Philon

Philon's of Byzantium problem

Given two intersecting lines e , d and a point A not on these lines, consider all lines through A and their intersection points B, C with e, d. To find the segment B₀C₀ realizing the minimum length of all such BC.
The problem can be reduced to that of finding the minimal segment B₀C₀ passing through a point A of a hyperbola and cut of by its asymptotes (see HyperbolaPropertyMiddles.html ). Below is a discussion of this "reduced" problem.
The minimal secant (called sometimes Philon's line) is defined by A and the other intersection point D₀ of the hyperbola with the circle having diameter OA, where O the center of the hyperbola.

To solve the problem we consider an arbitrary secant BC, and think of the angles at B and C depending on one parameter. Denoting the angles by the same letters and differentiating the relation B+C+O = pi => B'+C' = 0 (angle O is constant). On the other side we have to minimize BC = BA+AC = AB₁/sinB + AC₁/sinC, which differentiating gives: (AB₁/sin²B)*cosB*B' + (AC₁/sin²C)*cosC*C'. Equating this to zero and taking into account the relation B'+C'=0 we get:
(AB₁/sinB)*cotB - (AC₁/sinC)*cotC = 0 => AB*cotB - AC*cotC = 0.
But D being the projection of D on BC we have the relation OD=DB*tanB=DC*tanC => DB*cotC-DC*cotB = 0. Thus, if BC is minimal A, D divide BC in the same ratio but in opposite directions, hence we must have AC = BD and D must coincide with D₀.

Philon's of Byzantium problem

See Also