By inspecting the diagram we see that we have the following relations: A - 2n × OAK, B=2n × OMI, A'— 2n × OAI, B'= 2n × OAmI; hence, the ratios between the areas A, B, A', B', taken two and two, are the same as those between the triangles OAK, OMI, OAI, and of the quadrilateral OAmI; thus the whole is reduced to the determination of the ratios which exists between these last four figures. This being premised, we have immediately, in comparing the three triangles OMI, OAI, OAK, and observing that AK is parallel to MI, OMI: OAI::OAI: OAK (B. III., T. XXVI., C.); or, replacing these ratios of triangles by those of the polygons, we have that is, B: A':: A': A; In the second place, since in the triangle OMI, the straight line Om divides the angle O into two equal parts, we have the proportion Mm: mI:: OM: OI (B. III., T. XII.). Now, the two triangles OMm, OmI, having the same altitude OI, give OMm: OmI:: Mm: mI (B. III., T. XXIII., C. I.). Also, since the triangles OAI, OAK have the same altitude AK, we have OAI: OAK::OI: OK:: OM: OA:: OM: OI; hence, we have consequently, OMm: OmI:: OAI: OAK; OMm+OmI: OmI :: OAI+OAK:OAK; or doubling the consequents, and observing that OMm+OmI= OMI, and 2 × OmI = OAmI, we have OMI: OAMI:: OAI + OAK: 2 × OAK. Finally, substituting, in place of the three triangles OMI, OAI, OAK, and of the quadrilateral OAmI, the polygons of which they are like parts, we shall obtain Scholium. These two formulas show, at a glance, that A'> A and B'<B. Thus the areas of the regular inscribed polygons in the same circle will continue to augment as the number of sides, arising from doubling each time, increases. But the areas of the cir cumscribed polygons will decrease as the sides increase. However great may be the number of sides of the regular inscribed and circumscribed polygons, it is evident that the area of an inscribed polygon cannot exceed the area of the circle, and the area of a circumscribed polygon cannot be less than the circle. The same is true in regard to their perimeters as compared with the circumference of the circle. To prove this, it will be necessary only to consider the portion of the figure corresponding to the arc AB, since the same course of reasoning may be repeated for each of the corresponding portions BC, CD, etc. A L I K B Let the arc AB be divided into two, four, eight, etc., equal parts. Draw the chords of these new arcs. From the definition of a straight line, we evidently have chord AB <AI+IB < AL+LI+IK+KB <, etc., <arc AB. Hence, the perimeters of the regular inscribed polygons, in the same circle, will continue to augment as the number of sides increase, without being able to surpass the circumference of the circle. Now, let AM and BM be two half sides of the first regular circumscribed polygon corresponding to the arc AB; mn, and Am, Bn a side and half sides of the regular circumscribed polygon of double the number of sides. A m M I n B We shall have mM + Mn > mn; adding Am +nB to each member, we have Am+mM+Mn+nB, or AM+MB> Am + mn + nB. Hence, the perimeters of the regular circumscribed polygons will continue to decrease as the number of sides increase, without being able to become less than the circumference. This being premised, we give, in mathematics, the name of LIMIT to a constant and determinate magnitude, towards which a variable magnitude is constantly converging, either by increasing or diminishing. Hence, we say that: The area of a circle is the SUPERIOR limit of the areas of the regular inscribed polygons, and the INFERIOR limit of the areas of the regular circumscribed polygons. The circumference of a circle is the superior limit of the perimeters of the first polygons, and the inferior limit of the second. We feel authorized to regard the circle as a regular polygon of an infinite number of sides, infinitely small; these sides are called the elements of the curve. And this new definition of a circle, if it does not at first appear very rigorous, has at least the advantage of giving greater simplicity and precision to our demonstrations. THE DETERMINATION OF THE SIDES AND OF THE AREAS OF REGULAR POLYGONS OF A PARTICULAR KIND. THEOREM IX. The side of a regular inscribed hexagon is equal to the radius. Draw OA and OB; the value of the angle O, of the triangle OAB, is 4 = 3 of a right angle, hence the sum of the other two angles is 2 - 3 = 3 of a right angle, and since OA = OB, it follows that the angle OAB = angle OBA = of of 1⁄2 }=} E D F C L I K a right angle; hence the triangle OAB is equilateral, and gives AB=OA OB=R. Cor. I. The side AC of an inscribed equilateral triangle is to the radius as √3:1. If in formula (2), of Scholium to T. IV., we make a = R, we shall find hence, _ AC=R √3R2=R√3; Ꭱ AC: R:: √3:1. Cor. II. The side of a circumscribed equilateral triangle is double the side of the inscribed one. To show this, it is sufficient to substitute R√3 for a in formula (3) of T. VII., which thus becomes We may remark, that the altitude EL of the inscribed triangle is equal to R. For, we have EL=EO+OL=R+}R, since the figure OABC is a lozenge, consequently EL=3R. It follows, then, that the altitude of the circumscribed triangle is equal to 3R. Cor. III. If, in formula (2) of T. VII., we substitute R for a, we shall find AI = (2R2 — R√4R2 — R2)* = R √2 −√3, for the side of a regular inscribed dodecagon. Cor. IV. Finally, in formula (3), T. VII., substituting R for a, we find the radius and a side of a regular circumscribed hexagon, as follows: OM=3R √3; MN=3R√3. Scholium. We are now prepared, by the aid of the general formula (T. V.), to calculate the areas of these different polygons. Thus, for example, if we make n = 6, a=R, we shall obtain for the expression of the area of the regular inscribed hexagon, 8 A = 3R2√3. By making n = 3, a=R √3, we find A = 3R2 √3, for the area of the inscribed equilateral triangle. In a similar manner we proceed for the other cases. THEOREM X. The side of the inscribed square is to the radius in the ratio of √2 to 1. Draw the two diameters AB, CD perpendicular to each other, and draw the chords AC, CB, BD, DA. The figure ADBC is evidently a square; and we have AC2=AO2 + OC2 = 2R2; consequently AC=R √2, and AC:R::√2:1. C P A B M ע Scholium. In drawing through the points A, D, B, C, tangents, we form the circumscribed square; and we have MN=AB=2R. That is-The side of the circumscribed square is equal to the diameter of the circle. The areas of these two polygons are expressed respectively by 2R2 and 4R2. THEOREM XI. The side of a regular inscribed decagon is equal to the greater segment of the radius, when divided into mean and extreme ratio. Draw the radii OA, OB to the extremi ties of a side AB. The angle O of the tri 4 angle OAB is equal to of a right angle; there remains, then, 2-3= of a right angle, for the sum of the other two angles; and as OA = OB, it follows that the angle OAB=OBA = of a right angle. |