Cross Section Length of a Pentagon

[sjohnson89]

if we have a cube with side lengths of 2 inches each, what would have to be the side lengths of a regular pentagon fitted inside the cube (the cross section) Are there any formulas for this specific problem and where would we start first?

 

[pka]

if we have a cube with side lengths of 2 inches each, what would have to be the side lengths of a regular pentagon fitted inside the cube (the cross section) Are there any formulas for this specific problem and where would we start first?

What in the world does that mean? Have you tried to inscribe a regular pentagon in a cube?
If inscribe is the wrong word then what does "fitted in" mean?

 

[mmm4444bot]

Cross Section Length of a Pentagon ...

inside the cube (the cross section) ...

I don't understand 'cross-section of a pentagon', but the cross-section of a cube is a square -- as long as the cube is cut along a plane parallel to one of its faces.

Are you talking about a pentagon inscribed within a 2-inch square?

?

 

[sjohnson89]

I don't understand 'cross-section of a pentagon', but the cross-section of a cube is a square -- as long as the cube is cut along a plane parallel to one of its faces.

Are you talking about a pentagon inscribed within a 2-inch square?

?

What in the world does that mean? Have you tried to inscribe a regular pentagon in a cube?
If inscribe is the wrong word then what does "fitted in" mean?

Sorry I didn’t explain it that well but I think the image represents what I’m trying to show. If we have a pentagon inside the cube with side lengths of 2 inches each, what would be the side lengths for the pentagon if it had to be a regular pentagon.

 

[Dr.Peterson]

Why do you think it is possible for this to be a regular pentagon?

The site from which you appear to have taken the image says it is irregular.

You could describe the figure in terms of two variables locating the intersections with sides, and solve to make the sides of the "pentagon" equal, but you would also have to show that it is planar, and that the angles are equal. There are too many conditions.

 

[tfmcbride]

Since the edge of a regular pentagon inscribed in a unit cube is approximately 0.77437 76951 36495 67221 40842 57670 14333 23639, your answer is just twice this number.

The question concerning regular polygons inscribed in a cube is quite challenging. It's obvious how to place the largest equilateral triangle inside a cube and the larger possible hexagon is also well known. The placement for the largest square is not so obvious but the length of the edge of the largest square in a unit cube is 1.060660172=(sqrt(2*0.75^2)). The largest pentagon lies as shown in the diagram. Three of the edges lie in three of the cube faces; the other two edges are inside the cube and their intersect is on the top face of the cube. The center of this pentagon does not coincide with the center of the cube. As far as I know, the size of regular polygons with more than 6 edges that can be inscribed in a cube have not been calculated.

 

[tfmcbride]

Oops. There is a typo in my previous response. The largest square that can be inscribed in a unit cube is 1.06066017 but the formula should have been sqrt(2*0.75²)....(corrected)