.. the six s AOB, BOC, COD, DOE, EOF, FOA are

all equal;

.. the six arcs AB, BC, CD, DE, EF, FA are all equal;

.. the six chords AB, BC, CD, DE, EF, FA are all equal.

(2) To prove the hexagon equiangular.

III. 26

III. 29

Since the six arcs AB, BC, CD, DE, EF, FA are all equal, .. each is one-sixth of the whole Oce;

.. any four of them

=

four-sixths of the whole Oce.

Now the six s FAB, ABC, BCD, CDE, DEF, EFA stand each on an arc = four-sixths of the Oce; .. these six angles are all equal.

III. 27

COR.-The side of a regular hexagon inscribed in a circle

is equal to the radius.

1. If the points A, C, E be joined, ▲ ACE is equilateral.

2. The area of an inscribed equilateral triangle is half that of a regular hexagon inscribed in the same circle.

3. Construct a regular hexagon on a given straight line.

4. The area of an equilateral triangle described on a given straight line is one-sixth of the area of a regular hexagon described on the same straight line.

5. The opposite sides of a regular hexagon are parallel.

6. The straight lines which join the opposite vertices of a regular hexagon are concurrent, and are each || one of the sides.

7. How many diagonals can be drawn in a regular hexagon? 8. Prove that six of them are parallel in pairs.

9. The area of a regular hexagon inscribed in a circle is half of the area of an equilateral triangle circumscribed about the circle. 10. The square on a side of an inscribed regular hexagon is one-third of the square on a side of the equilateral triangle inscribed in the same circle.

11. What is the magnitude of the angle at the centre of a circle subtended by a side of an inscribed regular hexagon ?

12. Give the constructions for inscribing a circle in a regular hexagon; and for circumscribing a regular hexagon about a circle, and a circle about a regular hexagon.

PROPOSITION 16. PROBLEM.

To inscribe a regular quindecagon in a given circle.

A

B

E

Let ABC be the given circle:

it is required to inscribe a regular quindecagon in ABC.

IV. 2

Find AC a side of an equilateral triangle inscribed in the circle ; and find AB, BE two consecutive sides of a regular pentagon inscribed in the circle.

IV. 11

arc

CE(), or, of the O°°.

Hence, if CE be joined, CE will be a side of a regular quindecagon inscribed in the O ABC.

Place consecutively in the O chords equal to CE; IV.1 then a regular quindecagon will be inscribed in the circle.

1. How could the regular quindecagon be obtained, if, besides AC, a side of an equilateral triangle, only one side AB of the regular pentagon be drawn?

2. How could the regular quindecagon be obtained by making use of the sides of the regular inscribed hexagon and decagon?

3. In a given circle inscribe a triangle whose angles are as the numbers 2, 5, 8; and another whose angles are as the numbers 4, 5, 6.

4. Give the constructions for inscribing a circle in a regular quindecagon; and for circumscribing a regular quindecagon about a circle, and a circle about a regular quindecagon.

5. How many diagonals can be drawn in a regular quindecagon ? 6. Show that if a polygon have n sides, it will have n(n − 3) diagonals.

7. Show that the centres of the circles inscribed in, and circumscribed about, any regular figure coincide, and are obtained by bisecting any two consecutive angles of the figure.

NOTE 1.-The regular polygons of 3, 4, 5, and 15 sides, and such as may be derived from them by continued arcual bisection, were, till the time of Gauss, the only ones discovered by the ancient Greek, and known to the modern European, geometers to be inscriptible in a circle by the methods of elementary geometry. Gauss, in 1796, found that a regular polygon of 17 sides was inscriptible, and in his Disquisitiones Arithmeticæ, published in 1801, he showed that any regular polygon was inscriptible, provided the number of its sides was a prime number, and expressible by 2" + 1. (A good account of Gauss and his works is given in Nature, vol. xv. pp. 533-537.)

NOTE 2. The polygons of which Euclid treats are all of one kind, namely, convex polygons, that is to say, polygons each of whose angles is less than two right angles. There are others, however, called re-entrant, and intersectant (or concave, and crossed), such as ABCD in the accompanying figures. The reader will find

AA

B

D B

it instructive to inquire how far the properties of convex polygons (for example, quadrilaterals) are true for the others. Among the intersectant polygons there is a class called stellate or star, which

are obtained thus: Suppose A, B, C, D, E (see fig. to IV. 11) to be five points in order on the Oce of a circle. Join AC, CE, EB, BD, DA; then ACEBD is a star pentagon. If the arcs AB, BC, &c. are all equal, the star pentagon ACEBD is regular. Similarly, if 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 denote the vertices of a regular decagon inscribed in a circle, the regular star decagon (there can be only one) is got by joining consecutively 1, 4, 7, 10, 3, 6, 9, 2, 5, 8, 1. It will be found that if a regular polygon have n sides, the number of regular star polygons that may be derived from it is equal to the number of integers prime to n contained in the series 2, 3, 4, (n-1). (For more information on the subject of star polygons, see Chasles, Aperçu Historique sur l'Origine et le Développement des Méthodes en Géométrie, sec. éd. pp. 476–487, and Georges Dostor, Théorie Générale des Polygones Étoilés, 1880.)

...

APPENDIX IV.

PROPOSITION 1.

To describe a circle which shall touch three given straight lines.

(1) If the three straight lines be so situated that every two are parallel, the solution is impossible.

(2) If they be so situated that only two are parallel, there can be two solutions, as will appear from the following figure :

Let AB, CD, EF be the three straight lines of which AB is || CD.

Bisect 48 AEF, CFE by EI and FI, which meet at I; from I draw IH, IK, IL respectively 1 AB, EF, CD. Then As IEH, IEK are equal in all respects;

.. IH = IK.

Similarly, IK = IL;

.. IH = IK = IL.

Now since s at H, K, L are right,

I. 9

I. 12

I. 26

I. 26

.. the circle described with I as centre and IH as radius will touch AB, EF, CD.

III. 16 A similar construction on the other side of EF will give another circle touching the three given straight lines.

(3) If they be so situated that no two are parallel, then they will either all pass through the same point, in which case the solution is impossible; or they will form a triangle with its sides produced, in which case four solutions are possible.

Let AB, BC, CA produced be the three given straight lines forming by their intersection the ▲ ABC.

If the interior 48 B and C be bisected, the bisectors will meet at some point I, which is the centre of the circle inscribed in the triangle, as may be proved by drawing perpendiculars ID, IE, IF to the sides BC, CA, AB of the triangle.

IV. 4

If the exterior angles at B and C be bisected by BI1, CI, which