Charles Link's latest activity

Here is an aside item, and in a way a little simpler. The problem of finding integer solutions for ## x^2=(a^2+ab+b^2)/3 ## reminds me...

and a follow-on to the above: The case of ## m=1 ## gives the point ## (1,1) ## for ## (x,y) ##. It may be worth mentioning that I...

As others have noted, if the four sides of that quadrilateral have lengths a, x, b, and the diameter 2x, then 3x^2 = a^2+ab+b^2. As to...

I think I figured out now what @daverusin did in post 41. One begins with ## x^2+xy+y^2-3=0 ##, and to find the rational solutions...

I found the following video that might explain some of post 41. It seems to be a specialized topic, and not one that is found in the...

Could you elaborate on this a little? (post 41). My mathematics with calculus, etc. is reasonably advanced, but this looks like it is...

One more approach based on https://en.wikipedia.org/wiki/Cyclic_quadrilateral and the cosine theorem. ## \angle...

Good catch. To eliminate solution triples that are integer multiples of other triples I must verify that either ##b## or ##x## (or both)...

@kuruman Very good. :) Excellent derivation. Between you and @renormalize of post 32, we have a simpler expression for ## R=x ## than...

Here is the derivation mentioned in post #30. Refer to the figure on the right. It shows half the hexagon as a quadrilateral inscribed...

@kuruman You essentially have the same solution for ## x=R ## in post 30 that @renormalize has, but you should recognize that the...

@renormalize Thank you very much. :) You did verify the 3 combinations that I had previously found, plus the one in the Olympiad. You...

@Charles Link, given that ##a\geq1,b\geq1##, you can solve your formula ##(a+2b)^2=3(4x^2-a^2)## to get...

Very good. That reduces to ## R^2=(a^2+ab+b^2)/3 ##, and is in agreement with my post 24, but far simpler. Thank you @kuruman :)...

Using the law of cosines and reasonably simple algebra, I got the following expression for the radius-squared in terms of the given...