I don’t see why the two polygons have to have the same center. Consider the simpler case of packing one square inside one equilateral triangle. The most efficient packing has one side of the square correspond with (part of) one side of the triangle (cite). If we call the length of the side of the square 1, the side of the triangle is about 2.155. The altitude of the triangle is about 1.866 and the perpendicular distance from the side to the center of the triangle is one-third that or about 0.622. The perpendicular distance from the center of the square to the side is obviously only 0.5. Therefore the center of the square and the center of triangle do not correspond. I don’t see why the centers have to correspond in the case of a hexagon packed in a pentagon.
I wouldn’t expect the optimal nesting to ever be centered, when either one of the numbers of sides is odd (except for the trivial case where they’re both the same, of course). And it’s pretty easy to show that any optimal nesting will always have at least three points of contact, because with two points of contact you always have room to grow.
And I have to immediately correct myself: A hexagon and a triangle (either one larger) will always be centered. OK, then, let’s say when the numbers of sides are relatively prime.
While an analytic answer would be great to have I decided to fall back to brute force and found (perhaps) a numerical answer of:
r = 0.84247
with the following placement (the hexagon is indeed not centered but about 0.0024 units from the center):
What is the angle between hexagon and pentagon at the pentagon’s vertices?
If I start with the hexagon where two sides are parallel to the base of the pentagon, then rotate about \pi/10.
Thanks. You say “about” just because this us brute force, right? It could be exactly that analytically (or am I wrong?)?
It could be exactly \pi/10. Here it is the solution to about machine precision:
0.31415926255497995
But that could very well just be a local maximum.
It suggests a pattern for polygon fitting.
And I have to correct myself on this, too. For square-inside-a-hexagon, the optimal nesting only has two points of contact. Though I suppose you could argue that it’s four degenerate points of contact.
@Chingon , it’s not an absolute proof, but when I see an angle that comes out to equal pi/10 to six decimal places, as a physicist, yeah, I’m going to say that the true, exact answer really is pi/10. Though as you mention, it’s difficult to rule out that there might be a better global maximum somewhere.
Another feature of @Chingon’s cool diagram (thank you) is there are 4 points of contact between the 5-sided container and the 6-sided interior object. Is that 4-5-6 sequence a coincidence, or a feature of these sorts of packing problems? Does it apply to higher order problems?
Heck if I know, or have any clue how to find out.
Except in particularly symmetric cases (at the very least, where the numbers of sides have a common factor), it’s hard to see how one could, in general, ever have more than four points of contact. The problem has four degrees of freedom: The position of the center of the inner shape, the angle by which the inner shape is rotated, and the size of the inner shape. Roughly speaking, each degree of freedom can give you at most one point of contact.
Start with a regular pentagon, P, with vertices distance 1 from the origin with one vertex at (1, 0).
Then a regular hexagon, H, with a long diagonal along the x-axis centered on the point (d, 0) with sides length s is the largest hexagon that fits within P when:
d = \frac{(-23 + 22 \sqrt{5} - \sqrt{1275 - 330 \sqrt{5}})}{62}
s = \frac{(-25 + \sqrt{5} + \sqrt{4350 + 570 \sqrt{5}})}{62}
Lance_Turbo, is that the case where the hexagon touches the pentagon at four points?
Thanks for the calculation!
It is exactly what @Chingon depicted in post 24 which touches at four points (up to rotation).
Specifically, rotate Chingon’s diagram 54 degrees counter-clockwise. Then the x-axis passes through the two vertices of the hexagon that don’t touch the pentagon and the origin is equidistant from all five vertices of the pentagon.
You can show that the four vertices that are not on the x-axis do indeed lie on the pentagon with a little trigonometry and a lot of help from something like Wolfram Alpha. I can post the highlights if anyone is interested.