Friday puzzle -- Matchstick polygons

Re: Friday puzzle -- Matchstick polygons

Yes, if you choose a Hero and start with a rational approach.

Re: Friday puzzle -- Matchstick polygons

Maybe. I don't follow your answer at all, though. Let's give it a little more time, then explain your answer'

Solution

My "Hero" is the Greek mathematician, Hero of Alexandria. The key to the solution is a Heronic Triangle, the sides lengths and area of which are always rational numbers.

The best known Heronic triangle is the 3-4-5 triangle, which is ideally suited to this problem because the sum of the sides equals the number of matchsticks: 3 + 4 + 5 = 12. The area of this triangle is (3*4)/2 or 6. Two more than the problem requires. However, it is a simple matter to shift three of the matchsticks to subtract 2 unit squares from the area to get a total area of 4.

Re: Solution and followup question

Excellent. And thanks for the background.

It is obvious, following Bill's technique, how to get a polygon with an area of 3. Is it possible to get one of area 2 or area 1?

In theory

with 12 matchsticks, you should be able to construct polygons with areas just over zero to just over 11 - although I'm not sure I want to work out all the angles.

Correct

Bill, I think your solution to area=4, and the related one for area=3, have the advantage of being "constructable", which was not stated in the problem, but probably should be.

Your quick answer to the second made be think a little more about it and see how obvious it was -- just take a decagon (the maximum area) and "squash if" for any smaller area.

I saw another interesting solution, that had the interesting feature that small areas were attainable with no obtuse angles: start with the same decagon, and to get smaller sizes, move alternating points closer to the center. For small areas, this gives you a 5-pointed star with increasing degree of "pointiness".

Re: Correct

The original question was with 12 matches. Did you mean dodecagon and a 6 pointed star? That was my first thought.

For areas up to 9, the angle calculations are much much simpler if you use a rhombus with sides of length 3. For an area of A, the angle would be arcsin(A/9)

Yeah, meant dodecagon

and six-pointed star. I should have re-read (or even re thought about!) the original before typing the response. Thanks for correcting.

Re: Yeah, meant dodecagon

So can this be generalized to polygons with an even number of vertices? But not to polygons with an odd number of vertices?

With an even number of vertices it seems like you can always "push in" alternate vertices and approach an area of 0. You can "push in" some of the vertices of a polygon with an odd number of vertices, but you can not get arbitrarily close to an area of 0 (I think).

Even only?

I think you are right, Larry. You can get an arbitrarily small area by pushing in alternative points on an equal-sided polygon of even sides greater than 4, or by "squashing" any polygon with even number of equal sides.

I agree with you that this appears not to be doable for an odd number of matchsticks, but I don't have any proof of that.