<p>^^^A diagram is really needed here, imo. </p>
<p>So, you inscribe your eq. triangle within the circle. What I did from here is bisect all the angles of the triangle, and these bisectors all intersect at what should be the center of the circle (just something I assumed, it seems to make sense, but I didn’t want to prove it). So, the distance from the vertex of the triangle to this intersection thingy is 2 (the radius length). You should also see a right triangle formed, with the hypotenuse equal to this segment, the height equal to a littler segment, the base equal to 1/2 of the side length (the entire side length I’ll call x), and the angle between the base and hypotenuse of 60/2=30 degrees. So, the base of said triangle is x/2. The hypotenuse is 2. The angle is 30. So, the cos(30)=(x/2)/2=x/4. Remembering that cos(30) is root(3)/2, which you can use the unit circle to confirm, we have root(3)/2=x/4, and therefor x=2root(3) using some algebra. Since x is a side length, and we are looking for the perimeter of an equalateral triangle where all side lengths are the same, we know the perimeter is 3x=3*2root(3)=6root(3).</p>
<p>Edit: howon your diagram is hard to see fyi, at least on my computer.</p>