prob. of equolateral triangle

juantheron · 2011-12-14 05:59:48

from 8 vertex of a cube , 3 vertices are choosen at random. then find probability that these 3 vertices form an equilateral triangle

bobbym · 2011-12-14 06:33:04

Hi;

anonimnystefy · 2011-12-14 06:46:37

hi bobbym

i got that too

juantheron · 2011-12-14 06:49:08

thanks bobbym

bobbym can you explain it

bobbym · 2011-12-14 06:49:31

Hi;

Very good! I am glad at least someone else got the same answer.

Yes I can explain it. I broke the cardinal rule of combinatorics. I am sure that I will be hung from the neck until I am dead. Anyway, until that time here are all possible triangle distances made from a unit cube.

and

There are 56 and 8 of them are equilateral.

juantheron · 2011-12-14 07:20:27

great bobbym

as i have seen the answer, I am surprised how you count all that things

bobbym · 2011-12-14 07:23:07

Hi;

My computer did the counting. I wrote a small program to give the solutions. Do you have a method?

juantheron · 2011-12-14 07:37:25

no bobby actually i am sitting in a room , and in my room i have count it

bobbym · 2011-12-14 07:47:25

Hi;

It is easier if you take this example straight from a book. Give some time to post it.

TheDude · 2011-12-14 07:49:42

You can also solve it probabilistically. There are three possibilities for the length of a line segment connecting any two vertices of a unit cube: 1, √2, and √3. Line segments of length √3 only occur between vertices that are on opposite corners of the cube and pass through the center of the cube, so you cannot form a triangle solely with these lines. Similarly, you cannot form a triangle using line segments of length 1 since they all form 90 degree angles with each other. So an equilateral triangle must be formed out of line segments of length √2. These occur between pairs of vertices that are on opposite corners of a single side of the cube and the line segment passes through the center of that side.

WLOG assume that the first vertex chosen is the front-bottom-left vertex of the cube. Of the seven remaining vertices, three are length √2 from the first vertex: the front-top-right vertex, the back-bottom-right vertex, and the left-back-top vertex. Note that all four of these vertices are length √2 from each other, and no other vertices are √2 from any of these. So the probability that the next two vertices are from this group is 3/7 * 2/6 = 1/7.

Last edited by TheDude (2011-12-14 07:51:01)

bobbym · 2011-12-14 07:59:20

Hi;

Starting with vertex 1 there are ( 7 * 6) / 2 = 21 ways to select two more vertices from there. There are 1,3,5 and 1,5,7 and 1,3,7 that are all equilateral because they are all the diagonals of the three faces. There are no other possibilities and 3 / 21 = 1 / 7.

juantheron · 2011-12-14 08:03:02

thanks dude and bobbym

bobbym why we take total = 21

bobbym · 2011-12-14 08:09:21

There are (7 * 6) / 2 vertices left after you pick 1. There is no loss of generality in picking 1.

Math Is Fun Forum

#1 2011-12-14 05:59:48

prob. of equolateral triangle

#2 2011-12-14 06:33:04

Re: prob. of equolateral triangle

#3 2011-12-14 06:46:37

Re: prob. of equolateral triangle

#4 2011-12-14 06:49:08

Re: prob. of equolateral triangle

#5 2011-12-14 06:49:31

Re: prob. of equolateral triangle

#6 2011-12-14 07:20:27

Re: prob. of equolateral triangle

#7 2011-12-14 07:23:07

Re: prob. of equolateral triangle

#8 2011-12-14 07:37:25

Re: prob. of equolateral triangle

#9 2011-12-14 07:47:25

Re: prob. of equolateral triangle

#10 2011-12-14 07:49:42

Re: prob. of equolateral triangle

#11 2011-12-14 07:59:20

Re: prob. of equolateral triangle

#12 2011-12-14 08:03:02

Re: prob. of equolateral triangle

#13 2011-12-14 08:09:21

Re: prob. of equolateral triangle

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