Enigma 1376 - Square facts and faces

[Chappy]

Enigma 1376 - Square facts and faces
New Scientist magazine, 28 January 2006.
by Susan Denham.

I have just constructed a simple polyhedron,
ie a three-dimensional solid figure with
some vertices, straight edges, and faces.
The number of vertices is a square; the
number of edges is a square; the number of
faces is a square; the total number of
vertices and edges is a square; and over
half of the faces are squares! How many
faces are there, and how many of those are
triangles?

Ciao,
Chappy.

[Ilan Mayer]

SPOILER

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If there are F faces, V vertices, and E edges, then:
F+V = E+2
F = a^2
V = b^2
E = c^2
V+E = d^2
Where a, b, c, and d are integers.

The first solution is F = 9, V = 9, E = 16, and so there must be at
least 5 square faces. This can be achieved with the polyhedron formed
from a cube attached to the base of a square pyramid.

There next solution is F = 121, V = 25, E = 144, but it requires at
least 62 square faces, which is impossible with just 25 vertices.

Thus there are 9 faces, and 4 of them are triangles.

Please reply to drgmayer at hotmail dot com

__/\__
\ /
__/\\ //\__ Ilan Mayer
\ /
/__ __\ Toronto, Canada
/__ __\
||

[Dean Hickerson]

Chappy wrote:

> Enigma 1376 - Square facts and faces
> New Scientist magazine, 28 January 2006.
> by Susan Denham.
>
> I have just constructed a simple polyhedron,
> ie a three-dimensional solid figure with
> some vertices, straight edges, and faces.
> The number of vertices is a square; the
> number of edges is a square; the number of
> faces is a square; the total number of
> vertices and edges is a square; and over
> half of the faces are squares! How many
> faces are there, and how many of those are
> triangles?

Ilan Mayer described the simplest solution, a cube attached to a square
pyramid.

Here's another possibility: Start with a cylinder made of 2 regular
1060-gons and 1060 squares. Pick one of the square faces, with vertices
A, B, C, and D. Draw the diagonal AC. Introduce 481 points along this
diagonal and connect each of them to B and D. Thus we've divided the
square into 2*482=964 triangles. Move each of the 481 points slightly
out of the plane of the square so the triangles become actual faces of
the polyhedron. This polyhedron has V=2*1060+481=2601 vertices. It also
has 2 1060-gonal faces, 1059 square faces, and 964 triangular faces, for a
total of F=2025 faces. Hence it has E=V+F-2=4624 edges. As required,
V=51^2, E=68^2, F=45^2, and V+E=7225=85^2 are all squares, and the number
of square faces, 1059, is more than half of F.

Dean Hickerson
de...@math.ucdavis.edu

[Mark J. Tilford]

IIRC, you don't need the "at least half squares" to rule this out; any
simple polyhedron must have 2E/3 >= V > E/3 and 2E/3 >= F > E/3.

> Thus there are 9 faces, and 4 of them are triangles.
>
>
>
> Please reply to drgmayer at hotmail dot com
>
> __/\__
> \ /
> __/\\ //\__ Ilan Mayer
> \ /
> /__ __\ Toronto, Canada
> /__ __\
> ||
>

--
------------------------
Mark Jeffrey Tilford
til...@ugcs.caltech.edu

[Ed Murphy]

ObFollowup: How many topologically-distinct solutions are there?