MPMP: What is the optimal ellipse?

[@MrBrain4]

I submitted my answer by placing an ellipse directly into the answer box.

[@user-un7gp4bl2l]

Yes.

[@santoriomaker69]

you clever boyo

[@daniellebarker7205]

It's cool how often I see other scrabble players on this channel

[@justarandomdood]

0:25 He already sees it, he already knows that if he posts this video we'll notice it, but he posted the video with this intro anyways, what an absolute legend Anyways, Parker ellipse

[@PronteCo]

Not gonna lie, you had me in the first half

[@Robert256]

I don't get it

[@rewrose2838]

@@Robert256 Parker joke.. Parker Parker joke?

[@sk8rdman]

He gave it a go, and you've got to respect that.

[@lunasophia9002]

If it's a recap but something that we haven't seen yet, isn't it a precap?

[@petemagnuson7357]

Or a decap?

[@joshmyer9]

Hmm, whatever could the P in precap stand for?.. 🤔

[@lolerskates876]

Is the text crawl in Star Wars Episode IV a precap if you haven't seen the Star Wars Prequels first?

[@simonmultiverse6349]

I don't like to use the ratio a/b because that means an ellipse oriented horizontally has a different parameter (e.g. 2) to the same ellipse oriented vertically (e.g. 0.5). I have realised that I can use a Fourier series. If you invent a parameter phi (an angle) to describe the ellipse, then the ellipse semi-axis horizontally is cos(phi) and semi-axis vertically is sin(phi), the series for the perimeter is basically this: C0. + C4.cos(4.phi) + C8.cos(8.phi) + C12.cos(12.phi) etc. The terms decay quickly: C0=0.9580819 C4= -0.047013 C8= -0.006536 C12= -0.0020108 C16=-0.000114 I got this by messing around in a spreadsheet, but I think I can improve on the numbers by writing something more systematic in Python. Also, I integrate the ellipse perimeter length numerically, but I do it at two different resolutions, so I extrapolate to remove the leading error term, so the numerical integration gives me errors of about 10^-9 .

[@AgentM124]

I did not expect this video at this time of night/day. But I'm excited!

[@RussellSubedi]

Night/day? Must be a pretty confusing time you live in.

[@AgentM124]

@@RussellSubedi `Sorry. Floating in space it's kinda weird.

[@balping]

These puzzles vary vastly in difficulty. For some you need to think for tens of minutes, for some, like this one, you know the answer immediately.

[@filipsperl]

Right? I'm sitting here thinking I've missed something. Is it really that obvious?

[@Xeridanus]

@@filipsperl A few people thought the last one was obvious and got it wrong.

[@Mrsparky492]

@@filipsperl I literally sat and thought about it for 10 minutes trying to convince myself that it is that easy! The open challenge is very interesting though.

[@fozzzyyy]

I went and brushed up on notation for ellipses so I could differentiate the area twice to find a local maximum, and it turned out to be the easy answer

[@JamesCoutie]

@@fozzzyyy Sure is easier than just guessing 0 like lots of people seemed to have done :P

[@AlexSh789]

The actual formula for the perimeter/circumference of an ellipse is: The integral, from t=0 to t=2pi, of sqrt[ a^2*cos^2(t) + b^2*sin^2(t) ] dt, where a is the major radius, or half the major axis, and b is the radius perpendicular to the major axis. This formula makes sense for a circle as well. A circle is an ellipse with an eccentricity of 0, where a=b. If you set a and b equal to r, you can factor it out, so within the square root, you'll get r^2 * [cos^2(t) + sin^2(t)]. The sum within the brackets (i.e. square of sine + square of cosine), is 1, so it just goes away. So you're left with sqrt[r^2] dt, which simplifies to r*dt. If dt is from 0 to 2pi, the result is 2*pi*r, which is the formula for the circumference of a circle.

[@samburnes9389]

But now try to do the integral in general. I just wrote it down and it looks really messy, and if what Matt says is true, impossible to do.

[@AlexSh789]

​@@samburnes9389 - If we account for the four-part symmetry of an ellipse and the definition of the eccentricity, e (not to be confused with Euler's number) = sqrt[ 1 - b^2/a^2 ] (where a >= b), then the formula simplifies to: 4a times the integral, from t=0 to t=pi/2, of sqrt[ 1 - e^2*sin^2(t) ] dt. This integral is called the complete elliptic integral of the second kind, and is written as the function E(e). Thus, the formula for the circumference of an ellipse will fully simplify down to 4a*E(e). When e = 0, as in the case of a circle, E(0) is pi/2, or about 1.57. When e = 1, as in the case of an ellipse being squashed/stretched into a straight line, E(1) is 1. Every other value is somewhere in between.

[@sipkejorgmund753]

@@AlexSh789 If we're trying to find circumference as a function of the ratio of the major and minor axis lengths, and we find it is equal to an equally mysterious function in terms of eccentricity, it's not clear we've made any progress. :P

[@AlexSh789]

@@sipkejorgmund753 - The ratio of the major and minor axes pretty much *is* the eccentricity. I mean, you're right, we didn't make much progress. If you want to see the true extent of that progress, you can take a look at the main channel video. It's up now.

[@wrog7616]

I'm more excited about the approximations than the puzzle! :)

[@limegreenelevator]

Upside of being in America right now: this video isn't coming out shortly before midnight Downside of being in America right now: pretty much everything else. Eh, I'll take it.

[@AlexSh789]

@@justarandomdood - In the UK and Europe, it's midnight.

[@justarandomdood]

@@AlexSh789 oooooh wait I though that the guy was in the US 🤦‍♂️🤦‍♂️ Ignore me I guess

[@RowanAckerman]

Hopefully everything goes well in November.

[@NStripleseven]

I won't, I'm in Canada

[@telodemuestro7384]

I found an approximation with e and without pi, that when bigger the ratio a/b the more accurate it is. i established A as the biggest radius. The formula is 4a+((b^2)/a)(1-e^(-a^2/b))

[@bigJovialJon]

I intuitively came up with an answer and spent a bit of time proving to myself that it's correct. The key is noticing what happens to the length of the long and short axes as you vary the distance between the foci.

[@incription]

I'd just guess it was a circle since that's what I've learnt doing area maximization problems

[@davidalearmonth]

I did some math, and I'm happy that if I did things right, my immediate intuitive answer seems to be the correct one. :) (for the area problem, I mean)

[@jestongreenwood6815]

I got to your last puzzle just today. Very fun. It took almost 3 hours to figure out. Solution to the main puzzle for an ellipse. Start in polar coordinates. The equation for an ellipse is r = a cos θ + b sin θ dr/dθ = b cos θ - a sin θ The equation for the length of an arc in polar coordiantes is L = Int(0, 2π)[sqrt(r^2 + (dr/dθ)^2)]dθ. r^2 = a^2*(cos θ)^2 + 2*a*b*(cos θ)*(sin θ) + b^2*(sin θ)^2 and for (dr/dθ)^2 = b^2*(cos θ)^2 - 2*a*b*(cos θ)*(sin θ) + a^2*(sin θ)^2 Plugging it into our arc length formula we get L = Int(0, 2π) [sqrt(a^2*(cos θ)^2 + 2*a*b*(cos θ)*(sin θ) + b^2*(sin θ)^2 + b^2*(cos θ)^2 - 2*a*b*(cos θ)*(sin θ) + a^2*(sin θ)^2)]dθ ==> L = Int(0, 2π) [sqrt(a^2*(cos θ)^2 + b^2*(sin θ)^2 + b^2*(cos θ)^2 + a^2*(sin θ)^2)]dθ ==> L = Int(0, 2π) [sqrt((a^2 + b^2)((cos θ)^2 + (sin θ)^2))]dθ ==> L = Int(0, 2π) [sqrt(a^2 + b^2)]dθ ==> L = 2π*sqrt(a^2 + b^2) QED Better late than never.

[@eyondev]

Just yesterday i had an analytic geometry exam. What a nice timing

[@sk8rdman]

I'm amazed to learn that there exists no formula for finding the perimeter of an ellipse. It doesn't seem like that hard of a problem, and I can think of a few ways I would approach it, but apparently it's more complicated than it seems...

[@AaronSherman]

For the perimeter, the best estimate I could find is ((a/b-1)*c+2pi-sin((a/b)*d)/(e) with c, d and e being constants that you can optimize, but my current working values are 3.8645 , 0.249 and 1.645 that gives a stddev of the delta with your sample dataset of 0.1211906539 which is pretty reasonable, I think.

[@nathanfay1988]

I found the equation for the perimeter of an ellipse. I sent it to your think-maths submit page

[@dreznik]

the area of the ellipse is proportional to a b. w c being the distance from center to a focus, for an ellipse a^2-b^2=c^2 => b^2=a^2-c^2, then => (a b)^2 = prop to area^2 = a^2 (a^2-c^2) = a^4 - a^2 c^2. take the deriv wrt c and set it to zero to maximize area^2: -2 a^2 c = 0 => c = 0. i.e., for a given thread length the largest area occurs for c = 0, the circle.

[@u-kn]

Well, using Wolfram Alpha I figured out that the exact value (divided by 4) of the perimeter of an ellipse is: integrate sqrt(1+((b^2 x^2)/(a^2 (a^2 - x^2)))), x=0 to a which is lim{x,a}, (a Sqrt[1 - x^2/a^2] Sqrt[(a^4 - a^2 x^2 + b^2 x^2)/(a^4 - a^2 x^2)] EllipticE[ArcSin[x/a], 1 - b^2/a^2])/Sqrt[1 - x^2/a^2 + (b^2 x^2)/a^4] But I have no idea what to do with that information ^^'

[@diarya5573]

Sounds like this is one right up my alley yay

[@JohnDlugosz]

This sounds more like a problem from _Mind Your Decisions_ than one of MPMP's that are more conducive to fiddling and programming. The area is k(r1)(r2) where r's are the two semi-axis lengths, and k is a constant (unimportant here). We want to maximize the product of the two radii. Without looking up any formulas, we know that when the pen is in a line with the tacks, we are looking at r1 as the distance from one tack through the center past the second tack to the edge, and back to the second tack: (d) + 2 (extra) = 28. (extra) is (r1)-(d/2). When the pen is on the minor axis, half way between the pins, the thread forms a triangle with sides length (d), (14), and (14). Half that is a right triangle with sides (d/2) and (r2) with hypotenuse (14).

[@JohnDlugosz]

Math-free guess: if you start with a circle and move the pins apart a little, the height is reduced but the width is increased. But, not by equal amounts: it's a sin vs cos thing, so the height is reduced only slightly while the width goes up almost with the separation. When the pins are far apart it's the other way around, and clearly you're heading towards a zero-area flat tire. When the strings are at 45 degrees to reach the point on the bisector (the minor axis) is where the give/take between the wider vs taller is balanced. That must be the extremum. The hypotenuse of a 45 degree triangle is half the string, or 14 cm. So twice the side is 2 times14/sqrt(2) or approx 19.8 cm

[@DaviddeKloet]

The width of the ellipse actually doesn't change by moving the pins apart. It's always equally to the length of the string.

[@supermanvsgoldenunicorn4575]

Sorry to come late to the part. Nonetheless, how about a lazy approximation to the perimeter using an approximation for pi. p=3.1(6a/5 +3b/4)

[@pdpgrgn]

After 1 hour of frantically refreshing KZbin and switching between this and the other channel, I gave up thinking this was another blank week. Then I see this at this time, when the speed points have probably already been finished. And then, I spend 20 minutes after seeing the video on trying to verify whether the answer I had intuitively guessed is correct, forgetting the easiest method to do so on the first two tries.

[@JamesCoutie]

I wouldn't be so sure..... there's probably far less people doing it straight away because it came out so late.... and judging by the comments there are a lot of people getting the wrong answer too

[@Krebzonide]

I got lucky the second I looked at youtube this notification popped up so I was early. I had a guess from the start based on stuff I've done with surface area of a rectangle in calc, and after making a quick graph matched my guess, but I'm only like 60% sure I got the equation right to make the graph.

[@lukemejia9444]

I am a high schooler who loves math. I think your puzzles were the only thing keeping me sane this summer!

[@harikishanrakhade6108]

For the open problem: Perimeter = 2π * √( (a2 + b2) / 2 )

[@DoubleATam]

based on the delayed video, I think he means like, closer than that. More creative answers.

[@alan2here]

Someones going to submit an answer that's perfect for those 5 ratios, and awful for everything else. :-P :) Like a table.

[@rosiefay7283]

Yeah, fit a quartic.

[@alan2here]

Something that transitions between two different quartics around some value.

[@darktemp_de]

Finding the answer, made me question myself if I determined the semi-major axis correctly. But after verifying that, the answer makes even more sense and could even be found without any maths :D

[@samburnes9389]

I know, I was about to crack out a function and do some calculus, but then...

[@BeastOfTraal]

I wanted to program it in whitespace

[@efhiii]

You might be able to interweave your other code with the whitespace code so that either way, it works.

[@twojuiceman]

So since every equation to determine the perimeter is an approximation, are they all parker equations?

[@ThomasWinget]

On the one hand: I want to use the provided spreadsheet to compare my ideas to "proper" results. On the other hand: scipy.special.ellipe exists and the spreadsheet is xlsx. Yeah...I can't be bothered to parse or convert an xlsx when an easier solution exists. I cannot fathom why this table wasn't a csv file. tsk tsk, Matt. (I love your content, but if I have criticism I'm going to levy it.)

[@bjornmu]

Tip: ignore the number 28. Just call the string length 1, derive a solution, then multiply by 28. Or maybe 2 is even better.

[@ke9tv]

No, you have to multiply by a Grothendieck prime, like 57, and then Parker-square the result.

[@kongolandwalker]

why your method almost never works: take one and square it. Take 28 and square it. Does second divided by first give you 28?

[@jimmyh2137]

@@kongolandwalker B^2 / A^2 ≠ B It's not the "1" your problem here. 16/9 ≠ 4

[@kailomonkey]

Funnily enough I was thinking about this recently. The answer I had come up with is probably the worst you could hope to see. I figured when a is 0 the circumference is 4xb but with no area, when a=b the circumference is 2pixb with the difference being pi/2 - 4. So if we took a ratio a/b and x it by pi/2 - 4 in the equation we'd get: c=(4+(2*PI()-4)*(a/b))*b Let me know how very very wrong this is as I have no idea where to check it :)

[@kailomonkey]

I think I have something wrong with how I put the ratio even in but I'm sure it wouldn't be that simple and go in the wrong direction anyway :)

[@kailomonkey]

here we go: c=(4+(2*PI()-4)*(a/b))*b edited into original comment where originally it lacked the -4.

[@davidgould9431]

Before actually watching more than 0:30 of the video: the optimal one is one that's drawn with two pins and a loop of string (not one with the ends tied to the pins), so you can keep going round and don't have to scratch away backwards and forwards in a desperate attempt to draw something that doesn't look like a rectangle. I may be being a little harsh, but it wasn't an elegant ecli-- sorry, ellipse, was it?

[@mathization]

Better late than never: I made an adjustment to Ramanujan's formula. The adjustment factor to the value of (h) in Ramanujan's Ellipse Perimeter formula is: First calculate (h) from Ramanujan's formula. Then adjust its value: h = h + (h)(adjustment factor) adjustment factor = 0.00165231179593 (c) JPA 2021

[@donaldasayers]

"There is no equation for the perimeter of an ellipse" . That depends on your definition of what an equation is and what one means by "closed form". Sinx is not a closed form as it requires an infinite series to calculate it. However no one would think that y=sinx is anything but a simple equation. Helpfully though there is a rapidly converging polynomial approximation for sin and so there is a button on your calculator. But in the old days (my teens.) we only had tables of Sin. So a practical definition of "closed form" might be that an expression is closed if there is a rapidly converging series, leading to the easy implementation of a button on your calculator. The perimeter of an ellipse is an elliptical integral of second kind and there are books of tables (Jahnke & Emde, but not without errors...) for these and series approximations exist but they converge very, very slowly. Gauss found a neat solution to elliptic integrals of the first type using his "arithmetic-geometric mean." (AGM). This "mean" of two numbers, is not strictly a closed form, but it can be calculated using a very rapidly converging iteration, giving two decimal places per iteration. If people needed Gauss' AGM there would be a button on your calculator. So by extension one may consider elliptic integrals of the first kind to be expressible with a simple equation using practically closed forms. Thus there is a simple exact equation for the period of a simple pendulum. Elliptic integrals of the second kind, do not unfortunately yield to Gauss' method, but in a paper of 2012, Semjon Adlaj uses a modified AGM to provide a useful and rapidly converging solution to elliptic integrals of the second kind thus the assertion that there is no equation has been disproved.

[@effuah]

The trick with the AGM for the 2. Elliptical integral is older, see e.g. Borwein&Borwein Pi and the AGM

[@donaldasayers]

@@effuah Do you have a link for that? The only link I can find mentions, but does not solve for the second kind.

[@effuah]

@@donaldasayers You know libgen.rs? Nobody would click on this link and do a copyright infringement, it is a book, so it is not that easy to get legally for free in the internet

[@donaldasayers]

@@effuah OK but as with the abstract I read, whilst it mentions elliptic integrals of the second kind, I don't think they actually use the AGM to solve them, I could be wrong as this is at the edge of the math I was familiar with from my degree 30 years ago. The paper I cited uses a somewhat modified AGM and states explicitly that the AGM cannot be app;lied to the second kind.

[@effuah]

@@donaldasayers it is basically the calculation stated on Wikipedia: pi/2/AGM*(1-Series) and the terms series you get automatically when you calculate the AGM. If you allow the AGM, you already allow for rapidly convergent sequences.

[@scoutskylar]

If you plot the linear eccentricity (half the distance between tacks) compared to the area, it's a quarter of an ellipse! Not a coincidence, of course.

[@laremere]

I tried to find the answer, but I got nothing.

[@seanm7445]

I made the silly mistake of submitting the answers in inches, rather than centimetres.

[@bl4cksp1d3r]

My idea would be a distance of 0 to get a circle xS

[@johnchessant3012]

lmao nice

[@JohnDlugosz]

@@bl4cksp1d3r Yes, in general start looking at the extreme values. Clearly at the other extreme you get an area of 0. But if you move the pins a little farther apart starting at 0, you'll see that the height is reduced but the width is expanded. My guess would be when the string forms a 45 degree angle to each pin.

[@bl4cksp1d3r]

@@JohnDlugosz to my understanding the width should stay the same, but it's possible that I just overlooked smth.

[@cyrilio]

I love how mathematics can make it important about ridiculous issues.

[@rosiefay7283]

Mathematics isn't to blame! You think it's ridiculous, Matt shows it's worth making a puzzle about, let everyone else consider whether it's important enough to be worth some time puzzling over.

[@effuah]

I think the perimeters in your spreadsheet has larger errors than then length of the number would suggest

[@timh.6872]

This sounds like a fun thing to do over labor day weekend...

[@DaviddeKloet]

I don't understand the extended puzzle. My idea was to integrate over the Taylor expansion of sqrt(a sin2(theta) + b cos2(theta)) but it sounds like that's what you did to calculate the correct answers? So what's left to do for us? Are there rules about what kind of formula is allowed? I was going to wait for the main video in case it makes it clearer but it still didn't come.

[@alexismandelias]

Isn't there a formula for the perimeter of an elispe? Am I missing something?

[@heliocentric1756]

4(a-b)+2(pi)b

[@deskgo]

In row 1704 you accidentally copied the row number B1704 into the perimeter of the ellipse column. The B makes it NAN and screwed with my code for checking. You may want to fix the download.

[@foobargorch]

3:51 didi you say INTERCAL or unlambda? couldn't quite make it out

[@100dollarpie]

Cut out the ellipse. Cut out a square centimeter. Weigh both and divide. The answer is the square centimeters in your ellipse.

[@zenithparsec]

He didn't ask for the area though. But you could just take the cut out ellipse, mark some random point on the edge, and place it with the mark at 0 on a ruler. You could then roll along until you reach the mark again to get the perimeter. But that also doesn't solve the general question he asked.

[@Myckou]

At 11:39pm??? How do you expect me to work on it now?!

[@stevethecatcouch6532]

Change your past decisions so you are living in a different time zone.

[@Mystery_Biscuits]

Change your past decisions so your body clock operates in a different time zone

[@loreleihillard5078]

in my Time Zone, they usually come out around 1am, so be grateful it's just this one and not every time

[@JamesCoutie]

@@loreleihillard5078 Same here, I was up til 5am waiting for it to come out, then it ended up coming out at 7:30am, so I only got about 2 hours sleep

[@loreleihillard5078]

@@JamesCoutie I was up until about 1:30 waiting and then figured he must've skipped this week. Good thing I woke up at an appropriate time though

[@Huntracony]

That's unfortunate. When the video released I correctly intuited the solution, but did not have the mental capacity to prove it, so I decided to wait. I was here early too, probably lost out on quite a few speed points. Oh well...

[@Henrix1998]

I wonder if this can be just integrated

[@Qermaq]

So... been like a day and a half. Any updated timeframe?

[@Myckou]

0 < x < 0.28m Not close enough?

[@balping]

I think that is actually false

[@Blauefrucht]

Technically no. But you are infinitely close.

[@boom6766]

You are actually incorrect

[@loreleihillard5078]

@@balping spoilers, y'all

[@JamesCoutie]

@@loreleihillard5078 It's not really spoilers if it's wrong (and I mean the replies are wrong, the original comment is actually true)

[@akshat9282]

3:30am and I'm here watching this as it's uploaded

[@computerfis]

Just submitted my answer, hope it's not terrible wrong. I have a habbit of misunderstanding things =)

[@Questiala124]

Harmonica meme: 2pi(((1/a)+(1/b))/2). Haven’t testes it yet but it seems nice.

[@thrillscience]

I'm going to guess without even watching the video and seeing the question.

[@EumelHugo]

Grabbed some cardboard two pins and a piece of string and here we go.

[@alonshaltiel9919]

This seems to be more of a psychology test than a maths one... a few weeks of no puzzles and then suddenly one that seems to be so trivial! could it be...?

[@MCLegoboy]

Hmm I see... but could we make it better with more band?

[@thembushes1554]

Hey you should make a statement on the whole captions thing pls.

[@rosiefay7283]

What language do you want them in?

[@VaradMahashabde]

It is 3:14AM right now for me. π AM Clearly, God is on my side

[@VaradMahashabde]

Too easy if you know it/studied it recently tbh

[@t71024]

It's exactly 3:14:15.9265 in some time zone on some globular planet any time.

[@scoutskylar]

@@t71024 No, that's not how time zones work.

[@simonmultiverse6349]

@@t71024 The International Space Station experiences pi o'clock a lot more often than the rest of us do.

[@Veggie13]

The main channel video hasn't been posted yet, eh?

[@BryanLeeWilliams]

Is it just me or was there no video published on the main channel?

[@BryanLeeWilliams]

Oh. 5 more seconds on the video and it's answered

[@BryanLeeWilliams]

And either I completely missed the boat or the answer is trivial.

[@thomasschneeberger5792]

sooo, just using excel to fit something to the data does not count?

[@simonmultiverse6349]

I did that yesterday and I came out with some Fourier coefficients which are jolly useful. I'm setting the semi-axis of the ellipse to cos(phi) and the other semi-axis to be sin(phi). This gives an ellipse of area = pi.cos(phi).sin(phi). Then I put in phi and my spreadsheet draws it. Also I'm measuring the circumference as a function of phi, and a Fourier series does it very nicely, needing only cosine terms, and only every fourth one is non zero. The formula is C0 + C4.cos(4.phi) + C8.cos(8.phi) etc. and the terms get small very quickly.

[@Tfin]

I mean, it has to be that, right? I mean, it must be? So to check it anyway....

[@leachy3000]

This needed more band

[@Cloiss_]

more choir imo

[@JamesCoutie]

For people who have been commenting that it was really easy, it's possible to have an area >650cm^2, so perhaps check the area you get before submitting Edit: I stuffed up my calculations, I was wrong!

[@MrDannyDetail]

Do you mean 615 rather than 650? That seems to be about the upper bound to me.

[@JamesCoutie]

@@MrDannyDetail No, I mean 650. I'm assuming you get 615.75, but there's definitely ways to get a much larger area

[@MrDannyDetail]

@@JamesCoutie I've since realised that I don't even understand if the pen is attached to a fixed point on the string, or if if it is moving freely along the string despite being attached to it. It looks in the footage at the start of the video to be the latter, but I'm not 100% sure. If it was on fixed point on the string, and wasn't at the dead centre of the string, then would the presumably asymmetric result still count as an ellipse?

[@MrDannyDetail]

@@JamesCoutie And I get 615.7325 ish.

[@stephenbeck7222]

MrDannyDetail if the pen was attached to a fixed point on the string then I’m not sure what you could trace but it certainly would not be anything like an ellipse.

[@Joe_Payne]

I'm watching it too soon. :(

[@haalstra]

This feels way to easy to be correct.... just to be sure tried 4 different way's to solve this, but they all gave the same answer.

[@Septimus_ii]

Is it just a circle?

[@YassinElMohtadi]

@@Septimus_ii Yep

[@LeeSmith-cf1vo]

perimiter = 1 errorMaring=lots

[@t71024]

I was expecting a solution video today, but - zip, nada, zippo, squat, nought, nix, zilch, nothing, zero.

[@DaviddeKloet]

I'm waiting for it as well. The league table is updated though.

[@simonmultiverse6349]

You left out "diddly squat."

[@TheShanir]

A Parker Ellipse

[@AgentM124]

That's late today!

[@AlexSh789]

Not where I am, it isn't. 👍

[@ChongFrisbee]

There is no such thing as "too soon" to watch this

[@ChongFrisbee]

Just came in from the main channel to check... definetly not too soon

[@Petch85]

omg, i love this.... But I hate the perimeter of ellipse, way is this so hard, it should be easy f(a,b) = ? I think this is proof that we do not live in a simulation, cause who would come up with that? :-)

[@zenithparsec]

It's trivial to generate an ellipse though... perhaps the simulation doesn't need to calculate the perimeter using those parameters? If your simulation uses atoms which know what object they are part of and where they are relative to some point, as well the location of its nearest neighbors, you could simply ask the object how big it is, and it could send a message to its constituent parts and they could tell you. An advantage to this, is it would even work if the object wasn't a perfect ellipse... you know, like in the "real" world. Or you could count how many 'voxels' (volume-pixels) it passes through, and that would give you as accurate of a length as you probably need. Perhaps randomly perturb the offset/rotation of the ellipse and average the values over multiple samples to get a better approximation? Sometimes just because something is hard to work out one way doesn't mean it's actually hard to work out

[@Mrsparky492]

Pi * ( b+a) is extremely accurate for very small c. That's the best approximation I have.

[@peacockmoss1491]

Ha! I got here too early!

[@scanerang]

I don't know what the question is, but based on the title i'd say a perfect circle

[@banananaa]

You should get a new pen.

[@PetruRatiu]

I'm looking forward for the solution of the area puzzle, because I tried my hand at it and got zero, and that's too boring a result to be true, so I'm most likely missing something.

[@johnchessant3012]

standardized test question time... rectangle : square :: ellipse : _______

[@loreleihillard5078]

this is a massive spoiler tho...

[@zenithparsec]

The whole "come up with the same analogy that I did to prove you are smart" is a stupid thing to put in a test. I guess it does cause "standardized" thought though. Here are 5 answers that are not considered correct. - It's also 'square'! (you can define the area of both using no math more complex than calculating the 'square' of a number... multiplying the height by the width is like doing a square, and you do the same thing, basically, for an ellipse) - why not 'cube'? because for the ellipse, while my previous answer is basically true (multiplying the height by the width is basically the same as multiplying half the height by half the width by a constant), it's actually closer to calculating the cube of a number, except usually all three of the numbers will be different. - "a five pointed star"! (they are both shapes which start with the next letter in the alphabet) - 'an pair of parentheses pushed close to each other'? (The shape you get if you cut out the middle of each and push them together) - or an 'eyeipse'? (like an L-ipse, but with some of its length cut off )

[@VaradMahashabde]

18 views, 53 upvotes This makes sense

[@Xeridanus]

KZbin doesn't update the stats in real time.

[@zenithparsec]

@@Xeridanus So obviously they must use imaginary time to update the stats, which allows faster than light transfer of the data (and the video) between their servers around the world. (see special relativity). Checks out.

[@Grimlock1979]

Matt's first ASMR video...

[@CthulhusDream]

#toosoon

[@stevethecatcouch6532]

In the video, when you mentioned the major and minor axes, you traced the semi-major and semi-minor axes. Is that a British thing, like misspelling math?

[@abcdefg9213]

The fact that I've got an answer and there are less than 1000 views is promising Still, don't think it's the right answer

[@douglasbrinkman5937]

answer required in metric...i'm out.

[@jetison333]

Just pretend it's 28 inches instead, and perfect! :)

[@rebmcr]

Douglas Brinkman OK Boomer

[@andrewkepert923]

This will make it easier: 28cm = 11³⁄₁₂₈

[@olivier2553]

@@andrewkepert923 11 3/128? You mean 33/128? This notation of implying an addition when everything taught in maths says it should be a multiply is the mot confusing stuff possible, it is at least as bad as the imperial measures: how can you expect the kids to have a clear mind when you teach them two antagonist systems?

[@Xeridanus]

@@olivier2553 It's not multiplication, it's a single number pronounced with the word 'and' between ie: eleven AND three one hundred AND twenty eighths. I emphasised the second 'and' to show that this is used in other places as well. And also denotes addition btw same way times denotes multiplication.

[@alexismandelias]

Isn't there a formula for the perimeter of an elispe? Am I missing something?