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 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?

 Submitted by Erik O. Rating: 3.0000 (2 votes) Solution: (Hide) The first step in solving this problem is to figure out how many different triangles can be made using the vertices of a cube. Since there are 8 vertices, the first point can be picked from one of 8 vertices, the second point can be picked from the 7 remaining vertices, and the last point of the triangle can be picked from the 6 remaining vertices. There are, therefore, 8𡥚 = 336 ways to draw a triangle on (or inside) a cube, when you disregard the sequence of choosing points for the triangles. Taking the point sequence into account, we can divide 336 by 6 to come up with 56 unique triangles. Of those 56 triangles, only 8 are regular triangles. To draw a regular triangle in a cube, the first two points must fall on opposing diagonals of a single face. The last point is taken from one of the two points on the opposite face on a corner that does not share an edge with the first two points. The remaining triangles are all right triangles. Therefore the probability of getting a regular triangle is 8/56 = 1/7 = 0.14286..., and the probability of getting a right triangle is 48/56 = 6/7 = 0.85714...

 Subject Author Date Triangulars cubes REVISITED Ady TZIDON 2010-08-16 14:02:38 Answer K Sengupta 2008-03-15 03:42:16 re(2): i c it this way Ady TZIDON 2004-12-07 09:41:36 re: Same answer, different approach owl 2004-12-06 15:01:01 Same answer, different approach nikki 2004-12-06 14:16:01 The way I did it (statistical) Larry 2004-12-06 03:23:49 re: i c it this way Charlie 2004-12-05 15:15:44 i c it this way Ady TZIDON 2004-12-05 07:53:06 Geometry... It's been too long Dustin 2004-12-05 02:58:48 Solution Steve Herman 2004-12-04 17:25:13 Solution Bractals 2004-12-04 17:21:28

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