and circumscribed circles, is also called the center of the polygon; and the perpendicular from the center upon one of the sides, that is, the radius of the inscribed circle, is called the apothem of the polygon.

Since all the chords AB, BC, &c., are equal, the angles at the center, AOB, BOC, &c., are equal; and the value of each may be found by dividing four right angles by the number of sides of the polygon.

Scholium 2. To inscribe a regular polygon of any number of sides in a circle, it is only necessary to divide the circumference into the same number of equal parts; for, if the arcs are equal, the chords AB, BC, CD, &c., will be equal. Hence the triangles AOB, BOC, COD, &c., will also be equal, because they are mutually equilateral; therefore all the angles ABC, BCD, CDE, &c., will be equal, and the figure ABCDEF will be a regular polygon.

PROPOSITION III. PROBLEM.

To inscribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in it.

E

C

Draw two diameters AC, BD at right angles to each other; and join AB, BC, CD, DA. Because the angles AEB, BEC, &c., are equal, the chords AB, BC, &c., are also equal. And because the angles ABC, BCD, &c., are inscribed in semicircles, they are right angles (Prop. XV., Cor. 2, B. III.). Therefore ABCD is a square, and it is inscribed in the circle ABCD.

B

Cor. Since the triangle AEB is right-angled and isosceles, we have the proportion, AB : AE:2:1 (Prop. XI., Cor. 3, B. IV.); therefore the side of the inscribed square is to the radius, as the square root of 2 is to unity.

PROPOSITION IV. THEOREM.

The side of a regular hexagon is equal to the radius of the circumscribed circle.

Let ABCDEF be a regular hexagon inscribed in a circle whose center is O; then any side as AB will be equal to the radius AO.

Draw the radius BO. Then the angle AOB is the sixth part of four right angles (Prop. II., Sch. 1), or the third part of two right angles. Also, because the three an- A gles of every triangle are equal to two right angles, the two angles OAB, OBA are together equal to two thirds of two

B

D

right angles; and since AO is equal to BO, each of these angles is one third of two right angles. Hence the triangle AOB is equiangular, and AB is equal to AO. Therefore the side of a regular hexagon, &c.

Cor. To inscribe a regular hexagon in a given circle, the radius must be applied six times upon the circumference By joining the alternate angles A, C, E, an equilateral triangle will be inscribed in the circle.

PROPOSITION V. PROBLEM.

To inscribe a regular decagon in a given circle.

Let ABF be the given circle; it is required to inscribe in it a regular decagon.

Take C the center of the circle; draw the radius AC, and divide it in extreme and mean ratio (Prob. XX., B. V.) at the point D. Make the chord AB equal to CD the greater segment; then will AB be the side of a regular decagon inscribed in the circle.

A

F

B'

E

Join BC, BD. Then, by construction, AC: CD :: CD: AD; but AB is equal to CD; therefore AC: AB::AB: AD. Hence the triangles ACB, ABD have a common angle A included between proportional sides; they are therefore similar (Prop. XX., B. IV.) And because the triangle ACB is isosceles, the triangle ABD must also be isosceles, and AB is equal to BD. But AB was made equal to CD; hence BD is equal to CD, and the angle DBC is equal to the angle DCB. Therefore the exterior angle ADB, which is equal to the sum of DCB and DBC, must be double of DCB. But the angle ADB is equal to DAB; therefore each of the angles CAB, CBA is double of the angle ACB. Hence the sum of the three angles of the triangle ACB is five times the angle C. But these three angles are equal to two right angles (Prop. XXVII., B. I.); therefore the angle C is the fifth part of two right angles, or the tenth part of four right angles. Hence the arc AB is one tenth f

the circumference, and the chord AB is the side of a regular decagon inscribed in the circle.

Cor. 1. By joining the alternate angles of the regular decagon, a regular pentagon may be inscribed in the circle. Cor. 2. By combining this Proposition with the preceding a regular pentedecagon may be inscribed in a circle.

For, let AE be the side of a regular hexagon; then the arc AE will be one sixth of the whole circumference, and the arc AB one tenth of the whole circumference. Hence the arc BE will be — or, and the chord of this arc will be the side of a regular pentedecagon.

Scholium. By bisecting the arcs subtended by the sides of any polygon, another polygon of double the number of sides may be inscribed in a circle. Hence the square will enable us to inscribe regular polygons of 8, 16, 32, &c., sides; the hexagon will enable us to inscribe polygons of 12, 24, &c., sides; the decagon will enable us to inscribe polygons of 20, 40, &c., sides; and the pentedecagon, polygons of 30, 60, &c., sides.

The ancient geometricians were unacquainted with any method of inscribing in a circle, regular polygons of 7, 9, 11, 13, 14, 17, &c., sides; and for a long time it was believed that these polygons could not be constructed geometrically; but Gauss, a German mathematician, has shown that a regu lar polygon of 17 sides may be inscribed in a circle, by employing straight lines and circles only.

PROPOSITION VI. PROBLEM.

A regular polygon inscribed in a circle being given, to de scribe a similar polygon about the circle.

Let ABCDEF be a regular polygon inscribed in the circle ABD; it is required to describe a similar polygon about the circle.

L

E

A

D

Bisect the arc AB in G, and through G draw the tangent LM. Bisect also the arc BC in H, and through H draw the tangent MN, and in the same manner draw tangents to the middle points of the arcs CD, DE, &c. These tangents, by their intersections, will form a circumscribed polygon similar to the one inscribed.

M H

N

Find O the center of the circle, and draw the radii OG OH. Then, because OG is perpendicular to the tangent LM (Prop. IX., B. III.), and also to the chord AB (Prop. VI.

Sch., B. III.), the tangent is parallel to the chord (Prop. XX., B. I.). In the same manner it may be proved that the other sides of the circumscribed polygon are parallel to the sides of the inscribed polygon; and therefore the angles of the circumscribed polygon are equal to those of the inscribed one (Prop. XXVI., B. I.).

Since the arcs BG, BH are halves of the equal arcs AGB, BHC, they are equal to each other; that is, the vertex B is at the middle point of the arc GBH. Join OM; the line OM will pass through the point B. For the right-angled triangles OMH, OMG have the hypothenuse OM common, and the side OH equal to OG; therefore the angle GOM is equal to the angle HOM (Prop. XIX., B. I.), and the line OM passes through the point B, the middle of the arc GBH.

Now because the triangle OAB is similar to the triangle OLM, and the triangle OBC to the triangle OMN, we have the proportions

also,

AB: LM: BO: MO;

therefore (Prop. IV., B. II.),

AB: LM :: BC: MN.

But AB is equal to BC; therefore LM is equal to MN. In the same manner, it may be proved that the other sides of the circumscribed polygon are equal to each other. Hence this polygon is regular, and similar to the one inscribed.

Cor. 1. Conversely, if the circumscribed polygon is given, and it is required to form the similar inscribed one, draw the lines OL, OM, ON, &c., to the angles of the polygon; these lines will meet the circumference in the points A, B, C, &c. Join these points by the lines AB, BC, CD, &c., and a similar polygon will be inscribed in the circle.

Or we may simply join the points of contact G, H, I, &c., by the chords GH, HI, &c., and there will be formed an in scribed polygon similar to the circumscribed one.

Cor. 2. Hence we can circumscribe about a circle, any regular polygon which can be inscribed within it, and conversely.

Cor. 3. A side of the circumscribed polygon MN is equal to twice MH, or MG+MH.

PROPOSITION VII. THEOREM.

The area of a regular polygon is equivalent to the produc of its perimeter, by half the radius of the inscribed circle.

Let ABCDEF be a regular polygon, and G the center of

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the inscribed circle. From G draw lines to all the angles of the polygon. The polygon will thus be divided into as many triangles as it has sides; and the common altitude of these triangles is GH, the radius of the circle. Now, the area of the triangle BGC is equal to the product of BC by the half of GH (Prop. VI., B. IV.); and so of all the other triangles having their vertices in G. Hence the sum of all the triangles, that is, the surface of the polygon, is equivalent to the product of the sum of the bases AB, BC, &c.; that is, the perimeter of the polygon, multiplied by half of GH, or half the radius of the inscribed circle. Therefore, the area of a regular polygon, &c.

B

H

PROPOSITION VIII. THEOREM.

The perimeters of two regular polygons of the same number of sides, are as the radii of the inscribed or circumscribed circles, and their surfaces are as the squares of the radii.

Let ABCDEF, abcdef be

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D a

two regular polygons of the same number of sides; let G and g be the centers of the A circumscribed circles; and let GH, gh be drawn perpendicular to BC and bc; then will the perimeters of the polygons be as the radii BG, bg; and, also, as GH, gh, the radii of the inscribed circles.

B

C

The angle BGC is equal to the angle bgc (Prop. II., Sch. 1); and since the triangles BGC, bgc are isosceles, they are similar. So, also, are the right-angled triangles BGH, bgh; and, consequently, BC: bc:: BG: bg :: GH: gh. But the perimeters of the two polygons are to each other as the sides BC, bc (Prop. I., Cor.); they are, therefore, to each other as the radii BG, bg of the circumscribed circles; and also as the radii GH, gh of the inscribed circles.

The surfaces of these polygons are to each other as the squares of the homologous sides BC, bc (Prop. I., Cor.); they are, therefore, as the squares of BG, bg, the radii of the cir cumscribed circles; or as the squares of GH, gh, the radii of the inscribed circles.