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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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P

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 BỠ 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.

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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

B

circle. Describe circle round PAD, and also a circle

round PBC; then OP touches each of these circles (III. 37); therefore the angle OPA is equal to the angle PDA, and the angle OPB is equal to the angle PCB (III. 32). But the angle OPB is equal to the angles OPA and APB together, and the angle PCB is equal to the angles CPD and PDA together (I. 32). Therefore the angles OPA and APB together are equal to the angles CPD and PDA together; and the angle OPA has been shewn equal to the angle PDA; therefore the angle APB is equal to the angle CPD.

We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle.

For take any point Q which satisfies the required conditions. Describe a circle round QAD, and also a circle round QBC. These circles will touch the same straight line at Q; for the angles AQB and CQD are equal, and the converse of III. 32 is true. Let this straight line which touches both circles at Q be drawn; and let it meet the straight line containing the four given points at R. Then the rectangle RA, RD is equal to the rectangle RB, RC; for each is equal to the square on RQ (III. 36). Therefore R must coincide with O (34); and therefore RQ must be equal to OK. Thus Q must be on the circumference of the circle of which O is the centre, and OK the radius.

55. Required the locus of the vertices of all the triangles ABC which stand on a given base AB, and have the side AC to the side BC in a constant ratio.

If the sides AC and BC are to be equal, the locus is

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the straight line which bisects AB at right angles. We will suppose that the ratio is greater than a ratio of equality; so that AC is to be the greater side.

Divide AB at D so that AD is to DB in the given ratio (VI. 10); and produce AB to E, so that AE is to EB in the given ratio. Let P be any point in the required locus; join PD and PE. Then PD bisects the angle APB, and PE bisects the angle between BP and AP produced. Therefore the angle DPE is a right angle. Therefore P is on the circumference of a circle described on DE as diameter.

We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on DE as diameter; we shall now shew that any point on the circumference of this circle satisfies the assigned conditions.

Let be any point on the circumference of this circle, QA shall be to QB in the assigned ratio. For, take O the centre of the circle; and join QO. Then, by construction, AE is to EB as AD is to DB, and therefore, alternately, AE is to AD as EB is to DB; therefore the sum of AE and AD is to their difference as the sum of EB and DB is to their difference (23); that is, twice AO is to twice DO as twice DO is to twice BO; therefore AO is to DO as DO is

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