Ratio comparison of two arc tangents.

Using the given inequality, calculation of the mean value of function 1/(1+x²) allows series expansion of arc tan x in powers of x. _________________________ Gregory's Arc Tangent Series [b]Proposition: [/b][i]To determine the angles of a triangle from the sides without the use of tables.[/i] This is problem #17 in Heinrich Dorrie's [i]100 Great Problems of Elementary Mathematics[/i]

If the arms are fixed at a right angle

Notes: 1) The solution is my own (Geogebra practically solves it for you). 2) The positions of A, B, C are given by moving the red point. This point is not part of the solution, which means this construction goes backward. To solve the problem, I went in this order: Draw the axes of the two arms. Let A move freely on the first arm. Choose a means of fixing the segment length AB. Construct AB. B is dragged along the second arm as A moves. Determine the ranges of A and B on the two arms. Construct arbitrary C, rigidly connected to A and B. The solution must be an ellipse. There is no other possibility. Locus[] confirms the suspicion. Now, when A reaches either end of the arm, what direction will the triangle rotate? It depends on the motion of the system, and its arrangement in space. We could also choose a direction of rotation by applying force to B or C when A reaches the end of an arm. Or by any number of mechanical means. So, one option is to introduce mechanics. I think that is a beautiful choice.* Another, more convenient alternative is to rotate the triangle with a dial. :) *Like this: [url]http://www.geogebratube.org/material/show/id/71230[/url]

A more flexible set of axes. I will adopt this system for now.

Each axis slides along one meridian. Arrows indicate the direction of positive rotation. I adopt the following notation: Capital letters X, Y, Z denote the 3-dimensional vectors: X Y Z are mutually perpendicular unit vectors forming a right-handed coordinate system. Barred letters are projections: [math] \;\;\;{\rm \bar x}, {\rm \bar y}, {\rm \bar z}[/math] are the projections of X Y Z on the view plane (the coordinates of [math]{\bf \bar x}[/math] are the ordinary GGB coordinates.) I have begun with a coordinate-free representation. I submit without proof that the axes are bound by the following constraints: [math] \;\;\;1.\;|{\rm \bar x}|² + |{\rm \bar y}|² + |{\rm \bar z}|² = 2 [/math] [math] \;\;\; 2.\; |{\rm \bar x}×{\rm \bar y}|= \sqrt{1-|{\rm \bar z}|²}[/math] [math] \;\;\;\;\;\; |{\rm \bar y}×{\rm \bar z}|= \sqrt{1-|{\rm \bar x}|²}[/math] [math] \;\;\;\;\;\; |{\rm \bar z}×{\rm \bar x}|= \sqrt{1-|{\rm \bar y}|²}[/math] The two constraints can also be stated this way: Let [math]{\rm \bar x}[/math] determine an ellipse with center O, major axis the unit perpendicular to [math]{\rm \bar x}[/math], and minor axis of [math]\sqrt{1-|{\rm \bar x}|²}[/math]. Then [math]({\rm \bar y},{\rm \bar z})[/math] are conjugate radii on the ellipse. And likewise [math] ({\rm \bar z},{\rm \bar x}), ({\rm \bar x},{\rm \bar y})[/math] are conjugate on the ellipses determined by [math]{\rm \bar y},{\rm \bar z}[/math], respectively. [b]Update:[/b] So. I made that up, to respect freedom of motion. Today I find this is Gauss' Fundamental Theorem of Normal Axonometry. This was problem #74 in Heinrich Dorrie's [i]100 Great Problems of Elementary Mathematics.[/i]. ___________________ [b]Unit Sphere[/b] [list] [*]Setup: [url]http://www.geogebratube.org/material/show/id/101282[/url] [*][b]→ Trihedron:[/b] [*]Base Object: [url]http://www.geogebratube.org/material/show/id/105255[/url] [*] Spherical Coordinates {link} [*] Meridian, (Horizon Points) [*] Latitude, (Horizon Points)