Initial disposition of Archytas' construction, including the sphere. You can achieve the final disposition following the instructions below. The final disposition is achieved when A, M, K are aligned. The diagram will show also the trace of point K. How to manipulate left and right panels: [list] [*]On the left panel there is the three-dimensional canvas, with the initial disposition of Archytas' construction with the sphere and the Archytas' triangle ΑΜΔ. [*]On the right panel, there is a plane canvas with the same Archytas' triangle ΑΜΔ. [*]At the top there are some bars for each panel. [*]With the right button of the mouse, you can rotate the three-dimensional canvas to change the point of view: while clicking on the canvas, move the mouse to change the point of view. [*]You can select the point Δ with the left button of the mouse on the three-dimensional panel and move it along the semicircle of diameter ΑΔ, to find the correct disposition of elements to achieve Archytas' solution. NB: To move the point Δ, it must be selected the arrow in the top menu. [*]After clicking in one panel, you can Zoom In (Ctrl+) or Zoom out (Ctrl-) in the panel. [*]The three-dimensional canvas can have a uniform rotation from its center by clicking the play button surrounded by a circular arrow on the second bar of the left panel. [/list]

Construction of the [i]proportionatrix prima[/i] in a semicircle of diameter AG. Steps to find a point K of the curve from a point F of the semicircle: [list=1] [*]Semicircle of diameter AG and center D (see diagram). [*]Semicircle of diameter DG. [*]A point F on the large semicircle is selected and FG is drawn cutting the small semicircle in H. [*]Circle of center G and radius GH. [*]I is the intersection of AG with this circle. [*]Circle of center I and radius IG. [*]K is the intersection of this circle and FG. [*]CK is perpendicular to AG. [/list] The bottom bar of the canvas allows to follow these steps. Once the last step of the construction is arrived at, click on the 'Trace' button: the trajectory of the red point will trace the curve made by the point K when F moves on the semicircle. To stop the process, click on 'Stop Trace'. You can also draw the curve by selecting the point on the larger semicircle and moving it along the semicircle. To clean the canvas and get the initial configuration, click on the button with two circular arrows in the right upper corner of the canvas.

Once the curve has been drawn, it is easy to prove that AG/GF = GF/GK = GK/GC: it is enough to show that AGF, MFK and GKD are similar triangles, and that GF = GM. This statement is proved in Villalpando's proposition X [url]http://www.e-rara.ch/zut/content/pageview/3800973[/url]. Villalpando's [i]proportionatrix prima[/i] can be used to find two mean proportional between lines [i]a,b[/i], with [i]a>b[/i]. If we set AG = [i]a[/i] and the [i]proportionatrix prima[/i] is met by the straight line drawn from the extreme C of GC = [i]b[/i] perpendicular to AG, then GF and GK are the two mean proportional of AG and GC (Villalpando's actual use of this curve to find the two mean proportionals between two assigned straight lines is a bit different, still is equivalent to the one just expounded).