I take the following approach (from Dorrie's [i]100 Great Problems[/i]) Let X1, X2, X3, X4, X5 be the intersections of tangents (V, I), (I, II), (II, III), (III, IV), (IV, V). Now suppse X1... X5 define a hexagram, Mr. Penty, where two vertices coincide (pass into a tangent line). Working backward through Brianchon's and Pascal's theorems, we may [list=1] [*] Choose any point on the given tangents, and introduce it as a new vertex of a second hexagram inscribed about the same conic section, sharing four vertices with Mr. Penty [*] 2. Draw the sixth vertex of the new hexagram, giving us a new tangent line. [*] 3. Find the point of tangency on this new line. [/list] Here is a procedure for inserting a new point on the segment X1 X2:
Since we may choose the red point at pleasure, this is the same this as saying, [i]from five tangents we may draw the inscribed conic section.[/i] NOTES: 1. The red point can be freed to fall anywhere on tangent I. No changes to the construction are necessary. Only the numbering changes. 2. The red point may fall anywhere on any of the five tangents. 3. When drawing on paper, we will number the tangent lines so that the figure is easy to see and draw from. With a fixed numbering (above), the polygons can quickly become unruly and confusing (try moving the tangents around). Better, would be to let the computer rearrange the intersections X1, X2.... in a more natural way. Let me collect these changes into a more robust solution... Like this: