then draw the line DE parallel to the diameter AB; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the required rectangle. For, their sum is equal to AB; and their rectangle AFX FB is equivalent to the square of EF, or to the square of AD; hence, this rectangle is equivalent to the given square C Scholium. The problem is impossible, if the distance AD exceeds the radius; that is, the side of the square C must not exceed half the line AB. PROBLEM XVII. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line. Let C denote the given square, and AB the difference of the sides of the rectangle. Upon the given line AB, as a diameter, describe a circumference. At the extremity of the diameter, draw the tangent AD, and make it equal to the side of the square C; through the point D and the centre O draw the secant DOF, intersecting the circumference in E and , F D A E B F For, the difference of these sides is equal to the diame ter EF or AB; and the rectangle DE, DF is equal to AD3 (P. 30); hence, the rectangle DFX DE, is equivalent to the given square C PROBLEM XVIII. To find the common measure, between the side and diagonal of a square. G E Let ABCG be any square, and AC its diagonal. We first apply CB upon CA. For this purpose let the semicircumference DBE be described, from the centre C, with the radius CB, and produce AC to E. It is evident that CB is contained once in AC, with the remainder AD. The result of the first operation is, therefore, a quotient 1, with the remainder AD. This remainder must now be compared with BC, or its equal AB. A F B Since the angle ABC is a right angle, AB is a tangent, and since AE is a secant drawn from the same point, we have (P. 30), AD : AB : :: AB AE. Hence, in the second operation, where AD is compared with AB, the equal ratio of AB to AE may be taken instead : but AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore a quotient 2 with the remainder AD, and this must be again compared with AB. Thus, the third operation consists in comparing again AD with AB, and may be reduced in the same manner to the comparison of AB or its equal CD with AE; from which there will again be obtained a quotient 2, and the remainder AD. Hence, it is evident that the process will never terminate, and consequently that no remainder is contained in its divisor an exact number of times; therefore, there is no common measure between the side and the diagonal of a square. This property has already been shown, since (P. 11, c. 5), AB AC :: : 1 : √2, but it acquires a greater degree of clearness by the geometrical investigation. BOOK V. REGULAR POLYGONS-MEASUREMENT OF THE CIRCLE. DEFINITION. A REGULAR POLYGON is one which is both equilateral and equiangular. A regular polygon may have any number of sides. The equilateral triangle is one of three sides; the square, is one of four. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar figures. Let ABCDEF, abcedf, be two regular polygons. the number of sides; and the same is true of either angle of the other polygon (B. I., P. 26, c. 4); hence (4.1), the angles of the polygons are equal. Again, since the polygons are regular, the sides AB, BC, CD, &c., are equal, and so likewise the sides ab, bc, cd (D.), &c.; hence AB: ab :: BC: bc :: CD: cd, &c.; therefore, the two polygons have their angles equal, and their sides taken in the same order proportional; consequently, they are similar (B. IV., D. 1). Cor. 1. The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their surfaces are to each other as the squares of those sides (B. IV., p. 27). P. Cor. 2. The angle of a regular polygon, like the angle of an equiangular polygon, is determined by the number of its sides (B. I., P. 26, c. 4). PROPOSITION II. THEOREM. 4 regular polygon may be circumscribed by the circumference of a circle, and a circle may be inscribed within it. A Let HGFE, &c., be any regular polygon. Through the three points A, B, C, describe the circumference of a circle: the centre O will lie in the line OP, drawn perpendicular to BC at the middle point P (B. III., P. 6, s.) Then draw OB and OC. H B R E G F If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for, the side OP is common; the angle OPC=0PB, each being a right angle; hence, the side PC will apply to its equal PB, and the point C will fall on B: besides, the polygon being regular, the angle PCD=PBA (D.); hence, CD will take the direction BA; and since CD=BA, the point D will fall on A, and the two quadrilaterals will coincide. Hence, OD is equal to AO; and consequently, the circumference which passes through the three points A, B, C, will also pass through the point D. In the same manner it may be shown, that the circumference which passes through the three points B, C, D, will also pass through the point E; and so of all the other vertices; hence, the circumference which passes through the points A, B, C, passes also through the vertices of all the angles of the polygon, consequently, the circum'ference of the circle circumscribes the polygon (B. III., D. 7). Again, in reference to this circle, all the sides AB, BC, CD, &c., of the polygon, are equal chords; they are therefore equally distant from the centre (B. III., P. 8): hence, if from the point O as a centre, with the distance OP, a circumference be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon (B. III., D. 11). Scholium. The point 0, the common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and the angle AOB is called the angle at the centre, being formed by two lines drawn from the centre to the extremities of the same side AB. The perpendicular OP, is called the apothem of the polygon. E Cor. 1. Since all the chords AB, BC, CD, &c., are equal, all the angles at the centre are likewise equal (B. III., P. 4); and therefore, the value of any angle will be found by dividing four right angles by the number of sides of the polygon. Cor. 2. To inscribe a regular polygon of any number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides; for, when the arcs. are equal, the chords AB, BC, CD, &c., are also equal (B. III., P. 4); H A B hence, likewise the triangles AOB, BOC, COD, must be equal, because their sides are equal each to each (B. I., P. 10); therefore, by addition, all the angles ABC, BCD, CDE, &c., are equal (A.2); hence, the figure ABCDEH, is a regular polygon. PROPOSITION III. PROBLEM. To inscribe a square in a given circle. Draw two diameters AC, BD, intersecting each other at right angles; join their extremities A, B, C, D, the figure ABCD will be a square. For, the angles AOB, BOC, &c., being equal, the chords AB, BC, &c., are also equal (B. III., P. 4) : and the angles ABC, BCD, &c., being inscribed in semicircles, are right angles (B. III., P. 18, C. 2).. Scholium. Since the triangle BCO is right-angled and isosceles, B A C we have (B. IV., P. 11, c. 5), BC : BO D :: √/2 : 1, hence, the side of the inscribed square is to the radius, as the square root of two, to unity. |