I have a 2-Dimensional Closed Convex Compact Body $\mathbb{S}$(a set, for eg, a circular disc)). Assume, it has a non-zero intersection with the all-negative quadrant. Consider the following 2-D point $P_1$ \begin{align} x_{min}= \min_{(x,y)\in \mathbb{S}}x \\ y'= \min_{(x_{min},y)\in \mathbb{S}}y\\ P_1=(x_{min},y') \end{align}
and also the point $P_2$ \begin{align} y_{min}= \min_{(x,y)\in \mathbb{S}}y \\ x'= \min_{(x,y_{min})\in \mathbb{S}}x\\ P_2=(x',y_{min}) \end{align}

Simply put, $P_1$ is the left-most vertical edge ( west-most point) and $P_2$ is the bottom-most horizontal edge (south-most point). (its ok if you assume both edges are kind-of sharp). The question is

*Given $P_1$ and $P_2$, and also using properties of $\mathbb{S}$ (convex,compact,closed), how can we tell, whether the given body $\mathbb{S}$ have an intersection with $x=y$ line in the negative quadrant. *


You can't. Let $A=(-2,-3)$, $B=(-1,-4)$, $C=(-1,0)$. Both the line segment $AB$ and the triangle $ABC$ have the same $P_1=A$ and $P_2=B$, but only the latter intersects the line $x=y$.

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  • $\begingroup$ If $A$ was above $x=y$ line, (say $A=(-2,-1)$), does this ensure a sufficient condition for the intersection to happen? $\endgroup$ – dineshdileep Jan 30 '13 at 5:49
  • $\begingroup$ Yes, if $P_1$ and $P_2$ are on opposite sides of $x=y$, then the existence of an intersection immediately follows from convexity: $\mathbb S$ contains the line segment $P_1P_2$, which intersects the line $x=y$. $\endgroup$ – user856 Jan 30 '13 at 5:51

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