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bonanova
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It's trivial to ask the size of the largest point that will will fit on a unit line segment.

And it's not much harder to determine the largest line segment that will fit in a unit square.

So let's ask instead: What is the largest square that will fit in a unit cube?

The largest hexagon?

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It's trivial to ask the size of the largest point that will will fit on a unit line segment.

And it's not much harder to determine the largest line segment that will fit in a unit square.

So let's ask instead: What is the largest square that will fit in a unit cube?

The largest hexagon?

I'd say the largest square is one side of the cube?

The largest hexagon... sorry, maths down now.

Edited by bonanova
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I think this is optimal for the square:

The best I get has area 9/8 and edge length .75*sqrt(2) as follows: From a corner on one face place two corners of the square a distance d away from that corner along the two edges of that face that form the corner. So, the edge length will be d*sqrt(2). Next we go to the corner spatially diagonal to the corner we started from and place the two corners for the square d away from the cube corner on the face opposite our starting face. We calculate the spatial edge of the square to be sqrt(2*(1-d)^2+1) because the two spatial end points (The way I places the square in the coordinate system) are ((1,1,d) and (d,0,1). So, to be a square, the spatial edges must equal the facial edges. So, d*sqrt(2)=sqrt(2*(1-d)^2+1). solving for d, we get .75. The square, then, has edge .75*sqrt(2). The area is 9/8.

Edited by superprismatic
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I think this is optimal for the square:

The best I get has area 9/8 and edge length .75*sqrt(2) as follows: From a corner on one face place two corners of the square a distance d away from that corner along the two edges of that face that form the corner. So, the edge length will be d*sqrt(2). Next we go to the corner spatially diagonal to the corner we started from and place the two corners for the square d away from the cube corner on the face opposite our starting face. We calculate the spatial edge of the square to be sqrt(2*(1-d)^2+1) because the two spatial end points (The way I places the square in the coordinate system) are ((1,1,d) and (d,0,1). So, to be a square, the spatial edges must equal the facial edges. So, d*sqrt(2)=sqrt(2*(1-d)^2+1). solving for d, we get .75. The square, then, has edge .75*sqrt(2). The area is 9/8.

superprism has it. Nice. B))

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