@TomRocksMaths]

:)

[@garywalker8493]

I found a solution for N=3, but I don't have room in the margin to show the proof.

[@arnhelmkrausson8445]

I see what you did there, Fermat...

[@jayantjha3128]

Underrated

[@8Clips]

I used to always think about what scores it wasn't possible to have in Rugby, and never got further than 1, 2 and 4, but I never managed to figure out the chain of logic, linking it to once you have three consecutive scores, you can just add three over and over infinitely, until literally like second year problem solving. It seems so obvious now, but I remember thinking how elegant it was to prove that all numbers work infinitely after that with such a tiny bit of logic.

[@joaquinvigara1356]

I’m a simple man: I see Tom, I click

[@TomRocksMaths]

[@uy-ge3dm]

I paused it at 1 minute and solved it myself. Took a while but it was fulfilling. I believe that the answer is 2 and all numbers 5 and up. The way I proved it was as follows: I found a construction for n=5, n=6, and n=7, and then proved that if n works then n+3 works. The way we do this is to take a square and divide it into a 2x2 grid. In the top left square we put our solution for n scaled down by a factor of 2. In the top right square we divide it into two 2x1 rectangles. The bottom two squares are joined to form a big rectangle. Therefore, n+3 is possible. I'm looking forward to see what Numberphile's solution is.

[@Mrfrog2024]

now i kinda want a coaster with square seven

[@ianwrzesinski5676]

Looks like there are a ton of ways to get 7! I got it by trying to stack a vertical (x) on a horizontal (y) as the full length of one side, and solved 2y=x and 2x+y=1 for y=1/5, x=2/5. Then with a square of length 5 I used two 1.5x3 and three 2x1 to finish it.

[@robinder_]

yeah, same bro!

[@ambassadorkees]

Next question: HOW MANY WAYS to fill a square with x rectangles?

[@yaeldillies]

The corner remark (along with some arithmetical argument maybe) shows that there are only finitely many ways!

[@highpath4776]

@@yaeldillies I had not spotted the corner element, one could indeed split this into a vertices matching (topography?) method

[@0xfolling]

I solved the 7-rectangles version differently. Thinking about it I noticed that whatever I used as a solution I would at some point need to have a square that can be solved. Since we know a square can only be solved with 2, 5, and 6 rectangles I would need to end up at that square using either 5, 2, or 1 rectangle. 1 was clearly impossible, as was 2, so I tried it with 5 and it turned out that that's possible. This is interestingly also the solution you have, although you explained it differently. But the square for you is in the upper left side, but not on the edge. My solution goes as follows: imagine a 5x5 square, cut a 3x3 square off the lower left corner. That leaves 5 rectangles to fill the remainder. To the right of the square is now a 2x3 empty space, which you can conveniently fill with three 2x1s. That leaves the top bit which is now a 5x2 which can be filled with a 4x2 and a vertical 2x1. That uses 5 rectangles, leaving only the square which you can slice in half for the 2-rectangle solution, et voila: 7 rectangles with a 2x1 ratio that fill a square!

[@RafaelSolaPACalsaverini]

That's a freaking stylish mathematician. Dude's bright and handsome. Nice hairdo, brilliant tattoos. Awesome math.

[@TomRocksMaths]

[@RafaelSolaPACalsaverini]

@@TomRocksMaths

[@Aurora-oe2qp]

So, I went about this a slightly different way. I was trying to knock out all mod 3 categories. So n ≡ 0 (mod 3), n ≡ 1 (mod 3), and n ≡ 2 (mod 3). First we can divide it in 2, but for each rectangle we have we can divide it into 4 parts, which leaves us with 3 more ones. So that gives us 2 + 3k. For the second one I did the same as he did: put a square in the middle with sides 1/3, divide it in two, and then use 4 (2/3)×(1/3) rectangles to cover it up. This gives us 6 + 3k. Now, the last one is a bit harder, but I thought some, and came up with a solution that uses 1 (4/5)×(2/5), 2 (3/5)×(3/10), and 4 (2/5)/(1/5) rectangles. It's a bit complex, but, it looks a bit like this: _______________ | | | |___________|__| | | |_____| | | |_____| |____|____|_____| This means that we now have 7 + 3n, which means that we have a number for each congruence class of mod 3. The only ones we're left with that aren't proven possible are 1, 3, and 4. 1 is obviously not possible, and neither is 3 and 4 which is pretty easy to see. You can play around some, but you'll notice that one of the rectangles have to take up half of the square, and then there's just no way to tile 1:2 rectangles with 2 or 3 smaller 1:2 rectangles. So, that leaves us with all natural numbers except 1, 3, and 4. Aaaand, that's it, I guess?

[@michaelpennington6935]

Does anybody have an idea as to why a video about rectangles has brought out the nutters in the comments section?

[@mazetlionel]

There is another solution for the partition of the square in 7 parts: Layout the square in a 4x4 grid, with (A, B, C, D) columns and (1,2,3,4) rows. In column A : fill with 2 rectangles (1/4 x 1/2) In row 3 : fill in column B, C with 1 rectangles (1/2 x 1/4) In row 3, column D, fill with 2 rectangles (1/4 x 1/8) The last part is a 3x3 square, in columns B,C,D and rows 2,3;4 : divide it in 2 rectangles of (3/4 x 3/8) So to recap : 2 + 1 + 2 + 2 = 7

[@TheGarbageMann]

feels like it's oddly related to the cantor set

[@Ray25689]

In which way?

[@TheGarbageMann]

@@Ray25689 in divinding objects in half and the number of objects in the cantor set still being infinite

[@Melomathics]

The common bit is the fractal nature of both sets.

[@Ray25689]

@@TheGarbageMann yeah, but how does this occur here? All you do is dividing things up, but nothing with infinity.

[@TheGarbageMann]

@@Melomathics that's what I was thinking as well, but it feels like there's more than just the fractal nature, as the cantor set is mainly about dividing a whole into infinite fractions

[@hameedamathtuber]

That was an awesome solution. Excellent 👏

[@TomRocksMaths]

Glad you enjoyed it :)

[@camerongray7767]

This was very fun

[@TomRocksMaths]

:)

[@as-qh1qq]

I think 3n-1-k+\sum_{i=1)^{k}m_i^2 where m_i and k are any integer greater than or equal to 1, also covers almost all numbers. The 3n-1 came from recursively dividing while the k squares come from dividing ith rectangles into a m_i by m_i grid

[@TheHeroOfTomorrow]

Patrick: “RECTANGLES!!!”

[@ItachiUchiha-ns1il]

Came here looking for this comment lol

[@rhysknight8681]

This guy seems like the best dude

[@carloscontente]

Maths are quite interesting but I would like to see more information about this cute guy's tatoos.

[@jucom756]

My way of solving math problems is try out a few examples of the proposed scenario, try to find some sort of pattern, develop a ruleset for that pattern, try to prove that ruleset wrong, if i find a way it is definitely infinitely right i make a formula for clarity and then make an explanation that arrives at the formula. I had trouble fonding a definite formula for th8s since it is just a case by case sort of proof and in my problem solving method the case by case is the first step before pattern finding but i instinctively don't trust case by case which made me not try enough things to notice the patterns... :/

[@ExtendedDelta]

hahaha Thank you mathematicians for making us engineers seem normal.

[@racheline_nya]

for n=7, i found a symmetric way to fill it with all denominators being powers of 2: there are two 3/8 by 3/4 rectangles in opposite corners, and two 1/4 by 1/2 rectangles in the other two corners. to fill the gaps in the sides with 3/8 and 1/2, we use two 1/8 by 1/4 rectangles, and then one last 1/4 by 1/2 rectangle in the middle. i'm not sure if i explained it clearly, so here are the coordinates of the rectangles ((x1,y1,x2,y2) means (x1,y1) and (x2,y2) are two opposite corners of the rectangle): (0,0,3/4,3/8), (0,3/8,1/4,1/2), (0,1/2,1/4,1), (1/4,3/8,3/4,5/8), (1/4,5/8,1,1) (3/4,0,1,1/2), (3/4,1/2,1,5/8) i'm now curious about ways to fill a square with 1:2 rectangles such that if you remove any pair of rectangles, the hole left behind is not a rectangle (with any ratio of sides). i know it's possible with infinitely many rectangles - we can repeatedly break down the square into 4 rectangles and a square with a 3 times smaller side length, similarly to the "building up" part of the video. i'm not sure about any finite amounts of rectangles.

[@lelandshennett]

Punk rock look Big math brain Love it

[@TomRocksMaths]

[@vivek-battan]

What if we want to find the number of possible ways (Ratio of the rectangles/ratio of smaller or bigger side/ratio of area etc) to construct a square out of 7 rectangles. Here is what I think how we can do this: 2(a²+b²+c²+d²+e²+f²+g²)=A² [Where a to g are the smaller side of every rectangle and A² is area of the square constructed].