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the angles B and b; therefore the triangles ABO, a bo, are similar, as also the right-angled triangles ADO, a do;
AB: ab:: AO: ao:: DO: do;
consequently the perimeters of the polygons are to each other as the radii AO, a o, of the circumscribed circles, and also as the radii DO, do, of the inscribed circles.
The surfaces of these same polygons are to each other as the squares of the homologous sides AB, ab (221); they are therefore also as the squares of the radii of the circumscribed circles AO, a o, and as the squares of the radii of the inscribed circles DO, do.
283. Every curved line, or polygon, which encloses, from one Fig. 162. extremity to the other, a convex line AMB (fig. 162), is greater than the enclosed line AMB.
Demonstration. We have already said, that, by a convex line, we understand a curved line or polygon, or a line consisting in part of a curve and in part of a polygon, such that a straight line cannot cut it in more than two points (68). If the line AMB had re-entering parts or sinuosities, it would cease to be convex, because, as will be readily perceived, it might be cut by a straight line in more than two points. The arcs of a circle are essentially convex; but the proposition under consideration extends to every line, which fulfils the condition stated.
This being premised, if the line AMB be not smaller than any of those lines which enclose it, there is among these last a line smaller than any of the others, which is less than AMB, or at least equal to AMB. Let ACDEB be this enclosing line; between these two lines draw at pleasure the straight line PQ, which does not meet the line AMB, or, at most, only touches it; the straight line PQ is less than PCDEQ (3); consequently, if, instead of PCDEQ, we substitute the straight line PQ, we shall have the enclosing line APQB, less than APDQB. But, by hypothesis, this must be the shortest of all; this hypothesis, then, cannot be maintained; therefore each of the enclosing lines is greater than AMB.
284. Scholium. After the same manner, it may be demonstrated, without any restriction, that a line which is convex, and
returns into itself, AMB (fig. 163), is less than any line which Fig 163. encloses it on all sides, whether the enclosing line FGH touches AMB in one or more points, or whether it surrounds it without touching it.
285. Two concentric circles being given, there may always be inscribed, in the greater, a regular polygon, the sides of which shall not meet the circumference of the smaller; and there may also be circumscribed, about the smaller, a regular polygon, the sides of which shall not meet the circumference of the greater; so that, on the whole, the sides of the polygon described shall be contained between the two circumferences.
Demonstration. Let CA, CB (fig. 164), be the radii of the Fig. 164. two given circles. At the point A draw the tangent DE terminating, at the greater circumference, in D and E. Inscribe, in the greater circumference, one of the regular polygons, which can be inscribed by the preceding problems, and bisect the arcs subtended by the sides, and draw the chords of these half arcs; and a regular polygon will be described of double the number of sides. Continue to bisect the arcs until one is obtained which is smaller than DBE. Let MBN be this arc, the middle of which is supposed to be in B; it is evident that the chord MN will be further from the centre than DE, and that thus the regular polygon, of which MN is a side, cannot meet the circumference, of which CA is the radius.
The same things being supposed, join CM and CN, which meet the tangent DE in P and Q; PQ will be the side of a polygon circumscribed about the smaller circumference similar to the polygon inscribed in the greater, the side of which is MN. Now it is evident that the circumscribed polygon, which has for its side PQ, cannot meet the greater circumference, since CP is less than CM.
There may, therefore, by the same construction, be a regular polygon inscribed in the greater circumference, and a similar polygon circumscribed about the smaller, which shall have their sides comprehended between the two circumferences.
286. Scholium. If we have two concentric sectors FCG, ICH, we can likewise inscribe, in the greater, a portion of a regular polygon, or circumscribe, about the smaller, a portion of a similar GEOM.
polygon, so that the perimeters of the two polygons would be comprehended between the two circles. It is only necessary to divide the arc FBG successively into 2, 4, 8, 16, &c., equal parts, until one is obtained smaller than DBE.
By a portion of a regular polygon, as the phrase is here used, is to be understood the figure terminated by a series of equal chords, inscribed in the arc FG, from one extremity to the other. This portion has the principal properties of a regular polygon; it has its angles equal, and its sides equal; it is, at the same time, capable of being inscribed in, and circumscribed about a circle; it does not, however, make a part of a regular polygon, properly so called, except when the arc, subtended by one of these sides, is an aliquot part of the circumference.
287. The circumferences of circles are as their radii, and their surfaces are as the squares of their radii.
Demonstration. Denoting, by circ. CA and circ. OB (fig. 165), the circumferences of the circles whose radii are CA and OB, we say that circ. CA circ. OB:: CA: OB.
For, if this proportion be not true, CA will be to OB as circ. CA is to a fourth term either greater or less than circ. OB. Let us suppose that it is less, and that, if possible,
CA: OB:: circ. CA: circ. OD.
Inscribe, in the circumference of which OB is the radius, a regular polygon EFGKLE, whose sides shall not meet the circumference of the circle, whose radius is OD (285); inscribe a similar polygon MNPSTM in the circle whose radius is CA.
This being done, since the polygons are similar, their perimeters MNPSM, EFGKE, are to each other as the radii CA, OB, of the circumscribed circles (282), and we have
MNPSM: EFGKE:: CA: OB;
but, by hypothesis,
CA: OB: circ. CA: circ. OD; therefore MNPSM : EFGKE :: circ. CA : circ. OD. Now this proportion is impossible, because the perimeter MNPSM is less than circ. CA (283), while EFGKE is greater than the circ. OD; therefore it is impossible that CA should be to OB as circ. CA is to a circumference less than circ. OB; or, in other words,
it is impossible that the radius of one circle should be to that of another as the circumference of the first is to a circumference less than that of the second.
It follows, moreover, from what has been said, that CA cannot be to OB as circ. CA is to a circumference greater than circ OB; for, if this were the case, we should have, by inversion, OB: CA :: a circumference greater than circ. OB : circ. CA, which is the same thing,
OB: CA :: circ. OB: a circumference less than circ. CA; therefore the radius of one circle may be to the radius of another, as the circumference described upon the former is to a circumference less than the one described upon the latter, which has been shown to be impossible.
Since the fourth term of the proportion CA: OB:: circ. CA: X can be neither less nor greater than circ. OB, it must be equal to circ. OB; therefore the circumferences of circles are as their radii.
By a construction and course of reasoning entirely similar, it may be demonstrated that the surfaces of circles are as the squares of their radii.
We shall not enter into further details upon this proposition, which is indeed a corollary from the next.
288. Corollary. Similar arcs AB, DE (fig. 166), are as Fig 166. their radii AC, DO; and similar sectors ACB, DOE, are as the squares of their radii.
For, since the arcs are similar, the angle C is equal to the angle O (163); now, the angle C is to four right angles as the arc AB is to the entire circumference described upon the radius AC (122), and the angle O is to four right angles as the arc DE is to the circumference described upon the radius OD; therefore the arcs AB, DE, are to each other as the circumferences of which they are respectively a part; and these circumferences are as the radii AC, DO; therefore
arc AB arc DE :: AC : DO.
For the same reason the sectors ACB, DOE, are as the entire circles; but the entire circles are as the squares of their radii; therefore
sect. ACB: sect. DOE :: AC : DỔ.
289. The area of a circle is equal to the product of its circumference by half of the radius.
Demonstration. Denoting by surf. CA the surface or area of a circle whose radius is CA, we say that
surf. CACA × circ. CA.
Fig. 167 If CA × circ. CA (fig. 167) be not the area of the circle of which CA is the radius, this quantity will be the measure of a circle either greater or less. Let us suppose, in the first place, that it is the measure of a greater circle, and that, if it be possible, CA × circ. CA surf. CB.
About the circle, of which CA is the radius, circumscribe a regular polygon DEFG, &c., the sides of which shall not meet the circumference of the circle whose radius is CB (285); the surface of this polygon will be equal to its perimeter
DE + EF+FG+&c.,
multiplied by AC (280). But the perimeter of the polygon is greater than that of the inscribed circle, since it encloses it on all sides; consequently the surface of the polygon DEFG, &c. is greater than AC × circ. AC, which, by hypothesis, is the measure of the circle, of which CB is the radius; hence the polygon would be greater than the circle; but it is less, since it is contained within it; therefore it is impossible that
CA× circ. CA
should be greater than surf. CA, or, in other words, it is impossible that the circumference of a circle multiplied by half of the radius should be the measure of a greater circle.
Again, this same product cannot be the measure of a less circle; and, not to change the figure, I will suppose that the circle in question is that whose radius is CB; it is to be proved, then, that CB × circ. CB cannot be the measure of a less circle, of the circle, for example, whose radius is CA. Let us suppose, if it be possible, that CB × circ. CB= surf. CA.
The same construction being supposed as above, the surface of the polygon DEFG, &c. will have for its measure
(DE+EF+FG+ &c.) × CA;
but the perimeter DE+ EF+FG+ &c., is less than circ. CB, which encloses it on all sides; hence the area of the polygon is less than CA × circ. CB, and, for a still stronger reason, less