Agreed. *_A_* triangle "with as many integers as possible." Three integer angles = 60° Three integer sides = 1 Total of six integers to define one triangle -- this is "as many integers as possible" I didn't think of the equilateral triangle case until it was mentioned at 03:25. An engineering read of the question seemed to be answered there. I assume other 60-degree cases have two non-integer angles -- with an infinite number of all-integer-side possibilities.
You could extend it to rational angles provided you subdivide a full rotation into a rational number. That's equivalent to saying the lengths are rational and the angles are rational multiples of pi radians.
Radian angles of a triangle that are integers could only be 1 or 2 and must add up to pi so the closest u could do would be 2 angles that are 1 and one angle that is pi-2
(No clickbait) this video shows different angle measures, and i wonder if it is even possible to do the "all integer triangle" with these ruclips.net/video/9yfWkUpYuq8/видео.htmlsi=iJjs87pWIHrMq1td
A triangle's angles sum up to pi radians, and pi is approximately 3.1415, more than 3 and less than 4. Because triangle's can't have negative angles or 0s as angles, the minimum integer angle for any corner is 1 radian, which is less than pi. The next step would be to add 1 radian to 1 corner but that's more than pi. A simpler explanation that I just thought of is that the sum of integers is an integer, and if all angles are integers their sum is an integer, and pi must be the sum of the angles and it is not an integer therefore there are no integer-radian-angle triangles
I hadn’t seen the earlier video and was puzzled at 1:49 when he said that cosine only assumes three rational values. Of course cosine is onto [0,1] and assumes all rational values in that interval. But he meant what is relevant here and what was covered in the earlier video: cos(x) when x is rational only assumes three rational values.
I was coming here to say this. Clearly, the value of cos takes on all rational values in [-1, 1], so he must have had some sort of restriction on theta.
For a triangle with a 60 degree angle, the parametrization can be done with just 1 variable. Using b as the variable, the parametrization is 2b-1, 3b²-2b, 3b²-3b+1. The 60 degree angle is between the short and long sides of the triangle, or between 2b-1 and 3b²-2b. Note that when b=1, it forms an equilateral triangle with side length 1. When b>1, scalene triangles are formed.
7:40 There's a much earlier work that shows this technique. See Pocklington, H. C., Some Diophantine Impossibilities. Proceedings of the Cambridge Philosophical Society, 1913, 17, 108-121 at 109 ("Our second lemma..."). The entire paper is worth a read, as it solves several interesting classes of quadratic equations. It also includes a proof that a triangle with integral sides and exactly one angle of 60 degrees cannot be equal in area to an equilateral triangle (p. 112).
Since cosine is continuous, ALL of the rational numbers between 0 and 1 are in its range. This does not imply that cos(rational) is in this set however. At 14:15, that should be -qr.
The only rational values of cosine are 0, -1/2, 1/2, -1, and 1 (1:55)? That's ridiculous. Cosine can assume any rational value from -1 to 1. For any positive rational number b/c (with b and c positive integers and b
The classic formula for finding Pythagoriean triples is: a=q^2 - r^2 b=2qr c=q^2 + r^2 This can be generalized for any angle C : a = q^2 - r^2, b = 2q(q*cos(C) + r) c = q^2 + r^2 + 2qr*cos(C) a' = 2b*cos(C) - a for a 60-degree angle, cos(C) = .5, so the formulas reduce to: a = q^2 - r^2 (obtuse triangle) b = q^2 +2qr c = q^2 + qr + r^2 a' = 2qr + r^2 (acute triangle)
Before watching I am thinking the old 30-60-90 with 3 integer angles and 2 integer sides, then Pythagorean and Eisenstein triangles with three integer sides and one integer angle. Edit: After watching, how did I miss the equilateral triangle? But it was great to see the derivation of the parameterization of the Eisenstein triples.
Excellent work! △'s in the plain are a great start. I suppose an extension is to spherical △'s but these are also triangles on a curved plain surface rather than 3d space. Maybe if a planar triangle in 3d is a pyramid is there such a thing as a pyramid form made from spherical triangles and has it any redeeming qualities? And as before ... a good place to stop is also an excellent place to start!
My parametrization : a = (3t^2 + 1)^2s ; b = (3t^2 + 6t - 1)(3t + 1)(t - 1)s ; c = (3t^2 - 6t - 1)(3t - 1)(t + 1)s where t,s are rationals. Deduced from the parametrization of a 3-orbit in the elliptic billiard and the circle... (one of my observations : a 3-orbit with a 60° degrees internal angle always exists).
@@MarsAnonymous I think at least one of these may be linked to spherical triangles. If you have access to one of those round spinning globes of planet earth then a triangle can be formed by drawing a straight line from north pole to equator. Then along equator for some distance then back up to north pole. When measured the angles at each point of triangle do not sum to 180 degrees and there may even be cases of θ = 180°
@@Alan-zf2tt No, that's still about planar triangles, just degenerate ones. They all look like a line, and have the longest side (let's call it c) be the length of the sum of the smallest; a + b = c. Then the angle between a and b is 180° and the other two are 0°.
@@MarsAnonymous This is true but when plane is curved spherical or toroidal angles of 180° and sum of internal angles of triangle greater than 180° is possible. Sure I realize that in video it is sume of internal angles of a planar triangle is 180° but that is a specific case Just to say the horizon on topic ranges a bit wider and give context to triangle in math (I hope(
Not sure if the math checks out but I understood from the question it's the number of sides + the number of angles, in my mind I looked at Pythagorean triples and expect that they (with all integer sides) could have 4, 3 sides + 1 angle, they cannot have more than 4, I then started thinking that you can have a 5, I eventually thought about it and 5 can only occur with 3 angles and 2 sides, think 90 + 2x 45 (apologies English was not first language so not getting the name) not sure of other cases that there are 5 but my mind tells me there can be but this could be the exact opposite, I dismissed 6 initially but as soon as 3x 60 degree triangles were brought up, I believe it's the only time you can have 6, all 3 angles and all 3 sides as integers, anyhow, that's my rambling over, good way of stretching my mind.
@@ajety-- Please write sentences like an adult should. This is not texting. "I remember seeing you on Michael Penn's comment's section three years ago. It's impressive you are still going."
If we want to maximize the number of integers in a triangle without resorting to the trivial equilateral triangle, can we do 3 integer angles, 2 integer sides, and one non-integer side?
You're aware that your measurements being integers are totally dependent on your units, right? Although the "integer" statement is valid for lengths (as long as you're in 2D), it's unit-dependent: what you call "60", which is definitely an integer, I see it as π/3 mod 2π, which is not even a real number.
I was rather hoping the angles would be in radians, because the notion of integer seems kind of meaningless when you can pick an arbitrary measuring unit like degrees. I was wondering how you were going to pull a rabbit out of a hat.
@@jackychanmaths no it's not. A radian is a unit of measuring an angle, just like a degree is. It doesn't typically have a symbol to represent it, but it's still a unit.
There's only two possible integer numbers of radians that could be angles of triangles, and neither of them have a rational trig value of any trig function.
It's easy, just take all three sides = 1 and all angles = 60°
All integers 😊
Nehh
Agreed. *_A_* triangle "with as many integers as possible."
Three integer angles = 60°
Three integer sides = 1
Total of six integers to define one triangle -- this is "as many integers as possible"
I didn't think of the equilateral triangle case until it was mentioned at 03:25.
An engineering read of the question seemed to be answered there.
I assume other 60-degree cases have two non-integer angles -- with an infinite number of all-integer-side possibilities.
Exactly what I said.
what about just taking all three sides = 1n and all angles = 60° ? where n equals any integer greater than 1 ?
τ/6 is transcendental, not an integer
Integer angles in degrees, but this question and answer is perhaps not limited to this specific way of measuring angles?
You could extend it to rational angles provided you subdivide a full rotation into a rational number. That's equivalent to saying the lengths are rational and the angles are rational multiples of pi radians.
Radian angles of a triangle that are integers could only be 1 or 2 and must add up to pi so the closest u could do would be 2 angles that are 1 and one angle that is pi-2
(No clickbait) this video shows different angle measures, and i wonder if it is even possible to do the "all integer triangle" with these
ruclips.net/video/9yfWkUpYuq8/видео.htmlsi=iJjs87pWIHrMq1td
Perhaps it is fairer to consider rational multiples of pi radians...
A triangle's angles sum up to pi radians, and pi is approximately 3.1415, more than 3 and less than 4. Because triangle's can't have negative angles or 0s as angles, the minimum integer angle for any corner is 1 radian, which is less than pi. The next step would be to add 1 radian to 1 corner but that's more than pi.
A simpler explanation that I just thought of is that the sum of integers is an integer, and if all angles are integers their sum is an integer, and pi must be the sum of the angles and it is not an integer therefore there are no integer-radian-angle triangles
In radians, the task is trivial if we apply the fundamental theorem of engineering.
pi = 3?
@ That's the one.
@@isavenewspapers8890 oh okay, thanks! Never heard of that one before.
@ No problem!
I've been an engineer for 14 years, and never once have I used pi=3 for anything that mattered.
I hadn’t seen the earlier video and was puzzled at 1:49 when he said that cosine only assumes three rational values. Of course cosine is onto [0,1] and assumes all rational values in that interval. But he meant what is relevant here and what was covered in the earlier video: cos(x) when x is rational only assumes three rational values.
I was a bit arraid too !
Thank you for the comment !
I was coming here to say this. Clearly, the value of cos takes on all rational values in [-1, 1], so he must have had some sort of restriction on theta.
And of course I should have clarified 5 values in [-1,1] … clarification needed a clarification!
cosine of a rational multiple of pi assume only those values
On that note: did anybody actually find that video? He has hundreds of "previous videos", kind of an imprecise citation.
The smallest non-trivial one is (5,7,8). (8^2+5^2-7^2)/(2*5*8)=1/2
For a triangle with a 60 degree angle, the parametrization can be done with just 1 variable. Using b as the variable, the parametrization is 2b-1, 3b²-2b, 3b²-3b+1. The 60 degree angle is between the short and long sides of the triangle, or between 2b-1 and 3b²-2b. Note that when b=1, it forms an equilateral triangle with side length 1. When b>1, scalene triangles are formed.
7:40 There's a much earlier work that shows this technique. See Pocklington, H. C., Some Diophantine Impossibilities. Proceedings of the Cambridge Philosophical Society, 1913, 17, 108-121 at 109 ("Our second lemma..."). The entire paper is worth a read, as it solves several interesting classes of quadratic equations. It also includes a proof that a triangle with integral sides and exactly one angle of 60 degrees cannot be equal in area to an equilateral triangle (p. 112).
0:40 chess battle advanced
Since cosine is continuous, ALL of the rational numbers between 0 and 1 are in its range. This does not imply that cos(rational) is in this set however.
At 14:15, that should be -qr.
The only rational values of cosine are 0, -1/2, 1/2, -1, and 1 (1:55)? That's ridiculous. Cosine can assume any rational value from -1 to 1. For any positive rational number b/c (with b and c positive integers and b
The classic formula for finding Pythagoriean triples is:
a=q^2 - r^2
b=2qr
c=q^2 + r^2
This can be generalized for any angle C :
a = q^2 - r^2,
b = 2q(q*cos(C) + r)
c = q^2 + r^2 + 2qr*cos(C)
a' = 2b*cos(C) - a
for a 60-degree angle, cos(C) = .5, so the formulas reduce to:
a = q^2 - r^2 (obtuse triangle)
b = q^2 +2qr
c = q^2 + qr + r^2
a' = 2qr + r^2 (acute triangle)
Before watching I am thinking the old 30-60-90 with 3 integer angles and 2 integer sides, then Pythagorean and Eisenstein triangles with three integer sides and one integer angle. Edit: After watching, how did I miss the equilateral triangle? But it was great to see the derivation of the parameterization of the Eisenstein triples.
Excellent work! △'s in the plain are a great start. I suppose an extension is to spherical △'s but these are also triangles on a curved plain surface rather than 3d space.
Maybe if a planar triangle in 3d is a pyramid is there such a thing as a pyramid form made from spherical triangles and has it any redeeming qualities?
And as before ... a good place to stop is also an excellent place to start!
My parametrization : a = (3t^2 + 1)^2s ; b = (3t^2 + 6t - 1)(3t + 1)(t - 1)s ; c = (3t^2 - 6t - 1)(3t - 1)(t + 1)s where t,s are rationals. Deduced from the parametrization of a 3-orbit in the elliptic billiard and the circle... (one of my observations : a 3-orbit with a 60° degrees internal angle always exists).
I want integer area, integer side lengths, integer angles, and integer vertex coordinates. I want ALL OF IT!
cos(theta) = ±1, degenerates triangle
Yeah, he left out the cases where θ = 0° or θ = 180°.
@@MarsAnonymous I think at least one of these may be linked to spherical triangles. If you have access to one of those round spinning globes of planet earth then a triangle can be formed by drawing a straight line from north pole to equator. Then along equator for some distance then back up to north pole.
When measured the angles at each point of triangle do not sum to 180 degrees and there may even be cases of θ = 180°
@@Alan-zf2tt No, that's still about planar triangles, just degenerate ones. They all look like a line, and have the longest side (let's call it c) be the length of the sum of the smallest; a + b = c. Then the angle between a and b is 180° and the other two are 0°.
@@MarsAnonymous This is true but when plane is curved spherical or toroidal angles of 180° and sum of internal angles of triangle greater than 180° is possible.
Sure I realize that in video it is sume of internal angles of a planar triangle is 180° but that is a specific case
Just to say the horizon on topic ranges a bit wider and give context to triangle in math (I hope(
Not sure if the math checks out but I understood from the question it's the number of sides + the number of angles, in my mind I looked at Pythagorean triples and expect that they (with all integer sides) could have 4, 3 sides + 1 angle, they cannot have more than 4, I then started thinking that you can have a 5, I eventually thought about it and 5 can only occur with 3 angles and 2 sides, think 90 + 2x 45 (apologies English was not first language so not getting the name) not sure of other cases that there are 5 but my mind tells me there can be but this could be the exact opposite, I dismissed 6 initially but as soon as 3x 60 degree triangles were brought up, I believe it's the only time you can have 6, all 3 angles and all 3 sides as integers, anyhow, that's my rambling over, good way of stretching my mind.
depends what you mean by integers, angles (degrees, radians, etc.) , and if only Euclidean geometry
What do you mean "the only integer cosine is 0,1/2,1? There's infinitely many angles such that cosine is rational
He means that these are the only rational values the cosine can take when the angle is also constrained to be rational (in degrees)
I wonder if (m^2 - n^2 , m^2 -mn +n^2, 2mn - n^2) generates all primitive solutions.
17:43
i remember seeing u on michael penns comments section 3 years ago. impressive ur still going
We missed you
@@ajety-- Please write sentences like an adult should. This is not texting.
"I remember seeing you on Michael Penn's comment's section three years ago. It's impressive you are still going."
@ 👌
@@robertveith6383Why did you feel the need to say this?
But what about 3 angles being integers, there is a 90,30,60 degrees case where 2 sides can ne integers and third is not
If we want to maximize the number of integers in a triangle without resorting to the trivial equilateral triangle, can we do 3 integer angles, 2 integer sides, and one non-integer side?
Yes. a 30/60/90 triangle with sides 1,2, and root 3.
Previous video link??
You're aware that your measurements being integers are totally dependent on your units, right? Although the "integer" statement is valid for lengths (as long as you're in 2D), it's unit-dependent: what you call "60", which is definitely an integer, I see it as π/3 mod 2π, which is not even a real number.
How come quotient PI cosine only gives quotient outcome for these three angles :-/
it must be an equil. triangle.
TIL integer+integer+integer=pi
I was rather hoping the angles would be in radians, because the notion of integer seems kind of meaningless when you can pick an arbitrary measuring unit like degrees. I was wondering how you were going to pull a rabbit out of a hat.
cos(37), cos(53) are both rational right?
no
60° = pi/3 so 60° is not an integer
An integer should not bear a unit
all measurements have a unit.
@@phiefer3 but radian is dimensionless
@@jackychanmaths no it's not. A radian is a unit of measuring an angle, just like a degree is. It doesn't typically have a symbol to represent it, but it's still a unit.
Just ... wow :)
The answer in degrees is an infinite amount of equilateral triangles so... Probably the question is in radians, and that's impossible, right?
There's only two possible integer numbers of radians that could be angles of triangles, and neither of them have a rational trig value of any trig function.
Great CHANNEL PROF
Not all angles can be integers because there are no integers that add up to PI ;)
What the hell? Cosine is a continuous function between negative 1 and 1. All the rational numbers are there!
Yes, but for nearly all rational values of cos theta, theta itself is irrational, no matter if you use degrees or radians.
He misspoke. He meant those are the only rational angles (in degrees) that have rational cosines.