SX to PQ. On SX take SE=PQ and join E with the fourth point R. ER is a side of the square in position and direction, and the points first joined, P and Q, are on opposite sides of the required square.

Thence the square is readily constructed.

Since SE may be measured in two directions along the line SX, two squares can have their sides passing through the same four points P, Q, R, S, and having P and Q on opposite sides.

Also, since P may be first connected with R or S, two squares can be constructed fulfilling the conditions and having P and Q on adjacent sides.

Therefore, four squares can be constructed to have their sides passing through the same four given points.

CIRCLES FULFILLING GIVEN CONDITIONS.

The problems occurring here are necessarily of an elementary character. The more complex problems require relations not yet developed.

129°. Problem.—To describe a circle to touch a given line at a given point.

IM

P is a given point in the line L.

Constr.-Through P draw M

to L. A circle having any point C, on M, as centre and CP as radius touches L at P.

P

Proof.-L is to the diameter at its end-point, therefore L is tangent to the circle. (110°, Cor. 3)

L

Def.-As C is any point on M, any number of circles may be drawn to touch L at the point P, and all their centres lie on M.

Such a problem is indefinite because the conditions are not sufficient to determine a particular circle. If the circle

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 CQP=≤CPQ.

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

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 bisector, draw CA to L.

L

N

M

N

O

B

B

A

A

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

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, E1. the internal and external bisectors of the angle A; and I2, E2, those of the angle B.

LA +≤B is < 1, .'. LBAO+LABO is <

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

(79°, Cor.)

M

E1

A

*

C

10.

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

The four points 0, 01, 02, 03 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 I1 and E1 as their centre-locus (130°), and circles which touch N and L have I, and Eg 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, 01, 02, and 03,

.. O, 01, 02, and O, 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 I, 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

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 Isaid 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 ▲. 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 O1, O2, and O3 are the escribed or ex-circles of the triangle.

29

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

(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.

But

and (132°, Def. 2)

or

LOAB=LOBA=LOBC=etc.,

LOAB+LOBA=LABC.
LOAB+LOBA=1-LAOB,

LAOB-4 right angles

n

LABC= (2–4) right angles,

=(2-4)90°.

q.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 :—

=

A

Octagon,

Decagon,

LAOB LEOD=3,

LAOE=, and AOF=3
LAOB=LBOC=LCOD=etc.=

3135° 144°

Dodecagon, 150°

B

134°. Problem.-On a given line-segment as side to construct a regular hexagon.

A

B

Let AB be the given segment.

E

с

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.

D

C