@Priyaranjan: Suppose that the cube has side 1 and is placed in
"standard position at the origin." Then the triangle with vertices
(0,0,0), (1,0,1), and (1,1,0) is equilateral with side sqrt(2), and
therefore is not a right triangle.

Dave

On Feb 24, 8:49 am, Vikas Kumar <dev.vika...@gmail.com> wrote:
> @priyaranjan
> No,Your triangle will be right angle triangle.
>
> In fact any triangle chosen will be right angle triangle.
>
> Proof:
>
> suppose there are 2 planes kept at parallel with each other.(top bottom)
> join corresponding vertices of up to down and form  cube.
>
> Now 3 points chosen can be on same plane .(means 2 must be adjacent)
> so right angle triangle is only possible
>
> or at least 2 points are on same plane and 3rd point is other plane(pigeon
> hole as only 2 holes are there for 3 pegions)
>
> Now as 3 rd point always lies on perpendicular point so triagle formed will
> always be right angled.
>
> On Thu, Feb 24, 2011 at 5:23 PM, awesomeandroid
> <priyaranjan....@gmail.com>wrote:
>
>
>
> > The only way to get an acute triangle this way is to connect the
> > diagonals of three adjacent faces. You can select 3 adjacent faces of
> > a cube in (6*4*2)/(3*2*1) = 8 different ways.
>
> > Regards
> > Priyaranjan
> >http://code-forum.blogspot.com
>
> > On Feb 23, 12:10 am, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> > > answer is 8 ,,, duno how
>
> > > On Wed, Feb 23, 2011 at 12:31 AM, Sundi <sundi...@gmail.com> wrote:
> > > > Is it not 8C3 = 56
>
> > > > On Feb 22, 9:55 pm, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> > > > > Think of the 8 vertices of a given cube. You are allowed to join
> > three
> > > > > vertices to form a triangle. How many such unique acute triangles can
> > you
> > > > > make ??
>
> > > > > --
> > > > > With Regards,
> > > > > *Jalaj Jaiswal* (+919019947895)
> > > > > Software developer, Cisco Systems
> > > > > B.Tech IIIT ALLAHABAD
>
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> > > Software developer, Cisco Systems
> > > B.Tech IIIT ALLAHABAD
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