Please let me know if you prefer harder or easier problems. I cannot satisfy everyone with every puzzle but I can at least try to alternate difficulties.
I didn't read it that way, but I was wondering if the fact that the triangle is right-isosceles imposes some kind of additional condition that lets us find \theta.
Don't really have the time to work it further right now though. Geometry isn't really my forte.
\thetaappears to be 3*45°, but I don't know why.k=IItwiceI, the arc length?\theta.\theta, the inscribed angle theorem should help.chave equal lengths.\frac{45°}{2}.\theta = 3*45°.\frac{r}{k} = \sqrt{2-\sqrt{2}} \approx{0.76537}\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2}+1}which also\approx 0.76537if I'm not mistaken.