06 Jul '06 01:59

given a cube can you take a cross-section that is a regular pentagon.

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06 Jul '06 10:37

aginis asked:*Originally posted by aginis***why do you say that?**

"given a cube can you take a cross-section that is a regular pentagon."

I answered:

"No, but a hexagon."

"No", because a cross-section cannot give a pentagon.

"but a hexagon." because there is a hexagon to be found if you make a cross-section in the right way.

You asked:

"why do you say that?"

And I answer:

That's why I said that.

Am I wrong?- Joined
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06 Jul '06 11:041 edit

if your cube is is a cube of length one anchored at (0,0,0) and (1,1,1)*Originally posted by FabianFnas***aginis asked:**

"given a cube can you take a cross-section that is a regular pentagon."

I answered:

"No, but a hexagon."

"No", because a cross-section cannot give a pentagon.

"but a hexagon." because there is a hexagon to be found if you make a cross-section in the right way.

You asked:

"why do you say that?"

And I answer:

That's why I said that.

Am I wrong?

then the cross-section formed by the plane passing through (0,0.5,1) (0.5, 1, 1) and (1,0,0) will be a pentagon, therefore your solution is incorrect.

(although i agree that the cross-section defined by (0,0.5,1) (0.5, 1, 1) and (1,0.5,0) is a regular hexagon)

please note that i asked for a REGULAR pentagon

A regular polygon is an n-sided polygon in which the sides are all the same length and are symmetrically placed about a common center (i.e., the polygon is both equiangular and equilateral).

http://mathworld.wolfram.com/RegularPolygon.html- Joined
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06 Jul '06 11:43

A hexagon, i knew, a pentagon I didn't know.*Originally posted by aginis***if your cube is is a cube of length one anchored at (0,0,0) and (1,1,1)**

then the cross-section formed by the plane passing through (0,0.5,1) (0.5, 1, 1) and (1,0,0) will be a pentagon, therefore your solution is incorrect.

(although i agree that the cross-section defined by (0,0.5,1) (0.5, 1, 1) and (1,0.5,0) is a regular hexagon)

please note that i as ...[text shortened]... olygon is both equiangular and equilateral).

http://mathworld.wolfram.com/RegularPolygon.html

I have to get home and try myself, cutting a cheese or turnip or something.

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06 Jul '06 16:41

But FabianFnas -- the example aginis gives is not a regular pentagon.*Originally posted by FabianFnas***A hexagon, i knew, a pentagon I didn't know.**

I have to get home and try myself, cutting a cheese or turnip or something.

Very fascinating, indeed...

Doesn't aginis want a regular pentagon?

I would guess it's not possible to slice out a regular pentagon, but have no proof.

Of course the regular hexagon is easy -- you just slice perpendicular through the midpoint of a line joining opposite vertices.- Joined
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06 Jul '06 17:37

A regular pentagon, isn't that a polygon with 5 edges equal in length, and 5 angles equal in degrees?*Originally posted by SPMars***But FabianFnas -- the example aginis gives is not a regular pentagon.**

Doesn't aginis want a regular pentagon?

I would guess it's not possible to slice out a regular pentagon, but have no proof.

Of course the regular hexagon is easy -- you just slice perpendicular through the midpoint of a line joining opposite vertices.

I've tried to cut the d*rn cheese to get a pentagon but have not succeeded.

I'm about to think that it is impossible to cut a pentagon out of a cube.

A hexagon on the other hand ... easy.

I like cheese, especially the holes.- Joined
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06 Jul '06 18:141 edit

cut as if you want to make a regular hexagon but tilt the plane up so that it misses the bottom of the cube.*Originally posted by FabianFnas***A regular pentagon, isn't that a polygon with 5 edges equal in length, and 5 angles equal in degrees?**

I've tried to cut the d*rn cheese to get a pentagon but have not succeeded.

I'm about to think that it is impossible to cut a pentagon out of a cube.

A hexagon on the other hand ... easy.

I like cheese, especially the holes.

i like cheese too especially the cheese

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