Continous family of $n$-gons

Question

The statement of this problem asks to show that if $A$ and $B$ are two distinct convex $n$-gons there is a continous family of convex $n$-gons such that the first in that family is $A$ and the last is $B$. My first intuition is that convexity ought to be utilized at some point in this proof. Perhaps consider establishing a correspondence between the vertices and drawing the line segment in between. Then continuously translate the $n$-gon along the lines. Shape will clearly not be preserved but I think there is a way to do this preserving convexity. It is just not clear to me under which conditions this translation should be performed. Any thoughts?

Answer 1

For 3-gons one can consider the circumcircle, and move points along the circle until it is equilateral, without going through nonconvex 3-gons (since there are none). Since both 3-gons can be thus made into equilaterals, a composition of the two deformations takes the first into the second.

Suppose true for $n$ gons, and consider two $n+1$ gons, and take three adjacent vertices $a,b,c$ from one of them and $a',b',c'$ from the other. For each of these, one can move $b$ (and $b'$) until it coincides with the midpoint of $ac$ or $a'c',$ without destroying convexity. (E.g. move it on the line connecting $b$ to the midpoint of $ac.$)

Now you have essentially two $n$ gons and can finish by the inductive hypothesis.