Consider the following setup: three bounded, 'nice' convex sets $A \subseteq B \subsetneq C \subset \mathbb{R}^2$, and three points $x,y,z\in \partial A\cap \partial B\cap \partial C$ (see edit below). Imagine a triangle inscribed in a circle, with some convex curves between the edges of the triangle and the circle, 'clamped down' at the vertices of the triangle.


Additionally, there is a volume-preserving *affine* map $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that the perimeter of $f(A)$ and $f(C)$ are less than the perimeter of $A$ and $C$, respectively.


My question is this: is it true that the perimeter of $f(B)$ is smaller than the perimeter of $B$?


This seems true to me intuitively, but I've been wrong about similar things before; the problem I'm having is that the length in an individual 'sector' *can* increase, but this should be offset by the decrease in the other sectors. Any help would be very appreciated!


**EDIT**: As Wlodzimierz points out below, the question is not true in this generality. However, I would like to add the additional assumption that the points $x,y,z$ are each *extreme points* of *all three sets* $A,B$ and $C$. This would rule out the class of counterexamples for the original question.



Perimeter of a 'trapped' convex set
Let me add the required modification (as a separate "*answer*" to keep each text clean). I could simply replace $\ f\ $ by $\ f(x\ y):= (a\!\cdot\!x\ \,\frac ya)$, where $\ 0<a<3\ $ and $\ a\ $ is very close to $\ 3$. But I prefer to provide a specific example.


Let me repeat that the answer to the **Question** in general is: **NO**.


Let's consider triangles $A\subset B\subset C$ defined as follows:


* $A\ :=\ \triangle((0\ 0)\ \,(2\ 0)\ \,(0\ 6))$
* $B\ :=\ \triangle((0\ 0)\ \,(4\ 0)\ \,(0\ 6))$
* $C\ :=\ \triangle((0\ 0)\ \,(4\ 0)\ \,(0\ 12))$


Define $f$ as follows:
$$f(x\ y)\ \, :=\ \, (2\!\cdot\! x\ \ \frac y2)$$


The vertices of the image triangles are:


* $f(A)\ =\ \triangle((0\ 0)\ \,(4\ 0)\ \,(0\ 3))$
* $f(B)\ =\ \triangle((0\ 0)\ \,(8\ 0)\ \,(0\ 3))$
* $f(C)\ =\ \triangle((0\ 0)\ \,(8\ 0)\ \,(0\ 6))$


Let $\ \alpha\ \beta\ \gamma\ \alpha'\ \beta'\ \gamma'\ $ be the perimeters of triangles $\ A\ B\ C\ f(A)\ f(B)\ f(C)\ $ respectively. Then


* $\alpha\ =\ 2+6+\sqrt{40}\ >\ 14 > 12\ =\ \alpha'$
* $\beta\ =\ 4+6+\sqrt{52}\ <\ 8+3+\sqrt{73}\ =\ \beta'$
* $\gamma\ =\ 2\cdot\alpha\ >\ 2\cdot\alpha'\ =\ \gamma'$


That's all.


