No cuboid has an equal Volume, Surface Area and Perimeter. Here's why.
Вставка
- Опубліковано 22 тра 2024
- "The Impossible Cuboid"
In this video I discuss a proof of the fact that a cuboid can't have an equal volume, surface area and perimeter (at least in the real world!). We motivate the solution by considering a simpler version of the problem and go back into 3D.
As an aside, I'm extremely happy about how the animations have come out in this video (you can even watch in 4K). For those of you curious, I made the entire video using the free version of Davinci Resolve. The next improvement I plan on doing will be my microphone (as many of you have pointed out).
This video supplements my entry to the "Tom Rocks Maths essay competition"
(Yes I did miss a minus sign at 10:34, the equation really is infamous)
Also when the cubic discriminant = 0 it should say repeated real roots rather than 2 real roots (thanks to the comment pointing that out)
Discord Server: Discord: / discord
Chapters:
0:00 Introduction
0:49 Simplification
2:32 Roots of Polynomials
5:54 Main Problem
12:17 Final Thoughts
Music:
Yehezkel Raz - Ballerina
Yonatan Riklis - Tales of the Mind
Adrien de la Salle - Shimmer
Yes (for those of you who noticed) the quadratic formula really is infamous for that* reason!
Note: I realise I didn't make it super clear at the end of the proof when i showed Δ = -A²(3B²+C) that C is in fact positive, but in the final thoughts section I show C = 128/3 when you actually calculate its value. Hope that clears up any queries.
Thats is what was thinking!
Is your voice AI
Point. Has a surface of 0 a volume of 0 and a perimeter of 0 by definition, it is just impossible to distinguish from any other shape with side length of 0. And is therefore a solution to the equation (a,b,c)=(0,0,0)
(-0=+0, for N=0 you get A,B,C=0, F(x)=x^3 when x_1/2=+-0)
@@takuma359 No lol, I just sound like AI
Yeah the first term is -b not b.
Wait a minute, this is easy. I just set L, W, and H to zero. 😅
Edit: oopsie, I might have accidentally gotten top comment. Yes, this was a joke comment, because a cuboid with L, W, H of zero would be a single distinct point in space.
I don’t know if it could be considered a cuboid because that would imply 8 distinct points in space with each point having orthagonality to 3 other points (defining the dimensions). And positive infinitesimal arguments wouldn’t work because scaling from zero goes at different speeds for different exponents.
I guess so. But we are looking for *positive* solutions, which zero is not included in.
@@cosmicvoidtree r/wooosh
Then it's a dot, not a cuboid.
Well then it just doesn't exist
@@PaulRodrichSancti I disagree, you could get the same thing by setting each side equal to 1/x then watch as x gets closer to infinity. Guess what, Volume and Surface Area and Perimeter all approach 0! The sides of the cuboid still has space and it is approaching from the positive direction. Therefore it is not only still a cuboid but it even fulfills the positive real requirements.
An alternate solution using algebra:
Assume that abc = 2ab + 2bc + 2ca = 4a + 4b + 4c for some sidelengths a,b,c.
Then (abc)^2 = 4((ab)^2 + (bc)^2 + (ca)^2 + 2a^2bc + 2ab^2c + 2abc^2)
(abc)^2/abc = 8(a + b + c + other terms) > 4a + 4b + 4c
Hence they cannot be equal.
This is so underrated
It took me some time to get it but damn good proof.
This needs the assumption that a, b, and c are positive real values, which is kind of obvious given the question. But it does need to be stated.
@@Pfeg82 I did state that a,b,c were sidelengths
Without watching the video, here's my take on it :
Denote (x, y, z) to be the base, length, and height of the cuboid, with x, y, z > 0. If the volume, surface area, and perimeter are equal, they satisfy :
xyz = 2(xy + xz + yz) = 4(x + y + z) = k
where k is some arbitrary constant bigger than 0. We then construct a cubic in "t" such that the roots are (x, y, z) :
t^3 - (k/4)t^2 + (k/2)t - k = 0
The discriminant of such polynomial is D = (-3/64)k^4 + (7/4)k^3 - 27k^2, and because D < 0, the polynomial has always one real root and one pair of complex conjugates. This means that one of (x, y, z) will be a real number while the rest are complex conjugates, contradicting the assumption that x, y, z > 0.
I got a little bit lost on why that means that they can never exist, If there is one point where it crosses the x-axis doesn’t that mean there is one solution?
@@thedarthcader7055 You're right,there is one real solution if it crossed the X axis *once*, but you're forgetting about the other complex solutions. I showed that if a cuboid with an equal volume, surface area and perimeter exists,then one of the sides is real while the rests are complex conjugates. This however,is nonsensical .
@@integer9590 Ohhh I see, I was thinking that one cross of the x-axis meant that there was one cuboid, not one dimension, that would work. Thanks for explaining!
@@thedarthcader7055 One cross to the x-axis represents the root (sides)
Someone needs to get this guy a microphone to match his insane editing skills 🔥
Thank you Polyamaths for this video. It saved my marriage and life.
How exactly did it do that, I'm curious?
@@leif1075via complex solutions. ;)
@@leif1075helped him win an argument ofc
@@leif1075 We all are, although I’m pretty sure it was rather a sarcastic comment
@@leif1075the priest asked, will you prove that no real cuboid of equal volume, area and parameter exist?
before watching the video:
length: 0, width: 0, height: 0
@knutritter461 ??
@knutritter461 then there would be no rectangles with equal area and perimeter. as was stated in the video we're looking at the numerical value, not dimensions
this is what we might in mathematics call a "degenerate case" (not an insult, just terminology)
it wasn't quite stated in the video, but the idea seems to be a cuboid with positive side lengths. if you set everything to 0 then you just have a point.
That is a 0 dimensional point, not a 3D cube
Thats not a cuboid, just a singlular point/dot.
500 subs with these animations is wild
they're not that bad no need to be harsh
@@brownie3454 p sure hes saying that the animations are surprisingly really high quality for a channel with 500 subs
I for sure will recommend this channel to my Mathphile friends
Only 600 subscribers?! This channel is a gem! I'm gonna watch every video
wow your animations are smooth
love it!
These are some of the best animations I have seen in a while for math content; your video is so high quality, it reminded me immidiately of 3Blue1Brown. Please dont let a view / subscriber / like count demotivate you from making more vids; this is some great content, and Im sure people will realize it. Keep it up, here you have a new sub
I actually only started in the last few months. I'm as motivated as ever right now seeing everyone's positive reactions, and i hope to maintain this!
It is done using 3b1b library, that's why it looks the same
Holy cow! The animation for this video is top notch. I edit in davici resolve too and can't imagine the long hours in the fusion page to make this. Good job!
this channel (and those like it) is wonderful
i missed out on a lot of schooling due to medical and mental issues so i lost a lot of passion for math compared to when i was a kid but videos like these have reignited that interest
This is a gorgeous video. Thanks for making such an engaging and informative presentation.
You have done a really fantastic job with this video. The pacing, structure, and graphics of it all really were great. You should be proud of it. If you are going to keep working on this type of video, maybe invest in a higher quality mic, or similar audio system, nothing extreme, just something that will bring the audio quality up to par with your fantastic video.
Saw the thumbnail before going to bed last night and actually had the thought of “this looks like the expansion of (x+a)(x+b)(x+c)!” as well as using the relationship of them all being equal, but I didn’t connect it to solving a polynomial and finding the roots
Really fascinating video (and excellent visuals)
11:57 I would have liked if you had worked through completing the square to demonstrate that C is non-negative, because if it *is* negative then B^2 + C could possibly be zero or negative itself which would give us 2 or 3 real roots.
C *is* non-negative (it's 40 and a bit), but I had to work it out myself lol
If you go to 12:25 it shows C=+128/3 (although agreed this isn't clear)
Damn I was wrong too lol
How demanding! The guy does a great video but you don't give it a like because of some irrelevant detail.
@@samueldeandrade8535Bro is everywhere... Crazy!!!😂😂
@@harrymetu2746 hahahahahahahahahahahaha. Am I??? That actually would be nice.
Clean animations, clear narration. Nice!
so many lovely math channels these days! great video :) thanks
Great video, impressive animation and nicely explained .
Wow, just wow. This a gorgeous video presentation. Instant subscription from be 😁
This could easily be a contender for the next SOME grand prize. Great work!
I was actually considering holding off on this upload until SOME but Grant has said he's not running it again until 2025.
Of course you can't have equality of things in different dimensions. But if you are just talking about equality in the numbers, let me introduce a set of units:
Take a perfect cube with side lengths 1 m, now the volume is 1 m^3, the surface area is 6 m^2 and the perimeter is 12 m.
Instead of using metric units, we use the "perfect cuboid" units, where the length unit, called "perfect meters", denoted "pm" is equal to 12 meters, the area unit called "perfect square meters", denoted "psm" is equal to 6 square meters, and the volume unit is just normal cubic meters.
So the 1m×1m×1m cube has:
Volume: 1 m^2
Surface area: 1 psm
Perimeter: 1 pm
All three have the same numbers!
Unfortunately, the assumption, unstated, is that the numerical values are equal for a *consistent* set of length, area, and volume units - that is, if we define an 'Arbitrary Length Unit' (1 ALU) to be equal to some real world length, then the corresponding consistent Area unit (1 ALU^2) and Volume unit (1 ALU^3) follow directly from that, and we are basing our numerical equality directly on this. Alternately, the assumption can be taken to mean that we are using a Dimensionless Unit (as per a system of Natural Units, a fascinating system that sets some subset of the natural constants of the universe all equal to 1, dimensionless, and derives the other values from there through simple combinations of the natural quantities), which has the same end result, that the Natural Length, Natural Area, and Natural Volume are related in the same way as the ALUs above.
What you are describing is defining three distinct and unconnected units. An Arbitrary Length Unit (1 ALU), an Arbitrary Area Unit (1 AAU != 1 ALU^2), and an Arbitrary Volume Unit (1 AVU != 1 ALU^3). If you allow the definition of the AAU and AVU to be completely unrelated to the ALU, then yes, you can end up with any numerical arrangement you desire.
How can psm be different from pm * pm?
Once you have perfect meters pm, the units for area should be pm². You say that a “perfect square meter” is 1/6 m², but a 1m × 1m square should have area (1/24 pm) × (1/24 pm) = 1/576 pm².
@@theadamabrams Why should pm^2 = psm?
If you want to be consistent with your units, then yes, but it is very common to use e.g. liters as a volume measure, even though you use meters as a length measure...
@@Eknoma A liter is defined as 1000 cubic centimeters, and in science, you need to be consistent with your unit conversions, so in practice, when we're making measurements that convert between length and volume, we use m^3. You could also make a unit of area called the foot-meter. There's not really any reason to do this, unless your goal is just to piss off both Americans and everybody else at the same time.
By that logic any cuboid can have equal perimeter, surface area and volume. Not really an interesting observation
less than 1000 subs?!? it won't be that way for long. great video!
Great video - I look forward to checking out your channel. Thanks. Subscribed ... Cheers ...
Your music choice was impeccable
I loved the video and the interesting approach a lot. Along with that I want to share some feedback on showing the derivation of the vieta's formulas for quadratic case... Seeing the level of this channel it feels like to me that it you can assume the viewers are well versed with it.
I would love to hear thoughts of others on this
The derivation (even if most people know it) I think helped motivate the step of setting the sides as roots of a polynomial since that doesn't feel super obvious. (At least that was my aim)
@@Polyamathematics Thinking about it, I understand what you mean, tbh I too agree it is more important to motivate and make sure everyone is on the same page even if for some it may be obvious
Excellent video.
From a didactics point of view, I don't understand why you had to mention complex numbers at all. Stating simply that "there are no solutions" would have avoided the need for the awkward aside.
Also, arguably calculus is more advanced than complex numbers, so if you felt the need to refer the viewer to additional material on complex numbers, why not on calculus?
Finally, as a mathematician I'd have hoped for a more satisfying answer other than "this bunch of algebra reduces to delta < 0".
We have our variables a, b, c.
Hence,
A = a*b*c
P = 4*(a+b+c)
S = 2*a*b^+2*b*c+2*a*c
By equating the area to the perimeter we can isolate c as:
c = 4(a+b)/(ab-4)
And by equating the area and the surface we find that:
c = 2*a*b/(a*b-2*b-2*a)
Given that c has to be equal to itself, and setting a=1 without loss of generality, we have the equation for b as:
2*(1+b)*(b-2*b-2) = b*(b-4)
Rearranging and changing we find the equation:
3b^2+2b+4=0
Which has no real solution, only two imaginary solutions, therefore proving that there are no cuboids with equal volume, surface area and perimeter.
10:54 Mustn't there be -b instead of b?
Yeah you are correct
Actually, it depends. Maybe he is considering the quadratic equation
ax² - bx + c = 0
Not a big deal at all.
I am just a random viewer, I really appreciate your enthusiasm about this problem (and the nice animations!)
while at the same time I feel so disappointed after watching this video.
Taking a visual problem, projecting it onto polynomials and doing some grunt work to arrive at a negative number
is the complete opposite of "Here's why" for me. The simplification was a nice bit of "why", but then it became "Here's how" to prove.
Amazing !
Just to make sure I’m getting it right, the point of the proof is that presuming that all these measures are equal, there should be a constant that produces the sides that make them as real positive zeros in a cubic function (because the measures appear in its coefficients and imply the sides as zeros) right?
instead of cubic formula, you can:
1. take the derivative: 3x^2 - 2Nx + 2N
2. use quadratic formula to find local extrema: (-2N +- sqrt(4N^2 - 24N)) / 6
3. show that both extremas have to be for non-positive arguments. We can ignore the denominator and compare the two parts of the numerator. Squaring them makes the comparison obvious:
4N^2 >= 4N^2 - 24N
2N >= sqrt(4N^2 - 24N)
-2N +- sqrt(4N^2 - 24N)
Huh? A cubic function only has one inflection point. The zeroes of the first derivative are the local extrema, one local maximum and one local minimum. The inflection point is the second derivative, and would be the zero of 6x- 2N, so the inflection point is at x = n/3. But glancing at your math, I think everything is correct assuming you meant local extrema instead of inflection points.
@@iankrasnow5383 yeah, I meant local extrema. I've edited my comment.
Straight 🔥
gay 🔥
pan 🔥
@@fullfungo4476pan burning? I sure hope it doesn't /j
I never use type annotations. Partly because I'm self-taught, starting with 2.6, partly because I'm a one-man-band and therefore don't have to interact with any other programmers, partly because I mostly use Vim with no linting (although I'm moving to VS Code to use copilot), and partly because I rarely write anything in my code that isn't functional (including comments).
I'm also terrible at documentation. I really should start using more of these sorts of things.
Truly incredible
before watching, I took volume which is l*w*h, surface area, which is 2(lw+lh+wh), and perimeter, which is 4(l+w+h), and set them all equal.
Solved for h in terms of l and w, and then solved for l in terms of w, and then got a final monster equation that only holds true if w=-2. So as long as w, l, and h are positive real numbers, it shouldn't be possible.
The equations V=A=P are not homogeneous, so they will change by scaling. For example any box can be scaled so that V=P. For example, a cube of side root 12 has a volume and perimeter of 12 root 12.
Now the surface area is found to be a factor of root 3 too big. Indeed, for both a cube with edge 3 or 4 the area is bigger than both perimeter and volume.
For a very long solid with L=W=2, V and P are near 4H, but A is near 8H, twice as large.
For a flat panel with L=W large and H=8/L we have V and P are approximately 8L, but the area is 2L^2, a factor of L/4 too big.
Maybe this must always be the case. Let's try to express the idea that the area is too big when the perimeter and volume are scaled as a homogeneous equation.
From the assumption A^2 = PV, and both sides are homogeneous of degree 4.
Our conjecture suggests we expect A^2 - PV >= 0 (as all three 0 is possible).
But A^2-PV = 4 ( L^2W^2 + L^2H^2 + W^2H^2 + LWH^2 + LHW^2 + HWL^2 ), which must be strictly positive if L, W and H are.
Further, 2(A^2-PV) = A^2 + 4 ( L^2W^2 + L^2H^2 + W^2H^2 ), which can only be 0 if all terms are 0. This force at least 2 of L, W or H to be 0. But now P=0 gives us all three must be 0.
This also gives us intuition into WHY this fails GEOMETRIALLY.
If we scale any box with non zero volume so that the volume and perimeter are equal, the surface area will be larger than this value.
Hey question what font do your équations come in ? It's pretty!
It's Computer Modern, the main font of LaTeX and Manim (the latter probably being used to make the video). The italic typeface has a very special quality to it.
12:08 my question is, what if C is negative and has an absolute value greater than B²? Then that whole expression would be positive. Unless I missed something and C is greater than 0
C = +128/3. Its shown just after at 12:25, i think he said "of the form blah blah" so C was just a positive constant.
@@TheArizus alright thank you so much!
@@TheArizusI assumed C would be positive based off of the certainly in tone, but it was a bit sloppy to state it as such and leave it to the afterthought as proof. It would have been nice to clarify at that point instead of leaving it up to the viewer to verify in the hand written notes.
I wonder whether we can generalize the problem for higher dimensional cuboids.
Very cool!
What program did you use to make this video? The animations are so clean
Davinci resolve (free version)
@Polyamathematics I get using Vista since it tells you of the coefficient is positive pr negstive but why would anyone think to set the value b authentic root pf the polynomial? I'm not sure why anyone would do that
@Polyamathematics why is N There all of a sudden at 7:25 why not just use the coefficient from.the given equation 2 and 4..seems arbitrary and unclear...thanks for sharing
First off, they are all different units, and so regardless the specific formulas involved, the volume will be constructed by multiplying three lengths, the surface area will be constructed by multiplying two lengths, and the perimeter will be constructed using lengths but without multiplying any lengths together. Lengths can be picked such that the differences in the formulas between two of these are perfectly negated. However, since the third formula will necessarily be of a different polynomial degree, whatever corrective factor worked to equate the first two formulas will fail to equate the third. Simply put, assuming xyz = 2(xy + xz + yz) = 4(x + y + z) yields a contradiction for all nontrivial values of x y and z. They start out equal at x=y=z=0 and then grow at different rates.
This does beg an interesting question, though.. what 3d shapes would permit of having the same value for its volume, surface area, and perimeter? I imagine if they exist, they're something mathematicians like to 3d print.
a bit before 4:00, you need add a factor, as f(x) = 2 (x-1)(x-2) is also a quadratic formular
There's a note just after saying "this generalises by considering kf(x) but that would overcomplicate what is needed in the problem"
very nice
Your animation is amazing!
Though this problem can be solved much simpler like this:
Assume that abc=2(ab+bc+ca)=4(a+b+c). This condition on a, b, c is not homogenous so it is natural to try to homogenize it as follows:
4(a+b+c) × abc = (2(ab+bc+ca))^2
(a+b+c)×abc = (ab+bc+ca)×(ab+bc+ca). Expanding the right hand side obviously shows that the right hand side is bigger, so this is impossible.
This font looks nice
Does this generalize to higher dimensions? I.e. the perimeter, area, and volume of any n-dimensional rectanglular solid cannot be equal for n > 2?
Very interesting and logical question. I personally have no idea how you would even go around this, but certainly not with polynomials (as used here for n = 3), since it's been proven that for n >= 5 there is no closed form for roots, and even if there was, it would be a pain (and most likely impossible) to do the calculus for an n-degree polynomial describing an n-dimensional rectangular solid and showing that it certainly has at least one pair of complex roots.
Fam, what is the name of the application you use for animating these videos?
Davinci Resolve (Free version) - in the description btw
I am really bad at imaginary numbers, but have a question, the rectangle case have negative solutions that don't work because there is no such thing as a "negative side" on a rectangle. Would it be the same for the cuboid with "imaginary sides"?
It's more of a thought experiment than an actual tangible length. In reality only positive real lengths make physical sense
So what would the two complex solutions look like? Where are they going? Must not be in known 3D space, but somewhere.
Complex length does not make sense, norm is defined as a nonnegative real number. You can try to come up with another definition though.
wrong set volume surface area and perimeter to 0 boom done thank you for coming to my ted talk
(this is a joke please don't hurt me)
I am in your walls
Jokes will not be tolerated. Prepare for elimination.
the post below is me being pedantic, this is NOT a useful way to approach this problem
TLDR: pedantic shenanigans with defining new units of measurement
i have to disagree, you can have volume, surface area and perimeter have the same numerical value
it is pretty easy to do, if you are willing to be pedantic
you just have to use the right units of measurement
as in: take any rectangular cuboid, and let X = Volume in cm^3; Y = surface area in cm^2; Z = perimeter in cm
define 3 new units of measurement, "u_a" : 1 u_a = X; "u_b" : 1 u_b = Y; "u_c" : 1 u_c = Z
we now have a rectangular buboid with a volume of 1u_a, a surface area of 1u_b and a perimeter of 1 u_c
So it is possible if 2 sides have a complex length?
I don’t know if it’s an AI voice or a bad mic or an ailment, and I’m sort if it is the latter, but it’s extremely off-putting. Sorry again if it is an ailment and if it is I hope you get well soon. Besides that your voice is great for this style of video
So we need
lbh = 4(l+b+h)=2(lb+lh+bh)
We can further divide this into
lbh = 4(l+b+h) ...i
lbh = 2(lb+lh+bh) ...ii
We can immediately see that we have 3 variables and 2 equations, meaning no single solution
Lets go with the first one since that seems easier
4l+4b+4h-lbh=0
l(4-bh)+4b+4h=0
l = (-4(b+h))/(4-bh)
l = 4(b+h)/(bh-4) ...iii
Now,
lbh = 2(lb+bh+lh)
lbh = 2lb+2bh+2lh
Substituting value of l
4bh(b+h)/(bh-4) = (8b(b+h)/(bh-4))+2bh+(8h(b+h)/(bh-4))
(This got messy real fast)
4bh(b+h)/(bh-4) = [8b(b+h)+2bh(bh-4)+8h(b+h)]/(bh-4)
4bh(b+h) = 8b(b+h)+2bh(bh-4)+8h(b+h)
2bh(b+h) = 4(b+h)(b+h)+bh(bh-4)
2bh(b+h)-bh(bh-4) = 4(b+h)²
bh(2(b+h)-bh-4) = 4b²+4h²+8bh
bh(2b+2h-bh-4)-8bh = 4b²+4h²
bh(2b+2h-bh-12) = 4b²+4h²
2b²h+2bh²-b²h²-12bh = 4b²+4h²
This is the furthest i got before getting tired, you can still make it simpler (maybe) with some other algebra stuff, now im going to watch the video.
Btw, there are 3 equations, you forgot about 4(l+b+h) = 2(lb + lh + bh)
@@TheArizus if the first two that I mentioned are true, then the third one you mentioned will also be true so it's not worth considering
I think*
@@Kambyday hmm maybe you're right, not 100% sure how to make this approach work either.
11:21 when he mentions at delta = 0 there are only 2 real roots. I am confused because that is not possible for a cubic or is it only denoted like that because delta /= 0?
Delta is not the whole solution equation. It is the "determinant" of the solution equation and thus says how many roots real and complex roots there are.
Repeated roots (could technically be one root repeated 3 times too)
What about a tesseract? I think that gives x^4 - Nx^3 + 2N^2 - 4N + 8N
They can be equal “numerically” if you choose the right units
how did you animate?
What if all variables approach infinity or negative infinity?
Now my question becomes, what happens with a 4D+ hypercuboid?
12:00 This seems like a very awkward place to end the proof. If C
I thought that in polynomials with real coefficients, complex solutions had to come in conjugate pairs. If that's true, how do you get a cubic with two real solutions? What's the third solution - a duplicated real root?
you are correct, it must be real, and duplicate
Try proving it for n dimension cuboid
So what about higher dimensions?
I did actually think about this. Maybe a follow-up video?
does hyper dimensions change it too?
@@TheAdhdGaming my assumption was that say in 4D we'd look for solutions to abcd = 2(sum(abc)) = 4(sum(ab)) = 8(sum(a)) and i've thought of a decent argument for why this pattern continues in higher dimensions with "hyper volumes".
Then we'd prove / disprove whether positive real solutions exist (although its likely solutions don't exist since say in 4D we have 4 variables and 6 equations and in general we'll have n variables and n choose 2 equations, which is why it might not warrant a full video)
@@Polyamathematics Higher dimensions do sometimes subvert expectations - I have heard about a problem of a circle being fitted inside a square, and its higher dimensional equivalents of the same side lengths and radius; I'm pretty sure once you go up to a certain dimension, the n-sphere actually starts enveloping the n-cube or something weird like that!
@Polyamathematics why can't a cubic have 2 real roots and 1 complex? I don't see why ir couslbt? I'm assuming it's something that just "falls out"f the algebra somehow? Something to do with the co plex conjugates?
Actually, there DOES exist(n't) ONE single cuboid that has all 3 equal values: When all lengths are exactly Zero (it's kind of a singularity... but cuboid) ;P
What about the 4D solution?
I'm sorry but did I just get homework from a youtube video???? 10:15
Yes, yes you did!
Why did I thought the qn was "are there 2 different cuboids with the same volume, surface area and perimeter"
Same.
Actually one can answer that with the same idea as the video. If two cuboids have the same volume, surface area and perimeter, then they determine the same polynomial, so its roots are the same, and therefore the side lengths are the same. So there are no two cuboids with different side lengths and same volume, surface area and perimeter.
Just rig the units
Can someone explain why 4N is used in the 3D example? Why 4N?
Just to avoid having to deal with fractions. At the end of the day its just some positive real constant
why are "Nullstellen" (0- points, so where y=0) called roots in english? This is hella confusing.
11:56 B^2+C can be negative though?
C = +128/3 (if you look in that "final thoughts" bit)
I’m confused, isn’t the perimeter of a 3d object just the surface area?
My cubiod is the cuboid obtained taking the limit as n goes to zero if length width height all equal n
a length, an area and a volume being equal is kind of arbitrary though. what if we measure the length and area in inches/inches squared, but the volume in litres? i wonder if it is possible to pick up some unit conversion factors in that cubic to make it have solutions
Exactly. It's a meaningless question on it's own
@@TrailersReheard i didnt wanna say meaningless cus theres vector cross product for example
12:10 how do we know C isn’t negative?
C = +128/3 (its shown in the "final thoughts part")
(x-a)(x-b)(x-c)=0 where a,b,c >0 [V=volume, S=area and 2u=perimeter)
2x³ -ux²+Sx-2V=0
Δ= 72uSV+16V+4S²-8S³+(27*64)V³ if V=S=2u >0
Δ=36V³+16V+4V²-8V³+1728V³ (divide by 4)
439V³+V²+4V=0 --> V(439V²+V+4)=0 --> 439V²+V+4=0 --> Δ
If only V=S then for example a=b=c=6 one of the solution.
Am I the only person to notice that the quadratic formula is quoted wrong (it should be minus b, not b)?
(pinned comment 😅)
@@TheArizus It was a great video, by the way... and it's left me wondering about other ways to prove the point... and also what's the closest that you can get; if every case degenerates towards one side becoming zero (which I suspect is the case) then one problem might be what the closest you can get is whilst insisting that each side must be at least one unit long. Of course, you'd have to strictly define "closest"... I'm thinking that it should be something to do with the minimum square differences (the sum of the squared difference of area/volume area/perimeter and volume/perimeter)
I noticed.
❤
so what you're saying is, such a cuboid only meaningfully exists as a complex volume as seen by a creature of no less than six dimensions
When the quadratic equation popped up on screen, I had a brief existential panic.
Some part of my brain stated "that first b in the numerator is supposed to be negative." Then, a different part piped up to say "sure, maybe. But, you also know that your memory is especially fallible." Yet another Kool-aid manned through my skull to say "sure, but are any of ya'll even positive that any of this is consistent, let alone real?" And, finally, a particularly overworked set of neural pathways didn't even look up, but still remarked "typos are a thing, dude, and the weasels are obviously loose again."
Hear me out. Side lengths of Zero!
:o!
E32: No file name
Don’t write numbers in italic. Looks weird.
Generally speaking, most mathematical publications roughly follow ISO 80000-2, which recommends: Use italic text for variables, use upright text for constants: Trivial examples for constants are numbers 1, 2, 3, etc., but also π and e. Non-trivial examples are fixed functions such as sin, cos, tan, ln, log, exp, and many more. Variables are those things where you say “let” for introducing them, so not only real-number variables like _x_ are variables, but also functions _f_ introduced by something like: “Let _f_ : ℝ → ℝ be continuous.”
Also, the minus sign at 1:57 is probably a hyphen minus (U+002D), when it should be a real minus sign (U+2212).
Thank you, this is actually really useful (I spent a lot of time trying to get a font which looked nice but i still wasn't happy in the end, this is probably why)
You showed the cubic has one positive root. Why can't 4n be tht number?
EDIT: Oh wait, one of the length is that number, but the other two can't work out. Never mind.
4n isn't a root of the cubic but rather a, b and c are the 3 different (up to repetition) roots of the cubic so if 2 of those roots are complex, then the cuboid can't exist.
V=A and A=P provide 2 equations with three unknowns. Thus we expect a curve of solutions in real space. One component clearly goes through 0,0.
Even if this component avoids the positive octant, there might be another disconnected piece, as with the 2D case.
@@Utesfan100 Careful, there's 3 equations V=A, A=P AND V=P and 3 unknowns. (This is why generalising to n dimensions isn't that interesting cuz there's n choose 2 equations and n variables)
make it all positive infinity
ggez
0??
me with ±∞ iq and saying that 0, 0, 0 and ∞, ∞, ∞ are the solutions
I wonder if a cubic with side lengths of infinity is actually a valid solution.
Is the volume of multiple infinite planes actually equal to the same infinity?
I feel like if you actually did the math, you'd be able to prove that the volume of the object is a larger order of infinity, but that's just my intuition.
Now, I don't think there is anything stopping you from doing this with imaginary side lengths as well, aside from the fact that an object with complex side lengths would no longer be "3 dimensional."
physicists freaking out over the premise of this video (there is no way for a volume to ever be equal to any area)
Physicists know how to express things in dimensionless form. Try again, better luck next time 😂
Even a rectangle wouldn't have the same perimeter and area. It wouldn't be found, except if it's defined for one variable. It's the same with rectangular prism. There will be no exact number.
Last time i checked a rectangle with sides of 4,4,4,4 does indeed have a numerically equal area and perimeter
@@TheArizus It's the same as square.
@@wahyuadi35 a square is a type of rectangle
@@TheArizus Well, yes. But, actually, didn't look for "a square". It looks for the other solution, that is different and unique.
@@wahyuadi35 how about 3, 6, 3, 6
10:43 😬
QF should be -b not b!
Obviously not, since volume, area, and perimeter are three different units
10:45 negative b lil bro
x^3 = 6x^2 = 12x
Only real solution is x = 0 and the other 2 solutions are complex