Arcadian Functor

occasional meanderings in physics' brave new world

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Marni D. Sheppeard

Friday, April 13, 2007

Calabi's Triangle

Calabi found a unique non-equilateral isosceles triangle for which the largest inscribed square can be placed in the triangle in three ways. MathWorld gives us a neat picture: I wonder what Calabi was thinking about when he figured this out (see his proof here). MathWorld also tells us about the Biggest Little Hexagon. It is fun to think of uses for these special polygonal geometries.

5 Comments:

Anonymous a quantum diaries survivor said...

Hey this is quite cool. The kind of thing that may have no relevance whatsoever with whatever else in geometry or mathematics, but still is "unique" and thus worth cataloging.

You made me remember a thought I had a week ago, while looking at my two kids playing on a see-saw.
Imagine the see-saw is perfectly balanced: two kids of uneven weight will unbalance it. Now, given a gaussian distribution of
children weights, of mean M and width W, what is the most profitable unbalance the see-saw
should have when unloaded, in order to minimize on average the
unbalance once two random kids play on it ?
The problem is easy to solve, but even easier it is to know that there is indeed a unique solution, for a given single-modal distribution of weights. Or at least, that is what my intuition tells me.

Cheers,
T.

April 13, 2007 10:57 AM  
Anonymous a quantum diaries survivor said...

hmmm after posting the above comment, I tried solving the problem and did not manage it so easily... It is one in the morning and I need to sleep. I'll work at it tomorrow unless some kind soul here does it for me...

By intuition (the only thing still working at this time in my brain) the most profitable bias in the seesaw balance is the average difference between two children's weights. Am I right ?

Cheers,
T.

April 13, 2007 11:09 AM  
Blogger Kea said...

Tommaso, have you taken into account that one doesn't know which kid will sit on which end of the see-saw? I'm afraid my enthusiasm for this problem is waning already, but LOL. Whatever you're taking, it's some trip.

April 13, 2007 11:40 AM  
Anonymous a quantum diaries survivor said...

Hi Kea,

oh, you theorists... We need to specify even the goriest detail lest you'll argue the solution is undetermined. Of course the kids will try to balance the overall weight the best they can!

If I have time today I will try to solve the case of a flat prior weight distribution.

Cheers,
T.

April 13, 2007 7:05 PM  
Blogger Matti Pitkanen said...

A comment to the previous posting. Could one take the factor 3 out of the inverse of the Fermi distribution and add to the exponent -log(3) identified as chemical potential.

Matti

April 13, 2007 10:34 PM  

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