# Convex polygon is connected by broken lines with each chain parallel to one of two vectors

Given is convex polygon $P$ on the plane. Settle if there are two vectors $\vec{a}$, $\vec{b}$ such that any two points belonging to the polygon $P$ you can connect a broken line contained in the polygon and consisting of at most three sections, where each of sections is parallel to the vectors $\vec{a}$ or $\vec{b}$

I think that it's possible, for instance consider the following situations we have rectangle $ABCD$, our points are $H,I$ and we have $\vec{b}$ is parallel to sections $AH,IJ$ and $\vec{a}$ is parallel to $HI$, but I don't know how to prove it mathematically. • Do you want it proven for the displayed polygon or an arbitrary one (in the latter case I think it's impossible)? – Meow Dec 9 '13 at 17:42
• For an arbitrary one – Mark Dec 9 '13 at 17:47
• Okay, for arbitrary non-collinear $\mathbb{a,b}$? – Meow Dec 9 '13 at 17:48
• For the general case I think would be ok, in the content there's nothing mentioned about the collinear $a,b$ – Mark Dec 9 '13 at 17:56
• I'm genuinely lost .... what are you even asking?! What does "Given is convex polygon P on the plane." even mean? – Squirtle Dec 11 '13 at 20:26

Let $\ell$ be a diameter, i.e., a line segment in $P$ of maximal length. Then letting $\vec a$ be its direction vector and taking $\vec b$ orthogonal to $\vec a$ should suffice: To connect points $A$ and $B$ in $P$, connect each to its closest point on $\ell$; the maximality of $\ell$ guarantees these two points, call them $C$ and $D$, are inside $P$. So $AC$ is parallel to $\vec b$, $CD$ is parallel to $\vec a$, and $DB$ is parallel to $\vec b$, all inside $P$.