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# Illuminating a polygon

### #1

Posted 14 July 2014 - 04:11 PM

You are given a plane * n*-gon having no intersecting sides. That is, each segment of the perimeter has interior points on one side and exterior points on the other. We wish to illuminate the interior of the n-gon in its entirety by placing lamps at various points in its interior. Clearly, if the

*-gon is convex, one lamp will suffice. Concave points, however, may cast shadows in some interior regions. Since life here in the Den is never simple, we ask about the general case:*

**n**To illuminate the interior of a simple

*-gon, what is the smallest number of lamps that will always suffice? Does the answer change if we require the lamps to be placed at vertices?*

**n***Vidi vici veni.*

### #2

Posted 14 July 2014 - 09:44 PM

No proof, but I think the answer is

### #3

Posted 14 July 2014 - 10:57 PM

No proof, but I think the answer is

Spoiler for this

Your answer will surely work, but it is not the smallest number that will always suffice.

If you like, try drawing some 2p-gons, where p is 2 or 3 or 4, that are not convex.

Also, consider the fact that a triangle is always convex.

*Vidi vici veni.*

### #4

Posted 14 July 2014 - 11:53 PM Best Answer

**Women are definitely stronger. We are [Fe]males, after all...**

*Some of what makes me me is real, some of what makes me me is imaginary...I guess I'm just complex. ;P*

<3 BBC's Sherlock, the series and the man. "Smart

*is*the new sexy."

Chromatic Witch links now on my 'About Me' page! Episode 3 is finally here!

When life hands me lemons, I make invisible ink.

### #5

Posted 15 July 2014 - 02:58 PM

### #6

Posted 15 July 2014 - 06:34 PM

Spoiler for You need no more than

The idea that three is the fewest convex vertices is nice.

So far, y-san's approach gives the tightest upper bound.

*Vidi vici veni.*

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