I present here the philon line, as proposed by Philon, and two other ways of finding the same line to solve the duplication of the cube problem, one proposed by Apollonius and one by Heron.
Let AB and AC be two straight lines placed at right angles. Complete the rectangle ABDC (D is the point inside the angle through which the line shall be drawn). Let E be the center of the diagonal of the rectangle ABDC. Then a circle centered at E and going through D shall circumscribe the rectangle ABDC (note that the diameter will be AD).
Philon's wayPlace a ruler so that it passes through D and pivot it on D until it cuts AB and AC produced and the circle ABDC in points F,G,H such that the intercepts FD and HG are equal.Apollonius's way
Draw a circle centered in E and cutting the produced AB and AC in F and G respectively, but such that F,D and G are collinear.Heron's way
Place a ruler so that its edge passes through D, and move it about D until the edge intersects the produced AB and AC in points F and G respectively so that EF and EG are equals.
Obviously the three constructions compute the same points F and G. In Philon's construction, FD=HG. Then, the perpendicular from E on DH, which bisects DH, must also bisect FG, so that EF = EG. Therefore Philon's way is equivalent to Heron's way. We now proceed to proof that Apollonius's ways is equivalent to the other two. We will proof that (AF)(FB) = (AG)(GC), in other words: if the power of the point F with respect to circle AGDC is the same as the power of the point G and (AF)(FB) = (AG)(GC) then GH = FD.
Since the power of G and F are the same with respect to ABDC Then
(AF)(FB) = (FD)(FH) and (GC)(GA) = (GH)(GD)
if (AF)(FB) = (AG)(GC) then
(FD)(FH) = (GH)(GD)
But FH = FD + HD and GD = GH + HD
(FD)(FD + DH)=(GH)(GH + DH)
FD2-GH2 = HD(GH - FD)
(FD + GH)(FD - GH) = HD(GH - FD)
FD + GH = -DH
But two positive value can not add up to a negative value,
therefore (GH - FD) must be zero and we could not divide by it.
Hence FD = HG.
Proof that (AF)(FB)=(AG)(GC)::
a) Between Apollonius and Heron
Since K is the middle point of AB, then
(AF)(FB) + BK2 = FK2
since FK2 = (FB + BK)2 =FB(FB + 2BK) + FK2
Add KE2 to both sides, then
(AF)(FB) + BE2 = EF2
Also, by symmetry
(AG)(GC) + CE2 = EG2
But BE = CE and E F = EG
(AF)(FB) = (AG)(GC)
b) Between Philon and Apollonius
Since GH=FD, we have
But, since the circle GDHC passes through A, the power theorem tell us that
(AF)(FB) = (FD)(FH) and (GC)(GA) = (GH)(GD), therefore
(AF)(FB) = (AG)(GC)
With a few algebraic manipulation, we can obtain this interesting result, that will permit us to conclude that this construction is not feasible with Euclidean tools.
From our last result
AF / AG = CG / AG
By the similar triangle principle
FA / AG = DC / CG = FB / BD
DC / CG = CG / FB = FB / BD
AB / CG = CG / FB = FB / AC
We can use this last equation to set 3 equation that can relate to a solution from Menaechmus. The last equation can be rewritten a / x = x / y = y / b. From this we get 3 equation:
These represent the equation of two parabolas
and of one hyperbola.
solved the problem of the two
mean proportionals by means of the points of intersection of any two
of these conics.
But, if we add the first two equations, we have
which is a circle passing through the points common to the two parabolas x2 = ay, y2 = bx.
Therefore we can equally obtain a solution by means of the intersections of the circle x2 + y - bx2 - ay = 0 and the rectangular hyperbola xy = ab. This is exactly what Philon does, for if, AF and AG are the coordinate axes, the circle x2 + y - bx2 - ay = 0 is the circle BDHC, and xy = ab is the rectangular hyperbola with AF and AG as asymptotes and passing through D, which hyperbola intersects the circle again in H, a point such that FD = HG. This can also be seen by looking at the next picture. Let CP be the diameter of the circle going through C and P and let the parabola go through P and have CS and CR as asymptote. Then PQ produced is the Philon Line of P for angle RCS. For let PQ produced cut CR and CS in A and B, respectively. Then, by a well-known property of the hyperbola, AP = QB, and, since the circle has CP for a diameter, angle CQP is a right angle. Thus the CQ is the perpendicular from C on to PQ (see characterization). But their common intersection through P, the parabola and the circle are independent from one another. But it is known that in general it is impossible with euclidean tools to find another point of intersection of a conic and a circle of which only one point of intersection is given (see Menaechmus).