 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Triangular Cubes (Posted on 2004-12-04) Three points have been chosen randomly from the vertices of a cube. What is the probability that they form (a) an acute triangle; (b) a right triangle?

 See The Solution Submitted by Erik O. Rating: 3.0000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) The way I did it (statistical) | Comment 6 of 11 | I started by thinking about the total number of triangles:  8 choose 3, or 56.  That seemed too hard.

So I focused on what types of triangles were possible:
a) all 3 vertices on the same face, ie sides formed of 2 edges and a face diagonal, ie side lengths of 1 1 and sqrt(2) right triangle.

b) 2 of the vertices are opposite corners of the cube, ie sides formed of an edge, a face diagonal, and a "thru the air" diagonal, ie side lengths of 1 sqrt(2) and sqrt(3),   again right triangle

c) all 3 edges are face diagonals, ie sides are all sqrt(2) in length; so type "c" triangles are equilateral, ie acute.

OK, so statistically, say one vertex is already chosen.  To get an equilateral triangle, the other 2 vertices have to be along a face diagonal from my first point.  There are 3 such face diagonals going away from my first point.  So there are 3 vertices that are acceptable.  I have a total of 7 vertices to choose from.  To get an equilateral, the probablity of success (if I were picking at random) would be:  3/7 * 2/6 = 1/7.
 Posted by Larry on 2004-12-06 03:23:49 Please log in:

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