Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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. *

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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 $x=y$.

share|improve this answer
    
If $A$ was above $x=y$ line, (say $A=(-2,-1)$), does this ensure a sufficient condition for the intersection to happen? –  dineshdileep Jan 30 '13 at 5:49
    
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 $x=y$. –  Rahul Jan 30 '13 at 5:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.