0

I was solving this problem from the book, Putnam and Beyond.

Question: Prove that any convex polygonal surface can be divided by two perpendicular lines into four regions of equal area.

My problem: I have been able to prove it using the Intermediate value property of a continuous function but the main problem is the book says it's easy to see that when I am diving the polygon by a line or by rotating angle, all those operations are continuous functions.

How do I prove that those operations are continuous?

Sayantan
  • 187

1 Answers1

1

Continuity of area is a consequence of finiteness of polygon and properties of area.

Consider polygon $S$, split by line rotating around some point $O$. Let's denote one half of polygon as $P(\alpha)$ when line is rotated by $\alpha$. We want to show that $f(\alpha)=\mathop{\mathrm{Area}}(P(\alpha))$ is a continuous function of $\alpha$.

When we rotate the line by $\alpha$ some area will be added, and some will be subtracted. Consider following parts (backslash here is set difference operation): $$ P^+(\alpha)=P(\alpha)\setminus P(0),\qquad P^-(\alpha)=P(0)\setminus P(\alpha). $$ Since $\mathop{\mathrm{Area}}(P(\alpha))=\mathop{\mathrm{Area}}(P(0))-\mathop{\mathrm{Area}}(P^-(\alpha))+\mathop{\mathrm{Area}}(P^+(\alpha))$, we can conclude that: $$ |f(\alpha)-f(0)| \le \mathop{\mathrm{Area}}(P^-(\alpha))+\mathop{\mathrm{Area}}(P^+(\alpha)). $$

Let's consider now a circle with center at $O$ that covers polygon. Let its radius be $R$. Let $C^+(\alpha)$ and $C^-(\alpha)$ be sectors that are added and subtracted correspondingly to half-circle containing $P$ when line rotates by $\alpha$. We know that $P^+(\alpha)\in C^+(\alpha)$ and $P^-(\alpha)\in C^-(\alpha)$, so $$ |f(\alpha)-f(0)| \le \mathop{\mathrm{Area}}(C^-(\alpha))+\mathop{\mathrm{Area}}(C^+(\alpha)) = 2R|\alpha|. $$

Now it's easy to see that for every $\epsilon>0$, there is $\alpha=\frac13\epsilon/R$, that $ |f(\alpha)-f(0)| \le 2R\alpha < \epsilon $

Vasily Mitch
  • 10,129