Answer to the inscribed circle puzzle
First, we will prove that the inside circle is 1/4th the area of the outer circle. The rest becomes easy.
Draw a line from the center of the circle(s) (Point O, say) to the point where the inner circle touches the equilateral triangle’s bottom, horizontal side (point A, say).Draw another line from the center (Point O) to the right bottom vertex of the equilateral triangle (where it is touching the outer circle) (call it Point B).
Note that OA is the radius of the inner circle. OB is the radius of the outer circle. We have to prove OB = twice OA.
Look at the triangle OBA. Angle OAB is 90 degrees and Angle OBA is half the equilateral triangle’s angle = half of 60 deg = 30 deg.
Sin 30 deg = 1/2 = opp side / hypotenuse = OA/OB. Therefore OB = twice OA.
Since the radius is twice, the outer circle is 4 times the inner circle.
Therefore the area outside the inner circle but inside the outer circle is three times the full outer circle’s area. Note that the purple area is one third of that area (there are three such identical combination possible). So the purple ares is 1/3 of the area between the inner and outer circle = 1/3 of 3/4 of the outer circle = 1/4 of the outer circle.
Which is exactly what the inner circle is!