Homework vs. Research (Science Might Not Be What You Think)

I participated in an REU (Research Experience for Undergraduates) over the summer, and I've been thinking about how to do a video encapsulating how it felt. Being a researcher in Science is really completely different from being a student, and I've done the best I can to encapsulate that feeling with this video.
If you agree or disagree with me, please feel free to tell me in the comments!
Link to Preview Video: • Four Points Two Distan...
Resources
Blog Solution
www.solipsys.c...
Music
Reminisces of New York (Zhang Shuwen) www.google.com...
Other songs are listed by KZbin

[@berryesseen]

I have a PhD in probability theory (or in information theory to be more specific). You explained pretty well why research is different than doing homework. When I was a student, I had no trouble in doing any homeworks or exams. Before you start, the answer is either given to you, or you, at least, know that the problem is "solvable". You are supposed to be able to solve it in a few minutes. This is a huge relief. But in research, the story is very different. You are trying to do something nobody has done yet. There is no teacher to ask for answers. Your advisors usually know nothing about the problem. You are alone in your goal. You have to be self-motivating, persistent, and most importantly, creative. It can become very frustrating to get stuck at one technicality for months. This happens quite often, and it happens to even the best ones. But once it's solved, you feel very satisfied. Then you proceed to grind on another problem. The cycle continues. Good researchers are by definition good students. They can reproduce things that are already known. They are also expected to verify the validity of new solutions. But good students are not necessarily good researchers. Understanding and creating are two different things.

[@copywright5635]

Thank you for the comment. I admit that the puzzle in the video is not exactly analagous to the difference between being a student vs researcher. However, I wanted to provide a little taste of that experience for those unaware, and I hope that delivered. Research is inherently a creative process. Like you've stated, you have to be very self motivated. I think that often students think that they will make good researchers after doing well school, but are dissapointed/shocked after pursuing doctorates. This is why I think REUs and the like are so important.

[@DhavalBothra]

​@@copywright5635 Hi, in what way can I communicate with you a bit more easily?

[@musicplaylist6909]

not to mention the stage where whatever solutions or ideas you yourself validated will then be placed under intense scrutiny due to the nature of academia. All this before even going anywhere near publishing. An underrated skill of researchers should be the ability to take criticism and patience to go through multiple iterations of their methodologies to ensure the research itself is robust.

[@thuongthuong4457]

Thats why I make everything reseach

[@WilfNelson1]

The nice part about research is it is you versus the universe and you just have to figure out the problem. With homework unless it is just calculation there is a 'correct' answer defined by a teacher who can be wrong. This means you aren't really solving the problem you are trying to mindread and guess the answer the other person has prepared and it can be incredibly difficult to do this if you are bad at theory of mind. Research is wonderful when you have to figure out a problem, have it reviewed and work together. Homework is miserable and almost the opposite of how you teach a mind to solve real problems.

[@lih3391]

Rediscovering and rederiving things myself is always more fun for me than looking it up😁

[@copywright5635]

Totally agree!

[@lamMeTV]

its a shame that school doesnt care about fulfillment like that

[@somerandomboi8239]

​@@copywright5635Genuine question: Isn't it much more time efficient to look it up? Like there's other studying to be done, not much time to lose. Or are you two just that quick and brilliant? Genuine question about your experiences, not trolling you I swear. Bcs I feel completely stupid in the face of this. I rarely derive anything on my own without losing way too much time. Then I look it up when I'm like two steps close (bcs I feel a bit stuck and panic) just to save time. And that's only when I feel comfortable and confident enough to even try. Most of the time I'll just look it up to begin with. Am I stupid? P.s. what are your IQs? Idk mine I'm scared to check.

[@copywright5635]

@@somerandomboi8239 I wouldn't put very much stock into IQ at all. Never tested mine, and don't plan to, unless I had something other than vanity to gain. There's nothing wrong with looking things up, but if you want to learn something, you have to do it yourself. Often in life, that involves something physical. However, with math everything is purely mental. Essentially, looking something up is the same thing as getting someone else (or an AI) to do it for you. It works in the moment, but you don't learn nearly as much. It depends on priorities, sometimes it is efficient and convenient to look things up. But, for example for something esoteric like the example I showed, you don't really gain anything by just knowing the answer. However, the process of puzzling over it, trying to come up with a solution, that is a mental exercise that will be much more rewarding, regardless of it you get the answer correct

[@somerandomboi8239]

@@copywright5635 thank you very much for replying. TLDR at the end. I was recently learning centripetal motion and was thinking the whole day (bcs I wanted to draw them correct on my own) about the implications of friction vs frictionless (banked road) for the vectors on the car. I ended up with two ways that seemed possible, one of which was correct. But then I felt really reluctant to try it out with numbers or ponder any more to see if it makes sense - I just wanted the anxiety to go away and Iooked it up. I was happy I got it right, but also sad bcs I don't *quite* understand the vectors of friction in centripetal motion very deeply. I mean so deeply that it's just natural, you know... like you understand this solution and proof of yours. You know that feeling when you get and don't get something at the same time? Do you ever have it? I kind of do with the fact that friction disappears with ideal speed. I get the problems right every time, but it just seems so counterintuitive that friction would ever disappear on a tilted plane. TLDR I usually understand most formulas, but I feel like there is absolutely no way I could have ever discovered any of them on my own. What was the greatest thing you have ever derived/discovered on your own? I would appreciate a long reply.

[@sureashkumar1524]

"Comfort in math" a statement us pasents will never understand.

[@krishnachoubey8648]

anyone a bit well versed in math would understand. Math research is all about proof and is mostly left at that, while research in other sciences such as biology, medicines, etc. may need to also provide applications, it's occurrences (if any), etc. [you get the point]

[@gambrytube]

honestly I am terrible at math and I make mistakes AAAAAAAll the time but I still find a certain comfort in it, its quite simple when you get down to it I personally feel that we all associate it with difficulty due to how its taught

[@sureashkumar1524]

@@gambrytube I have nothing against math, in reality I quite enjoyed it until I got in class 11

[@users4007]

my genetically inferior ass cant comprehend

[@vitr1916]

In my opinion, homework is motivated for paying an attention and for learning ideas; it is really easy to get bored. Research is motivated by curiosity and starting an idea.

[@coco-ongelzela]

Homework teaches to bring your work at home for future employers

[@quantumgaming9180]

I always like when you post a new video. They feel like watching video essays sometimes; relaxing to listen

[@Kofhiliphus]

I think one of the most annoying but probably important parts of research is when you found an answer but you really have to spend a lot of time justifying that same answer even if you state the obvious because that's really how it should be. Ironically, this video came up at my recommendation as I am making my research paper.

[@geronimonm823]

Dots answer: 3, 6 or 7 depending on rules. Non overlapping: Square, Equilateral Triangle with dot on center, Rhombus made up of 2 Equilateral Triangles. Overlapping: 3 overlapping dots + another anywhere else, 2 sets of 2 overlapping dots any distance away, Equilateral Triangle with a dot overlapping one of it's points. Less than 2: 4 dots overlapping in space.

[@geronimonm823]

Apparently I sucked at this. I tried to get a rigorous proof but I'm sleep deprived so I gave up really easily in favor of finishing the video.

[@IustinThe_Human]

​@@geronimonm823 It was the same for me.

[@realityChemist]

Really great video! I'm working through a PhD in materials science right now, and I was just thinking about this uncertainty today actually. I'm about to write a paper and I think I've got pretty strong evidence resolving my research question, but I'm still not *sure* I'm not missing something, and I basically can't ever be. So I'll write it up and hope whatever mistakes I'm making are minor and not foundational!

[@alfredfeldt5694]

i have managed to find 4 so far 1.a normal square 2. a "diamond" that is made up from two equilateral triangles sharing one edge, with the last edge stretching between the two points 3.a kite, and equilateral triangle with a a line pointing from the tip through the middle of the base (all these egdges are same length) and the last two edges connecting to the point below the base. 4. a triangle with a point in the middle. Something i noted is that the diamond has 1 long edge, the square 2 long edges, the triangle 3 and the kite 4, perhaps this leaves a 5th shape with 5 long edges. the though of putting some points on top of each other also came to mind but seemed quite cheesy.

[@Noname-67]

It took me about 15 minutes to find all 6 arrangements and prove that there're no other cases. I used a graph theoretic approach like this. I splitted the problem into 3 cases based on the number of equal segments and analyse their graph before geometrically realising them. The first case has 5 equal segments. Any 5 segment from 4 points form 2 (possibly degenerate) triangles that share the same side. The only possible way they can all be equal in length is to form 2 equilateral triangles. That's the first arrangement. The second case has 4 equal segments. Any 4 segments either form a quadrilateral or a triangle with a tail. A quadrilateral with 4 equal sides and 2 equal diagonals is a square. This gives us another arrangement. There're 2 possible arrangements for the triangle with a tail: inward and outward pointing tail. So we have 4 in total. The final case has 2 triples of equal segments. This splits in 2 subcases: either a triple form a triangle or both triples are chains. The first subcase has a single arrangment: an equilateral triangle and its center. The second subcase can be geometrically realised in 2 different ways: a chain forms an U-shape and another self-intersect, or both chain form a Z-shape. The former gives an arrangement, an isosceles trapezoid with at a specific angle (which I'm too lazy to compute). The Z chains case gives no arrangement, because it would give a parallelogram such that each of its diagonals is equal to some of its sides, but that's impossible since the longer diagonal of a parallelogram is strictly longer than any of its sides. This gives 6 arrangement in total.

[@b_onkus2493]

I feel like this is also a good illustration of formulating good research questions, since the question as stated doesn't really discount the possibility of overlapping points with connectors of length 0. There could be 3 more configurations of the points where one of the lengths is zero, one of which appears as an equilateral triangle and two that appear as a line segment. From an inside perspective, it might be obvious that we only want configurations that have a quadrilateral boundary, but with only the question to guide me, an equilateral triangle was the first "solution" I found at all.

[@kwinvdv]

There are two more if you allow lengths of zero. The blog post does mention distinct points, which would prevent these solutions. Unless if you "scale up" the shapes, such that the zero length edges become finite and the initially nonzero length edges become infinitely long and one would allow that all infinite lengths to be equal.

[@basicusername1171]

I'm still a 1st year physics student and, while I know that I may not be good at research, I dedicate a lot of time to exploring pure mathematics on my free time because I really enjoy learning and then using that knowledge to go and try to think of a problem I want to know the answer of (not necessarily one that I find online or in my books) and then I try different approaches to proving it after I use my intuition to try and find an interesting pattern. It's actually very beautiful and my love towards mathematics grows exponentially with respect to time!❤ Thank you for this video, you reminded me why I chose to major in this: not only to do homework (although very fun) but to satiate the curiosity that I'm sure everyone has experienced throughout our journey, even if that implies being alone with your thoughts most of the time, haha.

[@Jernation]

This is a really great way of putting of it. The writing a lot part is so fr lmfaoooo.

[@copywright5635]

ikr 😭

[@vidal9747]

Me procrastinating doing my presentation on my research watching this video...

[@dank.]

Much like a math paper, I image no one will take the time to read through this, especially as I had to try to describe geometry without drawings lol. I had a fun time writing this out, anyway. I approached the problem by dividing it into casework of how many longer sides each shape has. 1 longer side: 1 shape. We can see this by starting with the longer side, then recognizing that since all the other sides are the same lengths, the other two points lie on the perpendicular bisector of the longer line. The points are uniquely determined since they must be also be equidistant from the first line and have a distance of one shorter side between them. This shape is a rhombus with 60, 120, 60, and 120 degree angles. 2 longer sides: 2 shapes, although we will have to further subdivide into casework. If the two longer sides are connected at a point or if they do not share any endpoints. This is a shape: When the two longer sides share an endpoint, we can at first imagine them attached at an angle, then complete the triangle with a shorter side. The fourth point must be on the perpendicular bisector of the first short side, either on the outside of the first triangle or on the inside. Whether it's on the inside or the outside, adding the fourth point must create a regular triangle. If the new point is on the outside of the first triangle, the only way the first point can be a shorter side length away from the intersection of the longer sides is if the longer sides are shorter than the shorter sides, which we will consider later, but for this case it is impossible. Thus the new point must be on the inside of the original triangle. This shape works, and it's uniquely identified to be a unit triangle with a fourth point on the outside one shorter length away from one of the points such that the connections to the other points are equal. In a coordinate system we can describe this shape as (-0.5,0), (0.5,0), (0,0.86), (0,1.86). This is a shape: When the two longer sides do not share an endpoint, they must be the diagonals of some shape. Since the other (shorter) sides must be equal, it is a rhombus. The two longer sides are equal, so it is a rhombus with diagonals of equal length, which must be a square. 3 longer sides: 2 shapes, and once again we further subdivide the casework. The three longer sides can all share an endpoint, or if they do not, we also have two cases: namely whether or not the three longer sides form a regular triangle. If the three longer sides do not form a regular triangle and do not share an endpoint, then there are two further cases, namely whether the longer sides form a z shape or if two of them cross each other. This can't be a shape: For the case where they form a z shape, we can start with one longer side in the middle, then add a longer side to each endpoint of this middle longer side, with their endpoints on opposite sides of the middle longer side (forming the z shape). We have that by symmetry the midpoint of a shorter side is in the middle of the middle longer side. This means that the middle of the middle longer side bisects a shorter side. We can construct circles from each endpoint of the middle longer side to imagine where we can place the other longer sides. However, the only way the shorter side bisected by the middle of the middle side can reach these circles is if half its length is at least half of the longer side. This is a contradiction, as it implies that the shorter side must be as long if not longer than the longer sides. This is a shape: For the case where they intersect themselves, we can imagine this shape being a trapezoid created by cutting an octagon in half (though annoyingly it isn't that simple). After some hellish trig (wasn't really that bad but took me a while lol) I found that It's an isosceles trapezoid with base angles of arctan(sqrt(5+2sqrt(5))). Cool. This is a shape: If the three longer sides do form a regular triangle, the fourth point must be equidistant from the first three, so it is simply on the circumcenter of the first triangle. The sides connecting this new point to the original points are shorter than the longer sides we started with, so this is a valid shape. This shape is formed by starting with a regular triangle and adding its center as the fourth point. This can't be a shape: If the three longer sides do share an endpoint, then we can connect the three non-shared endpoints of the longer sides to get a triangle. Since the three longer sides are equal, the first point is the circumcenter of this triangle. However, since this triangle is made of shorter sides, it is regular, so the circumcenter is inside the triangle and the circumradius is less than the side length. This implies that the longer sides are shorter than the shorter sides, which violates our assumptions. This is really just the previous shape, but so as not to overcount, it isn't valid in this case. 4 longer sides: 1 shape, and there are two cases to consider here: whether the shorter sides share an endpoint or not. This is a shape: If the shorter sides share an endpoint, then we can once more start our shape with a regular triangle made of longer sides. Fourth point must be chosen such that it's on the perpendicular bisector of one of the sides of the triangle and one longer side length away from the far point. This uniquely identifies a point. This can't be a shape: If the shorter sides don't share an endpoint, than the longer sides form a rhombus, and since the shorter sided diagonals are equal it's really a square. This violates our assumption that the longer sides are longer than the shorter sides so this case has no solutions. 5 longer sides: 0 shapes. Since 5 sides of the 6 possible connections must be longer sides, without losing generality we can start with a regular triangle of longer sides. The fourth point would have to be equidistant from one of the sides, forming another regular triangle. Since there is already a point on one side of the side, there's only one choice for the fourth side. However, this point leaves the far points further than the length of a longer side, which violates our assumption that we started with the longer sides, meaning there are no shapes with 5 longer sides. Thus my answer to the problem is there are 6 unique shapes which have only 2 possible lengths of sides/diagonals.

[@dank.]

After video note: hooray! I got it! I like proof based math better than competition math style problems, since I have some breathing room to make mistakes and explore. Cool video! I wish I saw the connection with the pentagon: as I was doing the math in desmos I was like "man this sure looks like a pentagon... and the trig I got sure reminds me of the trig of a pentagon..." lol. I can't wait to start doing math research

[@sphyncematry7267]

For some god forsaken reason i came up with only the sixth one. lol

[@forfucksakecomrade]

Lmao you are so galaxy brain for that

[@electrifiedbathbomb7383]

Love this channel, it makes math intuitive and more accessible. Though math is still a jumble of problems people make up to solve And as for the problem, I could only think of 2: the original square and 2 equalateral

[@StacyMcCabe]

OMG THE PENTAGON WAS THE ONLY ONE I THOUGHT OF!

[@thatoneguy7357]

I’ve always hated maths growing up but no I realise in maths, you’re either right or wrong and I’m slowly gaining passion for maths.

[@jenny_azoth]

thank for make this

[@nickm3694]

My approach started with equilateral triangles before I realized that every configuration of 4 points and 6 edges falls into either a configuration of 2 edges crossing or no edges crossing. Then, from there, I had 18 cases to check (after filtering the duplicate rotation, reflection, and short/long side inversion cases) which I could go down the line and figure out what was or wasn't possible to construct. Fun little problem!

[@codetoil]

The symmetry group is S4, the 4th permutation group. This is because switching pair of the dots will still result in all of the edges still existing. So, the solution to the problem is 4! or 24, equal to the size of S4.

[@juliannewman2ndchannelmusi475]

Results of my attempt at the question starting at 1:28: Up to enlargement and isometry, I could prove that there are exactly four solutions in which there is a connection that is either a shared leg of two isosceles triangles or a shared base of two isosceles triangle - namely, these are: a square; a kite whose opposite non-equal angles are 60 degrees and 150 degrees (leaving 75 degrees for each of the two remaining angles); a rhombus consisting of concatenating two equilateral triangles (i.e. the two distinct angles are 60 degrees and 120 degrees); and an equilateral triangle plus the point at the centre of the equilateral triangle. Using the intermediate value theorem, I think I could also prove that there exists at least one solution for which the convex hull of the four points is a quadrilateral for which the two interior connections are not perpendicular to each other. So there are at least five solutions. But I could see no way of determining the exact number of solutions outside of the first four examples, except to set up and try and solve a really complicated equation. After seeing the solution: I missed one of the ones where two isosceles triangles share a base, namely the one consisting of an isosceles triangle containing an equilateral triangle. The 6th solution is probably the one whose existence would have been constructed by my IVT argument, but IVT only guarantees existence of _at least_ one solution fulfilling a given requirement.

[@JacksonBell-ep6zw]

"Namely" instead of "mainly" at 1:45. Good point. People notice.

[@nile6076]

If you allow it, you can simply create a rectangle, then extend two vertices infinitely in any direction. The long side lengths will become equal to the diagonals at the limit, and the sides between the two points near the origin and the two points at infinity remain equal in length.

[@sirknightartorias68]

I came up with the trapezium first 😂. But never thought of the inside point one. Which I missed. I have been thinking about the same issue recently. About the uncertainty we are left over with despite having an answer to our problem. I too compared it with the H.W thing and the pre-existing solutions behind the books. But you put it up to a good perspective.

[@doobani3209]

great video keep it up 🔥

[@_samurai_x_7513]

English teaches you how to think. Research requires thinking.

[@TeilanLee]

The answer I got for the four dots problem was 6. The first thing I thought of was what would happen if you started with 2 dots; no matter how you arrange them, there will always only be one length. Then adding a 3rd dot would only be allowed to add 0 or 1 new length, i.e. an equilateral or isosceles triangle. From there, I looked for what ways a 4th dot could be placed while only using previously used lengths (or, in the case of the equilateral triangle, how many ways a 4th dot could be place while only adding one new length. However, the cases of the equilateral triangle end up being actually the same as other cases). I thought about how you could place the 4th dot in such a way that it was the same length away from the previous 3 dots as the previous 3 dots were from each other. There are a base length and a leg length that can be used, which means that you could figure out all the possible cases by making every combination of base and leg lengths. By this logic, with 2 lengths, and three dots, there would be 2x2x2=8 combinations. However, this overcounts, as the two dots at the base of the isosceles triangle act in identical ways, so the 2 cases with combinations of base leg and leg base for the two dots at the base are the same. So the calculation should be the 2 possibilities for the dot at the top of the isosceles triangle times the 3 possible combinations for the two base dots, which gives 6 cases. Edit: GG, but as some others are mentioning, you could have overlapping dots if you don't specify whether the points have to be distinct or not. With those cases, you would have 4 more cases: 4 overlapping dots, 3 overlapping dots + one more somewhere else, 2 pairs of overlapping dots, or and equilateral triangle with one point that's actually 2

[@copywright5635]

Thanks. Yeah idk why I didn’t say “distinct” points. Dumb oversight on my part. But at least, I think it’s also a lesson about formulating research questions.

[@philipm3173]

I got the trapezoid but missed the isosceles.

[@jeffreychandler8418]

This is actually pretty funny. My answer to the problem was "infinitely many" because my logic was being based on absolute positions, not distinct shapes. Essentially, my definition of configuration was different, so give my definition I would be correct, but under a different definition I would not be. In science I would present that and see how it stacks up to others beliefs. Then when you said six my brain instantly went "well I'm already in stretching absolutely coordinates mindset, so this must be 'distinct shapes'. Well you can make a rhombus a trapezoid a triangle..." lol.

[@michaelbraxton137]

Maths is hard. Failed maths and I'm going for lessions to do over maths.

[@copywright5635]

I pray for your success! You can do it!

[@JokerLover123]

The fact that you didn't quit when confronted with failure shows that you will succeed. Good luck and happy learning. 🫵💪

[@Tanvir1337]

this is so brilliant

[@crackersnip6505]

This video is actually really good

[@FPSIreland2]

So I didn’t put pen to paper, but I came up with the same idea in my head of using the three points but failed to notice the other four possibilities, so I just thought there were only two possibilities, a square and an equilateral triangle with the fourth point at the centre! Even though I barely got any of the answers, I’m still feeling pretty happy about myself that my intuitions seem to (kind of) work

[@abbasfadhil1715]

My guess is 5 not sure, but i played around with the dots. I figured that starting with an . Isosceles triangle then adding a forth dot would be easier since we know two sides are the same length . Edit: I was close and I wasn't able to find the sixth one re-doing the problem the ones I reasonably got were just 4 the 5th was more of a guess regardless it was a fun problem

[@copywright5635]

Thanks for actually pausing and trying! I always look forward to your comments on my vids. It's a tricky problem.

[@abbasfadhil1715]

@@copywright5635 no thank you for the amazing videos, I'm eager for the content more than ever u always have interesting problems that aren't too hard for a high-schooler like me to understand the Two vids that I liked the most were Rk4 and the domain problem from that pop anime. Also I wish u post more about electricity cuz the the spherical capacitor vid was phenomenal .

[@costlyhornet9994]

I just guessed 6 without really understanding the problem and am happy to be right lol, 6 just pops up a lot when it comes to 4 dots

[@koktszfung]

I just finished my MPhil degree on Physics. I realised very early on that I don't actually like doing research, because like what you said, research is for the most part aimless and don't have a clear goal. I don't want to admit it, but I actually like doing homework, the kind of works that I know for sure there is a question and an answer, and I just have to figure it out. For my research, there wasn't even a clear question I want to ask in the beginning, it was even more aimless than trying to solve some unsolved problems. Fortunately, after a little past the middle of the degree, I finally found the question I want to ask, and could begin finding ways to answer the question. The problem solving part is fun, which I enjoy the most out of the entire journey, but the part where I just play around with the topic and look for things to do is just painful and that is why I decided to stop doing research. Research seems fun at first glance because of all the freedom you are given, which could be true for some people, but it is exactly those freedom that can make you feel like you are wasting your time.

[@thanhtrungdo2520]

I think I am the same as you

[@user-xw4gn7do9x]

Before six solutions was revealed, I found the rhombus and triangle, after, I found all equilateral start solutions.

[@charlemagnethegreat3929]

If a length of 0 is allowed then technically you can create a line segment with 2 dots at each end, stacked on top of each other. Then there is only the distance 0 and the distance between each stack.

[@copywright5635]

I probably have been more thorough and said "non-trivial" configurations. But, regardless, the problem wants 2-distances, not 1.

[@tyruskarmesin5418]

@@copywright5635A distance of 0 and a non 0 distance does give two distances. If a distance of 0 is allowed, there are 9 configurations. 10 if stacking them all in one place is allowed.

[@blacklistnr1]

There's two ideas which I think you didn't give enough credit to: 1. measurement/checks 2. meticulous exploration Without 1 it's not science and sometimes the checks are pretty hard to figure out too, in this video they were pretty easy (measure the lines, count the different lengths, make sure it's 2). Good ones also come with the relief that what you did was right (whether it's a true positive or true negative). Lists (point 2) are how you make sure you don't miss anything. They often involve breaking down the problem into the space it exists(e.g break a picture down into a grid of rgb values) or something like in this video where you pick which lines of the shape are equal. Then you go through each case and check. Even if the list is not comprehensive due to various reasons, you can at least be confident that in the list's scope these are the results (Found x results in this region or found no results there) which can be a useful stepping stone for the next person because they can focus on a different region.

[@evanroderick91]

I think it's 4 ways: with a rhombus, trapezoid, square and kite (not certain if the kite is possible). Edit -- I see I was right about the kite, though I didn't think to think in terms of triangles as opposed to squares

[@newestski]

what if you had an equilateral triangle and one point infinitely far away. that way the only two lengths are the edges of the triangle and infinity.

[@tyruskarmesin5418]

I found 9 ways. Edit: three of them involved two or more points at the same spot, which he did not do. So I found all of them.

[@tyruskarmesin5418]

Also I guess if multiple points can be on top of each other then I should include all of them being at the same spot, so that’s ten ways.

[@meestyouyouestme3753]

If they all over lap then that’s just one distance (zero) not two.

[@meestyouyouestme3753]

Or is zero not a real distance?

[@AMARK1NG]

❤

[@thomaswright7325]

Yep - I HATED that aspect of writing classes. Even in college, professors have marked me down for explaining or saying something one way instead of another - despite perfect spelling and grammar. It made me irate.

[@lxtiani8135]

Buen video

[@jaredmccain7555]

This was a really good video but ngl I wasn't expecting that from the thumbnaail. But making a intresting clickbait thumbnail for this would be pretty hard tbf,

[@ArtArtisian]

For the puzzle: I'm thinking 4, two of which seem degenerate to me. We can do a convex hull of a square, equilateral triangle, line segment, and a single point. The last two require stacking points on top of each other.

[@rxeii8224]

I was searching for a video like this a few months ago since I am thinking of pursuing my career in the reasearch field, I don't know if it's the same here in my country though. :,(

[@georgekutras7063]

Not this damn thing again…

[@copywright5635]

hehe

[@MathsSciencePhilosophy]

I am not alone who don't like homework 😂

[@novygaming5713]

2:35 Infinite number of ways, as long as its sides are parallel with their opposites. (I seem to have overthought the problem)

[@potocats2274]

I have only been able to find two new ones. I got the first by shrinking one of the diagonals till it was equal to the sides.and I found another by elongating a pair of sides until they were the same lengths as the diagonals. Edit: I found the diamond and the trapezoid I tried the homework a bit but got kinda bored and wanted to see what all of them were

[@primordialious6945]

Homework has an answer, research doesn’t.

[@whtrbt]

Enjoyable little puzzle. What was the music that started around 3:54? It doesn't appear to be any of the listed songs.

[@mathics5913]

nice video

[@QSBraWQ]

I didn't quite get what distances mean, so I got one shape of a straight line... Also maybe all points in one point, but that would mean 0 distances and not 2, so idk Oh dang, with triangles it makes sense now

[@Boris.Becker.]

❤❤❤❤❤❤

[@Duncan3]

i thought i came up with all 6 but 3 of them were degenerate

[@yandere8888]

i read the task multiple times and somehow still misinterpreted it, i thought i was supposed to find a shape with all 6 lines having the same distance

[@demening7860]

i just feel really dumb now...

[@woudythesheep420]

5:27, what if the dot move infinitely right?

[@hedgehog3180]

English as a subject does actually have clearly defined goals and standards that you get graded on, humanities teachers are just terrible at communicating what their subject is about.

[@giacomocervelli1945]

Dont know about the first part, but the second one is definitely right.

[@pittyconor2489]

Couldn't I just prove that these are the only solutions? Mathematical research seems to be a "homework" in your explanation.

[@copywright5635]

Yes you could. However, you would be responsible for that. When I just tell you that there are 6 solutions, that burden of proof is lifted off of you. I admit that the analogy is not perfect, however, I think the contrast in how it feels to solve the problem before and after is relatively similar to the difference I felt when conducting research vs. doing homework.

[@pittyconor2489]

@@copywright5635 yes and im saying mathematical research doesnt have anything that makes you second guess what you are doing, everything is rigorous and proven, it could be different in other scientific research.

[@goldendome-l1l]

i like the video, but dont really understand it especially the trapezoid part. i do understand the underlying part of how it was made to be, but it just doesnt sit well with me (make sense), not quite sure why. ty though.

[@br4hem_M4th]

i hate the taste of research.

[@AidanLowder-z1u]

Do the dots not have three distances in a square? There are two lines connecting to adjacent vertices and the diagonal. So, each point has three distances. I find the wording either incorrect or confusing. I feel two distances would be a triangle. There is one line connecting to the other two verticies. For the four points, I imagined them in a straight line, and only the two points in the middle have two distance while the outer points have one. Then if you just move these dots around a little you realise there will always be more than two connections for other configurations. Even with the six answers you showed there are dots that have more than two connections. So I feel as though I just didn't understand the problem. You could say that each four points are a,b,c,d and you could count line segments. Say ab=ba then we're not overcounting, but then clearly there are only six lines in the square example so each point wouldn't have two line segments as 2×4=8. So even with this logic the problem couldn't have a square as an answer. Ah, after further watching the video I understand the actual question. So it is in what way can you arrange four points on a plane in such a way that the resulting line segments originating from these points contain only two unique segments. The uniqueness of segment is determined wholly by size and direction plays no factor. In this way we cam safely determine a square fits these rules. Also it should be noted that ab!=ba essentially meaning we count each segments from each point and a segment already counted can be reused for other points. Perhaps I'm being pedantic or obtuse, but I truly didn't understand the problem at first. I'll keep this comment for equally boorish people like me.

[@copywright5635]

I'm not exactly sure what you're saying here. We care about the distance of the connecting lines. That's all. For any quadrilateral with 4 distinct points, there are six connecting segments. We want arrangements of those points such that the lengths of those connecting segments fall into only 2 different distances. In the square, those two distances are the edge length, and the diagonal length

[@AidanLowder-z1u]

​@@copywright5635I edited my previous comment. Essentially I didn't understand the original question the way it was worded. As I considered distances equal in length to be unique. I didn't quite get the intuition that the distances should be unique from the original wording. I apologise for the confusion.

[@kylecow1930]

i dont know if this makes me a good or a bad researcher, but i saw the problem and kinda obvously you can case analysis your way to a full solution so i didnt bother just assume some length is one then do some geometry to find all the configurations for each choice of assigining lengths 1, r then you can do it. this probably makes me a bad one because when i see a solution i tend to get bored and not bother actually doing it lmao

[@kylecow1930]

maybe if there was more of a point to solving the problem than watching a youtube video i would bother but for me knowing how one could get a complete solution is good enough in this case

[@copywright5635]

yeah I changed the title from that as I think it's a bit misleading. The point of the way I showcased the problem is to give a little taste of how it feels to be in research

[@eduardokuri1983]

It's 4C2=6

[@rajatrathore4577]

its so simple... 4C2 where c is combination

[@NataliaPunko]

Wait, so why is it that a parallelogram isn't a viable configuration? Four points, no edge/point overlap, two distances total

[@copywright5635]

There's more than 2 distances (unless you do a rhombus, but that's in the video). The two diagonals would be different lengths, and the side lengths would be different as well.

[@NataliaPunko]

@@copywright5635 oh.. wait, you're right, the diagonals would be different in length. Thank you!

[@ssjzack4730]

Couldn’t . . be one too? . .

[@meestyouyouestme3753]

Can two points overlap? ._(:)_. Or (:)_(:)

[@copywright5635]

Well no they shouldn’t overlap. Even so, your first answer would have three distances.

[@meestyouyouestme3753]

@@copywright5635 but the (:) is d=0 for D1(first point to overlapping point) and D2(overlapping point to first point) and DLeft and DRight are spaced equally apart from center point (overlapping dots)

[@meestyouyouestme3753]

Sorry. Not point. First dot and second dot (are placed on same point)

[@copywright5635]

@@meestyouyouestme3753 close. There is a third distance. From the left point to the right point. There are 3 distances then.

[@meestyouyouestme3753]

@@copywright5635 noooooooooooooo, Thanks I would have never thought of that. I should have made a checklist/matrix lol

[@johncharles2357]

Interesting video, but half of it was just your excitement to present a solution to a fun problem; it had nothing to do with the actual topic of the video.

[@copywright5635]

Yes! I delivered a point AND showed a fun little math puzzle. I see it as a win-win, I’m sorry if you don’t though

[@find2hard]

Dis vid make I feel stupiderder.

[@lamMeTV]

I dont even understand most science one line tasks cause they use such specific and uncommon meaning for words. Wtf is a distance?

[@ferraneb]

I missed the second case for b = l (on the other side) 🥲

[@b_onkus2493]

I feel like this is also a good illustration of formulating good research questions, since the question as stated doesn't really discount the possibility of overlapping points with connectors of length 0. There could be 3 more configurations of the points where one of the lengths is zero, one of which appears as an equilateral triangle and two that appear as a line segment. From an inside perspective, it might be obvious that we only want configurations that have a quadrilateral boundary, but with only the question to guide me, an equilateral triangle was the first "solution" I found at all.

[@copywright5635]

Yes, and I should have been explicit that the points are distinct