A question I got asked I while ago: 


If $T$ is a triangle in $\mathbb R^2$, is there a function $f:T\to\mathbb R$ such that the integral of $f$ over each straight segment connecting two points in the boundary of $T$ not on the same side is always $1$?


(Of course, you can change $T$ for your favorite convex set... and the problem should really be seen as asking for what sets is the answer affirmative, mostly)



Egalitarian measures
In general, no. For the double integral $\iint\_T f(x,y)\,dx\,dy$ will be the height on any side, as is seen by turning the triangle with one side parallel to an axis and performing the integral. So at least, $T$ has to be equilateral. I don't know the answer in that case.


**Edit:** Wait, wait – the same trick works even if I turn the triangle at any angle, hence all heights (defined as $\sup\_{RT} y-\inf\_{RT} y$ where $R$ is a rotation) must be the same. That is never true for a triangle, and limits the number of convex sets seriously – but there are still non-circles that might satisfy the criterion. Once more, I don't know the answer, but for circles at least, it should in principle be straightforward to check if a radially symmetric function will do. And of course, if there is a solution, there is a radially symmetric one, as can be seen by rotating the solution and taking the average of all its rotated variants.


**Edit2:** For the unit disk (and radially symmetric $f$) the answer should in principle be obtainable by the Abel transform (which is really nothing but the Radon transform on radially symmetric functions). The required Abel transform $\Phi$ should be the characteristic function of the interval $[0,1]$ (we only use positive $x$ due to symmetry), and the inverse Abel transform provides the answer:
$$f(r)=\frac{-1}{\pi}\int\_r^\infty \frac{\Phi'(x)\,dx}{\sqrt{x^2-r^2}}=\frac{1}{\pi\sqrt{1-r^2}}$$
when $0\le r<1$. Being lazy, I checked the answer using Maple, and it seems right.


**Addendum:** Anton Petrunin pointed out in a comment that the above measure is the push-forward of the surface measure on the unit sphere on the unit disk under projection. It is well-known that the surface area of the portion of a disk between two parallel planes depends only on the distance between the planes (and is proportional to said distance), which ties in nicely with the desired property of $f$ on the disk.


