I like to think they have him go to the sink to put them on as a tease. "Oh good, he's at least going to rinse them." Then you remember this is the man who once was craving beer so much he licked the dirt under the bleachers at a baseball stadium.
@@ImDemonAlchemist The definition of a placebo is "A medicine or procedure prescribed for the psychological benefit to the patient rather than for any physiological effect." You could say that homer receiving the glasses or the scarecrow receiving his "brain" making them think they are smarter when in fact they are not as a placebo effect.
Oh man, I got that Homer's line was an homage to Wizard of Oz, and I could get that Homer got the Pythagorean theorem wrong, but I never noticed that the original line in Wizard of Oz was wrong!
I know this is a year old... One of main themes in Oz is that magic can't solve your problems. The wizard actually doesn't do anything in the world (allegory for false promises of politicians) and the work is left to the outsider Dorothy. Scarecrow thinks he's been fixed, but he was already "fixed," he just didn't know it.
I don't remember noticing that either! The lines go by so quickly it's hard to notice exactly which words they are using or have time to think about it!
@@itzmistz no, it's not. it's a green screen, he has a monitor to the side where he looks at a teleprompter or notes or a slideshow, and the edit is placed over later. no projector
I suppose if you take a cylinder at least 4 but less than 6 units in circumference and wrap the big side around and join it with the two shorter sides. I'd hesitate to call it a triangle though. It would be more like a letter C with the gap joined by a little v at right angles.
This line is also referenced in an episode of Hey Arnold, where Arnold’s grandpa goes back to elementary school to get his grade school diploma. Funny thing is Dan Castellaneta (who voices Homer) also voiced Arnold’s grandpa, whom recites this line to the principal in order to secure his diploma.
Estava pensando exatamente isso. Quando eu estava no ensino fundamental/médio não conseguia visualizar as equações dessa forma, era tudo muito abstrato, depois desse vídeo consegui compreender algumas coisas da época da escola.
David^2 - S^2 = Cohen^2 gives us a hint: the "D" from David stands for Donut, the S stands for Sign and the C stands for Colossal donut. When homer points at the colossal donut, we can see all of these 3 points (donut, sign, colosal donut) in one frame. If we connect these 3 points we get a triangle where a is the height of the sign including the colosal donut. You can also measure the angle of homers arms (alpha): 10, and the credits give us the number 24m as the length of b. We can now calculate the length of the hypothenose c: 24/cos(10) which is 24.3. Now we can calculate a: sin(10)*24.8 which is about ~4m. This means that the man holding the colosal donut plus the colosal donut is 4 meters high. They are about the same size so we can divide by 2 to get the size of the colosal donut: 2 meters!!!
Hold on: If b=0, then we have a line. Then, solve for a using the first equation, and you get (a)^(1/2) = - a^(1/2), so a=0. Thus, you are left with a point. That's the joke! They have a point! :)
Hold on, I don't think it'll work like that. We got b=0 for the second equation, so we can't use that in the first equation. These are not a set of equations, rather a matter of either/or . Also yeah ik this is meant as a joke lol :-)
That actually says '2+' on each power, which is actually probably better read as a contradiction of Fermat's last theorem, and if I remember my Simpsons correctly, is not the last such contradiction in the episode :) check the equations in the background when Homer is in the 3rd dimension...
+HerraTohtori Well, with more complex surfaces you first have to make up your mind what exactly you mean by a triangle. I've left a few comments earlier on in which I talk about this. Maybe have a look :)
I was thinking that would make sense considering Homer's favorite junk food! As for the math to prove it, I'll leave that to folks with more time and math training than me. If true, maybe Wizard of Oz screenplay writers (or Baum himself, if those exact words are found in the book) had donuts on the mind and/or knew something about tori.
They already did it! Season 10, Episode 22 "They saved Lisa's brain" Stephen Hawking: "Your theory of a donut-shaped universe is intriguing, Homer. I may have to steal it" (Dun dun duuuunn)
Any Lorenz geometry model usually works without triangle inequalities. Not sure now, but maybe homer's theorem holds true for minkowski space? Where triangle inequality is reversed?
I just want to say that I really enjoy this channel. It's difficult to find interesting videos about cool bits of mathematics, and so far I have found 2 channels that deliver this: numberphile and mathologer. Keep doing what you're doing!
You're missing 3blue1brown, especially if you're already somewhat advanced in your maths education. But even if you're not, there's plenty of cool stuff on that channel too, definetely check it out. (i know the comment is old af, but if you haven't seen it since then, GO FUCKING DO IT :p)
It works on a sphere when you ignore the any sides part. You can create a triangle were two of the sides equal a quarter of the circumference of the sphere and the other one spans around the equator. The angels between the equator line and the other two are always 90°, therefor the triangle is iscosceles. The third side can now vary from 0 to the circumference of the sphere. So if you subtract the other two sides (which equal half of the circumference) you still got the possibility to have half of the circumference left. Since in this example a equals b 2*(square root of a * square root b) equals 4*a. Since a equals a quarter of the circumference you get the solution when c spans the whole circumference. It doesn't look like a triangle but technically it is a triangle on a sphere I guess.
A man in the lightmode void talks about the mistakes Homer Simpson makes while looking at an omnipresent context and visual providing object that reacts to both his words and the content it showed previously.
In spherical geometry opposite points on the sphere are considered equivalent: this is because it changes the 5th postulate to say that parallel lines do not exist: lines can only be maximum circles (circles made by a place cutting the center of the sphere). All lines are perpendicular and cross at ONE point: so they consider the opposite points as one single point. So some of those points on your bigger side are part of the original triangle and the others are excedent like a side prolonged even ending on the same points. The distance in Riemannian geometry is given by the SMALLER maximum circle because a metric can not be a multivalued function and the metric by definition must obey the triangular inequality (or the hell will go loose and several contradictions arise because the metric should capture the intuitive notion of distance as being additive, and being symmetrical (in a loose sense that I am too lazy to explain). So your construction is not a triangle is a triangle with line segments added (my explanation is a little convoluted because I am lazy, maybe later I explain better).
There's the time homer solves fermat's last theorem. But they used an edge case where the answer is incorrect in decimals that a regular calculator doesn't show
There are examples in which an instance of that formula, sqrt(s)=sqrt(x)+sqrt(y) may be found. The triangle inequality is reversed in Minkowski space, so that's a candidate. Secondly, it might be possible to find instances of that on some surfaces, such as a variant of the pseudosphere or some other surface of revolution of some cusp-containing curve. Finally, something similar to it can be found in particular Lp spaces. For example, a space with a norm |s|^p = |x|^p + |y|^p will have something akin to the required formula for, say, p=1/2 What I find very intriguing about the last option is that circles, when drawn on a euclidean plane, will look like Lamé curves (with the power parameter being 1/2). In short it can be done in spaces with a quasi-norm.
Late to the party, I know. But my point is that sqrt(a) has TWO values. sqrt(b) + sqrt(a) = sqrt(a) might not work, but sqrt(b) - sqrt(a) = sqrt(a) could easily be true, as well as -sqrt(b) + sqrt(a) = sqrt(a)
@@Goldmos1 topology I believe, I'm taking my first course in it now (MATH 525). I'm in my final year as an undergraduate and the stuff in the post seemed like stuff I could probably start to grasp. And I'm in North America, you can learn any mathenatics you want. You just have to be passionate and eyeballs deep in student loans! (The later may be optional if your really gifted or driven, but scholarships are few and self study is difficult)
+John Galmann It does as long as all the lines are timeline - and that gives rise to the so-called twin "paradox" (not a paradox at all, of course, just the result of the triangle inequality in a Minkowski space).
Jasper Tan citation needed... That silly formula for the Minkowski metric doesn't make much mathematical sense, especially considering that the distance between two distinct simultaneous events is an imaginary number(?!)... Even assuming that is the case, imaginary numbers aren't comparable, so the hypotenuse is neither shorter nor longer. :-\
it's been a long time since I used any high level of math. mostly basic stuff, Pythagorean theorem always comes in handy, and geometry in general. I do grow increasingly fascinated with Eratosthenes. This guy was simply amazing. Kind of sad, put in all those endless hours of head splitting work, worry, study, panic, study more, obsess, and in the end I have to periodically give myself math test so I don't forget all of it. everything today is charts, computers, and more charts. I remember i started my job and could figure everything with mobil calculator, pencil, and paper. Co-workers were jealous I believe and said why figure it out like that it's in the tables. One professor I had said - I feel sorry for you if technology ever crashes. At The time I didn't care The exams were so damn long and hard that without a calculator I would have had a nervous breakdown trying to crunch it all before I ran out of time. Now I understand though. The most important stuff you will need in life is college algebra and geometry maybe some trig but probably not. However when you have that knowledge it feels good. In a job interview I got asked a math problem and immediately pointed out the flaw in the question and offered a math solution to solve it. The other mathlete in the room laughed and of course no job for me. However, it felt damn good.
+The Einhaender I’m afraid that’s it... 8:20 He said "it’s a tough one" and "there is a *hint* in the credit". Then at 8:38, they give the credit hint. I can’t believe the solution is this obvious. If it was all, they would say, the *solution* is in the credit, or something a bit more allusive I guess... David S Cohen is the math guy he must have done something clever in the sequence, not just adding a few squares in the credit... ;-) My comment was totally desperate, I know it can’t be about the dots on the donuts. I searched for triangles that could have some obvious ratios, I couldn’t find any right triangle! Or maybe some circle with a crossed diameter, no chance... I’m afraid I’ll just be disappointed in the end. In ... Anyways, the show is great.
If the triangle is inside of the sphere, the two shortest lines can split from the longest line right before it makes the full radios. it would be a weird shape. but it would have three corners and it would give the two short sides a opportunity to be infinitely shorter then the longest line. Also works for the outside of the triangle ofcourse :)
I wonder if the point was that the scare crow didn't really get a brain. He just had to think he did. The líon just had to think he had courage and the tin man had to think he got a heart. Which is kind of an interesting angle from when I was a kid and I literally thought they had somehow actually received these things.
But A+B < C does work on a sphere. You just have to go the long way around the sphere. So the Mercator projection would look like ____________/\______________Edit: I'm sure you've gotten this a thousand times. I tried to find a similar comment, but if it's not in Top Comments....
Not sure if I'm being an idiot and I would like more insight on this but wouldn't that Mercator projection make a hemisphere with a triangle missing instead of a triangle since the inside angles would exceed 180 degrees
At around 7 minutes, trying to make the Mutilated Pythagorean Theorem (MPT) work on a spherical triangle - the triangle you show won't satisfy it, but there are spherical triangles that do. If you put the apex at a pole, and _c_ along the equator, then _a_ and _b_ are ¼-great-circle arcs ( _a_ = _b_ = ½πR), and _c_ can be any length in the open interval, 0 < _c_ < 2πR. E.g., if R = 2/π, then _a_ = _b_ = 1, 0 < _c_ < 4. If you could make _c_ the entire equator, you'd have _(a,b,c)_ = (1,1,4), which satisfies the MPT; that is, for sides taken in the order _a,b,c_ ; √a + √b = √c. If you make _a_ = _b_ a little shorter than 1 and at slightly different "longitudes", then they can be adjusted so that the great circle joining them the long way, will be _c_ = _4a_ = _4b_ , and the MPT will hold. [Interesting to note: the MPT is homogeneous of degree ½, so it scales by any constant factor without changing.]
A world where a+b can be less than c can be gotten by taking that sphere diagram of yours, and having c go the LONG way around the circle instead of the short way. Boom, a+b
SCARECROW THEOREM SHOWN TO BE CORRECT! After the Scarecrow got his brains, this is what he actually said: "There is an inverse stereographic projection of an isosceles triangle onto a 3-D surface, such that the curve length of each of any two sides is equal to 1, and the 3-D surface can also be constructed such that the remaining side would have a projected curve length of 4. Thus, considering the projected sides... ...The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side." Unfortunately, most of the Scarecrow's mathematical statement ended up on the cutting room floor. All that is left is the abridged and inaccurate version in the movie. Hollywood had no appreciation of the Scarecrow's mathematical genius.
In my language, isosceles triage is called as "equal feet triage" and right angle triage is called as "bracket triage". Maybe that's why we not confuse with both.
For triangle inequality to fail, the space would not be a metric space. Minkowski space has pseudo Euclidean metric so it might work there. Or, stretch the definition of triangle so it does not have to be three points with the shortest paths between them, but any geodesic is allowed. Then there are such triangles on the sphere, by taking the greater arc instead of the lesser arc for the longest side.
@techtrashing: Play some old Super Nintendo RPGs that feature a world map. The world map will usually loop from "west" to "east" and "south" to "north", thus forming a donut shaped world.
You can't violate the triangle inequality, a+b>c, with some weird surface because no matter what surface and metric you're using, by definition the metric has to satisfy the triangle inequality. The only way is if you choose the sides of the triangle to be something other than geodesics (shortest paths), in which case you don't really have a triangle, just some 3 vertex shape.
Maybe it's a triangle on a cone. Varying 3D cones have different degrees, like cones that have 10 degrees or 37 degrees even 50 degrees. So the isosceles triangle is on a cone, where the remaining side cuts through the cone exactly and the first two same sides indicate the degrees of the cone. So maybe it's an "isosceles cone", and the formula is actually a way to measure the circumference of the bottom of the cone. It looks triangular from a certain angle, until you realize that it's 3D! So it's possible that it's the formula to calculate the circumference of the bottom of the cone. From there, maybe the cone height and even the cone volume can be calculated. And it we know the weight of the cone, we can use the formula "D=m/v" to calculate the cone density and then put it through the density experiment to see if it floats on oil or sinks in honey or floats on water or maybe floating in alcohol or lamp oil or sinking in galinstan liquid metal alloy. Or maybe it's a pac man cone. An incomplete cone with two sides that meet up in the bottom forming a pac man shape at the bottom of the pac man cone.
I think I have an easier proof for the isocele triangles that 2*sqrt(a) =/= sqrt(b). You can construct an other isoceles triangle with the equal sides which are still a and the remaining size which would be b' =/= b. Assuming the theorem mentionned is true, you have that sqrt(a) = sqrt( b )/2 = sqrt( b' )/2 which is a contradiction.
+Jelle (NL) Ah, yes, that's a nice one. There are actually two occurrences of "counterexamples" to Fermat's last theorem in the Simpsons. The one you mention is the second one. The first one pops up in Homer^3 (Homer cubed) where Homer stumbles into a 3d world. Very neat stuff. There is also one mention of Fermat's little theorem in the Futurama Simpsons crossover episode.
+Mathologer keep the counterexamples in quotation marks. As per the numberphile video those are only close to a solution, not exact. (even the parity of the sum is wrong). P.S. just to make sure no one thanks Ferma's last theorem is debunked.
7:00 The answer is *never* on any Riemannian manifold ... if "length" is defined as *geodesic distance* ... because the geodesic is the *shortest distance* between two points, which forces the triangle inequality. Now, on a *pseudo-Riemannian* manifold (even flat, like Minkowski space), that's another story. This leads naturally to a question for you: do the flight distances of New York, Miami, Chicago and Houston fit in *any* Euclidean geometry, if they are treated as straight lines? If not, then what's the minimum curvature they must have before they do? What about other sets of 4 cities on the Earth, like London, Tokyo, New York and Johannesburg? Which geometries will 4 cities fit on, as a function of how much curvature their flight paths are endowed with? (Yes, some cases require a 2+1 dimensional Minkowski Geometry). What about 5 or 6 cities? And since the Earth is *not* a sphere, what happens if you try to fit 6 cities, as a function of the curvature you give all the flight paths, assuming they're all given the same curvature? How much information can be said about the dimensions of the Earth - as well as the cities' *latitudes and relative longitudes* - on the assumption that the 6 cities fit on a ellipsoid? Try it with { New York, Miami, Chicago, Houston, Los Angeles, Seattle}, as well as {London, New York, Tokyo, Johannesburg, Melbourne, Rio de Janeiro}.
If we modify it just a little bit, so it says: c = sqrt(a) + sqrt(b) then some triangles do satisfy this, like a = 1, b= 2, c = 1+sqrt(2), but this is no longer Homer's triangle. Though, who knows. maybe the Scarecrow-Simpson triangle needs complex dimensions. I haven't tried that. :D
When I was riding in an elevator with a colleague, I asked him what he taught. He said "Math." He asked me what I taught and I said "Communications." He smugly replied "That doesn't count." He explained that math was the only subject that was worthwhile. Don't get too full of yourselves math people.
The problem is we conflate side with length. In Euclid's statement of the Pythagorean theorem he talks about a side's square, essentially referring to the area of the square built on a side. If we do that here, and recognize the existence of two square roots for each area, the correct theorem is: The sum of (one of) the square roots of any two sides (areas) of a degenerate triangle is equal to the square root of the remaining side (area). That means a triangle's vertices are three collinear points precisely when 4a^2b^2 = (a^2 + b^2 - c^2)^2 or equivalently (a^2 + b^2 + c^2)^2 = 2(a^4 + b^4 + c^4) These are fundamentally functions of the square areas built on the sides, not the side lengths. Homer's onto something.
There is no metric space where this equality could happen. In particular it is not true for any space with metric (Riemannian manifold: en.wikipedia.org/wiki/Riemannian_manifold), the sphere included. In such a strange world we would have a distance function which is does not satisfy the triangular inequality ...
The video suggested drawing a triangle on a sphere but a+b>c is still satisfied. If you draw a triangle on an ellipsoid you can get lines where a+b>c is now false. Think of a football cut in half from one pointed end to the other. the a and b lines come down from the center on the top to the center on the sides. The c line starts at the bottom of a, around the ellipse towards the small rounded end and back to the bottom of b. The real question is whether a triangle drawn on something other than a flat plane can still be considered a triangle.
Additional arithmetic proof that Scarecrow's theorem doesn't work: An isosceles triangle is two congruent right triangles that share a side. Let's call the length of this side h. So each right triangle has legs of length h and b/2, and hypotenuse a. h^2+(b/2)^2=a^2 This can be rearranged into: h^2=a^2-(b/2)^2 Since h is the height of a triangle, we know it must be a real, non-zero number. The square of any real, non-zero number is positive, so h^2 is positive. If h^2 is positive, then a^2-(b/2)^2 must also be positive, thus a^2>(b/2)^2 Take the square root of both sides and multiply out the 2, and we get 2a>b Substitute b=4a 2a>4a Divide out 2>4 Which is an invalid result. Thus, b=/=4a, thus Scarecrow's theorem is disproven by contradiction.
I know this is old but I'm going to take a crack at these Pythagorean clips. In the first clip, David S. Cohen's name is written as "David^2+S.^2 = Cohen^2", quite clever ;) Of course, the second time around, A^2+B^2 = C^2 is just on the "MATH BOOK"
I know this is a dumb question...I guess I'm just not nerdy enough, but I don't get the joke. how is "David squared plus S squared = Cohen squared" clever? Is there some hidden meaning? Is there some sort of language wordplay thing going on there? I understand the pythagorean theorem, I understand the reference, but I don't get the joke.
oh yeah, I looked back and I see the joke...it was just wordplay like on the halloween episodes they do that with the credits like "James Hell Brooks" instead of "James L Brooks" I just thought maybe it was some sort of like...higher mathematics joke like a reference to a famous equation or something.
I did the calculations thinking that David^2+S.^2 = Cohen^2 would correlate numerically, if, for example, each letter associated with a number value (A=1, B=2, C=3)... but I didn't find anything. Someone can check my math, but I got DAVID (4+1+22+9+14)=40^2= 1400 Plus S (19)=19^2=361, so together 1961 equals COHEN (3+15+8+5+14)=45^2=2025. So, all together, 1961=2025 which obviously doesn't add up.
It is possible on a sphere if the longest side almost stretches around the whole sphere and the rest is normal. It will make every side connected to each other, so I think it still counts as a triangle.
Take that same triangle on a sphere then rotate it up while looking straight at it till half the triangle is out of sight. The apparent lengths of the sides from your perspective now satisfy the equation.
Surely a + b < c would work on a spherical surface if c is greater than half a circumference. Eg. on a sphere of radius r, if c is 2r*pi-r/1000 then any combination of a, b would work if a+b > r/1000. An interesting special case is if c is exactly half an equator ie. c=r*pi and we have a and b meet at the north pole. There a + b = c exactly, and the three angles are 90,90,180 degrees respectively.
It could work with complex numbers where i*i=-1, then: a*i + b*i +2sqrt(a*b*i*i) = a*i + b*i - 2sqrt(a*b) = c*i might lead to some solutions. p.s. oh... its 5 years old - saw 5th of september and didnt noticed the year :D
