shmm wrote: Let $ABC$ triangle is inscribed into circle with radius $1$. If $ABC$ triangle's sides are integer numbers , then find that triangle's sides. We see the circle has diameter 2, so all the sides must be $\le 2$. Clearly they cannot all be 1, so we assume one side has length 2. Then we must have a $1-2-2$ triangle because $1-1-2$ is degenerate. However, a 1-2-2 triangle doesn't work...

Case 1: The circumcentre is inside the triangle. Use cosine rule, let angle AOB=x. Then c^2=2-2cosx, so c is bounded by (0,2) so c=1. Equally, we get a=1 and b=1. Case 2: One of the sides is the diameter Clearly impossible. Case 3: Circumcentre is outside the triangle More impossible: a=b=c=1. (Which is case 1). Hence (1,1,1) equilateral triangle.

As stated, that is a ridiculous question. Maybe again something has been lost in the translation ...