Perhaps my Google game is weak, but I’m trying to find a mathematical treatment (and ideally a visual representation) of the largest regular hexagon that fits inside a regular pentagon.
It’s not hard to find the other way around, or to find squares inside pentagons or vice versa, but I can’t find anyone addressing the pentagon around the hexagon.
Given a pentagon with side lengths s, The hexagon will be inscribed such that its vertices touch the sides of the pentagon.
The apothem (distance from the center to a side) of the pentagon can be found using:
apothem = s x COS (36 degrees).
This apothem will also be the distance from the center of the hexagon to one of its vertices. Using this distance, we can find the side length of the hexagon, h, using the relation 2 x apothem x TAN (30 degrees).
The side length h of the largest regular hexagon that can fit inside a regular pentagon with side length
s is given by:
h = ((1+sqroot(5)) x s) / (2 x sqroot(3))
There are two obvious orientations for the hexagon inside the pentagon, one where they share a side (incompletely) and two or maybe four vertices, the other and one where maybe four vertices touch four sides of the pentagon. Any hints which way this is or how your solution works before I screw up attempting to figure this out myself? I assume one is larger, but no sure if those side and points would connect as I described.
I’m not sure that I accept that a hexagon with distance-from-center-to-vertex the same as apothem-of-the-pentagon is the largest hexagon that can fit in the pentagon. If I skew the hexagon, can’t I make it larger?
I think that if the hexagon and pentagon both share a side, you get a smaller hexagon than if they only touch at vertices/sides.
Plus, the apothem of a pentagon in terms of the side is 0.5(tan 54)S, or 0.688S, not (cos36)S.
Yes, that’s what it looks like playing around in Paint. The distance between parallel sides of the hexagon seems to be the limiting factor. I was wondering about skewing also.
The location of the vertices for a regular hexagon and a regular pentagon can be calculated with simple trig (cos (360 i /6+skew) and sin (360i /6+skew) are the x and y coordinates for the hexagon (for i=0 to 5) and cos (360 i /5) and sin (360i /5) for the pentagon (you only need to skew one of them)). From that you can find the equation for the line segments that make up each figure, and go from there.
(I did this problem for circles, squares and triangles some years back. As I recall, there was a skew for the largest triangle in a square - they didn’t share a vertex or a side).
If the pentagon has a radius of 1, then the radius of the hexagon is \sqrt{\frac{5-\sqrt{5}}{2}}
Can you walk me through that, or give me a reference?
I based my solution on the assumption that the largest hexagon you can fit will be with three vertices touching three sides of the pentagon.
I think I made a mistake, let me double check my math.
It should be \frac{1+\sqrt{5}}{4}.
Consider a hexagon inscribed in the pentagon such that one of its sides is parallel to the base of the pentagon. The distance from the center of the pentagon to the midpoint of this side of the hexagon will be the radius of the hexagon.
You’ll notice that this is \cos{\pi/5}
Although we might need to take into account an non-centered hexagon…
I think that orientation is the largest regular hexagon that will fit inside a regular pentagon. What I see is the ‘height’ of the hexagon, IOW the distance between parallel sides must fit between to adjacent sides of the pentagon while one vertex touches the midpoint of one pentagon side. I don’t know how to do math with all those funny symbols so I’m trying to see how in your solution or any others here that works out.
Is that the largest hexagon that will fit though? The height of the hexagon oriented that way seems to be less than that with the orientation suggested above.
Using the formula I derived here’s the largest I’ve been able to fit, and the orientation:
Yes, the possibility of a non-centered hexagon is the crux of the problem.
It can be skewed, clearly. But I don’t see how a non-centered hexagon can ever be a solution.
I am pretty sure your are correct. It must be centered if it’s a regular hexagon.
Dorjan, Chingon I don’t think it has to be centered. Taking a look at Dorjan’s diagram, the red hexagon can clearly be larger. If you scale it up ever so slightly while keeping the SSE segment parallel to its current orientation (so maintaining contact with the SE and SW sides of the pentagon) until the westernmost corner of the hexagon touches the western side of the pentagon, you will have a non-centered hexagon making contact on three sides of the pentagon.
IOW if you rotate the hexagon around one of its corners to make it fit, you’ve uncentered it relative to the hexagon.