10°. Two parallel lines invert into two circles touching externally if the origin is between the lines; and internally if the lines are on the same side of the origin.

11°. If a quadrilateral ABCD inverts into a parallelogram from an origin 0; the pairs of circles BOC, AOD and COA, BOD touch at 0.*

SECTION III.

ANHARMONIC RATIOS UNALTERED BY INVERSION.

128. Theorem.-If A, B, C, D be any four concyclic points and A', B', C', D′ their inverses with respect to any circle of inversion, then

BC.AD:CA.BD:AB.CD=B'C'. A'D': C'A'.B'D': A'B'. CD'.

This property has been shown to hold for any four points and their inverses, and is therefore true in the particular case when they lie on a circle; hence the anharmonic ratios of four concyclic points are equal to the anharmonic ratios of their inverses with respect to any circle of inversion. Particular cases have been noticed in Art. 121, Cors. 1, 2.

129. Problem.-To invert a regular cyclic polygon ABC... from any origin P.

The circumcircle ABC... inverts into a circle aẞy...; the diameters AA', BB', CC'... into circles passing through the origin P and cutting aßy... orthogonally in aa', ßß, vy....

*

Hence a construction for the required centres of inversion.

They therefore pass through Q the inverse of P with respect to the inverse circle and thus form a coaxal system of the common point species. (Art. 127, 6°.) Also the chords aa', BB', yy... meet in a point K on PQ (Art. 72, Ex. 6).

1

On the primitive figure any side BC of the polygon and any diameter AA' meet the circle in a harmonic row of points; therefore (Art. 128) on the inverse figure Byaa' is an harmonic row; hence Balya = Balya', or, by Euc. III. 22, the diagonal aa' of the quadrilateral is the locus of a point such that its distances from either pairs of sides which meet at its extremities are proportional to the lengths of the sides; similarly for the quadrilaterals yoßß', etc. Therefore the distances of the point K from the sides of the polygon aßy... are proportional to the sides.

For an Harmonic Quadrilateral, K is evidently at the intersection of the diagonals; and the inverse of the regular polygon possessing, as has been shown, a corre

sponding and more general property has been termed by Casey an Harmonic Polygon.

Definitions. The point K is called the Symmedian Point of the Polygon; and if the ratio of any perpendicular from K to half the side on which it falls is tan w, then w is the Brocard Angle of the Polygon.

For the properties of harmonic polygons the reader is referred to Casey's Sequel to Euclid, Supplementary Chapter, Section VI.

130. Cosymmedian Triangles.--Let ABC be a triangle K, its symmedian point, and let the lines AK, BK, CK meet the circum-circle again in A', B', C'. If the circle of inversion be K, p where

ρ

KA.KA' KB. KB'= KC. KC' = — p2,

=

the vertices of ABC invert into A', B', C'.

Also since BCAA' is a harmonic quadrilateral, therefore B'C'A'A is harmonic, or A'A is a symmedian of the triangle A'B'C'; similarly the other symmedians are B'B and C'C.

It appears thus that the two triangles have the same symmedian lines, symmedian point, Brocard Circle, Brocard Angle, Brocard Points, etc. On account of these relations they have been termed Cosymmedian Triangles.*

*Their properties were first stated by Casey before the Royal Irish Academy in December, 1885. A further account of them will be found in Milne's Companion.

EXAMPLE.

1. If ABC be a triangle and G its centroid; AA', BB', CC' chords of the circum-circle passing through G; the symmedian point of A'B'C' is on the diameter which contains Tarry's point. (Vigarié.)

[Let the circle be self-inverted from G as origin and the points A, B, C invert into A', B', C' respectively. Let AA", BB", CC" be the symmedian chords meeting in K.

If a circle CGC" meet GK in the point I then

and similar relations hold for the circles AGA" and BGB"; therefore these three circles meet in a second common point L, which is the inverse of K', the symmedian point of A'B'C'.

Let J be the inverse of K with respect to the circum-circle ABC, and it follows that KO. KJ=KG. KL=the power of K with respect to the circum-circle. Hence OGJL is a cyclic figure, and

the angle GOK=L.

It has been shown (Art. 67, Ex. 18) that Tarry's point on the circum-circle corresponds to O the circum-centre on the Brocard Circle with respect to ABC and Brocard's first triangle, and that G is their common centroid; hence angle GNO=GOK and GRO=GKO=GF'O. Therefore OGKF" is a cyclic quadrilateral, and (Euc. III. 21) the points F, K, F' are collinear. Therefore KO. KJ KG. KL=KF. KF" or F, J, L are collinear, the line being the inverse of the circle OGKF" with respect to K as origin.

Now the circum-circles of ABC and GFL cut each other orthogonally since the angle OFG = L; hence the inverse of the latter from G is the diameter NR, and therefore L inverts into a point K' on it; therefore, etc.

This solution is due to M'Cay.*]

MISCELLANEOUS EXAMPLES.

1. The six circles that can be described to touch three given ones A, B, C, two externally and one internally and two internally and one externally, are in pairs the inverses of one another with respect to the common orthogonal circle of A, B, C.

[Invert with respect to the common orthogonal circle of A, B, C, and since A, B, C remain unaltered after inversion, three of the circles of contact invert into the remaining three; therefore, etc.]

2. The eight circles of contact with A, B, C have a common circle of antisimilitude.

[As in Ex. 1 they are in pairs the inverses of each other with respect to the common orthogonal circle of A, B, and C.]

3. Three circles are described touching the ex-circles of a triangle, two externally and one internally; prove that they each pass through the centre of Taylor's Circle.

[Invert with respect to Taylor's Circle and the circles in question invert into the remaining circles of contact, which in this case are the sides of the triangle; and since the circles invert into lines they each pass through the centre of inversion.]

4. If ABC be a triangle; C', p a circle of inversion, A' and B' the inverses of A and B ; to prove that

28= p2 sin Cr

where' is the radius of the in-circle of A'B'C.

[We have AC= p2/A'C, BC= p2/B'C and AB/A'B' = p2/A'C. B'C, hence by addition

*"Mathematical Questions with their Solutions," from the Educational p. 73.

Times, vol. lii.,