Let $S$ be a simply connected Riemannan surface . Suppose $\Delta ABC$ is a triangle on $S$. The Area of a triangle is denoted by $\mathcal{A}$. A point $P$ in the interior of $\Delta ABC$ is called "A point of half area" if $\mathcal{A}(\Delta PBC)=\frac{1}{2} \mathcal{A}(\Delta ABC)$. (In planar geometry, $S=\mathbb{R}^{2}$, the points of half area of a triangle is a straight line)


Assume that for every (small) triangle, the points of half area is a geodesic.


What can be said about the geometry of $S$?(Its curvature)


**Note** By small triangle I mean "each point of $S$ has a neighborhood $U$ such that all triangles in $U$ satify the above property.



The points of half area of a triangle
I believe that the answer is that the curvature of $S$ has to vanish, i.e., the surface is locally isometric to the plane.


I haven't checked all of the details, which are somewhat messy in my analysis, but here is the basic argument: If the Gauss curvature vanishes identically, then, obviously, the local property holds, so assume that the Gauss curvature is not identically zero, and, since everything is local, restrict to an arbitrarily small, strictly geodesically convex neighborhood $S$ of a point $p$ at which the Gauss curvature is not zero. In fact, one can assume that the Gauss curvature is either strictly positive or strictly negative throughout $S$.


First, the desired property does not hold when $S$ has constant nonzero Gauss curvature (either positive or negative), as one can easily check by working out what these curves are on the $2$-sphere or in the Poincaré metric.


Second, note that the desired property has a simpler formulation as follows: For any two distinct points $a,b\in S$, consider the condition on a point $c$ that the (oriented) area of the triangle $\Delta(a,b,c)$ be equal to some constant $t$. The desired property has the consequence that the points $c$ that satisfy the equation $\mathcal{A}\bigl(\Delta(a,b,c)\bigr) = t$ all lie on a curve $C\_t(a,b)$ that is, in fact, a geodesic. 


However, doing a (rather laborious) expansion about $t=0$ (since the curve $C\_0(a,b)$ is always the geodesic passing through $a$ and $b$), one finds that the geodesic curvature of $C\_t(a,b)$ has a nonzero coefficient in $t^2$ that is a (nonzero) universal constant times the Gauss curvature along $C\_0(a,b)$. In particular, this coefficient is nonzero (since the Gauss curvature does not change sign in $S$), so, for small, nonzero $t$, the curve $C\_t(a,b)$ is *not* a geodesic. 


Thus, surfaces with nonzero Gauss curvature do not have the desired property.


