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anonimnystefy
2012-11-13 05:06:52

Those are the ones. You are welcome.

anna_gg
2012-11-13 00:16:57

Yessss!! And there are only 3 such planes, because for every two sets of sides we get the same plane.
Thanks Stefy!!

bob bundy
2012-11-12 23:54:33

I think Stefy means this.

Thanks Stefy.

Bob

anna_gg
2012-11-12 22:16:33

#### anonimnystefy wrote:

Nope! There are three more, one for each pair of midpoints of opposite edges of the tetrahedron.

So 4 + 3 = 7 such planes?
Can you make a drawing for the 3 last planes?

Thanks!!

anonimnystefy
2012-11-12 22:11:34

Nope! There are three more, one for each pair of midpoints of opposite edges of the tetrahedron.

bob bundy
2012-11-12 21:50:30

Four would be my thinking too.

Bob

anna_gg
2012-11-12 20:03:05

Bob, I think this is it, although I do not understand your second drawing.

So I guess we have four such planes, correct?

Thanks,
Katerina

bob bundy
2012-11-12 05:21:52

Oh right.  Sorry I misunderstood.

In that case a horizontal plane half way up the height will do it.  (see picture)

The distance of a point from a plane has to be the perpendicular distance.  In the second part of the picture, BC not BA.

So all the points on the base are the same distance from the plane and, as it is half way up, so is the top.

Will that do?

Bob

anna_gg
2012-11-12 04:11:41

#### bob bundy wrote:

hi anna_gg

I have, of course, assumed that the triangular pyramid is regular; ie. all its faces are equilateral triangles.

If it is 'sitting' on one face with one vertex in the air, then, a line drawn straight down from that vertex to the base, will go through the centre of gravity (centroid) of the base.  Let's call that line (don't know the proper term for it) an axis.  Turn the pyramid so that a different vertex is on top and repeat the construction.  This new axis will intersect the first.  You can do this four times altogether and get four axes, all intersecting at the same point.  This point is the centre of gravity of the pyramid and it is the same distance from all four vertices.

Furthermore it is the only point that is the same distance from all four vertices.  But you want a vertical plane.

Ok let's try this.  I'll find a line in the base (ABC)  that is the same distance from all three base vertices.  Then make a vertical plane going upwards from this line.  See picture.  The base is ABC.  The line is shown dotted.  It's distance is the same from all three base vertices.

This plane is, therefore, (i) vertical  (ii) the same distance from all three base vertices.

Now suppose we are looking straight down from above on the pyramid.  Let D be the last vertex.  You can see it will be much closer to the plane.

Conclusion.  I think it is not possible to create the plane you want.

Bob

Just to clarify, I did not say that the plane must be vertical. I only clarified that the (vertical) distances of each vertex from this plane must be equal. Sorry for the confusion!

bob bundy
2012-11-12 01:18:24

hi anna_gg

I have, of course, assumed that the triangular pyramid is regular; ie. all its faces are equilateral triangles.

If it is 'sitting' on one face with one vertex in the air, then, a line drawn straight down from that vertex to the base, will go through the centre of gravity (centroid) of the base.  Let's call that line (don't know the proper term for it) an axis.  Turn the pyramid so that a different vertex is on top and repeat the construction.  This new axis will intersect the first.  You can do this four times altogether and get four axes, all intersecting at the same point.  This point is the centre of gravity of the pyramid and it is the same distance from all four vertices.

Furthermore it is the only point that is the same distance from all four vertices.  But you want a vertical plane.

Ok let's try this.  I'll find a line in the base (ABC)  that is the same distance from all three base vertices.  Then make a vertical plane going upwards from this line.  See picture.  The base is ABC.  The line is shown dotted.  It's distance is the same from all three base vertices.

This plane is, therefore, (i) vertical  (ii) the same distance from all three base vertices.

Now suppose we are looking straight down from above on the pyramid.  Let D be the last vertex.  You can see it will be much closer to the plane.

Conclusion.  I think it is not possible to create the plane you want.

Bob

anna_gg
2012-11-11 23:04:02

#### bob bundy wrote:

hi anna_gg

The picture I suggested goes through two vertices and the midpoint of the side joining the other vertices.

So it is the same distance from only two of the vertices, and it is a plane of symmetry.

There is one point only that is the same distance from all four vertices, (see picture below), so you could draw any plane that all that goes through this point.  It would not be a plane of symmetry.

Bob

Hi Bob,
Actually we are looking for a plane that has equal (vertical, obviously) distance from ALL FOUR vertices, so I guess this plane must not pass from any of the sides (because in this case, for the 2 of them, distance will be zero).

Anna

bob bundy
2012-10-22 22:55:59

hi anna_gg

The picture I suggested goes through two vertices and the midpoint of the side joining the other vertices.

So it is the same distance from only two of the vertices, and it is a plane of symmetry.

There is one point only that is the same distance from all four vertices, (see picture below), so you could draw any plane that all that goes through this point.  It would not be a plane of symmetry.

Bob

anna_gg
2012-10-22 22:12:27

#### bob bundy wrote:

hi anna-gg

See picture below.  Is that what you are after?

Don't understand your word "apices"  .

Did you mean vertices ?

Bob

Yes, vertices

bob bundy
2012-10-22 21:41:53

hi anna-gg

See picture below.  Is that what you are after?

Don't understand your word "apices"  .

Did you mean vertices ?

Bob

anna_gg
2012-10-22 20:20:36

The question is to find a plane that "cuts" (passes through) the triangular pyramid and which has equal distance from all its apices.