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Geometry math

asked 2021-05-02 14:44:12 +0100

Buk gravatar image

updated 2021-05-02 15:18:46 +0100

I can't wrap my brain around this today.

If you have hexagons attached to the edges of a pentagon in 2D, how far must you rotate two of the hexagon around their common edge with the pentagon, to bring their common edge into alignment? image description image description

That is, how do you calculate it. It might be squares around a triangle, or pentagons around a square etc.

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answered 2021-05-02 17:53:43 +0100

slelievre gravatar image

updated 2021-05-04 21:00:27 +0100

Baskets of polygons

Baskets

Fix two integers m ≥ 3 and n ≥ 4 and consider, in the horizontal plane in three-space, a regular m-gon surrounded by m regular n-gons with the same side length, glued to it edge to edge.

Depending on m and n, the outer n-gons can

  • overlap with their neighbours (think of m and n both large)
  • tile perfectly, that is, intersect with each of their neighbours along an edge
  • leave out some space, so that two neighbouring n-gons only share a vertex

The second case happens only for finitely many pairs (m, n) and is well studied.

The third case happens for every (m, 4) and for finitely many other pairs.

In that third case, if we rotate each n-gon by a same angle around its common edge with the central m-gon, as the rotation angle increases, the gap between neighbouring n-gons decreases, until for some special value alpha (depending on m and n), it vanishes and each n-gon shares a full edge with each of its neighbours.

We obtain a "basket" ; let's call that an (m, n)-basket.

Link with the soccer ball polyhedron

The soccer ball polyhedron is the polyhedron made of 12 regular pentagons and 20 regular hexagons, all with the same side length, glued edge to edge, each pentagon adjacent to 5 hexagons, each hexagon adjacent to 3 pentagons and 3 hexagons.

One of the pentagons together with its five adjacent hexagons form a (5, 6)-basket.

Basket angle

The question is: for each m and n, what is the angle alpha to use?

Modelisation and solution

To determine the angle alpha, we restrict our attention to one m-gon and one adjacent n-gon.

For convenience we take half the side length as our length unit.

We set up some notation. Call

  • M the m-gon, E its centre,
  • N the n-gon,
  • T, U, V, W four consecutive vertices of N, such that [UV] is the common edge with M
  • A the midpoint of [UV]
  • C the midpoint of [TW]
  • P the plane that contains M
  • L the line through E perpendicular to P
  • p the orthogonal projection map from the whole space to P (parallel to L)
  • B = p(C) the orthogonal projection of C to P
  • tau the angle of a full turn, that is, tau = 2*pi
  • r the rotation map by tau/m (an m-th of a turn) around L sending U to V and T to W;
  • for each integer k, r^k the rotation by (k/m)*tau around L, (the k-th composition power of r),

Notice the rotates r^k(T) for k from 0 to m - 1 (including T and W) form a regular m-gon parallel to M.

Call R that m-gon, and F its centre. Notice that p(F) = E.

Since the planes containing M and R are parallel, p induces an isometry between them, so FC = EB (both as vectors and as lengths).

Notice E, A, B are aligned, so, as lengths, EB = EA + AB

Right-angled triangles, and a trapezoid

We notice some right-angled triangles:

  • (A, B, C) has a right angle at B and an angle alpha at A, so

    cos(alpha) = AB / AC
    

    (if n = 4 that triangle is degenerate but the formula still holds)

  • (E, A, V) has a right angle at A and an angle pi/m at E, so

    tan(pi/m) = AV / EA
    cot(pi/m) = EA / AV
    

    and notice that by our choice of length unit, AV = 1

  • (F, C, W) has a right angle at C and an angle pi/m at F, so

    tan(pi/m) = CW / FC
    cot(pi/m) = FC / CW
    

Furthermore, calling H the orthogonal projection of U to the line (TW),

  • (T, H, U) has a right angle at H and an angle tau/n at T, so

    TH = 2 * cos(tau/n)
    UH = 2 * sin(tau/n)
    

In the trapezoid (T, U, V, W), C is the midpoint of [WT], the points C, H, T are aligned, and UACH is a rectangle, so

    CW = CT
    CT = CH + HT
    AC = UH = 2 * sin(tau/n)
    CH = AU = 1

Combining our equations

Starting from the equation

FC = cot(pi/m) * CW

and substituting

FC = EB = EA + AB = cot(pi/m) + 2 * sin(tau/n) * cos(alpha)

and

CW = CT = CH + HT = AU + TH = 1 + 2 * cos(tau/n)

we get

cot(pi/m) + 2 * sin(tau/n) * cos(alpha) = cot(pi/m) * (1 + 2 * cos(tau/n))

which simplifies to

2 * sin(tau/n) * cos(alpha) = cot(pi/m) * 2 * cos(tau/n)

and gives

cos(alpha) = cot(pi/m) * cot(tau/n)

How can Sage be useful here?

Define a function and try it out:

def basket_angle(m, n):
    return arccos(cot(pi/m) * cot(2*pi/n))

Pentagon surrounded by hexagons:

sage: a = basket_angle(5, 6)

sage: a
arccos(1/15*sqrt(3)*sqrt(10*sqrt(5) + 25))

sage: a.n()
0.652358139784368

sage: (a/pi).n()
0.207652045225832

sage: (a * 180 / pi).n()
37.3773681406497

Triangle surrounded by octagons:

sage: a = basket_angle(3, 8)

sage: a.n()
0.955316618124509

In degrees:

sage: (a * 180 / pi).n()
54.7356103172453

Going further

It would be a fun project to draw pictures and animations to illustrate the problem using Sage. For example, one could produce

  • a 3d animated scene to illustrate folding the hexagons from an initial position spread out in the plane of the pentagon to a final folded up position

  • a 3d animated scene drawing each of the steps of the reasoning above, highlighting the relevant right angled triangles one after the other.

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Comments

Solving the problem by using a coordinate system may have been misguided..

I have replaced my initial answer with a geometric solution.

slelievre gravatar imageslelievre ( 2021-05-03 23:45:13 +0100 )edit
1

Solving the problem by using a coordinate system may have been misguided..

I have replaced my initial answer with a geometric solution

Presenting both solutions would have illustrated the differences of approach between synthetic and analytic geometry which is a point not absolutely devoid if interest for budding mathematicians...

I would, however, have started wit a few hints, leaving the (obvious) student his/her (obvious) homework...

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2021-05-04 09:14:42 +0100 )edit

Obvious can be tricky.

slelievre gravatar imageslelievre ( 2021-05-04 10:18:20 +0100 )edit

Thank you for taking a second crack at this.

This method closely resembles the construction method I use, based on the simple observation that the coincident apexes of the rotated polygons form a polygon similar to the base polygon. animation*

In the case of the hexagon around a pentagon, the subscribing circle of the formed hexagon is simply twice that of the base pentagon.

The close correspondence between the methods gave me great hope for your final equation: n(deg( arccos(tan(pi/6)/tan(pi/5)) )) = 37.3773681406497

However, for other combinations of sides -- eg, from the OP: "It might be squares around a triangle, or pentagons around a square etc" -- it no longer holds true.

Eg.octagons and triangle

Buk gravatar imageBuk ( 2021-05-04 10:47:13 +0100 )edit

@Emmanuel Charpentier I'm 65+, and definitely not a "student" and this is definitely not homework.

Retired engineer turned programmer, but not a mathematician; looking to use SageMath as a tool to help me use math to solve engineering tasks.

Buk gravatar imageBuk ( 2021-05-04 10:50:28 +0100 )edit

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Asked: 2021-05-02 14:44:12 +0100

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Last updated: May 04