'Constrained morphing' of planar convex regions

Question

Morphing may be defined as a continuous transition of one shape to another. This post is about modifying planar regions continuously from one form to another under some constraints.

Qn: If $C_1$ and $C_2$ are planar convex shapes (not necessarily polygonal) with equal area, can one of them be morphed into the other such that all intermediate shapes are convex and area remains constant right through the transition?

Guess: The answer to the above seems "yes". But I don't know if the answer remains "yes" if more functions such as perimeter, diameter,... are equal for $C_1$ and $C_2$ and also need to remain constant during the entire morphing from $C_1$ to $C_2$.

Answer 1

A construction for Question 1:

Suppose $C_0$ and $C_1$ are planar convex shapes with equal area, which is without loss of generality $1$. For $t\in[0,1]$, let $$B_t:=(1-t)C_0+tC_1$$ and $$C_t:=\frac{B_t}{|B_t|^{1/2}},$$ where $|B_t|$ is the area of $B_t$.

Then $(C_t)_{t\in[0,1]}$ is a family of convex shapes of constant area continuously interpolating between $C_0$ and $C_1$.

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Details: Any planar convex set of area $1$ must be bounded. Therefore, the set $B_t$ is continuous in $t$ (say, with respect to the Hausdorff distance). Also, by the mixed volumes formula, $|B_t|$ is a (quadratic) polynomial in $t$ and hence continuous in $t$. Moreover, $|B_t|\ge\max(|(1-t)C_0|,\,|tC_1|)\ge1/4>0$ for all real $t$. So, the set $C_t$ is continuous in $t$ (with respect to the Hausdorff distance).

That $C_t$ is convex and is of area $1$ is easy to see.