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**Math Student****Guest**

Hi, I've worked these amounts of planes, but I'm not sure if they are correct, could you help me confirm them?

Cuboid - 3

Triangular Prism - 4

Square-based Pyramid - 4

Cylinder - 2

Tetrahedron - 3

Thanks you!

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

I'm not sure I quite understand what is meant by "plane of symmetry." If it is just "a plane that passes through the object which cuts it into two symmetrical pieces," then a cube would have an infinite number of such planes. Just pass a vertical plane through the center, and rotate it by infinitesimal amounts. It will always be symmetrical, regardless of the angle!

El que pega primero pega dos veces.

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**Math Student****Guest**

No, that isn't right, pretend the plane is a mirror line, if you just turn it then you will not always get a cube.

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

For a cube (not a cuboid) there would be the x, y and z planes, but also the planes that pass through the corner points, am I right?

I am not referring to anything other than my imagination, but here goes:

Cuboid (not cube): 3 seems right

Triangular Prism: the cross-section in the middle, and if it is regular then each symmetrical plane of the triangle. 1+3=4

Square-based Pyramid - x,y,x=y,x=-y, so 4

Cylinder - the cross-section in the middle, plus if it is a circular cylinder wouldn't there be infinite planes thru the circle?

Tetrahedron - you could run a plane through one edge and bisect the face opposite. A tetrahedron has 6 edges, so 6.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Math Student is right. In the same way that squares only have 4 lines of symmetry, cubes will have a finite amount as well.

However, ryos's reasoning is valid for the cylinder, so it would have infinite planes.

I'm not entirely sure on this, but I think a tetrahedron would have 6.

Edit: Rarrrrggh.

Why did the vector cross the road?

It wanted to be normal.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

I agree, except that I believe that a cuboid may have 9 planes of symmetry...

3 that run perpendicular to the sides (in each direction)

and then 6 that cut through the diagonal for each side.

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

Oh, yes!

This is an interesting mental exercise ... I guess it depends on the cuboid ...

If the cuboid is a **cube** (all faces square) then any face can be cut diagonally giving 2 × 3 (x,y,z) = 6, and then there are the 3 cross sections parallel to faces, giving a total of 9 planes

At the other extreme, if no face is square, then the diagonals won't work, giving only 3 planes.

Maybe ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's right. And in the middle extreme, with a square prism, you can get 5 planes.

But a cuboid is defined as having rectangular faces. They can also be squares, but they don't have to be. So, Math Student's original answer of 3 is correct.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

"Middle extreme" ... good concept

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**zena****Guest**

well i found out a formula that helps oyu determin the planes of symmerty of any 3d shape here its n+1 which means the number of sides of the 2d surface area +1

**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

hi anna-gg

See picture below. Is that what you are after?

Don't understand your word "apices" .

Did you mean vertices ?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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!

*Last edited by anna_gg (2012-11-11 05:13:07)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

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

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

Four would be my thinking too.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

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

Here lies the reader who will never open this book. He is forever dead.

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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!!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,097

I think Stefy means this.

Thanks Stefy.

Bob

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 105

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

Thanks Stefy!!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,813

Those are the ones. You are welcome.

Here lies the reader who will never open this book. He is forever dead.

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