NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17445

Internet Message Format | 1992-12-27 | 2.7 KB

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From: [e-mail] (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Two (I think) interesting Problems
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Date: 28 Dec 92 01:38:45 GMT
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Organization: U of Cambridge, England
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1. To fold a sheet of paper into a regular pentagon: If it's an A4 sheet
of paper (or any other of the A... sizes; what matters is that the ratio
of the sides should be root(2)) there is a fairly elegant way to do this
with perfect accuracy. It goes like this:
 - Find the centre of the piece of paper by folding both pairs of opposite
   sides together.
 - Fold all four corners to the centre, producing a sort of hexagonal
   shape.
 - Fold this in half along what was the shorter of the two "orthogonal"
   folds in the first step.
 - You now have a sort of five-sided greenhouse shape. Fold the two walls
   of the greenhouse in so that the top of each wall is on the line of
   symmetry of the shape, and so that the line joining that to the corner
   newly created by this fold is perpendicular to that line of symmetry.
 - You now have a regular pentagon.
The last step sounds intimidating, but if you've got the earlier stages
right it's easy because you can see it's producing a regular pentagon.
This produces a pentagon with a couple of "flaps" folded back; if you fold
the pentagon in the right way (in the second stage, fold two opposite
corners and then the other two opposite corners) you can then stick 12 of
these things together and make a quite rigid dodecahedron!

2. You cannot draw an equilateral triangle all of whose corners are at points
of a square grid. (Proof: Set up coordinates so that the "grid points" are
exactly the points with integer coordinates. Now, on the one hand the area
of the triangle is root(3)/4 times square of side length = root(3)/4 times
an integer; on the other, if the vertices are (x1,y1) (x2,y2) and (x3,y3)
you can check that the area is (plus or minus) the determinant of the matrix
(1 1 1   /   x1 x2 x3   /   y1 y2 y3) over 2, which is half an integer. Hence
root(3) is rational. But it isn't: contradiction.)
You can get pretty good approximations, of course: for instance, put two
vertices 8 squares apart along a grid line, and the other one half-way between
but displaced by 7 squares. I'm sure you can do better than this without too
big an increase in size...

--
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
[e-mail]  Cambridge University, England.    [Research student]