Follow-up: British Flag Theorem

UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle. Take two pythagorean triples: (u, v, w) and (x, y, z); Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz). It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5). I don't know if there are other ways to do it. Richard Holmes wonders whether this is the only way to construct the rectangles, but found a counterexample: (25, 60, 65), (25, 312, 313), (91, 312, 325), (60, 91, 109) These are four genuinely different pythag triples. I don't know how to make other examples like this. Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring. -- Here is the original video https://youtu.be/scZMQEHEt1w And here is the description from that video, copied over to this one: 1. Is there a 3d version of this? There is. First of all, the point can be above/below a rectangle, and if we connect the four corners to the point (now in 3d space), it is still true. But also, I've just checked for a cuboid and it's still true. If AB is a space diagonal (for example, from the bottom left corner of the cuboid to the top right corner of the opposite face), and CD the other space diagonal (for example, bottom right to top left of opposite face), then a^2 + b^2 = c^2 + d^2. You can prove that by splitting the height, width, and depth into u, v, w, x, y, z and doing 3d pythagoras on that. The two sides are equal to u^2 + v^2 + w^2 + x^2 + y^2 + z^2. 2. Can this be done for a parallelograms? There is a version for parallelograms, although a^2 + b^2 does not equal c^2 + d^2. Instead, the two sides differ by a value that is independent of the choice of point. (I will leave that as a challenge for you, I might do the answer in a follow-up video). 2a. ironpencil observes that if we place two copies of the rectangle, side-by-side, then the diagonals a, b, c, d form a quadrilateral with orthogonal diagonals. In that case a and b are opposite sides, c and d are opposite sides and a^2 + b^2 = c^2 + d^2. The diagonals would have length (w+x) and (y+z), and the area of the quadrilateral will be (w+x)(y+z)/2. 3. Can we make w, x, y, z and diagonals a, b, c, d all integers? You can! If (w, z, a) are all integers, it is called a pythagorean triple. We need to find four pythagorean triples, (w, z, a), (x, y, b), (w, y, c) and (x, z, d) so they can fit together to make a rectangle. That's how I made my example, with pythagorean triples (280, 210, 350), (72, 96, 120), (280, 96, 296) and (72, 210, 222), making a 352 by 306 rectangle. UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle. Take two pythagorean triples: (u, v, w) and (x, y, z); Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz). It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5). I don't know if there are other ways to do it. Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring. 4. What other flag theorems could we have? Here is the Dutch National Flag Problem: https://en.wikipedia.org/wiki/Dutch_national_flag_problem And someone suggested pythagoras is the Trinidad and Tobago flag theorem https://en.wikipedia.org/wiki/Flag_of_Trinidad_and_Tobago 5. How can pythagoras be a special case of the British Flag Theorem, when you use pythag to prove the British Flag Theorem? We could prove the British Flag Theorem the same way we prove pythag, without using pythag itself. There are a few hundred ways to do that, take your pick. In other words, Pythag and BFT are equivalent theorems. 6. Is there some sort of British celebration going on? There is. The Queen has been queen for 70 years. For the record, I'm not bothered about that, but there will be lots of flags about and I'm using that as an excuse to talk about maths.