another way to show na+nb = ab has a contradiction is that we can observe that n needs to be a multiple of a and b using the similar argument shown in the video, so it must be a multiple of ab. Let n = mab where m is an integer, then we can rewrite the LHS into mab(a+b) = ab, since neither a nor b is 0, we can cancel out the ab to get m(a+b) = 1, but m is at least 1 and a+b clearly larger than 1, so na+nb = ab is impossible
Regarding the "implicit" assumption that x is an integer in the last proof. I think it comes to play when infering a|x. That is only true if x is also an integer. Otherwise, you have "c^2*x|a", which could be true if x=a/c^2.
The type 1 inscribed square case may be treated without considering the hypotenuse. Just x/(a-x)=(b-x)/x => x^2=ab-bx-ax+x^2 => ab=bx+ax => ab=(a+b)x and finally x=ab/(a+b).
This fascinates me. One consequence is that the length of the perimeter of either an inscribed circle or square is irrational. (EDIT: square perimeter is rational....) In the case of the circle, it is transcendental, but I haven't taken the time to determine if this is also the case for square. It makes me wonder if the perimeter of the square is somehow also related to pi as for the circle. Another thing for me to think about: does the restriction gcd(a, b) = 1 always hold for a "minimal" similar pythagorean triangle? I wasn't sure where that stipulation came from. Anyway, thank you for another thought-provoking video!
@@heliocentric1756 Thank you, I had forgotten that gcd(a, b) = 1 and incorrectly concluded that the length was irrational by assuming I could scale up the original triangle to remove denominators, but that I see now that is not true.
if a and b have a common factor. you have (a'*k)^2 + (b'*k)^2 = c^2 with a' and b' having gcd of 1. you can factor the k out.which shows us that the triangle is just an integer triangle scaled by k. now, since our x' (the side length of the square for the a', b' triangle) was not an integer, k*x' is not an integer either EDIT: unless k is exactly a multiple of the denominator... so it is possible
Take a Pythagorean triangle where an and b do not have a common factor (e.g., 3-4-5), and scale it by some multiple of a + b. Then you will get an integer inscribed square.
Why do side lengths of Pythagorean triangles need to be positive integers? Surely, you can scale the sides by any rational number (e.g., 3/2, 4/2, 5/2)?
A Pythagorean triangle is essentialy represented by the relationship of its sides (eg one side is 1 unit another is 6), as you suggested it could have two sides of length eg 3/2 and 11/5, this could however be rewritten as a similar triangle with side lengths 15 and 22. This would have the same ratio between side lengths just a scaled version showing that every rational scaled triangle has a similar triangle with integer side lengths.
I don't understand why we can safely assume gcd(a,b)=1. Couldn't we just scale the triangle to another similar Pythagorean triplet so the side of the square x is an integer?
If we select 21, 28, 35 as our right angle triangle, then the side of the first kind of inscribed square becomes 21 x 28 / (21 + 28) = 12. So, yes you can scale it like that. (I chose 21, 28 and 35 as it was a 3-4-5 triangle scaled by a factor of the sum of the shorter sides, 3+4)
Wait, what?!? The fact that a circle inscribed in a Pythagorean triangle necessarily has an integer radius... blows my mind! 🤯
I know, this is sooo cool.
@@philstubblefield by the same token, the side length of a square inscribed inside a Pythagorean triangle is always rational
4:24 good place to sneeze
4:24 bless you
another way to show na+nb = ab has a contradiction is that we can observe that n needs to be a multiple of a and b using the similar argument shown in the video, so it must be a multiple of ab.
Let n = mab where m is an integer, then we can rewrite the LHS into mab(a+b) = ab, since neither a nor b is 0, we can cancel out the ab to get m(a+b) = 1, but m is at least 1 and a+b clearly larger than 1, so na+nb = ab is impossible
nice!
Regarding the "implicit" assumption that x is an integer in the last proof. I think it comes to play when infering a|x. That is only true if x is also an integer. Otherwise, you have "c^2*x|a", which could be true if x=a/c^2.
The type 1 inscribed square case may be treated without considering the hypotenuse. Just x/(a-x)=(b-x)/x => x^2=ab-bx-ax+x^2 => ab=bx+ax => ab=(a+b)x and finally x=ab/(a+b).
Three very nice results!
This fascinates me. One consequence is that the length of the perimeter of either an inscribed circle or square is irrational. (EDIT: square perimeter is rational....) In the case of the circle, it is transcendental, but I haven't taken the time to determine if this is also the case for square. It makes me wonder if the perimeter of the square is somehow also related to pi as for the circle. Another thing for me to think about: does the restriction gcd(a, b) = 1 always hold for a "minimal" similar pythagorean triangle? I wasn't sure where that stipulation came from. Anyway, thank you for another thought-provoking video!
In the case of the inscribed square, x is not an integer but it's necessarily a rational number. So the perimeter of that square is rational.
@@heliocentric1756 Thank you, I had forgotten that gcd(a, b) = 1 and incorrectly concluded that the length was irrational by assuming I could scale up the original triangle to remove denominators, but that I see now that is not true.
Square no, pentagon no, hexagon no, heptagon no, ..., circle YES!
How much of this is still true if you relax the gcd constraint? Does it work for pythagorean triples that aren't reduced?
if a and b have a common factor. you have (a'*k)^2 + (b'*k)^2 = c^2 with a' and b' having gcd of 1. you can factor the k out.which shows us that the triangle is just an integer triangle scaled by k. now, since our x' (the side length of the square for the a', b' triangle) was not an integer, k*x' is not an integer either EDIT: unless k is exactly a multiple of the denominator... so it is possible
Take a Pythagorean triangle where an and b do not have a common factor (e.g., 3-4-5), and scale it by some multiple of a + b. Then you will get an integer inscribed square.
Why do side lengths of Pythagorean triangles need to be positive integers? Surely, you can scale the sides by any rational number (e.g., 3/2, 4/2, 5/2)?
A Pythagorean triangle is essentialy represented by the relationship of its sides (eg one side is 1 unit another is 6), as you suggested it could have two sides of length eg 3/2 and 11/5, this could however be rewritten as a similar triangle with side lengths 15 and 22. This would have the same ratio between side lengths just a scaled version showing that every rational scaled triangle has a similar triangle with integer side lengths.
Well, a Pythagorean triple is defined as three positive integers a, b and c such that a² + b² = c².
There's no getting away from that really.
That's just the definition of Pythagorean triangle
I don't understand why we can safely assume gcd(a,b)=1. Couldn't we just scale the triangle to another similar Pythagorean triplet so the side of the square x is an integer?
I think he meant b is not multiple of a or vice versa otherwise triangle wont be right angled
Yeah you're right. He just wanted to consider the smallest triangle from each equivalence class of similar Pythagorean triangles.
@@puneetbajaj786 No, that's not what gcd(a,b) = 1 means.
If we select 21, 28, 35 as our right angle triangle, then the side of the first kind of inscribed square becomes 21 x 28 / (21 + 28) = 12. So, yes you can scale it like that. (I chose 21, 28 and 35 as it was a 3-4-5 triangle scaled by a factor of the sum of the shorter sides, 3+4)
@@hoodedR Oh ok ty
No
I love this comment
Yes
Indeed there was no good place to stop...
@@nyancat2143 You say yes, I say no 🎶
In any pithagorean triangle abc are all even or two odd and the third even!
Could be all even because you can double any pythagorean triangle to get another
Or he could have just said "primitive Pythagoras triples"
Cool video
and that's a good place to stop
He forgot to say the thing!
I think one day AI will chew up all of his videos and teach itself math.
AI cannot teach itself or even learn like us
No