less than the triangle CAB; for, as the perimeter of every circumscribed polygon is greater than the circumference AB (I. 10. Scholium), and therefore, as before shown, the polygon itself greater than the triangle CAB, to which, however, it may be made to approach within any given difference, because its perimeter may be made (31. Cor. 1.) to approach to AB within any given difference; so, because similar polygons may be inscribed and cicumscribed approaching to one another more nearly than by any given difference (31.) a polygon may be inscribed approaching to the triangle CAB within any given difference, that is greater than the circle, if the circle be supposed to be less than the triangle CAB; which is absurd. Therefore the circle is neither greater nor less than the triangle' C A B, that is, it is equal to it. Therefore, &c. Cor. Any circular sector is equal to half the rectangle under the radius of the circle, and the arc upon which it stands for it is less than the circle in the ratio of that arc to the circumference (13.). PROP. 33. (Euc. xii. 2.) The circumferences of circles are as the radii, and their areas are as the squares of the radii. Let R, r, represent the radii of two circles, C, c their circumferences, and A a, their areas: then Cc: Rr, and A: a: R2 : 72. For, in the first place, there may be inscribed (31. Cor. 2.) two similar polygons, the perimeters of which approach more nearly to the perimeters C, c of the two circles, than by any the same given difference; and the perimeter of the one polygon (30.) is to the perimeter of the other, always in the same ratio, viz. as R to r: therefore, C:c::R:r (II. 28.). And, in the same manner, because there may be inscribed in the circles two similar polygons, the areas of which (31. Cor. 2.) approach more nearly to the areas A, a of the circles, than by any the same given difference; and because the area of the one polygon (30.) is always to the area of the other in the same ratio, viz. as R to r2, A: a: R2: r2 (II. 28.). Therefore, &c. ratio, viz. that of the angle at the centre (13.) to four right angles, are to one another, alternando, as the whole circumferences (or circles), that is, by the proposition, as the radii (or the squares of the radii.) Cor. 1. Hence, similar arcs of circles are as their radii; and similar sectors are as the squares of their radii. For such arcs (or sectors) being to the whole circumferences (or circles) in the same Cor. 2. Similar segments of circles are as the squares of the radii (II. 22.). For they are the differences of similar sectors, and similar triangles, (def. 13. and II. 32.), which sectors (Cor. 1.) as also the triangles (II. 42.) are as the squares of the radii. PROP. 34. If K and L represent two regular polygons of the same number of sides, the one inscribed in, and the other circumscribed about, the same circle, and if M and N represent the inscribed and circumscribed polygons of twice the number of sides; M shall be a geometrical mean between K and L, and N shall be an harmonical mean between L and M. Let C be the centre of the circle, AB a side of the inscribed polygon K, CD an radius drawn perpendicular to, and therefore (3.) bisecting A B in the point I. Then, if EF be drawn, touching the circle in D, and terminated by CA and CB produced; EF will (27. Cor. 2.) be a side of the circumscribed polygon L of the same number of sides. Also, if AD be joined, and at the points A and B tangents be drawn to meet E F in the points G and H; AI and G H will be sides of the polygons M and N of twice the number of sides (27.). Now, because the triangles CAI, CED, CAD, CG H are severally contained in the polygons K, L, M, N, the same number of times, viz. as often as the angle ACD, or G C H, is contained in four right angles, the polygons are one to another as those triangles(II.17.). But the triangle C AI is to the triangle CAD as CI to CD, (II. 39.) that is (because A I, ED are parallels), as CA to CE, that is, again, as the triangle CAD to the triangle CED (II. 39.). And, because CAI is to CAD as CAD to CED, K: M:: M: L (II. 17. Cor. 2.) and M is a mean proportional between K and L. Again, because the triangle C GH is double of C G D, and, therefore, equal to the quadrilateral C A GD; the tri angle A EG is equal to the difference of the triangles CED, CGH, and the triangle AGD to the difference of the triangles C GH, CAD. But A EG is to AGD as EG to G D, that is, (II. 50.) because the line GC bisects the angle ACD, as EC to CD or CA, that is again, as the triangle CED to the triangle CAD (II. 39.). And, because CED is to CAD as the difference of C ED, CGH to the difference of C GH, CAD, L: ML-N: N-M (II. 17. Cor. 2.); that is, N is an harmonical mean between L and M (def. 17.). Therefore, &c. Scholium. The proposition which has been just demonstrated, affords one of the most simple methods of approximating to the area of the circle to which purpose it may be applied as follows. Let the diameters AB, D E be drawn at right angles to one another: the straight lines joining their extremities will include an inscribed square, and the tangents drawnthrough the same a circumscribed square. Now it is plain that the circumscribed square is equal to 4 times the square of the radius, and the inscribed to half the circumscribed, that is, to twice the square of the radius. Therefore, if the radius be assumed for the linear unit, the inscribed square will contain 2, and the circumscribed 4 units of area. But the inscribed figure of eight sides is a mean proportional between them: therefore, the number of units of area which it contains, will be a mean proportional between 2 and 4,2x4, -2.8284271247 &c. to the tenth decimal place inclusively. And the number which is an harmonical mean between 4 and 2.8284271247 will, in like manner, be the number of superficial units in the circumscribed figure of 8 sides. Now, to find such a mean x between two numbers m, n, we have this proportion, m: n::m-x : x − n (II. def. 17); whence, multiplying extremes and means, m x (x−n) = n × (m−x); transposing, mx + nx = 2 m n ; and dividing by m+n,x= ; that is, an harmonical mean between two numbers is obtained by dividing twice their product by their sum. Thus we find the cir 2 m n D 5 Of a calculation which has attracted so much atten tion, it is not impossible that the student may be curious enough to revise the steps, or even push it to a still greater degree of approximation. In doing this by the method here given, his labour will be considerably abridged by attending to the 1°. Annex one more to the decimal places which are required to be exactly ascertained, and with this additional place, use the abbreviated modes of multiplying, dividing, and extracting the square root, viz. by inverting the multiplier, cutting off successively the figures of the divisor, and dividing out when the root is obtained to half the required number of places (See Arith, art, 167. 185.) 20. When the calculation has proceeded so far that ( being the difference of the two preceding poly.. H polygons: therefore, the area of the circle is correctly denoted by 3.1415926 535 as far as the tenth decimal place inclusively. This number is commonly represented T4 1663 gons, and b the lesser of the two), the quotient of the fraction when expressed in decimals, has no significant figure in the first ten decimal places, the harmonical mean may be found by taking half the sum or arithmetical mean, and subtracting therefrom x2 Since b=3.14, &c., this rule may be used when 46° 24 does not appear in the last place but three. is not found 3 3°. And in like manner, when 16 62 in the last decimal place (or which is the same thing nearly, 73 in the last place but three), the geometrical mean may be obtained by taking the arithme tical mean and subtracting therefrom x2 8 b x 2 [4°. When is not found in the last decimal 4 b by the Greek letter, being the first letter of the Greek word which signifies circumference.* For the same number which represents the area of a circle when the radius is taken for unit, re place, (or 82 in the last but two) neither the harmonical nor the geometrical mean will differ apparently from the arithmetical, which may therefore be taken for them. Or, when this comes to be the case, instead of computing the intermediate polygonal areas, the area of the circle may be directly found to the required number of places by the following rule. "Let an inscribed polygon be the last computed; take the difference between its area and that of the divide this difference preceding circumscribed: (considered as a whole number) by that powerof 2, say 2m, which is next less than it; multiply the quotient m-1 2 by 3 the inscribed polygon, placing the units of the product under the last decimal place of the area; the sum shall be the circular area required." Thus, in the preceding table of areas, the difference between the inscribed polygon L and the circumscribed polygon K is 36962; the power of 2, which is next less, is 32768; the quotient of 36962 divided by 32768 is 1.128; the number by which this is to be 1 32768 multiplied -1) or 5461; the product to 3 2 the nearest unit 6160; and 14215, together with the double of this product, is 26535, which has the remaining digits in question. The second, third, and fourth of these rules may be established by the assistance of the binomial theorem: the last is derived from the algebraical form of a series of quantities, each of which is an arithmetical mean between the two preceding. 1 , and add twice the product to the area of The letter is, however, more generally understood to represent the semicircumference of a circle whose radius is unit; this being evidently the same number which represents the circumference when the diameter is assumed for unit. as the former, and an equal perimeter; m shall be an arithmetical mean between k and 1, and n a geometrical mean between 1 and m. To demonstrate this: Let AB be a side, and C the centre of any regular polygon; let CD be drawn perpendicular to AB, and join CA, CB: then CD is the radius, k, of the inscribed circle, and CA the radius, 1, of the circumscribed circle. From DC produced cut off CE equal to CA or CB, and join EA, EB: from C draw CF perpendicular to EA, and therefore (I. 6. Cor. 3.) bisecting EA, and through F draw FG parallel to AB, and therefore (I. 14.) perpendicular to ED, which it cuts in the point H. In fact represents (1), the superficial area of the circle where the unit of superficies is the square of the radius; (2) the linear value of the circumference, where the diameter is the unit of length; and (3) the linear value of the semicircumference, where the radius is the unit of length. The last of these is the meaning most commonly attached to the symbol. In the method of approximation which is adopted in the text, although the principle is perhaps more obvious, the computation is not so concise as in another method, which may be derived from the following elegant theorem. If k and represent the radii of the circles which are inscribed in any regular polygon, and circumscribed about it; and if m and represent these radii For a regular polygon which has twice as many sides F Then, because the angle AEC is equal to half the angle ACD (I. 19.), the angle AEB, or FEG, is equal to half the angle ACD: also, because EF is equal to the half of EA, FG is equal (II. 30. Cor. 2.) to the half of AB: therefore FG is the side of a regular polygon, which has twice as many sides as the former, E its centre, EH the radius, m, of the inscribed circle, and EF the radius, n, of the eircumscribed circle. But, because EF is equal to half EA, EH is (II. 30.) equal to half ED, or to half the sum of CD and CA; that is, m is an arithmetical mean between k and . And, again, because from the right angle F of the triangle EFC, FH is drawn perpendicular to the hypotenuse EC, EF is a mean proportional between EC and EH (II. 34. Cor.); that is, a is a mean proportional between and m. Therefore, &c. Hence, beginning with the square or hexagon, we may proceed, by alternate arithmetical and geometri cal means, to determine these radii for a regular polygon, the number of whose sides shall exceed any given number; in which process it is evident that the values of the radii will continually approach to one another, and, therefore, to the intermediate value of the radius of a circle which has the same given perimeter. There is yet a third theorem, nearly related to the preceding, which may be applied to the purpose of this approximation. Ifk and 1 represent the radii of the circles which are circumscribed about any regular polygon, and inscribed in it, and m an arithmetical mean between them; and if k' and ' represent these radii for a regular polygon which has twice as many sides as the former, and an equal area, k' shall be a mean propor tional between k and 1, and ľ a mean proportional between land m. To demonstrate this: Let AB be a side, and C the centre of any regular polygon; let CD be drawn perpendicular to AB, and join CA, CB: then CA is the radius, k, of the circumscribed circle, and CD the radius, 1, of the inscribed circle. Draw the straight line CE bisecting the angle ACD; in CD produced take CF a mean proportional between CA and CD; from F draw FG perpendicular to CE, and produce it to meet CA in H. H E ID Then, because CG bisects the angle FCH, and FG is perpendicular to CG, the triangle FCH is isosceles (I. 5.); and, because CHXCF is equal to CAX CD, the triangle CFH is equal to the triangle CAD (II. 40. Cor.): therefore FĤ is the side of a re presents also the circumference when the diameter is taken for unit, because the area of a circle, being equal to the rectangle under the semicircumference and radius (32.), bears to the square of the radius the same ratio which the semicircumference bears to the radius, or the circumference to the diameter. And hence if R be the radius of any circle, its circumference (greater than 2 R in the ratio of: 1) is 2 R: and its area (greater than R in the ratio of : 1) is = R2. 7 It remains to observe that the circumference of a circle is incommensurable with its diameter, for which reason their ratio can never be exactly represented by numbers. This was for the first time demonstrated in the year 1761 A. D. by Lambert. During the long period for which it was only matter of conjecture, the quest of the exact numerical ratio (and that by methods not more expeditious than the above) occupied many laborious calculators. Could they have assigned any such, it is evident that they might likewise have assigned the exact value of the area of a circle, whose radius is given, and vice versâ ; because that area is (32.) equal to half the product of the radius and circumference. But the hope of arriving at a term of the approximation is now demonstrated to have been vain, and accordingly an exact solution of the celebrated problem of squaring the circle, that is, of finding a straight line, the square of which shall be exactly equal to a given circle, impracticable. At the same time, the gular polygon which has twice as many sides as the former, and an equal area, C its centre, CF the radius, k', of the circumscribed circle, and CG the radins, l', of the inscribed circle But, by the construction, CE is a mean proportional between CA and CD; that is, k' is a mean proportional between k and l. And, again, because the triangle CGF is similar to CDE, the triangle CGF is to the triangle CDE as CG to CD2 (11. 42. Cor.); but, because the triangle CGF is equal to half CHF, that is to half CDA, CGF is to CDE as half DA to DE (H. 39.), or,because CE bisects the angle ACD as half CA+CD to CD (II. 50.); therefore (II. 12.) CG2: CA+CD. 2: CD, and (II. 37. Cor. 2.) CG is a CD2 :: 2 : mean proportional between CD and CA+CD that is, is a mean proportional between / and m. Therefore, &c. This theorem is applied in the same manner as the preceding. It is necessary to observe that CG is greater than CE, and not equal to it, as is wrongly represented in the figure: for, if P be taken a third proportional to EC and ED, it may be shown that CG is greater than CE2 by a square which is to P2 as CA to CD. approximate solution exhibited in the number 3.1415926535 &c. is sufficient for every useful purpose. If the ratio be considered as expressed by the integer and first ten decimal places, the error committed will bear a less propor tion to the whole circumference than an inch to the circuit of the earth. Instead of the number 3.1415 &c. the fractions and may also be conveniently used in cases not requiring a great degree of approximation. The first (discovered by Archimedes) will be found to fail in the third decimal place: the other (due to Metius, and remarkably made up of the odd numbers 1, 3, 5) fails in the seventh decimal place only. SECTION 5.-The circle a maximum of area, and a minimum of perimeter. In the present section it is proposed to show that of all plane figures having equal perimeters, the circle contains the greatest area; and consequently, of all plane figures containing equal areas, has the least perimeter; in other words, as it is announced in the title of the Section, that the circle is a maximum of area and a minimum of perimeter. PROP. 35. Of equal triangles upon the same base, the isosceles has the least perimeter; and, of the rest, that which has the greater vertical angle has the less perimeter. Let the triangles ABC, D B C be upon the same base B C, and between the same parallels AD, BC (I. 27.), and let the triangle ABC be isosceles: the triangle ABC shall have a less perimeter than the triangle D B C. B K From B draw BE perpendicular to AD, and produce it to F, so that EF DF. Then, because the triangles BE A, may be equal to EB: and join A F FEA have two sides of the one equal to two sides of the other, each to each, and the included angles BEA, FEA equal to one another, A F is equal to A B (I. 4.) and the angle FA E to the angle B A E, that is, to ABC (I. 15.) or (I. 6.) AC B. But the angles AC B, EAC are together equal to F E 100 two right angles (I. 15); therefore FAE, EAC are likewise equal to two right angles, and (I. 2.) FÁ, A C are in the same straight line. And because the triangles BED, FED have the two sides BE, ED of the one equal to the two FE, ED of the other, each to each, and the included angles equal to one another (I. 4.) DF is equal to DB; and it was shown that A F was equal to AB. But DF, DC are greater than FC (I. 10.); therefore DB, DC are greater than AB, A C; and, BC being added to each, the perimeter of the triangle DBC is greater than the perimeter of the triangle A B C. In the next place, let G B C be another triangle upon the same base B C, and between the same parallels, but having the angle B G C less than BDC: the perimeter of the triangle G B C shall be greater than the perimeter of the triangle D B C. Bisect B C in K, and join A K. Then, because ABC is an isosceles triangle, AK is (I. 6. Cor. 3.) at right angles to B C. And, because AK bisects B C at right angles, it passes (3. Cor. 2.) through the centre of the circle which is circumscribed about the triangle DBC (5. Cor. 2.) Take A d equal to AD. Then, because A K passes through the centre of this circle, and bisects the chord BC, it bisects also the chord which passes through the point A parallel to BC (3. Cor. 1.); and therefore the point d is in the circumference of the circle. Now, because the angle BGC is less than BDC, the point G must lie without the circle, (15. Cor. 3.) that is, G must be some point in the line D d produced, and does not lie between the points D, d. But if it lie upon the same side of FC with the point D, FG, GC together must be greater (I. 10. Cor. 1.) than FD, DC together; and therefore, because FG is equal to BG, and FD to BD, (I. 4.) the perimeter of the triangle GBC must be greater than the perimeter of the triangle DBC. And if it lie upon the other side of FC, FG, GC together will be greater than Fd, dC together. But because the diagonals FC, Dd bisect one another (1.22.) the figure FDCd is a parallelogram, and (I. 22.) the sides Fd, d C together are equal to the sides FD, DC together. Therefore FG, GC together are greater than FD, DC together, and, as before, the perimeter of the triangle G BC is greater than the perimeter of the triangle D B C. Therefore, &c. PROP. 36. If a rectilineal figure ABCDE have not all its sides equal and all its angles equal, a figure of equal area may be found, which shall have the same number of sides and a less perimeter. For, in the first place, if it have not all its sides equal, there must be at least two adjacent sides which are unequal. Let these be AB, AE, and join BE: and let a BE be an isosceles triangle of equal area, and upon the same base B E. Then the whole figure a BCDE is equal to the whole ABCDE; and because (35.) a B, a E together are less than AB, AE together, the figure a BCDE has been found of equal area with the figure ABCDE, and having a less perimeter. Next, if it have not all its angles equal, there must be two adjacent angles A, B, which are unequal. And, first, let the sides AE, BC, meet one another in a point P. Take Pa a mean proportional (II.51.) between PA, PB, and make Pb equal to Pa. Then, in the first place, if one of these points, as b, lie in the corresponding side BC, that is, between B and C, join ab: the figure ab CDE shall be of equal area with the figure ABCDE, and shall have a less perimeter. For, because PA is to Pa as Pa or Pb to PB, a B joined is parallel to Ab (II. 29.). Therefore (I. 27.) the triangle a A b is equal to the triangle BAb, and the figure abCDE is equal to the figure ABCDE. And because the triangle Pab is isosceles, the angle Eab is equal to the angle Cba (I. 6. or I. 6. Cor. 2.); but the two Eab, Cba are together equal to the two EAB, CBA (I. 19.), of which one, viz. EAB, is the greater; therefore the angle Eab is greater than the other CBA: And these latter angles are the vertical angles of the equal triangles a A b, B A b, which stand upon the same base Ab: therefore (35.) the sides a A, a b together are less than the sides B A, B b together; and the figure abCDE has a less perimeter than the figure ABCDE. E a A E E P 3 |