TITLE: Balanced cutting of a convex polygon
QUESTION [7 upvotes]: Given a convex polygon $C$ and a number $R\geq 1$, say that a point $x$ is an $R$-balance-point of $C$ if every line through $x$ divides $C$ to two parts $C_1,C_2$ such that:
$$1/R \leq Area(C_1)/Area(C_2)\leq R$$
Some polygons have a 1-balance-point, e.g. the centroid of a rectangle or an ellipse, since every line through it cuts $C$ to two parts of equal area.
Initially I thought that every convex polygon has a 1-balance-point, but then I found a counter-example. Consider the unit right-angled isosceles triangle, whose total area is 0.5:

Suppose by contradiction that it has a 1-balance-point, H. Then, the verical line through H must cut a triangle of area 0.25, so it must have $x = 1-\sqrt{0.5} \approx 0.29$. Similarly, the horizontal line through H must have $y = 1-\sqrt{0.5} \approx 0.29$. This means that H must be the point (0.29,0.29). But, the line through H at angle $135^\circ$ from the x axis cuts a triangle of area $\approx 0.17$.
So, my question is: what is the smallest $R$ such that every convex polygon has an $R$-balance-point? 
(and what is the standard name of this point?)

REPLY [2 votes]: I believe the answer given by @orangeskid is correct, and at least to see the balance point is the center of mass for triangles, note that the balance point is preserved under rotations, and also scaling axes (because these carry lines to lines, preserve incidence of points and lines, and scale the area of any region by a common factor). You can transform any triangle to an equilateral triangle with these operations. For an equilateral triangle, the balance point is the center of mass. One way to see the center of mass is the answer is by noting that the balance point must be the same when the triangle is rotated by 120 or 240 degrees around the center of the triangle. And finally, center of mass is preserved under rotations and scaling axes, so center of mass is the balance point for any triangle.