Conference rusure::math

Actually, as of Aug 1990 that set of tiles belongs to Home Equity and I am
in a new state. The new bathroom floor has a classical pattern: octagons 
alternating with squares. Something like:

	  >----<   >----<
	 /      \ /	 \
	|	 |	  |
	|	 |	  |
	 \      / \      /
	  >----<   >----<
	 /      \ /	 \
	|	 |	  |
	|	 |	  |
	 \      / \      /
	  >----<   >----<

where '*' is in one of the squares, which is at 45 degree angles.

The puzzle I see here is: What lengths of sides of the octagon makes the
area S of the square 1/2 the area O of the octagon? How about 1/n, where n
is an integer? [As the sides of the square increase from 0, with the sides
of the octagon fixed in position, the ratio S/O runs from 0 to 1, so every
1/n is feasible.] 

Lynn Yarbrough 

> The puzzle I see here is: What lengths of sides of the octagon makes the
> area S of the square [some fraction] the area O of the octagon?

Easy.  Let the width of the octagon be 1.  Now a little right isoceles
triangle with sides x, x, and x*sqrt(2) is clipped from each corner of
a square to form the octagon; and 0 < x < 1/2.  The area of this triangle
is � x�, so the area of the octagon is 1 - 2x�, and the area of a square
is (of course) 2x�.

You want (area of square)/(area of octagon) = (some fraction), so
2x�/(1-2x�) = f, or (ignoring the negative root) x = sqrt( � f / (1+f) ).
For f=1/2, x=1/sqrt(6); for f=1/3, x=1/sqrt(8); for f=1/n, x=1/sqrt(2n+2),
which is rational if n=2m�-1 : for f=1/(2m�-1), x=1/|2m|.  For both f and x
to be rational, simply let x be rational.

I think one can prove that >=4 is not possible, although I have not worked 
out the details. Consider any set S of tiles, with three centers of 
rotation C1, C2, C3. (Call that a 3-set.) Since any 4-set is also a 3-set,
it should be possible to find a 3-set such that the 4th center is implied.
So given any candidate 3-set, consider the restrictions placed on C4 by
certain points on the convex hull of S, and show that the existence of C4
requires the existence of another point in S not already there, e.g.
outside the convex hull. And that's a contradiction. 

For example, if C4 is outside the convex hull of S, this implies that C4
forms an equilateral triangle with the two nearest points on the hull, and 
also with the two farthest points on the hull. Also, min/max(dist(Ci, Sj))
must each be acheived for two distinct j, and the distance between them
must be that min/maximum, which implies a certain roundness of S. I'm
hand-waving by now, but I think the proof is there somewhere. 

Lynn Yarbrough 

Find the centers of the following:

		� � � � � � � � � � � 
		 � � * * � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � � * � � 
		 � � � � � � � * * � �
		� � � � � � � � * � � 
		 � � * � � � � � � � �
		� � * * � � � � � � � 
		 � � * � � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � * * � � 
		 � � � � � � � � � � �

		� � � � � � � � � � � 
		 � � 1 1 � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � � 2 � � 
		 � � � � � � � 5 4 � �
		� � � � � � � � 2 � � 
		 � � 3 � � � � � � � �
		� � 4 5 � � � � � � � 
		 � � 3 � � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � 6 6 � � 
		 � � � � � � � � � � �

None so far. Each number above lacks a partner when the other cell of the 
same number is center, and other points seem to have even worse problems.
You may be on the right track, though. How about:

		� � � * � � � � � � � 
		 � � * * � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � � * � � 
		 � � � � � � � 1 0 � �
		� � � � � � � � * � � 
		 � � * � � � � � � � �
		� � 0 1 � � � � � � � 
		 � � * � � � � � � � �
		� � � � � � � � � � � 
		 � � � � � � � � � � �
		� � � � � � � * * � � 
		 � � � � � � � * � � �

That's closer: now the 0's work, but still the 1's don't. Moving the new 
points inside will make the 1's work but breaks the 0's. Hmmm.

Oh dear.  That looks like a typo, but my notes are at home so I can't
correct it right now.  Here's a pretty one, by way of atonement.

		� * * * * �
		 * * � * *
		� * � � * �
		 * * � * *
		� * * * * �

There are FIVE centers of rotation in this pattern!


			       * * * *     * * * *
			      * * * * *   * * * * *
			     * * * * * * * * * * * *
			    * * * * * * * * * * * * *
			     * * * * * * * * * * * *
			      * * * * * * * * * * *
			   * * * * * * * * * * * * * *
			  * * * * * * * * * * * * * * *
			 * * * * * * * * * * * * * * * *
			* * * * * * * * * * * * * * * * *
			 * * * * * * * * * * * * * * * *
			  * * * * * * * * * * * * * * *
			   * * * * * * * * * * * * * *
			          * * * * * * *
			           * * * * * *
			            * * * * *
			             * * * *


The FIVE centers of rotation in the lobster below are marked with an `8' and
four `Z's; the `8' is also a colored tile, while the Zs are not.

			 . . . . . . . . * * * . . . . . . .
			. . . . . . . * * * * * * . . . . .
			 . . . * . * . * * * * * * . . . . .
			. . * * * . * * . * * * * * . . . .
			 . * * * * . . . * . * * * . . . . .
			. * * * . * . . . . * * * * * . . .
			 . * * * . . . . * * * * * * * * . .
			. * * * . . . . . * * * * * * * * .
			 . . . . . . . Z Z * * * * * * * * .
			. . . . . . . . . 8 * * * * * * * .
			 . . . . . . . Z Z * * * * * * * * .
			. * * * . . . . . * * * * * * * * .
			 . * * * . . . . * * * * * * * * . .
			. * * * . * . . . . * * * * * . . .
			 . * * * * . . . * . * * * . . . . .
			. . * * * . * * . * * * * * . . . .
			 . . . * . * . * * * * * * . . . . .
			. . . . . . . * * * * * * . . . . .
			 . . . . . . . . * * * . . . . . . .