varies its radius while fulfilling the conditions of the problem, the centre moves along M; and M is called the centre-locus of the variable circle. Hence the centre-locus of a circle which touches a fixed line at a fixed point is the perpendicular to the line at that point. Cor. If the circle is to pass through a second given point Q the problem is definite and the circle is a particular one, since it then passes through three fixed points, viz., the double point P and the point Q. (109°, 4) In this case LCQP=LCPQ. But LCPQ is given, since P, Q, and the line L are given. .. LCQP is given and C is a fixed point. N B N 130°. Problem.-To describe a circle to touch two given non-parallel lines. Let L and M be the lines intersecting at O. Draw N, N, the bisectors of the angle between L and M.(121°, Cor. 2) From C, any point on either bi sector, draw CA I to L. The circle with centre C and radius CA touches L, and if CB be drawn I to M, CB=CA. (68) Therefore the circle also touches M. As C is any point on the bisectors the problem is indefinite, and the centre-locus of a circle which touches two intersecting lines is the two bisectors of the angle between the lines. M 131°. Problem.—To describe a circle to touch three given lines which form a triangle. L, M, N are the lines forming the triangle. Constr.-Draw 11, E2, the internal and external bisectors of the angle A; and 12, E2, those of the angle B. LA + LB is <1, :: LBAO+LABO is < 7 .. I, and I, meet at some point o (799) and are not I to one another and therefore E, and E, meet at some point Oz. му (79°, Cor.) 3 Also I, and E, meet at some point 01, and similarly 1, and Ey meet at 0., The four points 0, 01, 02, 0, are the centres of four circles each of which touches the three lines L, M, and N. Proof.---Circles which touch M and N have I, and E, as their centre-locus (130°), and circles which touch N and L have 12 and E, as their centre-locus. :. Circles which touch L, M, and N must have their centres at the intersections of these loci. But these intersections are 0, 01, 02, and O3, .. 0, 01, 02, and O2 are the centres of the circles required. The radii are the perpendiculars from the centres upon any one of the lines L, M, or N. Cor. 1. Let Iz and E, be the bisectors of the LC. Then, since () is equidistant from L and M, Iz passes through O. (68°) ... the three internal bisectors of the angles of a triangle are concurrent. Cor. 2. Since 03 is equidistant from L and M, 13 passes through 03. (68°) .. the external bisectors of two angles of a triangle and the internal bisector of the third angle are concurrent. Def. 1.—When three or more points are in line they are said to be collinear. Cor. 3. The line through any two centres passes through a vertex of the AABC. .. any two centres are collinear with a vertex of the A. The lines of collinearity are the six bisectors of the three angles A, B, and C. Def. 2 -With respect to the AABC, the circle touching the sides and having its centre at O is called the inscribed circle or simply the in-circle of the triangle. The circles touching the lines and having centres at 01, 02, and O2 are the escribed or ex-circles of the triangle. REGULAR POLYGONS. 132°. Def. 1.-A closed rectilinear figure without re-entrant angles (89', 2) is in general called a polygon. They are named according to the number of their sides as follows: 3, triangle or trigon; 8, octagon; 10, decagon ; 12, dodecagon ; etc. The most important polygons higher than the quadrangle are regular polygons. Def. 2.-A regular polygon has its vertices concyclic, and all its sides equal to one another. The centre of the circumcircle is the centre of the polygon. 13.3o. Theorem.If n denotes the number of sides of a B LAOB=4 right angles regular polygon, the magnitude of an internal angle is (2-4) right angles. Proof.—Let AB, BC be two consecutive sides of the polygon and O its centre. Then the triangles AOB, BOC are isosceles and congruent. LOAB=LOBA=LOBC=etc., LOAB+LOBA=LABC. But LOAB+LOBA=|-LAOB, and (132°, Def. 2) LABC=(2-4) right angles, 9.e.d. Cor. The internal angles of the regular polygons expressed in right angles and in degrees are found, by putting proper values for n, to be as follows : Equilateral triangle, 60° Octagon, 3 135° Decagon, 144° 108° Dodecagon, $ 150° Hexagon, $ 120° .. or I 90° . . B F с 134°. Problem.—On a given line-segment as side to construct a regular hexagon. Let AB be the given segment. Constr.-On AB construct the equilateral triangle AOB (124°, Cor. 1), and with O as centre describe a circle through A, cutting AO and BO produced in D and E. Draw FC, the internal bisector of LAOE. Then ABCDEF is the hexagon. Proof. LAOB=LEOD=7, LAOE= 1), and AOF=37 E And the chords AB, BC, CD, etc., being sides of congruent equilateral triangles are all equal. Therefore ABCDEF is a regular hexagon. .. the side of a regular hexagon is equal to the radius of its circumcircle. R 135o. Problem.—To determine which species of regular polygons, each taken alone, can fill the plane. That a regular polygon of any species may be capable of filling the plane, the number of right angles in its internal angle must be a divisor of 4. But as no internal angle can be so great as two right angles, the only divisors, in 133°, Cor., are }, 1, and $, which give the quotients 6, 4, and 3. Therefore the plane can be filled by 6 equilateral triangles, or 4 squares, or 3 hexagons. It is worthy of note that, of the three regular polygons which can fill the plane, the hexagon includes the greatest area for a given perimeter. As a consequence, the hexagon is frequently found in Nature, as in the cells of bees, in certain tissues of plants, etc. Ex. I. Let D, E, F be points of contact of the inR, circle, and P, P', P", R, R', R", etc., of the ex-circles. (131°) Then AP=AP',CP'=CP", and BP=BP", (114°,Cor.1) .. AP' +AP =AB+BC + AC =a+b+c, and, denoting the perimeter of the triangle by 2s, we have AP=AP'=s, A 3 R E В D P |