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.

In this case CQP=4CPQ.

(109°, 4)

But CPQ is given, since P, Q, and the line L are given. .'. 4CQP is given and C is a fixed point.

130°. Problem.-To describe a circle to touch two given non-parallel lines.

L

N

B

A

A

N

B

M

C

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 bisector, draw CA L to L.

The circle with centre C and radius CA touches L, and if CB be drawn L to M, CB=CA.

Therefore the circle also touches M.

(68°)

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.

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 I1, E, the internal and external bisectors of the angle A; and I2, E2, those of the angle B.

LA+≤B is < 1, .'. LBAO+LABO is <7.

1

.. I, and I meet at some point O (79°) and are not to one another and therefore E1 and E2 meet at some point O3.

1

Also I and E, meet at some point O1, and similarly I, and E1 meet at O2.

3

The four points O, 01, 02, Og are the centres of four circles each of which touches the three lines L, M, and N.

1

Proof.---Circles which touch M and N have I, and E1 as their centre-locus (130°), and circles which touch N and L have I2 and E2 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 O, O1, O2, and O3,

... O, 01, 02, and Og are the centres of the circles required.

3

The radii are the perpendiculars from the centres upon any one of the lines L, M, or N.

3

3

Cor. 1. Let Ig and E, be the bisectors of the ≤C. Then, since O is equidistant from L. and M, I, passes through O. (68°) .. the three internal bisectors of the angles of a triangle

are concurrent.

3

3

Cor. 2. Since O, is equidistant from L and M, I, passes through O3.

(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 ▲ABC.

... 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 ABC, 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 O1, O2, and O3 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 ;

4, quadrangle, or tetragon, or quadrilateral;

5, pentagon; 6, hexagon; 7, heptagon ;

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.

133°. Theorem.—If n denotes the number of sides of a

regular polygon, the magnitude of an internal angle is

Cor. The internal angles of the regular polygons expressed in right angles and in degrees are found, by putting proper

134°. Problem.-On a given line-segment as side to con

struct a regular hexagon.

Let AB be the given segment.

Constr.-On AB construct the equi- F

lateral triangle AOB (124°, Cor. 1), and with O as centre describe a circle through

A, cutting AQ and BO produced in D

and E. Draw FC, the internal bisector of LAOE. Then ABCDEF is the hexagon.

Proof.-

LAOB=LEOD=37,

LAOE and AOF=37,

LAOB=LBOC=LCOD=etc.=7

And the chords AB, BC, CD, etc., being sides of congruent equilateral triangles are all equal.

Therefore ABCDEF is a regular hexagon.

Cor. Since AOB is an equilateral triangle, AB=AO ; .. the side of a regular hexagon is equal to the radius of its circumcircle.

135°. 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

and, denoting the perimeter of the triangle by 2s, we have

AP=AP'=s,