that of the inscribed polygon of double the number of sides; and the area of a circumscribed polygon is greater than that of the circumscribed polygon of double the number of sides.

PROPOSITION II.-THEOREM.

5. A circle may be circumscribed about any regular polygon; and a may also be inscribed in it.

circle

Let ABCD... be a regular polygon; then, 1st. A circle may be circumscribed about it. For, describe a circumference passing through three consecutive vertices A, B, C (II. 88); let O be its centre, draw OH perpendicular to BC and bisecting it at H, and join OA, OD. Conceive the quadrilateral

=

AOHB to be revolved upon the line OH (i. e., folded over), until HB falls upon its equal HC. The polygon being regular, the angle НВА HCD, and the side BA CD; therefore the side BA will take the direction of CD and the point A will fall upon D. Hence OD = OA, and the circumference described with the radius OA and passing through the three consecutive vertices A, B, C, also passes through the fourth vertex D. It follows that the circumference which passes through the three vertices B, C, D, also passes through the next vertex E, and thus through all the vertices of the polygon. The circle is therefore circumscribed about the polygon.

2d. A circle may be inscribed in it. For, the sides of the polygon being equal chords of the circumscribed circle, are equally distant from the centre; therefore, a circle described with the centre O and the radius OH will touch all the sides, and will consequently be inscribed in the polygon.

6. Definitions. The centre of a regular polygon is the common centre, O, of the circumscribed and inscribed circles.

The radius of a regular polygon is the radius, OA, of the circumscribed circle.

The apothem is the radius, OH, of the inscribed circle.

The angle at the centre is the angle, A OB, formed by radii drawn to the extremities of any side.

7. The angle at the centre is equal to four right angles divided by the number of sides of the polygon.

8. Since the angle ABC is equal to twice ABO, or to ABO+ BAO, it follows that the angle ABC of the polygon is the supplement of the angle at the centre (I. 68).

PROPOSITION III.-THEOREM.

9. Regular polygons of the same number of sides are similar. Let ABCDE, A'B'C'D'E', be regular polygons of the same number of sides; then, they are similar.

For, 1st, they are mutually equiangular, since the magnitude of an angle of either polygon depends only on the number of the

E

sides (7 and 8), which is the same in both.

H

B'

B

H'

CE

2d. The homologous sides are proportional, since the ratio AB: A'B' is the same as the ratio BC: B'C', or CD : C'D', etc. Therefore the polygons fulfill the two conditions of similarity. 10. Corollary. The perimeters of regular polygons of the same number of sides are to each other as the radii of the circumscribed circles, or as the radii of the inscribed circles; and their areas are to each other as the squares of these radii. For, these radii are homologous lines of the similar polygons (III. 43), (IV. 24).

PROPOSITION IV.-PROBLEM.

11. To inscribe a square in a given circle.

Draw any two diameters AC, BD, perpendicular to each other, and join their extremities by the chords AB, BC, CD, DA; then, ABCD is an inscribed square (II. 12), (II. 59).

B

12. Corollary. To circumscribe a square about the circle, draw tangents at the extremities of two perpendicular diameters AC, BD.

13. Scholium. In the right triangle ABO, we have AB'

2

OA2 + OB2 = 204', whence AB = OA. V/2, by which the side of the inscribed square can be computed, the radius being given.

PROPOSITION V.-PROBLEM.

14. To inscribe a regular hexagon in a given circle. Suppose the problem solved, and let ABCDEF be a regular inscribed hexagon.

B

F

E

Join BE and AD; since the arcs AB, BC, CD, etc., are equal, the lines BE, AD, bisect the circumference and are diameters intersecting in the centre O. The inscribed angle ABO is measured by one-half the arc AFE, that is, by AF, or one of the equal divisions of the circumference; the angle AOB at the centre is also measured by one division, that is, by AB; and the angle BAO ABO; therefore the triangle ABO is equiangular, and AB OA. Therefore the side of the inscribed regular hexagon is equal to the radius of the circle.

=

=

Consequently, to inscribe a regular hexagon, apply the radius six times as a chord.

15. Corollary. To inscribe an equilateral triangle, ACE, join the alternate vertices of the regular hexagon.

2

=

16. Scholium. In the right triangle ACD, we have AC2 AD2- DC2 (240) — AO2 = 3A0'; whence, AC = AO. √3, ᎪᎠ by which the side of the inscribed equilateral triangle can be computed, the radius being given.

The apothem, OH, of the inscribed equilateral triangle is equal to one-half the radius OB; for the figure AOCB is a rhombus and its diagonals bisect each other at right angles (I. 110).

The apothem of the inscribed regular hexagon is equal to one-half ΑΟ the side of the inscribed equilateral triangle, that is, to V3; for 2

the perpendicular from O upon AB is equal to the perpendicular from A upon OB, that is, to AH.

The angle at the centre of the regular inscribed hexagon is of 4 right angles, that is, of one right angle = 60°.

The angle of the hexagon, or ABC, is of a right angle

120°.

The angle at the centre of the inscribed equilateral triangle is of one right angle

=

120°.

PROPOSITION VI.-PROBLEM.

17. To inscribe a regular decagon in a given circle. Suppose the problem solved, and let ABC.... L, be a regular inscribed decagon.

Join AF, BG; since each of these lines bisects the circumference, they are diameters and intersect in the centre 0. Draw BK intersecting OA in M.

The angle AMB is measured by half the sum of the arcs KF and AB (II. 64), that is,

H

E

by two divisions of the circumference; the inscribed angle MAB is measured by half the arc BF, that is, also by two divisions; therefore AMB is an isosceles triangle, and MB

=

AB.

Again, the inscribed angle MBO is measured by half the arc KG, that is, by one division, and the angle MOB at the centre has the same measure; therefore OMB is an isosceles triangle, and OM : MB = AB.

=

The inscribed angle MBA, being measured by half the arc AK, that is, by one division, is equal to the angle AOB. Therefore the isosceles triangles AMB and AOB are mutually equiangular and similar, and give the proportion

that is, the radius OA is divided in extreme and mean ratio at M (III. 73); and the greater segment OM is equal to the side AB of the inscribed regular decagon.

Consequently, to inscribe a regular decagon, divide the radius in .extreme and mean ratio (III. 74), and apply the greater segment ten times as a chord.

18. Corollary. To inscribe a regular pentagon, A CEGK, join the alternate vertices of the regular inscribed decagon.

19. Scholium. By (III. 75), we have

E

by which the side of the regular decagon may be computed from the radius.

The angle at the centre of the regular decagon is of one right angle = 36°; the angle at the centre of the regular pentagon is of one right angle 72°.

=

The angle ABC of the regular decagon is of one right angle 144°; the angle ACE of the regular pentagon is g of one right angle = 108°.

PROPOSITION VII.-PROBLEM.

20. To inscribe a regular pentedecagon in a given circle. Suppose AB is the side of a regu

lar inscribed pentedecagon, or that the arc AB is of the circumfer

ence.

Now the fraction

ference between

and

B

; therefore the arc AB is the difof the circumference. Hence, if we

inscribe the chord AC equal to the side of the regular inscribed hexagon, and then CB equal to that of the regular inscribed decagon, the chord AB will be the side of the regular inscribed pentedecagon required.

21. Scholium. Any regular inscribed polygon being given, a regular inscribed polygon of double the number of sides can be formed by bisecting the arcs subtended by its sides and drawing the chords of the semi-arcs (4). Also, any regular inscribed polygon being given, a regular circumscribed polygon of the same number of sides can be formed (3). Therefore, by means of the inscribed square, we can inscribe and circumscribe, successively, regular polygons of 8, 16, 32, etc., sides; by means of the hexagon, those of 12, 24, 48, etc., sides; by means of the decagon, those of 20, 40, 80, etc., sides; and,