46. In a given circle it is required to inscribe a triangle so that the sides may pass through three given points.

Let A, B, C be the three given points. Suppose PMN to be the required triangle inscribed in the given circle.

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Draw NE parallel to AB, and determine the point F as in the preceding problem. We shall then have to describe in the given circle a triangle EMN so that two of its sides may pass through given points, F and C, and the third side be parallel to a given straight line AB. This can be done by the preceding problem.

This example and the preceding are taken from the work of Catalan already cited. The present problem is sometimes called Castillon's and sometimes Cramer's; the history of the general researches to which it has given rise will be found in a series of papers in the Mathematician, Vol. III. by the late T. S. Davies.

ON LOCI.

47. A locus consists of all the points which satisfy certain conditions and of those points alone. Thus, for example, the locus of the points which are at a given distance

from a given point is the surface of the sphere described from the given point as centre, with the given distance as radius; for all the points on this surface, and no other points, are at the given distance from the given point. If we restrict ourselves to all the points in a fixed plane which are at a given distance from a given point, the focus is the circumference of the circle described from the given point as centre, with the given distance as radius. In future we shall restrict ourselves to loci which are situated in a fixed plane, and which are properly called plane loci.

Several of the propositions in Euclid furnish good examples of loci. Thus the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same area, is a straight line parallel to the base; this is shewn in I. 37 and I. 39.

Again, the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same vertical angle, is a segment of a circle described on the base; for it is shewn in III. 21, that all the points thus determined satisfy the assigned conditions, and it is easily shewn that no other points do.

We will now give some examples. In each example we ought to shew not only that all the points which we indicate as the locus do fulfil the assigned conditions, but that no other points do. This second part however we leave to the student in all the examples except the last two; in these, which are more difficult, we have given the complete investigation.

48. Required the locus of points which are equidistant from two given points.

Let A and B be the two given points; join AB; and draw a straight line through the middle point of AB at right angles to AB; then it may be easily shewn that this straight line is the required locus.

49. Required the locus of the vertices of all triangles on a given base AB, such that the square on the side terminated at A may exceed the square on the side terminated at B, by a given square.

Suppose C to denote a point on the required locus; from C draw a perpendicular on the given base, meeting it, pro

duced if necessary, at D. Then the square on AC is equal to the squares on AD and CD, and the square on BC is equal to the squares on BD and CD (I. 47); therefore the square on AC exceeds the square on BC by as much as the square on AD exceeds the square on BD. Hence D is a fixed point either in AB or in AB produced through B (40). And the required locus is the straight line drawn through D, at right angles to AB.

50. Required the locus of a point such that the straight lines drawn from it to touch two given circles may be equal.

Let A be the centre of the greater circle, B the centre of a smaller circle; and let P denote any point on the required locus. Since the straight lines drawn from P to touch the given circles are equal, the squares on these straight lines are equal. But the squares on PA and PB exceed these equal squares by the squares on the radii of the respective circles. Hence the square on PA exceeds the square on PB, by a known square, namely a square equal to the excess of the square on the radius of the circle of which A is the centre over the square on the radius of the circle of which B is the centre. Hence, the required locus is a certain straight line which is at right angles to AB (49).

This straight line is called the radical axis of the two circles.

If the given circles intersect, it follows from III. 36, that the straight line which is the locus coincides with the produced parts of the common chord of the two circles.

51. Required the locus of the middle points of all the chords of a circle which pass through a fixed point. Let A be the centre of the given circle; B the fixed

point; let any chord of the circle be drawn so that, produced if necessary, it may pass through B. Let P be the middle point of this chord, so that P is a point on the required focus.

The straight line AP is at right angles to the chord of which P is the middle point (III. 3); therefore P is on the circumference of a circle of which AB is a diameter. Hence if B be within the given circle the locus is the circumference of the circle described on AB as diameter; if B be without the given circle the locus is that part of the circumference of the circle described on AB as diameter, which is within the given circle.

52. O is a fixed point from which any straight line is drawn meeting a fixed straight line at P; in OP a point Q is taken such that OQ is to OP in a fixed ratio: determine the locus of Q.

We shall shew that the locus of Q is a straight line.

For draw a perpendicular from O on the fixed straight line, meeting it at C; in OC take a point D such that ŎD is to OC in the fixed ratio; draw from O any straight line OP meeting the fixed straight line at P, and in OP take a point Q such that OQ is to OP in the fixed ratio; join

QD. The triangles ODQ and OCP are similar (VI. 6); therefore the angle ODQ is equal to the angle OCP, and is therefore a right angle. Hence Q lies in the straight line drawn through D at right angles to OD.

53. O is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P; in OP a point Q is taken such that OQ is to OP in a fixed ratio: determine the locus of Q.

We shall shew that the locus is the circumference of a circle.

For let C be the centre of the fixed circle; in OC take a point D such that OD is to OC in the fixed ratio, and draw any radius CP of the fixed circle; draw DQ parallel to CP meeting OP, produced if necessary, at Q. Then the triangles OCP and ODQ are similar (VI. 4), and therefore OQ is to OP as OD is to OC, that is, in the fixed ratio. Therefore is a point on the locus. And DQ is to CP in the fixed ratio, so that DQ is of constant length. Hence the locus is the circumference of a circle of which D is the centre.

54. There are four given points A, B, C, D in a straight line; required the locus of a point at which AB and CD subtend equal angles.

Find a point in the straight line, such that the rectangle OA, OD may be equal to the rectangle OB, OC (34), and take OK such that the square on OK may be equal to either of these rectangles (II. 14): the circumference of the circle described from O as centre, with radius OK, shall be the required locus.

[We will take the case in which the points are in the following order, O, A, B, C, D.]

For let P be any point on the circumference of this