Convex polygon and circle

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Convex polygon and circle

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Source: A. Slinko

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#1 h Mar 8, 2005, 7:03 AM • 13 Y

Y by Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Adventure10, Mango247

1.Prove that every convex polygon of area S and perimeter P contains a circle of radius greater than S/P

2.A convex polygon of area S and perimeter P contains a convex polygon of area T and perimeter Q,

Prove that 2S/P > T/Q.

(Use part 1 to prove part 2)

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#2 h Mar 9, 2005, 1:04 AM • 2 Y

Y by Adventure10, Mango247

It's easier to deduce the second one from the first one:

Let $\ell_i,\ i\in\overline{1,n}$ be the sides of the polygon. If $O$ is a point inside, then the radius $r$ of a circle centered at $O$ and contained inside the polygon is $\le\min_i(d(O,\ell_i))$ . This means that $r\le \frac{\sum_i d(O,\ell_i)\cdot \ell_i}{\sum_i \ell_i}=\frac {2S}P$ . Now, since the second, smaller polygon contains, according to the first problem, a circle of radius $r>\frac TQ$ , we get $\frac {2S}P\ge r>\frac TQ$ , and we are done.

Now for the first problem:

Using arguments similar to those above, we can see that all we need to do is show that there is a point $O$ inside the polygon s.t. $d(O,\ell_i)>\frac SP,\ \forall i\in\overline{1,n}$ . First draw lines $\ell_i'\|\ell_i$ s.t. the distance between $\ell_i$ and $\ell_i'$ is $d$ , and $\ell_i'$ lies on the same side of $\ell_i$ as the interior of the polygon, for small $d$ . It's easy to see that the area enclosed by $\ell_i,\ell_i'$ is smaller than $P\cdot d\ (*)$ . At first, there's still some space inside the polygon formed by $\ell_i'$ . We want to show that there's still space there when $d=\frac SP$ . From $(*)$ we see that we can increase $d$ until it becomes $\frac SP$ , because when it's $\frac SP$ , the area enclosed by $\ell_i,\ell_i'$ is $<P\cdot \frac SP=S$ , so we haven't yet covered the initial polygon.

Maybe it's not that rigorous, but I think something like that definitely works.

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#3 h Mar 9, 2005, 7:18 AM • 2 Y

Y by Adventure10, Mango247

For the first one (which is Grobber's way, but maybe a bit clearer

)

On each side of the given polgon, we draw internally a rectangle with height $\frac S p.$ Since the polygon is convex some of these rectangles have some common interior point, and the sum of the areas of these rectangles is $S$ . It follows that at least one point, say $M$ , of the interior point of the polygon is not covered by the rectangles, which means that $M$ is at distance at least $\frac S p$ from each sides so that the disk with center $M$ and radius $\frac S p$ gives a solution.

Pierre.

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#4 h Mar 9, 2005, 7:20 AM • 2 Y

Y by Adventure10, Mango247

Definitely nicer than what I said!

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#5 h Mar 9, 2005, 2:03 PM • 2 Y

Y by Adventure10, Mango247

It is an old russian problem by the way

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Myth

#6 h Mar 9, 2005, 3:19 PM • 2 Y

Y by Adventure10, Mango247

Moreover, Pierre's solution is, actually, the official one.

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