Wikipedia:Reference desk/Archives/Mathematics/2015 April 27

= April 27 =

Pythagorean theorem generalization
Draw a tetrahedron whose faces have ratio 3 square units, 4 square units, 5 square units, and 6 square units. Also try different tetrahedrons whose faces have ratio a:b:c:d where a^3 + b^3 + c^3 = d^3. Will tetrahedrons of this kind have any special property?? Georgia guy (talk) 15:56, 27 April 2015 (UTC)

StuRat (talk) 16:58, 27 April 2015 (UTC)
 * It's not a homework question; it's a question I would like to know the answer to. Georgia guy (talk) 17:03, 27 April 2015 (UTC)
 * You came up with the language "Draw ... Also try" on your own? —Tamfang (talk) 08:12, 28 April 2015 (UTC)


 * Is the "square" and "^3" (cube) distinction intentional? Or should it have been "^2"? —Quondum 18:27, 27 April 2015 (UTC)
 * The ^3 is intentional. The Pythagorean theorem is about triangles where the squares of the lengths of the 2 short sides add up to the square of the length of the longest side. I'm asking you to generalize the theorem to 3 dimensions so that a tetrahedron whose face areas are a quadruple where a^3 + b^3 + c^3 = d^3. Georgia guy (talk) 18:30, 27 April 2015 (UTC)
 * De Gua's theorem generalizes the Pythagorean theorem to tetrahedrons, but it involves the square of the areas. -- ToE 19:10, 27 April 2015 (UTC) For those who wish to follow this aside to the possible improvement of a couple of article: while we have Pythagorean theorem, there is no mention in De Gua's theorem whether or not its converse is necessarily true, and there is no mention of Pythagorean quadruple in De Gua's theorem or vice versa. -- ToE 19:40, 27 April 2015 (UTC)
 * "3 square units" is the equivalent of 3m^2, rather than 3^2 units, which would be equivlant to 9m, i.e. a "square unit" is a unit of area MChesterMC (talk) 08:22, 28 April 2015 (UTC)
 * One generalization of the Pythagorean identity for simplices is the Cauchy-Binet formula, I believe. For right tetrahedra, this says that the sum of the squares of the areas of the faces meeting at the right vertex is equal to the square of the area of the face opposite the vertex.  (But this does not involve sums of cubes in any obvious way.)   Sławomir Biały  (talk) 20:18, 27 April 2015 (UTC)

Specifying three sides uniquely determines a triangle (up to isometry), described as the SSS Postulate in our Triangle. Does specifying the areas of four faces similarly uniquely determine a tetrahedron? -- ToE 13:54, 28 April 2015 (UTC)
 * No. Start with a regular tetrahedron and stretch along a line through the midpoints of two opposite edges. Or just take coordinates of vertices of a tetrahedron to be (a, 0, b), (a, 0, -b), (-a, b, 0), (-a, -b, 0). All sides have the same area but the tetrahedron is not regular. Another way of looking at it, the number of degrees of freedom in a tetrahedron up to congruence is 6, which is more than the number of sides of the tetrahedron. There is a 3-dim version of SSS, but it says if all six corresponding edges are equal then the tetrahedrons are congruent. For dimension n, you would compare n(n+1)/2 edges. --RDBury (talk) 23:30, 28 April 2015 (UTC)
 * Thanks RD! I thought that there were too many degrees of freedom, but I couldn't quite visualize it.  Your simple counterexample is great.  I suspect that the converse of the De Gua's theorem is likewise false but I need more time to work on that. -- ToE 16:54, 2 May 2015 (UTC)