Euclid no essential advance was made in the problem of dividing the circle until the year 1796, when Gauss * proved the possibility of dividing the circle into n equal parts, if n is any prime number that exceeds a power of 2 by unity. The first four numbers that satisfy this condition are 3, 5, 17, 257. Gauss further proved that the division can be performed if n is the product of any two or more different numbers of this series; the first four numbers that satisfy this condition are 15, 51, 85, 255. Gauss gave the complete analysis for the case of 17 parts, and proved that the problem can be reduced to simpler ones that depend ultimately on the postulates of construction; but the method of proof is beyond the range of elementary geometry.

INSCRIBED AND CIRCUMSCRIBED CIRCLES

The following theorem and its corollaries furnish the basis for the two succeeding problems, which relate to the construction of the inscribed and circumscribed circles of any given regular polygon.

Concurrence of angle-bisectors.

130. THEOREM 36. The lines that bisect the angles of any regular polygon all meet in a point.

Outline. Let A, B, C, D be consecutive angles of any regular polygon. Bisect the angles A and B; and prove that the bisectors meet at that side of the line, AB, at which the polygon itself is (I. 124). Let the bisectors meet at the point O, and draw OC. Prove that OC bisects the angle C. (This is done by proving that the angle OCB equals OAB, which equals half of the angle A, and hence equals half C.) Prove similarly that OD bisects the angle D; and so on.

131. Cor. I. In any regular polygon, the segments of the angle-bisectors intercepted between the vertices and the point of concurrence are all equal.

132. Cor. 2. In any regular polygon, the perpendiculars from the intersection of the angle-bisectors to the sides are all equal.

*"Disquisitiones Arithmeticæ," published in 1801.

Circumscribed circle.

133. PROBLEM 15. To circumscribe a circle about a given regular polygon. (Use 130, 131.)

Inscribed circle.

134. PROBLEM 16. To inscribe a circle in a given regular polygon. (Use 130, 132.)

135. Cor. The inscribed and circumscribed circles of a regular polygon are concentric.

Ex. 1. If two regular polygons are equal, then their inscribed circles are equal, and so are their circumscribed circles.

Ex. 2. If two regular polygons of the same number of sides are inscribed in equal circles, then the two polygons are equal.

136. Definitions. The common center of the inscribed and circumscribed circles of a regular polygon is called the center of the regular polygon. The angle at the center subtended by any side of the polygon is called the central angle of the regular polygon.

In a regular polygon, a line joining the center to any vertex is called a radius, and a perpendicular from the center to any of the sides is called an apothem. Thus a radius of a regular polygon is a radius of its circumscribed circle, and an apothem is a radius of its inscribed circle.

Ex. 1. If any two regular polygons have the same number of sides, then their central angles are equal.

Ex. 2. If two regular polygons have the same number of sides, and if the radius of one is greater than the radius of the other, then the apothem of the first is greater than the apothem of the second, the side of the first is greater than the side of the second, and the surface of the first is greater than the surface of the second.

[Place the polygons with their centers in coincidence, and so that each radius of the first may fall on a radius of the second; then prove that each side of the first is parallel to a side of the second; etc.]

Ex. 3. If two regular polygons have the same number of sides, then the following pairs of magnitudes are in the same order of size :

The bases (b, b'); the radii (r, r'); the apothems (a, a'); the surfaces (s, s'); the perimeters (p, p'). 、

Equivalent rectangle.

137. THEOREM 37. A polygon circumscribed about a circle is equivalent to the rectangle contained by the perimeter and half the radius of the circle.

Ex. A regular polygon is equivalent to the rectangle contained by the perimeter and half the apothem.

MUTUALLY EQUILATERAL POLYGONS

138. THEOREM 38. If two circles are equal, and if two mutually equilateral polygons are inscribed in them, then the polygons are equal.

Outline. Compare the respective triangles whose vertices are at the centers and whose bases are corresponding sides of the polygons.

139. THEOREM 39. If two mutually equilateral polygons are each circumscribable by a circle, then the circles are equal, and the polygons are equal.

Outline. Suppose the radii unequal. Compare the respective central angles subtended by corresponding sides; see ex. 42 at end of Book I. Reduce to absurdity.

140. Cor. If in two semicircles are inscribed two broken lines, and if the segments of one broken line are respectively equal to those of the other, taken in order, then the semicircles are equal, and the two figures are superposable. [Prove as in 139.]

EXERCISES

1. Two regular polygons of the same number of sides circumscribed about equal circles are equal.

2. The center of the inscribed circle of a triangle is the orthocenter of the triangle formed by the centers of the escribed circles.

3. Show how to cut off the corners of an equilateral triangle so as to leave a regular hexagon; also of a square to leave a regular octagon.

4. If a parallelogram is circumscribed to a circle, then it is a rhombus.

5. Show that it is possible to trisect the central angle of a regular n-gon, when n is any one of the first or third series of Euclid's numbers (111).

6. In the same cases show that it is possible to trisect the interior or exterior angle of the regular n-gon.

7. Show that it is possible to divide a right angle into n equal parts, when n is any of Euclid's numbers.

8. Show that it is possible to divide the angle of an equilateral triangle into n equal parts, when n is any number belonging to the first three series of Euclid's numbers.

NOTE ON EXS. 5-8. The general problem to trisect a given arbitrary angle is one of the famous problems of antiquity, and has never been solved by methods permitted in elementary geometry. Modern mathematicians have demonstrated that this general problem cannot be analyzed into simpler ones that require only the drawing of straight lines and circles.* Thus the construction cannot be performed by means of only a pair of compasses and an unmarked straightedge. Several general solutions are known which overstep these limitations to a greater or less degree. One of the simplest employs the sliding motion of a straightedge on which two points are marked. There are, however, certain special angles that can be trisected by the methods of elementary geometry. (See exs. 5-8 above.)

The still more general problem of dividing a given arbitrary angle into n equal parts can be solved only when n is one of the first series of Euclid's numbers (111; I. 73); but in the case of certain special angles the problem can be solved for some other values of n (exs. 7, 8).

MAXIMA AND MINIMA †

141. Certain principles of maxima and minima relating to triangles were considered in Book II. 92-108. Similar principles can now be extended to polygons in general, subject to certain given conditions. The theorems here considered fall into five groups according to the nature of the assigned conditions.

* See Klein's "Vorträge über ausgewählte Fragen der Elementar Geometrie." (Translated by Professors Beman and Smith.)

This topic is discussed here on account of its intimate connection with the properties of the circle, and of inscribed and circumscribed polygons. It may, however, be postponed without inconvenience.

GIVEN SIDES

This group of two theorems with their corollaries will show how to make the surface of a polygon a maximum subject to various assigned conditions relating to the magnitude of the sides. In each case the additional condition is to be proved both necessary and sufficient for a maximum; and accordingly each theorem is accompanied by its converse (II. 93).

Greatest polygon with one arbitrary side.

142. THEOREM 40. Among the polygons that have all the sides but one equal respectively to given lines taken in order, any one that is a maximum is circumscribable by a semicircle having the undetermined side as diameter.

Let the polygon ABCDEF be a maximum subject to the condition that the sides AB, BC, CD, DE, EF are respectively equal to given lines taken in

D

E

the semicircle does not pass through C; and draw CA, CF. Then the angle ACF is not a right angle (55, 56).

F

Hence, by rotating the figures ABC and FEDC about the point C until ACF becomes a right angle, the triangle ACF could be increased (II. 94); and therefore the whole polygon ABCDEF could be increased without changing any of the given sides. This is contrary to the hypothesis that ABCDEF is a maximum under the given conditions.

Hence the semicircle described on AF passes through the point c. Similarly it passes through the other vertices.