For $n$-gons $A$ and $B$, is it possible for $A$ to have shorter (or equal) corresponding edges but longer corresponding diagonals than $B$?

Question

Let $A$ and $B$ be two $n$-gons with sides $i$, $i=1,...,n$, and diagonals $ij$, denoting the diagonal between $i$ and $j$. Is it possible that the length of each side in $A$ is smaller than or equal to the length of corresponding side in $B$, and the length of each diagonal in $A$ is bigger than the length of corresponding diagonal in $B$?

I think using the law of cosines for sides and diagonals may work. But, is there easier solution?

Answer 1

If both polygons are convex that is not possible. If $$ A_iA_{i+1}\le B_iB_{i+1},\quad A_{i+1}A_{i+2}\le B_{i+1}B_{i+2},\quad A_{i}A_{i+2}> B_{i}B_{i+2}, $$ if follows from elementary geometry that $\angle A_iA_{i+1}A_{i+2}>\angle B_iB_{i+1}B_{i+2}$, but this cannot hold for all $i$, as the sum of all angles is $(n-2)\pi$ for both polygons.

Answer 2

If we allow non-convex, non-simple polygons, then you can compare a regular pentagon with a regular five-pointed star. If the star is inscribed in the pentagon then each figure's sides are the diagonals of the other figure.