Comments on: Tidbits

By: David Grayson

David Grayson — Mon, 08 Mar 2010 06:43:50 +0000

Actually we DO need to assume there are a finite number of paraellelograms. Otherwise, any tiling that has a rectangle can be modified by replacing the rectangle with an infinite number of non-rectangle parallelograms, to yield a tiling without rectangles that contradicts the problem statement.

By: David Grayson

David Grayson — Mon, 08 Mar 2010 00:52:25 +0000

While I’m thinking about this problem, I am going to assume that there are a finite number of parallelograms, but hopefully I’ll discover later that this assumption is not necessary!

By: David Grayson

David Grayson — Mon, 08 Mar 2010 00:51:07 +0000

This is an interesting question. I’d like to know the answer! I found a math problem set on the web that asks the same question for an octagon instead of a 400-gon, and then it asks the students to generalize their answer to the octogon. So we should probably think about the octagon first. lukas’s octagon drawing shows three rectangles but we could easy eliminate one of them, leaving only two.

Here is that problem set I mentioned:
http://math.uci.edu/~krubin/oldcourses/08.194/ps8.pdf

By: Weekend Roundup at Steven Landsburg | The Big Questions: Tackling the Problems of Philosophy with Ideas from Mathematics, Economics, and Physics

Sat, 06 Mar 2010 07:02:43 +0000

[...] started the week with a pointer to a hilarious recipe for salted water. If you didn’t follow the link then, you should follow [...]

By: Al V.

Al V. — Thu, 04 Mar 2010 16:05:52 +0000

@Bennett Haselton, I’m not clear on why “… every parallelogram in the path leading from the north edge to the south edge, has a width equal to the edge lengths of the 400-gon…” I would think that parallelogram sides of 1/2 edge length would work equally well.

As a test, I tiled an octogon with 1/2 edge length parallelograms, and found it easy to tile the octogon, and wound up with 8 squares.

By: Dan Rosenberry

Dan Rosenberry — Tue, 02 Mar 2010 01:53:03 +0000

Lukas is right, as I was driving to work, I realized that the could split. The sum of the widths (in a direction parallel to the original edge) of the split paths is at least as long as the original edge.

Each crossing is a parallelogram or a decomposition into multiple and hence has at least one rectangle. Pushing the conclusion a bit further, a 4n-gon has rectangles/2 + squares >= n, so each of the 100 rectangles is either “special” and is also square or there’s an additional rectangle somewhere in the figure.

By: dave

dave — Tue, 02 Mar 2010 01:41:01 +0000

hehe. interesting that the recipe was at 3.5 forks with a 78% ‘would make this again’. i was sorely tempted to leave my own review but thought it better to leave it here at the origin.

the puzzle is as usual far to difficult for me. it needs a dash of salt.
your optical illusionesque picture didnt help at all either, lukas.

By: lukas

lukas — Mon, 01 Mar 2010 21:14:48 +0000

Well I guess I'll have to link to it.

By: lukas

lukas — Mon, 01 Mar 2010 21:13:20 +0000

Let’s try this:

By: lukas

lukas — Mon, 01 Mar 2010 20:59:31 +0000

Aw, Steven’s blog doesn’t like my

 tags :(