Defs. The point E is called the Director Point, and S', S., S, the Adjoint Points of F„, F„ F„

104. Theorems.*-In any three similar figures there exists an infinite number of triads of homologous points C1 С, C, which are collinear. 2°. The loci of these points are circles passing through E. 3°. The variable line CCC turns around E. Neuberg.

2) 3

3

2 3'

2

The triangles S ̧ ̧ ̧ S¿C ̧C1 S ̧CC, are constant in species (Art. 97, 2°); hence the angles S,CS, S,C,S,, SC2S2 are given, and therefore the loci of the points are circles passing through each pair of double points.

Again, since SCC, is a constant angle, the variable line CC, meets the locus of C1 in a fixed point, and similarly it meets the loci of C, and C, in fixed points. Therefore the fixed points are coincident; that is to say, the circular loci have a point in common.

In the particular case of the collinear triads S'S11, SSS SSS, it has been proved (Art. 103) that their lines of collinearity pass through E; therefore, etc. The points S, S, S' are on the corresponding circles.

3

105. Particular Cases.-Let the three similar figures F, F, F, be described on the sides of a triangle ABC. It has been shown that the middle points of the symmedian chords of the circum-circle † are the common vertices of directly similar triangles described on the sides, taken in pairs (Art. 25, Ex. 2), and they are therefore the three double points. Hence,

1o. Brocard's second triangle is the triangle of simili

* Mathesis, 1882, pp. 76-8.

+ The middle points of the symmedian chords of the circum-circle are the vertices of the triangle known as Brocard's Second Triangle.

tude, and the Brocard circle the circle of similitude, of three directly similar figures described on the sides of a triangle.

2o. Brocard's first triangle is their invariable triangle, Art. 29, Ex. 3.

3°. Brocard's second triangle and the given one are in perspective at a point on the circum-circle of the former whose distances from the sides of ABC are in the ratios of their lengths (Art. 100, 2°). See also Art. 16, Ex. 2. 4°. The centre of perspective is the symmedian point of ABC.

5o. The locus of the intersection of concurrent triads of homologous lines is the Brocard Circle, Art. 100, 4°.

6°. Brocard's first and second triangles are in perspective (Art. 101), and their centre of perspective E, or director points, is the centroid of ABC. (Art. 53, Ex. 6.)

7°. All collinear triads of homologous points lie on a variable line passing through E, and each point describes a circle passing through two vertices of Brocard's second triangle and the centroid of ABC.

M'CAY'S CIRCLES.

106. The loci in 7° of the previous Article are fully described by M'Cay in his memoir "On Three Circles related to a Triangle."* Amongst many other properties they possess those given in this and the following Article. The notation employed is as follows:- ABC is the given triangle; A‚‚ф „С, Brocard's first and second triangles; E centroid; A', B', C' three homologous collinear points; M middle point of AB; H circum-centre;

2 2 2

* Transactions of the Royal Irish Academy, vol. xxviii.-Science.

2

A, B, C, the homologues of A, B, C, respectively as double points of F1, F2, F. Pac the c correspondent of P regarded as an a point, and Lac and Lab the c and b

correspondents of any line L regarded as an a line; the circular loci the "A," "B," and "C" circles of the triangle. 1o. The mean centre of any three collinear homologous points is at E (Art. 53, Ex. 6).

2o. If one of them C' coincides with E, A'B' is a tangent to the "C" circle and EA' - EB' or EEca=EEcb; simi

=

larly we have EEab=EEac and EEbcEEba

=

3°. If one of them coincides with a double point A,, the line of collinearity is A,ЕA,A, (Art. 103) and EA,=2EA,

Similarly the lines B,B,B, and C,C,C, each pass through E, which is the common point of trisection of the segments

2 3'

4°. The circles cut each other at angles A, B, and C.

5°. Their centres are on the perpendicular bisectors of the sides.

This is proved for the "C" circle as follows:

On the sides of ABC construct three directly similar triangles BCA', CAB, ABC', each inversely similar to

ABC. Their centres of gravity are therefore corresponding points. But they lie on a parallel through E to AB; hence E, the centroid of ABC, is on the "C" circle and E and E' are reflexions with respect to the perpendicular bisector of AB.

107. Problem.-To find the Centres and Radii of M'Cay's Circles.

This is done by finding where the circles again cut the corresponding medians. We take, for example, the "C" circle and require to find C. Let L denote the median CM, and take it an a line. Since it makes an angle BCM with the side a, we draw the corresponding b and c lines by making angles CAB' and ABC' equal to BCM.

From similar triangles MBC and MCB we have MC. MC' MB2= MC. MI; hence MCMI. This also follows, since the triangles ABI and BAC are similar.

=

Again, the triangle CBC, is inversely similar to ABC, but it is (hyp.) directly similar to BAC, Hence BAC and ABC' are inversely similar; therefore C' is the

reflexion of C, with respect to the perpendicular bisector of the base.

The connection between three collinear points A', B', C on the median to the side c of the given triangle and C2, C, C has thus been established.

The triangles BCA', CAB', ABC are similar to one another, and to CBC, ACC,, and BAC,; and therefore A', C; B, C; C', C, are reflexions of one another with respect to the corresponding perpendicular bisectors of the sides of ABC.

2

3

It follows that if the median and symmedian cut the circum-circle in I and J, and these points be joined to M, the lines MI and MJ produced through M pass through C and C, respectively; MJ= MC, and MI= MC, or C and C, are the reflexions of I and J with respect to the base AB.

Let d be the distance of the centre of the "C" circle from AB, m the median, and the angle it makes with the base, t the tangent from M to the circle.

Then c2

.(1)

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