# For any convex polygon there is a line that divides both its area and perimeter in half.

I already proved that for any convex polygon and any slope there is a line with this slope that divides the polygon's area in halves. I guess this should help me with the main problem. While proving the lemma I showed that a function of area above the line, when the line moves vertically is continuous and used the intermediate value theorem.

I guess the main problem will be solved in a similar manner. My idea is to take any line dividing the area in halves and start going through all the lines of this property, but changing their slopes. When I'll have rotated 180 degrees I must have gone through a stage when the perimeter was also halved.

If the above reasoning is correct I only need to prove that the function that gives perimeter above the line is continuous when the slope changes.

How do I do this?

• This should be a result of the ham-sandwich theorem. – Michael Burr Sep 27 '15 at 12:08
• It is unclear to me which "circumference" we are trying to bisect together with the area of the polygon. Do you mean the perimeter of the convex polygon? – Jack D'Aurizio Sep 27 '15 at 12:12
• It's similar but I can't see how I could use this theorem here. And I want to prove everything from scratch and I don't know how to prove this general theorem. – I want to make games Sep 27 '15 at 12:13
• Yes, I meant the perimeter of the same polygon. Sorry – I want to make games Sep 27 '15 at 12:13

## 2 Answers

You could take the dual approach of starting with two points on the perimeter with distance half the perimeter between them, then moving them in sync until they're swapped. As in your approach, the difference between the areas on both sides of the line connecting the points changes signs and hence must vanish at some point. That the area depends continuously on the points is perhaps a bit more obvious than that the perimeter depends continuously on the slope in your approach. (Also you don't need the lemma for this version.)

• Nice, now proving the continuity is much easier. Thank you! – I want to make games Sep 27 '15 at 13:26
• I think this is a beautiful argument. ^_^ – Vectornaut Sep 27 '15 at 18:35

Your problem is very similar to an application of Bolzano's theorem given by Courant and Robbins in "What is Mathematics?" (see picture below). They don't bother to give an argument to justify continuity, but I think, as you do, that it is not so obvious. • This is not just very similar but in a sense a generalization, since the perimeter of $A$ can be seen as a limiting case of a second area $B$. – joriki Sep 27 '15 at 13:52