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More triangles in circles


Best Answer BMAD, 30 March 2014 - 02:36 PM

Spoiler for I should have considered two fixed points

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#11 BMAD

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Posted 01 April 2014 - 12:03 AM

are we assuming a regular pentagon?


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#12 bonanova

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Posted 01 April 2014 - 03:35 AM

Yes. Basically I'm wondering whether the number of sides of the containing (regular) polygon matters.

I'm working on it, don't have the answer yet.


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#13 BMAD

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Posted 01 April 2014 - 04:42 AM

it would seem that the answer is that it matters since a pentagon can't easily be divided into four identical pieces.


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#14 plainglazed

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Posted 01 April 2014 - 04:12 PM

 

Spoiler for > too

 


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#15 bonanova

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Posted 01 April 2014 - 07:21 PM

 

Spoiler for > too

 

 

I think for triangles with vertices either inside or on a circle the quadrant analysis

applies and the result is the same. The only insight needed is to see that a line

connecting points in opposite quadrants can be on either side of center with equal likelihood.

 

I'm going to run simulations for other polygons (an equilateral-triangle containing

shape seems interesting for starters) to see whether the result holds in every case.

I started by dividing it into sixths and inspecting the pairs of opposing sixths.

 

It certainly holds for a square container.


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#16 BMAD

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Posted 01 April 2014 - 07:24 PM

I think polygons are different because they are not equal distance to the center. Which is important in my approach and really matters in the forming of possible triangles across the center
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#17 bonanova

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Posted 01 April 2014 - 07:36 PM

Possibly true so long as they do not have 4-fold rotational symmetry.

I should have the triangle container simulated soon.


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#18 BMAD

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Posted 01 April 2014 - 10:14 PM

Probably only shapes that can be properly circumscribed
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#19 bonanova

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Posted 02 April 2014 - 08:47 AM

Spoiler for Results for triangle


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#20 plasmid

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Posted 03 April 2014 - 06:17 AM

A general case could in principle be solvable with an integral, but in practice it would be so messy that simulation would still be more feasible.
Spoiler for
Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center.

Edited by plasmid, 03 April 2014 - 07:16 AM.

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