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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.

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  • $\begingroup$ what role do the points $x,y,z$ play? $\endgroup$ Apr 14, 2014 at 0:36
  • $\begingroup$ Thanks for pointing that out! The three common points are actually in the intersection of the boundaries of the convex sets. I edited the question. $\endgroup$
    – quick_q
    Apr 14, 2014 at 0:50
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    $\begingroup$ Is $\ \subsetneq\ $ the same as $\ \subset\ $? $\endgroup$ Apr 14, 2014 at 1:32
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    $\begingroup$ My of my examples features sharp inequalities. A small modification will keep the inequalities intact while all boundary points of all three convex sets can be extreme. Thus the general answer to the question is still NO. $\endgroup$ Apr 14, 2014 at 15:54
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    $\begingroup$ Correction: One of my ... (instead of My of my ..., and certainly not My oh my). $\endgroup$ Apr 14, 2014 at 16:02

2 Answers 2

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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.

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  • $\begingroup$ Good example - thanks! I realize now that in the situation I'm in, the three common points are all extreme points of each set. This would force their boundaries to 'move in the same direction'. Is there still no hope with this additional assumption? I will edit the question to add this in. $\endgroup$
    – quick_q
    Apr 14, 2014 at 15:20
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    $\begingroup$ As I've explained in a comment under the Question, the above additional assumption does not make any essential difference. A small modification can have ALL boundary points extreme, while the sharp inequality will stay intact. $\endgroup$ Apr 15, 2014 at 19:55
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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 $\ B'$.

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  • $\begingroup$ The exact answer is given below. However the present version is nicer, even if it's not exactly in the required format. $\endgroup$ Apr 14, 2014 at 8:06
  • $\begingroup$ The exact answer is given below or above--it's not possible to know around MathOverflow. $\endgroup$ Apr 14, 2014 at 10:02

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