Last Time | Next Time |
If you think that you'd like to protest a change in climate in our world this Friday, something that matters to you, please join us! Talk to me, or email.
What should Danielle's political logo be? How about some Borromean rings with Integrity, Honesty, Devotion, or the like?
![]() |
![]() |
If you want to draw the Borromean rings, you'd draw three circles, as in the first figure above: | but you'd want to indicate, somehow, that one ring is below another ring (aka the "Irish Trinity"): |
Let's draw those. Kamora and I were drawing these yesterday, and saw that it's not too bad:
This is the most beautiful piece of mathematics I know:
It turns out that there are golden rectangles in the heart of an icosahedron, interlocked as Borromean rings!
We'll make some using 3x5 (or 4x6) cards -- which are not quite golden. Their corners are the 12 vertices of the icosahedron. The tricky part is locking them together!
But we can easily figure out which is which, because the knot is a single continuous piece of material, whereas the link is two separate pieces of material.
It's all about overs and unders!
![]() |
![]() |
![]() |
Type I | Type II | Type III |
The third images on that page stems from this picture from a recent Science issue:
![]() |
![]() |
In order to consider the picture on the left a knot, we have to know what its ends are doing. In the figure at right, I assumed that they are just connecting to each other in the simplest way.
The succession of steps then go on to show that the knot is actually an unknot! That's good news for the hyperbusy people!
Here's another picture of an unknot, which could trick you -- but knowing the R1 move saves you:
We'll just try a few with a string, to see what we can learn.
Reidemeister Move I is tricolorable. | Reidemeister Move II is tricolorable. | Reidemeister Move III is tricolorable. |
---|---|---|
So, for example: if you've got your picture of a knot down to three crossings, and it's not tricolorable, then it's the unknot.
Because the above written pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second montth, and thus there are in the second month 3 pairs; of these in one month two are pregnant and in the third month 2 pairs of rabbits are born, and thus there are 5 pairs in the month; ...
there will be 144 pairs in this [the tenth] month; to these are added again the 89 pairs that are born in the eleventh month; there will be 233 pairs in this month.
To these are still added the 144 pairs that are born in the last month; there will be 377 pairs, and this many pairs are produced from the abovewritten pair in the mentioned place at the end of the one year.
You can indeed see in the margin how we operated, namely that we added the first number to the second, namely the 1 to the 2, and the second to the third, and the third to the fourth and the fourth to the fifth, and thus one after another until we added the tenth to the eleventh, namely the 144 to the 233, and we had the abovewritten sum of rabbits, namely 377, and thus you can in order find it for an unending number of months.