Gosper, re your Brama-tangent remark below: looking at it another way, both triangles & quads satisfy sum of tans = sum of products of 3 tans. This looks like it could generalize further, to something like sum of tans - sum of products of 3 tans + sum of products of 5 tans -+ ... = 0 Of course, the LHS is the numerator of the tangent addition formula. And the sum of the interior angles of a polygon is ... Rich ------ Quoting Bill Gosper <billgosper@gmail.com>:
Fred>On 4/7/12, rcs@xmission.com <http://gosper.org/webmail/src/compose.php?send_to=rcs%40xmission.com> <rcs@xmission.com <http://gosper.org/webmail/src/compose.php?send_to=rcs%40xmission.com>> wrote:> ...> The final comment about splitting up the sides other than> 1 vs N-1 should give a very simple 2-2 formula for quadrilaterals,> and could lead to an angle-free formula relating the lengths> of the sides and the diagonals. Presumably this would give> the 2nd diagonal as a quadratic in the 4 sides and the first> diagonal. And there's likely something pretty involving the> sides of a pentagon and the "inscribed star".>> Rich>
The relation between the sides and diagonals of a plane quadrilateral is that the volume of the tetrahedron with those edge-lengths vanishes. The volume is given by the Cayley-Menger determinant for 3-space, a cubic in the squares of the edge-lengths. See eg.
http://en.wikipedia.org/wiki/Distance_geometry
[I know most of you have heard all this before several times --- Rich obviously wasn't paying attention.]
WFL
See also: http://en.wikipedia.org/wiki/Brahmagupta%27s_formula#Extension_to_non-cyclic...
--rwg
Something is wrong with either my math-fun archive or Apple's Finder|Spotlight. I can't find
any references to "Brahmag" despite, e.g., my (2008?)
subj: any noun can be verbed "Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0. --rwg
which I found in a scratch file.
"extremely ingenious" erratum: I miscopied more modern values for Euler's Bessel roots,
which were considerably less accurate. Simply manually evaluating Bessel functions must
have been so tedious as to discourage Newton's method. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun