Denote the three different arc lengths as A,B,C. The lengths of the edges of the polygons correspond to either P=A+B or Q=B+C or R=C+A. For the same perimeter and same area, it is obviously sufficient if the red and blue polygons have the same number of edges whose length is P, and the same number of edges whose length is Q, and the same number of edges whose length is R. Consider the sequence of arcs around the circle, - a sequence of A's, B's and C's. Now this is the key step: Any subsequence of the form ABA can be reduced simply to A, since it contributes an edge of length A+B = P to (say) the blue polygon, and an edge of the same length B+A = P to the red polygon. This means that the whole sequence eventually must reduce to ABCABC.... cyclically around the circle. In this sequence, the subsequence ABC must appear an even number of times (since we started off with an even number of elements, and the reduction process reduced this number by 2 at each step). Now it is trivial to see that the final ABCABC... cycle meets the required condition. - Wherever we start counting pairs, the totals of P, Q, and R are always the same.