Perimeter of a 'trapped' convex set 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.
 A: 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.
A: The answer in general is: NO.
Let's consider triangles  $A\subset B\subset C$ defined as follows:


*

*$A := \triangle((0\ 0)\ \,(1\ 0)\ \,(0\ 3))$

*$B := \triangle((0\ 0)\ \,(2\ 0)\ \,(0\ 3))$

*$C := \triangle((0\ 0)\ \,(2\ 0)\ \,(0\ 6))$


Define $f$ as follows:
$$f(x\ y)\ \, :=\ \, (3\!\cdot\! x\ \ \frac y3)$$
Then the perimeter of $\ f(A)\ $ is the same as of $\ A$, and of $\ f(C)\ $ as of $\ C$, while the perimeter of $\ f(B)\ $ is strictly larger than that of $\ B$.
REMARK: The sharp and not sharp inequalities are not exactly as required in the question but it is easy to fix it (trivial, I may do it later).
On the other hand, this example has the following symmetric feature:
Consider $\ g := f^{-1}\ $ -- the inverse function, $\ g:f(A)\rightarrow A$; also let
$$A' :=f(A)\qquad B':=f(B)\qquad C':=f(C)$$
Function $\ g\ $ is still volume preserving. Also:
the perimeter of $\ g(A')\ $ is the same as of $\ A'$, and of $\ g(C')\ $ as of $\ C'$, while the perimeter of $\ g(B')\ $ is strictly smaller than that of $\ B'$.
