51. Def.-If a point is taken on the radius of a circle and another point on the same radius produced, so that the product of their distances from the centre is equal to the square of the radius, each is the pole of the line (its polar) drawn through the other perpendicular to the radius.

Thus if OP.OQ= R2 the point P is the pole of the line QS, and the line QS is the polar of the point P with respect to the circle X.

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52. If a line passes through a given point, the pole of the line is on the polar of the point.

Hint.-Let P be the given point, PC the line, QS the polar of P.

Draw OC perpendicular to PC. Since OC.OB=OP.OQ, B is the pole of

PC.

53. Cor.-The line joining two points is the polar of the intersection of their polars; and the point of intersection of two lines is the pole of the line joining their poles.

It follows that the poles of lines which meet in a point are in a straight line, and the polars of points which are in a straight line meet in a point.

M

FIG. 53 (1)

1

FIG. 53 (2)

B

Thus if PM and PN [Fig. 53 (1)] are the polars of A and B, AB is the polar of P, and if A and B are the poles of PM and PN, P is the pole of AB.

If several lines meet in a point P [Fig. 53 (2)] their poles are in a straight line AB, and vice versa.

54. The locus of the intersection of tangents to a circle, drawn at the extremities of a chord which passes through a given point, is the polar of the point.

Hint.— Let P be the given point, Q a point on OP such that OP.OQ=R2, and B the intersection of the tangents at the extremities of TT'.

Hence Cinverts into B with respect to the centre of inversion O. But the locus of C is a circle on OP as diameter.

Therefore the locus of B is the straight line QS perpendicular to OQ. § 25 55. If four points on a straight line form a harmonic system, their four polars form a harmonic pencil.

Hint.-Let ABCD be harmonic points on the line QS, and P the pole of QS with respect to the circle X.

0.4, OB, OC, OD form a harmonic pencil. Also the polars of the four points A, B, C, D pass through P and are respectively perpendicular to the rays of this harmonic pencil.

Hence the four polars form a pencil which is equiangular with the pencil (0.ABCD) and therefore harmonic.

56. A line cutting a circle and passing through a fixed point is cut harmonically by the circle, the point, and the polar of the point.

Hint.-Let P be the fixed point, LC its polar, and PM the line cutting the circle.

Since PO.PC=PA.PB=PM.PN, a circle may be circumscribed about the quadrilateral OCNM.

Hence angle OCM=OMN=PCN; then CP and CL are the external and internal bisectors of the angle MCN. Therefore P, K, N, M are harmonic points.

§ 334, p. 151

57. A method of drawing the polar of a given point follows from

$56.

Hint.-Draw the secants PA, PM; also draw AM, AN, BM, BN.
The line through L and J is then the polar of P.

For LMAJNB is a complete quadrilateral; and A, B, C, P, and M, N, K, P, are therefore two systems of harmonic points.

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NINE POINTS CIRCLE

58. The circle through the middle points of the sides of a triangle passes through the feet of the perpendiculars from the opposite vertices, and through the middle points of the segments of the perpendiculars included between their point of intersection and the vertices. Hint.-Let ABC be the triangle, L, M, N the middle points of the sides, O the intersection of perpendiculars, X the middle point of CO. MX is parallel to AP and consequently perpendicular to ML. circle on LX as diameter passes through M. For a similar reason it passes through N.

Hence a

Since LSX is a right angle, the circle passes through S. The circle on MY as diameter must coincide with this circle since it passes through the points L, M, N. Hence the circle also passes through P, etc.

Therefore the circle passes through L, M, N, P, R, S, X, Y, Z. 59. Def.—This circle is the nine points circle of the triangle.

60. The circumscribing circle of a triangle can be inverted into che nine points circle of the triangle formed by joining the points in which the inscribed circle of the original triangle touches the sides. The centre of inversion is the centre of the inscribed circle; the constant of inversion is equal to the square of its radius.

Hint. Since OP.OB= OC", etc., the vertices A, B, C invert into the middle points of the sides of the triangle.

Hence the circle through A, B, C inverts into a circle through the middle points of the sides of the triangle A'B'C'.

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PERSPECTIVE

61. Def.-Two figures are in perspective if the lines joining their corresponding points meet in a common point, the centre of perspective. If the figures are in the same plane, and if, when the lines of the figures are indefinitely produced, the lines joining the corresponding points of intersection meet in a common point, the figures are in plane perspective.

Thus if in Fig. (1) lines Aa, Bb, Cc meet in a point O, the triangles 4 BC, abc are in perspective.

62. If two triangles are in perspective their corresponding sides intersect in points which are in a straight line.

Hint.-Let O be the centre of perspective of ABC, abc.*

Since AB and ab are both in the plane AOB they must meet; since AB is in the plane MN and ab in the plane M'N, the point of meeting must be in LN the line of intersection of these planes.

(2.) If the two triangles are in the same plane.

Hint.-Draw any line 00'0" not in the plane of the triangles through the centre of perspective. From any two points O', O" on this line draw lines through the vertices of the triangles.

O'A and O'a meet in a point A' because both are in the plane O'OA ; Thus both the triangles ABC and abc are projected into A'B'C'; hence, their corresponding sides meet on the line of intersection of the plane MN with the plane of A'B'C'.

63. Exercise. If two polygons are in perspective their corresponding sides meet in points which are in a straight line.

* If MN be a transparent plane and a point of light be at 0, the shadow cast upon the plane M'N by the triangle ABC is the triangle abc.