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$begingroup$

I'm reading through this paper and in the part Proof of Theorem 2 (page 4, at the bottom) I stumbled upon this problem.

Basically, we have a unit square partitioned into several districts $D_i$ with the area denoted by $|D_i|$ and perimeter by $|partial D_i|$. We take some constant $psi$ and make a "$psi$-thickened $partial D_i$", which I guess means constructing some kind of a belt around the perimeter, which has a constant width of $psi$ (but not constant distance from $delta D_i$). They then casually claim that this belt has an area bounded from above by
$$psileft|partial D_iright| + pi psi^2.$$

That makes intuitive sense when I imagine $D_i$ being a polygon, and I can even prove it for a $n$-sided polygon. The first part of the formula corresponds to rectangles with exactly $psi$ height drawn above every side of the polygon and the second part is an area of the circle which is a sum of all the circle-parts above the vertices.

The problem is, it can be any general curve that doesn't cross over itself. I'm not sure how to generalize this idea to curves.

geometry

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asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A more current terminology is the dilated set. The formula you give works only for convex shapes with a regular boundary.
  $endgroup$
  – Jean Marie
  Mar 15 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great to hear that, you can't imagine how hard it is to google "constant-thickened perimeter". I'll look it up; however, they don't state or even mention any of those conditions.
  $endgroup$
  – Sh4rP EYE
  Mar 15 at 18:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Rectification : I had read too quickly : for an inequality, you do not need a convex set. It is for the equality : area of the dilated set by $psi $ = initial area + $psi times $ perimeter +$pi psi^2$ that you need a convex set.
  $endgroup$
  – Jean Marie
  Mar 15 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Other keywords for extension to 3D for example (that you hopefuly do not need) : Steiner's formula, Minkowski functionals
  $endgroup$
  – Jean Marie
  Mar 15 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I can't seem to find anything about dilated sets beyond 7th grade material, though. Could you please provide some further pointers, especially for the 2D version? Please excuse my inability to google properly.
  $endgroup$
  – Sh4rP EYE
  Mar 16 at 9:26
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

I'm reading through this paper and in the part Proof of Theorem 2 (page 4, at the bottom) I stumbled upon this problem.

Basically, we have a unit square partitioned into several districts $D_i$ with the area denoted by $|D_i|$ and perimeter by $|partial D_i|$. We take some constant $psi$ and make a "$psi$-thickened $partial D_i$", which I guess means constructing some kind of a belt around the perimeter, which has a constant width of $psi$ (but not constant distance from $delta D_i$). They then casually claim that this belt has an area bounded from above by
$$psileft|partial D_iright| + pi psi^2.$$

That makes intuitive sense when I imagine $D_i$ being a polygon, and I can even prove it for a $n$-sided polygon. The first part of the formula corresponds to rectangles with exactly $psi$ height drawn above every side of the polygon and the second part is an area of the circle which is a sum of all the circle-parts above the vertices.

The problem is, it can be any general curve that doesn't cross over itself. I'm not sure how to generalize this idea to curves.

geometry

share|cite|improve this question

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A more current terminology is the dilated set. The formula you give works only for convex shapes with a regular boundary.
  $endgroup$
  – Jean Marie
  Mar 15 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great to hear that, you can't imagine how hard it is to google "constant-thickened perimeter". I'll look it up; however, they don't state or even mention any of those conditions.
  $endgroup$
  – Sh4rP EYE
  Mar 15 at 18:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Rectification : I had read too quickly : for an inequality, you do not need a convex set. It is for the equality : area of the dilated set by $psi $ = initial area + $psi times $ perimeter +$pi psi^2$ that you need a convex set.
  $endgroup$
  – Jean Marie
  Mar 15 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Other keywords for extension to 3D for example (that you hopefuly do not need) : Steiner's formula, Minkowski functionals
  $endgroup$
  – Jean Marie
  Mar 15 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I can't seem to find anything about dilated sets beyond 7th grade material, though. Could you please provide some further pointers, especially for the 2D version? Please excuse my inability to google properly.
  $endgroup$
  – Sh4rP EYE
  Mar 16 at 9:26
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I'm reading through this paper and in the part Proof of Theorem 2 (page 4, at the bottom) I stumbled upon this problem.

Basically, we have a unit square partitioned into several districts $D_i$ with the area denoted by $|D_i|$ and perimeter by $|partial D_i|$. We take some constant $psi$ and make a "$psi$-thickened $partial D_i$", which I guess means constructing some kind of a belt around the perimeter, which has a constant width of $psi$ (but not constant distance from $delta D_i$). They then casually claim that this belt has an area bounded from above by
$$psileft|partial D_iright| + pi psi^2.$$

That makes intuitive sense when I imagine $D_i$ being a polygon, and I can even prove it for a $n$-sided polygon. The first part of the formula corresponds to rectangles with exactly $psi$ height drawn above every side of the polygon and the second part is an area of the circle which is a sum of all the circle-parts above the vertices.

The problem is, it can be any general curve that doesn't cross over itself. I'm not sure how to generalize this idea to curves.

geometry

share|cite|improve this question

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

$endgroup$

share|cite|improve this question

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

share|cite|improve this question

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945

asked Mar 15 at 18:00

Sh4rP EYESh4rP EYE

945