It is easy to see that an equilateral triangle can be cut into 2 identical 30-60-90 degrees right triangles which can then be patched together to form a 30-30-120 degrees triangle. So, via 2 intermediate pieces, we can dissect an equilateral triangle into the 30-30-120 triangle of the same area.
Question: Find two equal area, non-congruent triangles T1 and T2 such that T1 can be cut into 3 (and not less than 3) pieces which can be reassembled into T2. Not sure if such a T1-T2 pair exists.
One can readily generalize and ask for equal area {T1-T2} pairs which can be dissected into each other via n (and not less than n) intermediate pieces (n cannot be arbitrarily large, from the Wallace-Bolyai-Gerwein theorem).
Further question: What can one say about equal area pairs of triangles which are the worst for mutual dissection - ie for which the number of intermediate pieces is the highest possible? Guess: a triangle pair with equal area and equal perimeter is bad for mutual dissection.
Further Note - Going to Pairs of Quadrilaterals: the problem is trivial for rectangles - indeed, for any rectangle R and any n, there is a different rectangle with same area as R and n times as long which can be got by cutting R into n equal, length-parallel strips and attaching them end to end. For general quadrilaterals, things seem less trivial. For any cyclic quad Q (each pair of opposite vertices of Q add to 180 degrees), it appears that we can cut Q into 2 pieces and patch these to get another quad Q' different from Q. But for non-cyclic Q's, one is not sure of finding such a Q' even for n =2. For general n, things appear harder.
