Let $\{\bar{x},\bar{y},\bar{z},\bar{t}\}$ be 4 points in $\mathbb{R^2}$, such that $conv\{\bar{x},\bar{y},\bar{z} \} \cap conv\{\bar{x},\bar{y},\bar{t} \} \cap conv\{\bar{x},\bar{z},\bar{t} \} = \{\bar{x} \}$, show $ \bar{x} \in conv\{\bar{y},\bar{z},\bar{t} \} $
"$conv$" denotes the convex hull.
I think I miss a trivial conclusion from the condition that $conv\{\bar{x},\bar{y},\bar{z} \} \cap conv\{\bar{x},\bar{y},\bar{t} \} \cap conv\{\bar{x},\bar{z},\bar{t} \} = \{\bar{x} \}$, when can these three convex hulls intersect at one point?
Can anyone give me a hint? Any help would be appreciated.
It's more heuristic than rigorous probably, but I hope it helps. Convex hulls of three points in $\mathbb{R}^{2}$ are just triangles (unless the points lay on a line). Fix three points $x,y,z$ that do not lay on a line. Then if you want to choose a fourth point $t$ (which doesn't lay on the same line as $x,y$ or $x,z$ or $y,z$) such that the triangles with vertices $x,y,z$, - $x,y,t$ - $x,z,t$ intersect in the single point $x$ you're forced to choose $t$ in such a way that the triangle with vertices $y,z,t$ contains $x$. This becomes clear once you draw the picture. One should then handle the degenerate cases (in which three points lay on the same line), but the drawing should solve the problem again.