I present here three characterization of the philon
line and the proof that they do are characterizations. Two of
them are old characterizations of unknown origin and the third one was
developed by B.
Bhattacharya and G.
Characterization 1 and 2:
Let RCS be the angle of discussion and let P be the philon point, then AB is the philon line.
First characterization:The Bhattacharya and Toussaint characterization:
Let Q be the foot of the perpendicular from C on AB. Then P and Q must isotomically divide the segment AB (the philon line); that is we must have AP = QB.Second characterization:
If AB is the philon line, then the perpendiculars to CR, CS, AB at A, B, P respectively, must be concurrent in a point D.
We now have to prove that these condition arrive when AB is the philon line. We will first show that these two characterization are equivalent, and then, we will proceed to prove the first characterization. Suppose that the three perpendiculars to CR, CS, AB at A, B, P are concurrent in a point D. Then CD is a diameter of the circle.
If you look at Fig. 2, CAD and CBD are 90°, therefore ADBC can be inscribe in a circle, since a quadrilateral can be inscribe in a circle if it as two opposite angles which sums up to 180°. Since CAD is 90°, it support an arc of 180, this arc support a cord, which is CD, which is the diameter. Therefore CD must be a diameter of the circle CADB.
Furthermore, we can complete the rectangle CfDe by producing CQ and DP until they intersect the circle again. Since Cf and De are parallel and equal (because they emerge from the vertices of a diameter with equal angle), then CfDe is a parallelogram. Therefore its diagonals meet at their center which is O, the circle's center. Therefore M, the foot of the perpendicular from O onto AB, bisect the segment PQ, since OM is parallel to CQ and to DP and that is equidistant to both CQ and DP. Therefore characterization 2 imply characterization 1.
Conversely, suppose the midpoint M of AB bisects the segment PQ. Then the perpendicular to AB at P meets the circle ABC in a point D diametrically opposite to point C (use again of the rectangle DfCe). It now follows that the perpendiculars to CR, CS, AB at A, B , P are concurrent in point D. Therefore the two characterization are equivalent.
We now have to prove that the first characterization is a valid one.
Let AB be a philon line, and let p and q denote the lengths of the perpendiculars dropped from P on CA and CB, respectively.
p/sin A + q/sin B = c, A + B + C =
Differentiating, and recalling that we must have dc = 0, we find
(p*cos A / sin2 A )dA + (q*cos B / sin2 B)dB = 0, dA + dB = 0,
(p / sin A)*cot A = (q / sin B)*cot B,
AP cot A = PB cot B
AP / PB = (cot B)/(cot A)
BQ*tan B = CQ = QA*tan A,
BQ/QA = (cot B)/(cot A) = AP/PB
AP = QB
Note that those two characterizations are implicit in the computation of Q from the intersection of the hyperbola and the circle (see Hyperbola and circle)
A characterization for a general case:
Let ROL be the angle of study, and let let KH be the philon line through P. Let PT be perpendicular to OR and let PS be perpendicular to OL. Finally, let T' and S' be points on the extensions of PT and PS respectively, such that S'H and T'K are both orthogonal to HK. The length KH is minimum when S'H = TK.
l() = PK + PH
l() = PT*sec( )+ PS*sec()
dl() / d = PT*sec( )*tan( ) - PS*sec()*tan( )
Now since l() is convex, dl() / d = 0 has one nontrivial solution and l() attains its minimum at this extreme point. Therefore, the minimum value is attained when
PT*sec( )*tan( ) - PS*sec()*tan( ) = 0
PK*tan( ) - PH*tan() = 0
S'H = T'K
We might generalize by replacing the sides of the given angle by any two arbitrary curves and , and at the same time the point P by a third arbitrary curve, and then consider the problem of drawing the minimum segment AB tangent to and having A on and B on . It can be shown that for this minimum position the normal to at A, the normal to at B, and the normal to at the point of contact of AB with , must be concurrent. This generalization seems to be due to Isaac Newton.