Where Do Points Sum to Two-Thirds the Perimeter of a Rectangle?

[Numeriprimi]

I found one interesting example...

We have a rectangle ABCD with his perimeter o. Where is the set of points when their (for each point) sum of distance lines AB, BC, CD, DA is 2/3o ?

I tried it, but geometric problems is quite hard for me and I don't know how do it. So, have you got any idea?

 

[tiny-tim]

Where is the set of points when their (for each point) sum of distance lines AB, BC, CD, DA is 2/3o ?

Sorry, Numeriprimi, I don't understand.

 

[Vargo]

I think the problem is to identify the set of points X such that
$d(X,AB)+d(X,BC)+d(X,CD)+d(X,DA)= \frac{2}{3}*\text{Perimeter of Rectangle ABCD}.$
The distances on the left hand side are lengths of perpendiculars dropped from the point X to the corresponding line. For example, if X lies inside the interior, then the left hand side would equal (1/2) the perimeter. So the locus cannot include points inside the rectangle. Without doing an explicit calculation, I bet the answer is an octogon.

 

[Numeriprimi]

Sorry for my English... And Vargo, yes, I mean this, thanks for wrote it in better form :-)

 

[tiny-tim]

Thanks, Vargo!

Numeriprimi, divide the exterior of the rectangle into 8 sections (by extending the sides),

and consider each section separately …

you should be able to find an easy equation for each section. ​

(and yes, it's an octagon!)