So we got the Parker Square; is this the beginning of the Grime Triangle? A metaphor for being so amazed by maths that you wholeheartedly ignore the quality of real-world applications.
My wife died two weeks ago. She was a huge "Hitchhiker's Guide" fan, and used 42 in some way whenever possible. She would have LOVED this video. That she also had a bit of a crush on Dr. Grime wouldn't have hurt. I've purchased the 42 Superhero Triangle shirt, to commemorate my own beloved superhero. Thank you for this.
Yes I saw that, and yes he says it's a cheat... But because it can be made from two RA triangles when actually it's one itself. Not the same thing at all
@@E1craZ4life They have done at least one video on Matt Parker's channel, with one of them proving something (I can't remember what) with geometry, and the other proving the same thing with algebra. But I would enjoy them coming up with something together, reminiscent of the Parker Square, or the Grime Triangles, so it can be called the Parker-Grime [what-ever-it-is]. Or Grime-Parker [thingamajig].
As stated at 12:50 in video, for a given q, there are a finite number of Hero triangles with Area=q*Perimeter. An intuitive way to think about this is to note that as you you scale up the size of an object, without changing its shape, its perimeter grows in linear proportion to the scaling factor, whereas its area grows proportional to the square of the scaling factor. This suggests that, for a fixed value of q, if you increase the required area sufficiently, there will be no Hero triangles satisfying Area=q*Perimeter for that particular value of q.
I think scaling up will change the proportions since we're also tracking into account area: perimeter is in 1 dimension and area is in 2 dimensions. If you take a triangle and double all the sides, the perimeter doubles, but the area quadruples.
@@gavin5410 That is for his "Superhero" triangles. For the Hero triangles they just both need to be an integer. Scaling up a triangle with an integer length and area by an integer multiple will result in both still being integers.
9:10 it's quite easy to work out by hand as well since it's the square root of a product and it's an integer. 32 = 2^5 27 = 3^3 which means 32x27x3x2 = 2^6 x 3^4 and then the square root is simply 2^3 x 3^2 = 8x9 = 72
Was feeling down this morning, thought to myself I need to see some numberphile with James. Went to youtube and immediately a new video, love you guys you are my comfort channel for the past 5 years xD
The framed brown paper about Graham's Number behind James, written and signed by Ron Graham himself, really is something like the holy grail of Numberphile!
Something I would like to see you do some day: Take one of these ancient mathematical truths and do the arithmetic using the ancient system of numerals. Some possible ideas: Calculate pi using Roman numbers. (Roman numerals were used for over a thousand years. How does one add, subtract, multiply, divide, take square roots with Roman numbers? ) As you know, the Pythagorean theorem predates Pythagoras. How did those earlier mathematicians find the length of the hypotenuse using the numerals available to them?
What I find so astounding, is that with the infinity of whole numbers, all the super hero triangles use such small numbers. Yes I realize that the fact that the perimeter will scale linearly while the area scales to the square so one could intuit that they all should be relatively small, but the fact that even the areas and perimeters are still only two digits (at least in base 10) I still find surprising.
I found an intriguing connection between nearly-equilateral Heronian triangles, the Pell numbers, and sequences of consecutive integers whose standard deviation is an integer, but a TH-cam comment is too small to contain the proof.
14:13 Thank you, Prof. Grime for making a connection between this interesting phenomenon and something I'd already been studying. The latter is triples of integers {x_i, y_i, z_i}, with the same sum and the same product. Following your notation at 10:28, suppose that for various indices i we have Hero triangles {a_i, b_i, c_i} with the same perimeter a_i+b_i+c_i=2s and the same area A. Then define x_i=s-a_i, y_i=s-b_i, z_i=s-c_i. Then x_i+y_i+z_i=s as desired. Moreover, x_i y_i z_i=P=A^2/s. So if you can find triples with the same sum s and the same product P, then you have triangles with the same perimeter 2s and the same area sqrt(sP). And if sP is a square, then your triangles are Heronian.
Here's a further nice feature. Define q=s/2, the quarter-perimeter. Then if we define d_i=a_i-q, e_i=b_i-q and f_i=c_i-q, then all of the (d_i, e_i, f_i) have the same sum q and the same cube-sum q^3-3P. Now suppose that you managed to find two such Hero triangles, and in one triangle the shortest side c_2 was q, the quarter-perimeter. Then f_2=0. (I'm using _1 and _2 to index the two triangles.) Now define sets L={d_1, e_1, f_1, -d_2, -e_2} and R the same but with all five signs flipped. Then L and R each have sum 0 and cube-sum 0 (by the previous paragraph). More trivially, L and R agree in square-sum and fourth-power-sum, because a number's even powers remain the same when you flip its sign. So L and R are a solution to the Tarry-Escott problem for powers up to the 4th. For example, triangles (18, 17, 5), (19, 11, 10) with perimeter 40, and 10=40/4 as required. Then L={8, 7, -5, -9, -1} and R={-8, -7, 5, 9, 1}.
12:49 "If you want your area to be a multiple of the perimeter" A nice thing about all triangles (not just Heronian ones) is that A=rs where r is the inradius.
I had an aha! moment with the rational height -- so simple yet so efficient! It means that every one of these is two Pythagorian triangles stuck together, but possibly with a rational scaling factor. How cool is that!
Is it proven that those are the only superheros? I understand that as you scale things the area goes faster than the perimeter, but there could hypothetically be a monster triangle with massive side lengths and angles of about 0, about 0, and about 180 degrees. Or to put it another way, there are clearly infinite triangles with massive side lengths who's areas are smaller than their perimeters, and also infinite triangles with area larger than their perimeter, and with how many massive integers there are, it seems surprising none of those massive triangles fall on the line where area equals perimeter. If such a proof exists I'm sure it's neat.
I was thinking about triangles with equal perimeter and area but not necessarily integers, but I realised all triangles can be scaled to such a triangle: Any triangle has a perimeter p and an area a. If we scale the triangle (lengths) by p/a, the new area is a(p^2/a^2)=p^2/a and the new perimeter is p(p/a)=p^2/a.
Oh this is phenomenonal, easy enough to understand for almost anyone and a great introduction to the concept of Mathematical proof for people who haven't studied it.
Primitive Heronian triangles always decompose to two rational-sided right triangles, but if you scale them up you will get two Pythagorean right triangles, primal or not. I like how if you glue a 9-12-15 and a 5-12-13 together, say, to get a 13-14-15 Heronian triangle, then flip the triangle so 13 or 15 is now the base, you can decompose this into one of the original Pythagorean triangles and also a 36-53-65. I’d love to see Dr Grimes explore where the new triangle comes from.
In school if I said something like: "The length is 5", I got asked "What 5? Five eggs? Five apples?" we were always demanded to name the unit. Even if the actual unit was not needed, we were demanded to call it. Lets say "30 length units" or "30 length units to the square" and so on. And honestly this is so much inside me, I was cringing alot in this video. "It has the same perimeter as it has area" and such things simply feel very wrong from the beginning. - Of course I like the video, it was very cool as always! :-)
From the studio that brought you the Parker Square and the stunning cinematic sequel of the Parker Circle comes the exciting new new series of the Grimes Triangle.
I'm Polish and the only moment I realised that your English 'Hero of Alexandria' is Polish 'Heron z Aleksandrii' was when you showed the equations that we covered in High school. It seems as math's language is more universal than national languages, who would have guessed😉
I scrolled through most of the comments and didn't see anyone point this out: the 5,29,30 triangle can be obtained by taking a 5,5,6 triangle (Heronian) and a 6,25,29 triangle (also Heronian) and gluing them along the edge of length 6. So there should be some "prime" Heronian triangles that cannot be obtained this way...
He’s back!!! We’ve missed you, James - it’s just not the same without you. I haven’t even watched the video, yet, and I’ve already upvoted it. Please don’t take so long to make your next Numberphile video. edit: James is the click bait 😊
After watching this video, I had a dream where The Moody Blues were on Sesame Street performing a song about perfect numbers with Cookie Monster. The song included some sort of nonsensical visual proof in high dimensions about their properties. It did not sound much like a Moody Blues song, and Cookie Monster sounded like he had a cold, so he was a particularly unconvincing lead when they did "Nights in White Satin."
I propose another category of triangles called Super Villain triangles. They are Super in the same way as the heroes, in that the perimeter is the same number as the area, but they are not heroes, meaning that their side lengths do not have to be whole numbers! Let me introduce you to their leader: 12sqrt(3); He is an equilateral triangle with all sides = 4sqrt(3), as well as perimeter and area = 12sqrt(3).
6:03 You don't even need to drop a perpendicular. You already have a right angle at the top. Your triangle is 3, 4, 5 scaled up by 5. 10:37 This is why I like James Grime. Real zeds, and real "less than or equal to" signs.
"The way I was told to work out the area in school requires using Pythagoras occasionally. Let me instead show you this other method, which ALWAYS uses Pythagoras (or, actually, Ptolemy, who Pythagoras' theorem is a special case of) in order to avoid occasionally having to use Pythagoras." If that isn't the biggest troll ever then I don't know what is. 😂
However, there are infinitely many Hero triangles with rational side lengths. Take any triangle, say with perimeter p and area a. Then scale it by the factor p/a, so that its perimeter and area are both p^2/a.
Lay the equal sides of two of those superheros, the ones with a common side length of 10, against each other and you get another right angled triangle, 8 15 17. Invert one of them but keep the equal sides together and you get a Grime quadrilateral.
Seeing James in the Thumbnail gives me a feeling of Nostalgia of 10 Years ago and the beginnings of NP, ahhh good times...
Polynomial vs numberphile - greatest unsolved problem in computing
The man was genetically engineered to make math videos.
Seeing James just made my day! Numberphile is never really Numberphile without James.
There is a P vs NP joke hiding somewhere around here...
@@LowellMorgan my ex was attracted to him. While we were together. xD
Recreating the horrible misshapen triangles on the animation made me burst out laughing. Sometimes, it's the little details that makes a difference.
He really Parker Squared them.
HEY, they are the lovely ones, don´t mock them hahahaha
I was laughing so hard when the numbers actually appeared next to the janky triangles
So we got the Parker Square; is this the beginning of the Grime Triangle?
A metaphor for being so amazed by maths that you wholeheartedly ignore the quality of real-world applications.
Hey! Those are jazz triangles, man!! They're totally cool.
Fred
My wife died two weeks ago. She was a huge "Hitchhiker's Guide" fan, and used 42 in some way whenever possible. She would have LOVED this video. That she also had a bit of a crush on Dr. Grime wouldn't have hurt.
I've purchased the 42 Superhero Triangle shirt, to commemorate my own beloved superhero. Thank you for this.
Sorry for your loss...
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41, 43,44,45,46,47,48,49,50,51,52,53,54
That’s beautiful
My condolences.
??
5:49 this is a 3-4-5 triangle sized up. It's a right triangle as well!
Yes I saw that, and yes he says it's a cheat... But because it can be made from two RA triangles when actually it's one itself. Not the same thing at all
I'm glad someone else has spotted that too, I thought I was going mad!
he probably drew up something quickly to illustrate his point without noticing it was a poor example
I'm glad to have James Grime back. Wonderful surprise Numberphile.
james is no square.
why? was he ever gone or something?
??.
James Prime is back!
James Prime > Optimus Prime
i think you clicked on the wrong video
@@Bardigrade what?
@@Bardigrade oh, James Prime is another TH-camr
13:43 This lovely drawn triangles should be called Grime triangles.
Usually mathematicians get recognized for their accomplishments, but on Numberphile for their mistakes!
@@supermarc *cough* Parker Square *cough*
@@ob3vious not to mention the parker circles
when a numberphile guest poorly draws something we shall call it a Grimes drawing, it's only fair to Matt
I was surprised to see there were not shirts and mugs with those
"These triangles are really interesting - you can see where my printer ran out of ink"
LOL
True story!
"Parker Super Triangle" hahahahah
Yeah, that made me laugh
Was gonna say the same lol
What we need now is Matt and James cooperating on something, so we can have Parker-Grime [something]s. That could go down in the history books!
the ch community says hello
xpqr12345 They were already together in a video about dividing by zero.
@@E1craZ4life They have done at least one video on Matt Parker's channel, with one of them proving something (I can't remember what) with geometry, and the other proving the same thing with algebra. But I would enjoy them coming up with something together, reminiscent of the Parker Square, or the Grime Triangles, so it can be called the Parker-Grime [what-ever-it-is]. Or Grime-Parker [thingamajig].
As stated at 12:50 in video, for a given q, there are a finite number of Hero triangles with Area=q*Perimeter. An intuitive way to think about this is to note that as you you scale up the size of an object, without changing its shape, its perimeter grows in linear proportion to the scaling factor, whereas its area grows proportional to the square of the scaling factor. This suggests that, for a fixed value of q, if you increase the required area sufficiently, there will be no Hero triangles satisfying Area=q*Perimeter for that particular value of q.
This guy is a living legend
Anti-hero triangle: the area is equal to the perimeter written backwards
That is a cool idea. What would you call a triagle with area that the reciprocal of the perimeter for example, a=2/5 and p=5/2?
@@alexandertownsend3291 Villain
We've got ourselves a complete triangle cinematic universe
less cool since it becomes base-10 dependent.
@@missphase8127 that's why it is not a true hero :)
Can we have a t-shirt of Grime’s Lovely Triangles? 13:49
Your wish is my command: bit.ly/Griangle
Numberphile thank you! 😁
NOPE
5:51 that is a right triangle
15, 20, 25 triangle is also a cheat because it is a scaled up 3, 4, 5.
I think scaling up will change the proportions since we're also tracking into account area: perimeter is in 1 dimension and area is in 2 dimensions. If you take a triangle and double all the sides, the perimeter doubles, but the area quadruples.
@@gavin5410 yes, the 15, 20, and 25 is not a superhero either (but the 6, 8, 10 is)
@@gavin5410 That is for his "Superhero" triangles.
For the Hero triangles they just both need to be an integer.
Scaling up a triangle with an integer length and area by an integer multiple will result in both still being integers.
9:10
it's quite easy to work out by hand as well since it's the square root of a product and it's an integer.
32 = 2^5
27 = 3^3
which means 32x27x3x2 = 2^6 x 3^4
and then the square root is simply 2^3 x 3^2 = 8x9 = 72
Glad that James Grime is back. Nostalgia brought back. BTW love your channel
more videos of James Grime please! I really like the way he explains math. He makes it very interesting!
This man's voice us so nice to listen to. And he's SO WHOLESOME. I STAN James
0:46 James we have got you on the thumbnail!!That is the best clickbait though this one isn't
Welcome back James!
12:47 the left triangle is an impossible triangle. The third side cannot be longer than the sum of the two first.
Guessing that the left side is a 2-digit number with the left digit cut off by mistake.
Yeah looks like 75
Seems like the second longest side is 72, the 7 got cut off the screen
@@ar_xiv Yeah, that works: Perimeter 162, Area 324, ratio 2
If the cut-off number is 72, how come the number shown is 5?
Was feeling down this morning, thought to myself I need to see some numberphile with James.
Went to youtube and immediately a new video, love you guys you are my comfort channel for the past 5 years xD
The framed brown paper about Graham's Number behind James, written and signed by Ron Graham himself, really is something like the holy grail of Numberphile!
I could watch James Grime talk about numbers all day, bring him back soon pls!
"Now I've given you permission to put clickbait thumbnails on this video."
Numberphile goes self-aware.
Something I would like to see you do some day:
Take one of these ancient mathematical truths and do the arithmetic using the ancient system of numerals. Some possible ideas: Calculate pi using Roman numbers. (Roman numerals were used for over a thousand years. How does one add, subtract, multiply, divide, take square roots with Roman numbers? ) As you know, the Pythagorean theorem predates Pythagoras. How did those earlier mathematicians find the length of the hypotenuse using the numerals available to them?
15-20-25 triangle: why drop a verticle to "cheat"? It's a 3-4-5 scaled up by a factor of 5, no?
What I find so astounding, is that with the infinity of whole numbers, all the super hero triangles use such small numbers. Yes I realize that the fact that the perimeter will scale linearly while the area scales to the square so one could intuit that they all should be relatively small, but the fact that even the areas and perimeters are still only two digits (at least in base 10) I still find surprising.
Yeah!!!! Brady does read and takes heed of the comments. Nice to see Dr.James.
I love the Graham's Number paper framed in the background
Looks to me like he had it signed.
where? time frame?
13:43 is my new favorite thing Brady has ever done
I am so happy Grimes is still here :) I have to find his banans channel again.
I still don't know but this channel give me an insight of how amazing and elegant mathematics really is ...thank you numberphile... love from India...
I found an intriguing connection between nearly-equilateral Heronian triangles, the Pell numbers, and sequences of consecutive integers whose standard deviation is an integer, but a TH-cam comment is too small to contain the proof.
Funnily enough, James’s drawings of the ”P = 70; A = 210” -pair would slot together perfectly, like pieces of a jigsaw 😅.
It is quite nice to see James again! I rushed when I saw him and 6 hours ago
13:38 Grimes triangles, akin to Parker circles? When’s the tshirt coming out?
Qermaq's Shape. It is the Parker Square
At 5:52 the triangle is already a right triangle with pythagorean ratio sides (3:4:5); there is no need to split it up to reveal the "cheat."
Parker-Squares and Grime-Triangles...
which polygon will be next?
Following the progression, we need a 2D shape with two sides. Annulus?
@@Tombsar Interesting idea. Now there's only the need for an eponym the annulus will be named after.
A Zvezda Circle (a straight line with "Ptolemy>Pythagoras" graffitied over it)
Wow, I love these pretty easy to follow proofs.
James has more math enthusiasm than anyone is enthusiastic about anything. Great vid.
14:13 Thank you, Prof. Grime for making a connection between this interesting phenomenon and something I'd already been studying. The latter is triples of integers {x_i, y_i, z_i}, with the same sum and the same product. Following your notation at 10:28, suppose that for various indices i we have Hero triangles {a_i, b_i, c_i} with the same perimeter a_i+b_i+c_i=2s and the same area A. Then define x_i=s-a_i, y_i=s-b_i, z_i=s-c_i. Then x_i+y_i+z_i=s as desired. Moreover, x_i y_i z_i=P=A^2/s. So if you can find triples with the same sum s and the same product P, then you have triangles with the same perimeter 2s and the same area sqrt(sP). And if sP is a square, then your triangles are Heronian.
Here's a further nice feature. Define q=s/2, the quarter-perimeter. Then if we define d_i=a_i-q, e_i=b_i-q and f_i=c_i-q, then all of the (d_i, e_i, f_i) have the same sum q and the same cube-sum q^3-3P.
Now suppose that you managed to find two such Hero triangles, and in one triangle the shortest side c_2 was q, the quarter-perimeter. Then f_2=0. (I'm using _1 and _2 to index the two triangles.) Now define sets L={d_1, e_1, f_1, -d_2, -e_2} and R the same but with all five signs flipped. Then L and R each have sum 0 and cube-sum 0 (by the previous paragraph). More trivially, L and R agree in square-sum and fourth-power-sum, because a number's even powers remain the same when you flip its sign. So L and R are a solution to the Tarry-Escott problem for powers up to the 4th. For example, triangles (18, 17, 5), (19, 11, 10) with perimeter 40, and 10=40/4 as required. Then L={8, 7, -5, -9, -1} and R={-8, -7, 5, 9, 1}.
12:49 "If you want your area to be a multiple of the perimeter" A nice thing about all triangles (not just Heronian ones) is that A=rs where r is the inradius.
wtf is the inradius
@@leo17921 the radius of the inner circle. For right triangles you can use the formula b + c - a, let a be the hypotenuse and b, c the catheti
16:01 - oooh the banter..
I had an aha! moment with the rational height -- so simple yet so efficient! It means that every one of these is two Pythagorian triangles stuck together, but possibly with a rational scaling factor. How cool is that!
Reminds me of this (though not as amazing)
A square with all sides of 4, has area 16, and perimiter 16.
A circle of radius 2 has perimeter 2*tau and area 2*tau.
*sees James*
Never clicked on a video so fast
Yeah, he is the best Clickbait!
@FullTimeSlacker so? 😂
same!
That semi-perimeter formula was taught in my school in 8th grade in India.
Was waiting the whole video for mention of a Parker triangle and you did not disappoint!
Is it proven that those are the only superheros? I understand that as you scale things the area goes faster than the perimeter, but there could hypothetically be a monster triangle with massive side lengths and angles of about 0, about 0, and about 180 degrees. Or to put it another way, there are clearly infinite triangles with massive side lengths who's areas are smaller than their perimeters, and also infinite triangles with area larger than their perimeter, and with how many massive integers there are, it seems surprising none of those massive triangles fall on the line where area equals perimeter. If such a proof exists I'm sure it's neat.
i love this joke continuation about "well" drawn triangles..
I was thinking about triangles with equal perimeter and area but not necessarily integers, but I realised all triangles can be scaled to such a triangle:
Any triangle has a perimeter p and an area a. If we scale the triangle (lengths) by p/a, the new area is a(p^2/a^2)=p^2/a and the new perimeter is p(p/a)=p^2/a.
Oh this is phenomenonal, easy enough to understand for almost anyone and a great introduction to the concept of Mathematical proof for people who haven't studied it.
This one is for Kobe. He’s a hero. Days after he died a video about 24 comes out. The world just works like that
13:43 I was going to comment about the other mathematician, but it's James Grime, so...
Happy to see James Grime back, I just love the guy
As soon as I saw Dr.James, I click so so fast .
oh, i thought this was uploaded years ago. it was actually only uploaded 12 minutes ago.
E
I could listen to Dr James Grime talk Geometry all day. 😍
Love this guy. I don’t even like math but love this channel.
5:57 this is a right-angle triangle (3-4-5 scaled up), so it's not only a cheat by dividing it in two right-angle triangles.
HE’S BACK
Primitive Heronian triangles always decompose to two rational-sided right triangles, but if you scale them up you will get two Pythagorean right triangles, primal or not. I like how if you glue a 9-12-15 and a 5-12-13 together, say, to get a 13-14-15 Heronian triangle, then flip the triangle so 13 or 15 is now the base, you can decompose this into one of the original Pythagorean triangles and also a 36-53-65. I’d love to see Dr Grimes explore where the new triangle comes from.
I love how you troll the presenters so ruthlessly 😂
My favorite presenter is back. Welcome back, James!
In school if I said something like: "The length is 5", I got asked "What 5? Five eggs? Five apples?" we were always demanded to name the unit. Even if the actual unit was not needed, we were demanded to call it. Lets say "30 length units" or "30 length units to the square" and so on. And honestly this is so much inside me, I was cringing alot in this video. "It has the same perimeter as it has area" and such things simply feel very wrong from the beginning. - Of course I like the video, it was very cool as always! :-)
From the studio that brought you the Parker Square and the stunning cinematic sequel of the Parker Circle comes the exciting new new series of the Grimes Triangle.
With a special crossover spinoff: the Parker Triangle
The 5,29,30 circle does decompose into two Pythagorean triangles if you scale it up by a factor of five: a 7,24,25 and a 24, 143, 145.
The 15-20-25 triangle was a right triangle to begin with!
I'm Polish and the only moment I realised that your English 'Hero of Alexandria' is Polish 'Heron z Aleksandrii' was when you showed the equations that we covered in High school.
It seems as math's language is more universal than national languages, who would have guessed😉
15:24 - so now Parker geometry has taken on a life of its own.
I scrolled through most of the comments and didn't see anyone point this out: the 5,29,30 triangle can be obtained by taking a 5,5,6 triangle (Heronian) and a 6,25,29 triangle (also Heronian) and gluing them along the edge of length 6. So there should be some "prime" Heronian triangles that cannot be obtained this way...
He’s back!!! We’ve missed you, James - it’s just not the same without you. I haven’t even watched the video, yet, and I’ve already upvoted it. Please don’t take so long to make your next Numberphile video.
edit: James is the click bait 😊
Yay, Mr.James is back!!!
5:45
That's already a Pythagorean triangle. 5 * 3-4-5.
7:36 It's quite surprising to see that Dr. Grime wasn't taught Hero's Formula at school. We learnt it in class 7, i.e, when we were 12 years old.
I was schooled in the UK and I too was never taught Hero's formula. Maybe it's just lacking from the British curriculum.
Yeah, same, that surprised me too
There is a related, very similar looking, area-formula by Brahmagupta - maybe that is a reason, why this is part of the Indian curriculum.
arcanics1971
It’s not in the Norwegian curriculum either. It feels related to the sine area rule, though, which I was taught at 16.
@@ragnkja It was taught to us in Hungary. I was in special maths class though, but I'm pretty sure they at least mention it to everyone.
Non-Euclidian Superhero triangles? Semi-SuperHero cones? (Surface area is ½ volume?) Lovely to see and hear you, James!
After watching this video, I had a dream where The Moody Blues were on Sesame Street performing a song about perfect numbers with Cookie Monster. The song included some sort of nonsensical visual proof in high dimensions about their properties.
It did not sound much like a Moody Blues song, and Cookie Monster sounded like he had a cold, so he was a particularly unconvincing lead when they did "Nights in White Satin."
Superheronian (octave reduced) scale from the 5 unique Pythagorean triples whose areas=perimeters:
0: 1/1 0.000000 unison, perfect prime
1: 25/24 70.672427 classic chromatic semitone, minor chroma
2: 18/17 98.954592 Arabic lute index finger
3: 15/14 119.442808 major diatonic semitone
4: 13/12 138.572661 tridecimal 2/3-tone
5: 10/9 182.403712 minor whole tone
6: 29/25 256.949766
7: 20/17 281.358304 septendecimal augmented second
8: 6/5 315.641287 minor third
9: 29/24 327.622193
10: 5/4 386.313714 major third
11: 13/10 454.213948 tridecimal semi-diminished fourth
12: 4/3 498.044999 perfect fourth
13: 7/5 582.512193 septimal or Huygens' tritone, BP fourth
14: 10/7 617.487807 Euler's tritone
15: 3/2 701.955001 perfect fifth
16: 20/13 745.786052 tridecimal semi-augmented fifth
17: 8/5 813.686286 minor sixth
18: 48/29 872.377807
19: 5/3 884.358713 major sixth, BP sixth
20: 17/10 918.641696 septendecimal diminished seventh
21: 50/29 943.050234
22: 9/5 1017.596288 just minor seventh, BP seventh
23: 24/13 1061.427339 tridecimal neutral seventh
24: 28/15 1080.557192 grave major seventh
25: 17/9 1101.045408 septendecimal major seventh
26: 48/25 1129.327573 classic diminished octave
27: 2/1 1200.000000 octave
I see James, I press the like button!
I see people liking James, I like their comment!
I propose another category of triangles called Super Villain triangles. They are Super in the same way as the heroes, in that the perimeter is the same number as the area, but they are not heroes, meaning that their side lengths do not have to be whole numbers!
Let me introduce you to their leader: 12sqrt(3); He is an equilateral triangle with all sides = 4sqrt(3), as well as perimeter and area = 12sqrt(3).
James on thumbnail - instant click
Liking for Hero’s formula, holy heck that would’ve made geometry so much easier
There seem to be triangles with areas that are different ratios of P (2/3, 3/4, 6/7), but none smaller than P/2.
Holy crap that triangle area formula is amazing! I can't believe this isn't taught in every school
Heron's formula was taught in my high school in the mid 1980s. Maybe grade 10 ??
James Grimes utilising superheros to become the most liked Numberphile person again!
Finally , James Grime again(not that others aren't acceptable but James is unbeatable!)!! Clicked instantly
"Putting captain america on one side in the tumbnail"
*puts James Grime*
6:04 The 15-20-25 is already a right triangle, tho
I want merchandise of "these really valid, lovely drawn triangles" (along with quote underneath lol)
6:03 You don't even need to drop a perpendicular. You already have a right angle at the top. Your triangle is 3, 4, 5 scaled up by 5.
10:37 This is why I like James Grime. Real zeds, and real "less than or equal to" signs.
You need a T-shirt with James's deformed triangles on it.
I love that you used his 'terrible' triangle drawings!
5:57 It is even a right triangle itself ! How did you not see that it was an expension x5 of the normal 3,4,5 right triangle ?!?
"The way I was told to work out the area in school requires using Pythagoras occasionally. Let me instead show you this other method, which ALWAYS uses Pythagoras (or, actually, Ptolemy, who Pythagoras' theorem is a special case of) in order to avoid occasionally having to use Pythagoras."
If that isn't the biggest troll ever then I don't know what is. 😂
However, there are infinitely many Hero triangles with rational side lengths. Take any triangle, say with perimeter p and area a. Then scale it by the factor p/a, so that its perimeter and area are both p^2/a.
Now do the same with tetragons, pentagons, hexagons, higher dimensional shapes!
Lay the equal sides of two of those superheros, the ones with a common side length of 10, against each other and you get another right angled triangle, 8 15 17. Invert one of them but keep the equal sides together and you get a Grime quadrilateral.
The triangle at 5:53 is actually a right angled triangle itself. You don't need to drop a perpendicular