Can two non-congruent quadrilaterals have the same four sides and same four angles?
Give an example or prove the task is never possible.
As usual, fundamental questions that are easy to ask yet hard to answer are especially intriguing. Having spent a day reviewing triangle congruence (SSS,SAS) with my mathematically-minded
11 year old, I am searching for an avenue of attack here.
My intuition says that any pair with all the same angles and sides must be congruent.
My thought is to prove by contradiction. First, reduce the problem: Assume (I think it can be shown) that to work, the sides must be out of order, Start in each case with a common side A. Then if the sides in the original shape that attach to A are ordered BCD, then they must be ordered BDC (without loss of generality) in the other shape. Call the endpoints of A: a1 and a2. BCD and BCD using the same 4 angles must then connect a1 to a2 in each case. Is this possible?
I don't think so. Since some of some of B, C, D are different lengths, I think it is impossible to land exactly back at a2 with any permutation of angles.
Another approach is to require the diagonals match and cross at the same point. One diagonal will be produced by a congruent triangle SAS on one side of the two quadrilaterals, but on the other side, not. This depends on the parity of the scrambled sides interlaced with the parity of the scrambled angles. A bit of a mess to enumerate.....
I was wondering if there were a hint. The problem has got to be solved and published somewhere....
Edited on July 26, 2022, 4:34 pm
looking
