Wikipedia:Reference desk/Archives/Mathematics/2022 September 11
| Mathematics desk | ||
|---|---|---|
| < September 10 | << Aug | September | Oct >> | Current desk > |
| Welcome to the Wikipedia Mathematics Reference Desk Archives |
|---|
| The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
September 11[edit]
Tetrahedron inside an octagon[edit]
About a week ago I posted a question to the Octagon talk page, here, concerning whether the eight sides of an irregular octagon could subtend a regular tetrahedron. Could anyone help please? Thank you. Sandbh (talk) 06:50, 11 September 2022 (UTC)
- An octagon is by definition a polygon, which is by definition a plane figure. Any figure it subtends, including the "internal" quadrilateral, remains in the plane and is not three-dimensional. A tetrahedron is by definition a polyhedron, which is by definition a three-dimensional figure. --Lambiam 11:57, 11 September 2022 (UTC)
- Are you asking whether a tetrahedron can cast an octagonal shadow? No, a shadow cannot have more vertices than the body casting it. —Tamfang (talk) 01:20, 12 September 2022 (UTC)
- Thank you Lambian and Tamfang. I've now included the picture of the irregular octagon. The green quadrilateral looks to me like the 2D skeleton of a tetrahedron. Could it be construed as that of a regular tetrahedron? Sandbh (talk) 01:28, 12 September 2022 (UTC)
- The quadrilateral can be viewed as the projection of a tetrahedron. Confining ourselves to the orthographic projection that maps point in three-dimensional space to the point in the plane, I think it is not possible in general to construct a regular tetrahedron of which the quadrilateral is the projection. The positions of the 4 vertices of the quadrilateral are each determined by 2 coordinates, giving 8 equations to be satisfied. A regular tetrahedron in three-dimensional space has 7 degrees of freedom, which means we have only 7 unknown variables we can use to construct it. One degree of freedom is even useless, since the tetrahedron can be shifted "vertically" without altering the projection, so there are only 6 "usable" degrees of freedom, in general not enough to solve the system of equations. --Lambiam 06:58, 12 September 2022 (UTC)
- And since the two marked diagonals must be equal, the corresponding opposite edges of the projected regular tetrahedron would both have to be at the same angle to the plane. Having rotated the first edge to match the direction on the plane (one degree of freedom), the angle of the first edge away from the plane is set (one degree of freedom), then the angle of the second has to match it by rotating around the first edge (no freedom, but allow for reflection). Clumsily explained, but I think it only has two degrees of freedom. -- Verbarson talkedits 09:34, 12 September 2022 (UTC)
- The quadrilateral can be viewed as the projection of a tetrahedron. Confining ourselves to the orthographic projection that maps point in three-dimensional space to the point in the plane, I think it is not possible in general to construct a regular tetrahedron of which the quadrilateral is the projection. The positions of the 4 vertices of the quadrilateral are each determined by 2 coordinates, giving 8 equations to be satisfied. A regular tetrahedron in three-dimensional space has 7 degrees of freedom, which means we have only 7 unknown variables we can use to construct it. One degree of freedom is even useless, since the tetrahedron can be shifted "vertically" without altering the projection, so there are only 6 "usable" degrees of freedom, in general not enough to solve the system of equations. --Lambiam 06:58, 12 September 2022 (UTC)
- Thank you Lambian and Tamfang. I've now included the picture of the irregular octagon. The green quadrilateral looks to me like the 2D skeleton of a tetrahedron. Could it be construed as that of a regular tetrahedron? Sandbh (talk) 01:28, 12 September 2022 (UTC)
