1. ## 3D Geometry Problem

for some reason, i just cannot seem to solve this problem. Does anyone know a way to solve it and maybe even draw a diagram for it? i can't even picture this image let alone draw it out.

Through each edge of a cube, draw outside the cube the planes making 45 degree angles with the adjacent faces. Compute the surface area of the polyhedron bounded by these planes, assuming that the edges of the cube have length a. Is this polyhedron a prism?

2. The polyhedron you create by that process is kind of hard to visualize. But if you look at just one face of the cube, you have a square pyramid.

The surface area of the pyramid (not including the base) is 4 times the area of one triangle. The area of a triangle is $\displaystyle \frac{1}{2}bh$, where b (the base) is just a, and you can calculate h from the fact that the angle at the midpoint of a side is 45 degrees. The midpoint of a side, the center of the square base, and the apex of the pyramid form a 45-45-90 triangle, so you can calculate h.

So you have the area $\displaystyle \frac{1}{2}bh$ of 1 triangle times 4 to give the surface area of the pyramid (minus its base), times 6 faces on a cube gives you the surface area of the whole polyhedron.

If you can imagine 4 rhombi meeting at a vertex pointing straight up, there are 4 points where 2 rhombi meet and 4 points where there is only 1 rhombus. Now connect 4 more rhombi, oriented vertically, with the "upper" level of vertices now finished with 3 rhombi and the "middle" level also having 3 rhombi, but the smaller angles. Now make another group of 4 rhombi meeting at a vertex pointing straight down, and it connects so that the "middle level" is complete with 4 rhombi meeting at a vertex, and the vertically-oriented rhombi fit into the "lower level", with 3 rhombi per vertex.

Post again if you are still having trouble.

3. Hello mathwizard325
Originally Posted by mathwizard325
for some reason, i just cannot seem to solve this problem. Does anyone know a way to solve it and maybe even draw a diagram for it? i can't even picture this image let alone draw it out.

Through each edge of a cube, draw outside the cube the planes making 45 degree angles with the adjacent faces. Compute the surface area of the polyhedron bounded by these planes, assuming that the edges of the cube have length a. Is this polyhedron a prism?
The polyhedron is a regular octahedron, which has 8 equilateral triangles as its faces. If you join the mid-points of its edges you get a cube - see my attempt at drawing one!

I think it's pretty obvious what the relation is between the lengths of the edges of the two respective solids, and therefore you should have little difficulty in working out the surface area of the octahedron in terms of $\displaystyle a$, the length of the edge of the cube.

4. Cool picture! How do you generate something like that?

But you do not have a plane going through the four vertical edges. The pyramid on your top and bottom faces should be on the four sides, too.

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6. Originally Posted by mathwizard325
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Through each edge of a cube, draw outside the cube the planes making 45 degree angles with the adjacent faces. Compute the surface area of the polyhedron bounded by these planes, assuming that the edges of the cube have length a. Is this polyhedron a prism?
The cube has 12 edges thus the new polyhedron must have 12 faces.

If the surface planes of the new polyhedron and the faces of the cube include an angle of 45° the face of the new polyhedron consist of 12 rhombii with the side length $\displaystyle s=\frac a2 \sqrt{3}$

I've attached a sketch of only 3 faces of the new polyhedron. The heights of the added pyramids (drawn in red) have equal length.

EDIT: The 2nd sketch is a view of the complete solid.

7. Originally Posted by earboth
The cube has 12 edges thus the new polyhedron must have 12 faces.

If the surface planes of the new polyhedron and the faces of the cube include an angle of 45° the face of the new polyhedron consist of 12 rhombii with the side length $\displaystyle s=\frac a2 \sqrt{3}$

I've attached a sketch of only 3 faces of the new polyhedron. The heights of the added pyramids (drawn in red) have equal length.

EDIT: The 2nd sketch is a view of the complete solid.
Quite right! My answer was complete nonsense! Not one of my better efforts!

If the surface planes of the new polyhedron and the faces of the cube include an angle of 45° the face of the new polyhedron consist of 12 rhombii with the side length $\displaystyle s=\frac a2 \sqrt{3}$