szimmetria-airtemmizs - Tumblr blog

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Now with rotations.

I made a gif not long ago (see below). Many of you expressed that it was "deeply uncomfortable" and "upsetting". I hope this new one will calm your feelings.

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Mathematical fun fact of the day 4:

You can cut a circular cake into seven parts using three straight cuts, but you cannot cut it into seven parts of equal area with three cuts.

And it gets worse, if the shape of the cake is any convex set in the plane, you still can not cut it into seven parts of equal area with three cuts.

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Here is a nice open problem:

Take a regular 2k-gon and draw all the diagonals. This breaks the plane into regions, let A_k denote how many sides does the region with the most sides have. For example, in this image we have a regular 10-gon, and we only have triangles and quadrangles, so A_5=4.

Question: Is it true that A_k is bounded by an absolute constant? For example, is there a k such that A_k>100?

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Mathematical fun fact of the day 3:

You can tile a regular 30-gon using a pentagon and rhombus.

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Mathematical fun fact of the day 2.

The maximum number of diagonals of the regular n-gon that meet at a point other than the center is 7. This can only be achieved when n is a multiple of 30.

For more details see the paper: The number of intersection points made by the diagonals of a regular polygon by Bjorn Poonen and Michael Rubinstein

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Balance

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Sliding and rotation. I combined two ideas:

Sliding,

and rotating:

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Recursive slide.

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Slide.

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Some of you requested a proof. I don't want to spoil it, so you can read it after clicking:

We will use vectors. A nice property of vector addition is that the order of the terms is not important (associative). For example, a+b+c=c+a+b for any vectors.

We start at the inner vertex of one of the 2k+1-gon and follow the colored path:

We consider each colored segment as a vector directed along the path. For example, the two red edges give us vectors that point in opposite direction.

Since the order is not important, we can calculate the sum of the vectors grouped by the colors. In each color we have two vectors with opposite direction and same length, so they sum to 0. Hence, the sum of all vectors is clearly 0. But summing the vectors is the same as putting them one after the other. Therefore, following these vectors we get back to the same point, i.e the inner vertices of the two 2k+1-gons are the same.

We need a 4k+2-gon so that we can do this pairing of the vectors, for other values this doesn't work.

Mathematical fun fact of the day:

If you draw a regular 4k+2-gon and add a regular 2k+1-gon inside it sharing a side, then the innermost vertex of the 2k+1-gon will be the center of the 4k+2-gon. For example, you can add two 2k+1-gon so that they meet in the middle:

[szimmetria-airtemmizs]

Mathematical fun fact of the day:

If you draw a regular 4k+2-gon and add a regular 2k+1-gon inside it sharing a side, then the innermost vertex of the 2k+1-gon will be the center of the 4k+2-gon. For example, you can add two 2k+1-gon so that they meet in the middle:

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Here is a fun theorem: Consider 31 points evenly spaced on a circle, and color each of them arbitrarily blue or red. Then we can always find 5 points with the same color that divide the circle into arcs proportional to 1 : 2 : 4 : 8 : 16. The arcs need not be in the order suggested by the proportion. That is, 1 : 4 : 8 : 2 : 16 counts as a success!

One of my favorite open problem is whether the generalization of this holds:

Stromquist's conjecture: For any k>= 3, consider (2^k)-1 points evenly spaced on a circle, and color each of them arbitrarily blue or red. Then we can always find k points with the same color that divide the circle into arcs proportional to 1:2:4: ... :2^(k-1), but not necessarily in this order.

I managed to prove this up to k=7 using a computer. Can you push it further?

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A little infinite puzzle for you.

We have an infinite hotel, that is a hotel, where the rooms are numbered by 1, 2, 3, ...

In each room there is a guest. Nothing interesting happened for a while, so the guests got very bored. They decide that each guest moves to a new room simultaneously. Furthermore, they want to move in such a way that for any nonempty finite set of rooms S there is a guest who did not live in the rooms of S but now moves into one of them. Can they achieve this?

For example, the guests in 1 and 2 are not allowed to switch places as there would be no new guest in S={1,2}.

Note that the condition implies that no room is left empty, as we could pick S to be that single room.

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In the math camp I also received a marvelous gift. A large set of Penrose tiles made from wood!

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Last week I participated in a math camp, where I taught students about tilings. Here are some photos:

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Fun fact: The union of a yellow and an orange shape gives a spectre tile (the chiral aperiodic tile, see here: https://arxiv.org/pdf/2305.17743.pdf). Of course, when I made this drawing the idea of finding an aperiodic tile based on this didn't even enter my head.

A tiling based on the wheel tiling. (See here: https://tilings.math.uni-bielefeld.de/substitution/wheel-tiling/)

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Nice! Unfortunately, some of them have the same size, so this is not a solution to the original question. But still, very nice!

Did you know that there is a way of cutting an equilateral triangle into seven similar triangles that have pairwise different sizes, and they are not right triangles? (With right triangles it is easy to cut into any number of similar pieces.)

One of the angles in each triangle is 120 degrees and the other two are roughly 40.67915375798 and 19.32084624 degrees. You can read the details in the paper:

A note on perfect dissections of an equilateral triangle by Andrzej Zak