Balanced cutting of a convex polygon 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?)
 A: What you have for one triangle is valid for all triangles. The best point for triangles is the center of gravity. If I remember correctly, in general, for a convex region in the plane the best $R$ $\le$ the one for a triangle , with equality only for triangles. I remember seeing this in a book about combinatorial geometry, maybe by Yaglom. 
${\bf Added:}$ I've found the reference, it's called Winternitz theorem, and says that the worst $R$ is the one for the triangle, which is $\frac{5}{4}$. I've found it in the book by Yaglom, Convex figures, Ex 3.10. All right!
Did not check the proof, but here is another result that could insure some $R$, not as good though. Kovner theorem (Ex 3.7) says that every convex region contains a symmetric convex region with area at least $2/3$ of the original. From this one we can conclude that any line passing through the center of that smaller region will give ration $\le \frac{2/3}{1/3} = 2$. 
A: 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.
