Most valuable rectangle contained in an equilateral triangle A measure $V$ (for "value") is defined on an equilateral triangle. $V$ is absolutely-continuous with respect to the Lebesgue measure, and the value of the entire triangle is $1$.
What is the largest $r$ such that, for every $V$, the triangle contains a rectangle with a value $\geq r$?
One special case is that $V$ is the Lebesgue measure, so the value of a rectangle is its area. The largest area of a rectangle is $1/2$, so $r\leq 1/2$:

However, there is a worse case: if the measure $V$ is spread evenly in a thin strip along the perimeter of the triangle, then any rectangle can contain a value of at most $1/3$, so $r\leq 1/3$:

Is there a measure $V$ for which $r$ is smaller?
P.S. the question is related to both geometry and measure theory. Is there a field that deals with such questions?
 A: Place $1/4$ of the mass at the centre of the triangle.  Place $1/8$ of the mass along each of the six green segments near the corners.   
Any rectangle that contains the centre contains no more than two points of green, total value $1/4+0+0=1/4$.
Any rectangle that contains more than one point of any side must be parallel to that side and contain no more than one point of any other side.
$1/4<1/3$, so this is an improvement on the OP.  

.
$r=0$.
Draw a circular arc in the triangle, joining one corner to another.
Pick any three points on the arc, give them $V$ weight $1/n$ each.  They are not co-linear, so there is a minimum non-zero width for any rectangle containing all three.  So there is a finite gap near the corners free of all these rectangles.   Pick a fourth point on the red arc in this gap, give it $V$ weight $1/n$.
Now consider any rectangle that contains at least three of the four points.  There is a finite number of triples, so there is still a minimum non-zero width for all of these rectangles; and a finite gap near the corners.  Add a fifth point in this gap, and repeat until you have $n$ points.  No rectangle has total weight more than $3/n$.

