=

DEF. A system of pairs of points Aa, Bb, &c. on a straight line such that XA. Xa = XB. Xb = XP2 = XQ2 is called a system in Involution, the point X being called the centre, P and Q the foci of the system, and any two corresponding points A, a, conjugate points.

PROBLEM 13. Two pairs of conjugate points A, a and B, b, being given, to find the centre and foci of the involution.

The existence of a focus is only possible when both points of a pair are on the same side of the centre, and hence two cases arise, 1st, in which one pair of points lies within the other, and 2nd in which each pair lies wholly outside the other.

Case 1. (Fig. 12.) Let ab be less than AB. Through a the extreme point of the range draw any line ac, and through B the

more distant from a of the second pair of points draw a parallel line Bd. Make acab, Bd=BA, then de will intersect ABba in

therefore by definition X is the centre of the system.

Take a mean proportional between either XA and Xa or XB and Xb, which determines the distance XP and XQ from X of the foci.

E.

18

CENTRE AND FOCI OF POINTS IN INVOLUTION.

Case 2. (Fig. 13.) Through the extreme points of the system draw any two parallel lines as bc, Ad. Make bc ba the distance

from 6 of the nearer point of the opposite pair and make Ad AB the distance from A of the similar point, then cd will cut Ab in X the required centre-for

or

i.e. XA : AB :: Xb : ab,

XA : AB-XA :: Xb : ab – Xb,

XA : XB :: Xb : Xa.

The foci must be determined as in Case 1.

XA- XP : XA + XP :: XP – Xa : XP + Xa,

i.e. AP: AQ :: Pa : aQ,

or each pair of conjugate points forms, with the foci of the system, a harmonic range.

It follows of course that if APaQ be an harmonic range and X the centre point of PQ,

XA. Xα= XP2 = XQ2.

The following relations between two pairs of conjugate points Aa and Bb, and their centre X and foci P and Q are sometimes useful.

This determines the ratio in which Bb is divided by P.

PROBLEM 14. (Fig. 14.) Through a given point P to draw a line meeting two given lines AB and CD in B and D so that PB=PD.

Through P draw any line meeting one of the given lines as at A. On AP produced make Pa= PA and draw aD parallel to BA

meeting the other given line in D. The line DPB will be the line required, i.e. PB=PD (by the similar and equal triangles APB, aPD).

20

TRIANGLE WITH VERTICES ON GIVEN LINES, &C.

PROBLEM 15. To draw a triangle with its sides passing through three given points A, B, C, and with its vertices on three given concurrent lines OD, OE, OF (Fig. 15).

Take any point (as E) on any one of the given lines and from it draw lines to any two of the given points (as EA, EB) meeting

the other lines in a and b. Let the lines AB and ab meet in M. Through M draw a line MC passing through the remaining point (C) and meeting the lines Oa and Ob in P and Q. PQ will be one side of the required triangle which can be completed by drawing the lines PA, QB which will intersect in R on the third given line. There are generally six solutions as lines can be drawn through each point terminated by either pair of lines.

PROBLEM 16. To draw a triangle with its vertices on three given lines AP, BQ, CQP, and with its sides passing through three given points A, B, C one on each line (fig. 16).

Let two of the given lines (as AP, BQ) meet in O; the third line

meets the others in P and Q. Draw the lines AQ and BP intersecting in D, and draw OD intersecting PQ in E. Take a mean

proportional PM between CP and PE (Problem 5), and a mean proportional QN between CQ and QE. With centres P and Q and radii respectively equal to PM and QN describe arcs intersecting in K. Draw a line bisecting the angle PKQ, intersecting PQ in Z. Z will be one of the vertices of the required triangle which can be completed by drawing BZ intersecting AP in X and AZ intersecting BQ in Y. X and Y are the other vertices and XY will pass through C.

PROBLEM 17. To determine the locus* of the vertex of a triangle on a given base AB and with sides BP, AP in a given ratio a : b. (Fig. 17.)

On the given base AB describe any one triangle with sides

Bisect the angle APB by PD meeting AB in D and draw PC perpendicular to PD meeting AB in C.

On DC as diameter describe a circle, which will be the required locus of the vertex.

For definition of locus, see p. 29 post.