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From: Alois Steindl on 7 Jul 2010 07:02
Fred Nurk <albert.xtheunknown0(a)gmail.com> writes:

> Alois Steindl wrote:
>> <snip>
>> If you look at the Tetrahedron from above, where do you think that you
>> see the top vertex?
>> <snip>
>
> At the centre of the base triangle.
How far ist that from the edge?
(You may try to draw it and compare the measurement with your
calculation.)


Alois
From: Rob Johnson on 7 Jul 2010 07:50
In article <ZIYYn.298$FH2.87(a)viwinnwfe02.internal.bigpond.com>,
Fred Nurk <albert.xtheunknown0(a)gmail.com> wrote:
>Alois Steindl wrote:
>> <snip>
>> If you look at the Tetrahedron from above, where do you think that you
>> see the top vertex?
>> <snip>
>
>At the centre of the base triangle.

Which center?

<http://en.wikipedia.org/wiki/Encyclopedia_of_Triangle_Centers>

For an equilateral triangle, many of these notions of center yield
the same point. The description that seems to apply best to this
situation is the circumcenter, the center of the circle which passes
through all three vertices.

In this case, the positions of the vertices relative to the center
are, with proper rotation, (-1/2,sqrt(3)/2), (-1/2,-sqrt(3)/2), and
(1,0), and the edges have length sqrt(3). Where does the vertex lie
that is not in the base plane?

Rob Johnson <rob(a)trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
From: Frederick Williams on 7 Jul 2010 09:09
Fred Nurk wrote:
>
> VABC is a regular tetrahedron with base triangle ABC. (All faces are
> equilateral triangles.) Find the magnitude of the angle between a sloping
> edge and the base.
>
> Let x be the length of any edge.
> Why doesn't acos{[(x ^ 2 - (x / 2) ^ 2] ^ (1 / 2) / x} yield the right
> answer?

It can't depend on x. A regular tetrahedron with side length 1 cm will
have the same angles as one with side length 1 m.

--
I can't go on, I'll go on.
From: Alois Steindl on 7 Jul 2010 10:29
Frederick Williams <frederick.williams2(a)tesco.net> writes:

> Fred Nurk wrote:
>>
>> VABC is a regular tetrahedron with base triangle ABC. (All faces are
>> equilateral triangles.) Find the magnitude of the angle between a sloping
>> edge and the base.
>>
>> Let x be the length of any edge.
>> Why doesn't acos{[(x ^ 2 - (x / 2) ^ 2] ^ (1 / 2) / x} yield the right
>> answer?
>
> It can't depend on x. A regular tetrahedron with side length 1 cm will
> have the same angles as one with side length 1 m.
The x cancels,
but it seems to me that he used some wrong length (the height of the
base triangle instead of the distance of the base point of the top edge
to some edge on the base.
Alois
From: Jim Heckman on 7 Jul 2010 18:50

On 7-Jul-2010, Fred Nurk <albert.xtheunknown0(a)gmail.com>
wrote in message <E1YYn.292$FH2.172(a)viwinnwfe02.internal.bigpond.com>:

> VABC is a regular tetrahedron with base triangle ABC. (All faces are
> equilateral triangles.) Find the magnitude of the angle between a sloping
> edge and the base.
>
> Let x be the length of any edge.
> Why doesn't acos{[(x ^ 2 - (x / 2) ^ 2] ^ (1 / 2) / x} yield the right
> answer?

I have no idea, but why are you even using the length of an edge in
the first place?

Hint: Clearly the sum of the vectors from the center of the
tetrahedron to each of its four vertices is 0. What does this tell
you about the angle subtended at the center by two vertices? And
what does that in turn tell you about the angle you're looking for?

--
Jim Heckman
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