NOTES TO BOOK XII. The first comparison of rectilinear areas is made in the first book of the Elements by the principle of superposition, where two triangles are coincident in all respects ; next, comparison is made between triangles and other rectilinear figures when they are not coincident. In the sixth book, similar triangles are compared by shewing that they are in the duplicate ratio of their homologous sides, and then by dividing similar polygons into the same number of similar triangles, and shewing that the polygons are also in the duplicate ratio of any of their homologous sides. In the eleventh book, similar rectilinear solids are compared by shewing that their volumes are to one another in the triplicate ratio of their homologous sides. In the twelfth book a new principle is introduced, called “the method of exhaustions,” which is founded on the principle of exhausting a magnitude or the difference of two magnitudes, by successively taking away a certain part of it. The method of exhaustions was employed by the Ancient Geometers and was strictly rigorous in its principles; but it was too tedious and operose in its application to be of extensive utility as an instrument of investigation. It is exemplified in Euc. XII. 2, where it is proved that the areas of circles are proportional to the squares on their diameters. In demonstrating this truth, it is first shewn by inscribing successively in one of the circles, regular polygons of four, eight, sixteen, &c, sides, and thus tending to exhaust the difference between the areas of the circle and polygon, that a polygon may be found which shall differ from the circle by an area less than any magnitude that can be assigned : and then since similar polygons inscribed in circles are as the squares on their diameters (Euc. xII. 1), the truth of the proposition is established by means of an indirect proof. " The method of exhaustions” may be applied to find the circumference and area of a circle. A rectilineal figure may be inscribed in the circle and a similar one circumscribed about it, and then by continually doubling the number of sides of the inscribed and circumscribed polygons, by this principle, it may be demonstrated, that the area of the circle is less than the area of the circumscribed polygon, but greater than the area of the inscribed polygon; and that as the number of sides of the polygon is increased, and consequently the magnitude of each diminished, the differences between the circle and the inscribed and circumscribed polygons are continually exhausted. In a similar way the principle is applied to the volumes and surfaces of the sphere, cone, &c. The Second Proposition of the twelfth book is perhaps retained merely as an example of the method employed by the Ancient Geometers. This method has been replaced by the method of prime ard ultimate ratios, which is now employed in the proofs of such propositions as were formerly effected by the method of exhaustions. THEOREM I. - If semicircles ADB, BEC be described on the sides AB, BC of a rightangled triangle, and on the hypotenuse another semicircle AFBGC be described, passing through the vertex B; the lunes AFBD and BGCE are together equal to the triangle ABC. It has been demonstrated (XII, 2) that the areas of circles are to one another as the squares on their diameters; it follows also that semicircles will be to each other in the same proportion. Therefore the semicircle ADB is to the semicircle ABC, as the square on AB is to the square on AC, and the semicircle CEB is to the semicircle ABC, as the square on BC is to the square on AC, hence the semicircles ADB, CEB, are to the semicircle ABC as the squares on AB, BC are to the square on AC; but the squares on AB, BC are equal to the square on AC, (1. 47.) therefore the semicircles ADB, CEB are equal to the semicircle ABC. (v. 14.) From these equals take the segments AFB, BGC of the semicircle on AC, and the remainders are equal, that is, the lunes AFBD, BGCE are equal to the triangle BAC. THEOREM II. If on any two segments of the diameter of a semicircle, semicircles be described, all towards the same parts, the area included between the three circumferences (called äpßnlos) will be equal to the area of a circle, the diameter of which is a mean proportional between the segments. Let ABC be a semicircle whose diameter is AB, B and on AD, DC let two semicircles be described on the same side; also let DB be drawn perpendicular to AC. Then the area contained between the three semicircles, is equal to the area of the circle whose diameter is BD. Since AC is divided into two parts in C, the square on AC is equal to the squares on AD, DC, and twice the rectangle AD, DC; (II. 4.) and since BD is a mean proportional between AD, DC; the rectangle AD, DC is equal to the square on DB, (VI. 17.) therefore the square on A C is equal to the squares on AD, DC, and twice the square on DB. But circles are to one another as the squares on their diameters or radii, (XII. 2.) therefore the circle whose diameter is A C'is equal to the circles whose diameters are AD, DC, and double the circle whose diameter is BD; wherefore the semicircle whose diameter is AC is equal to the circle whose diameter is BD, together with the two semicircles whose diameters are AD and DC: if the two semicircles whose diameters are AD and DC be taken from these equals, therefore the figure comprised between the three semi-circum ferences is equal to the circle whose diameter is DB. THEOREM III. There can be only five regular solids. If the faces be equilateral triangles. The angle of an equilateral triangle is one-third of two right angles; and six angles, each equal to the angle of an equilateral triangle, are equal to four right angles : and therefore a number of such angles less than six, but not less than three are necessary to form a solid angle. Hence there cannot be more than three regular figures whose faces are equal and equilateral triangles. If the faces be squares. Since four angles, each equal to a right angle, can fill up space round a point in a plane. A solid angle may be formed with three right angles, but not with a number greater or less than three. Hence, there cannot be more than one regular solid figure whose faces are equal squares, If the faces be equal and regular pentagons. Since each angle of a regular pentagon is a right angle and a fifth of a right angle: the magnitude of three such angles being less than four right angles, may form a solid angle, but four, or more than four, cannot form a solid angle. Hence, there cannot be more than one regular figure whose faces are equal and regular pentagons. If the faces be equal and regular hexagons, heptagons, octagons, or any other regular figures; it may be shewn that no number of them can form a solid angle. Wherefore there cannot be more than five regular solid figures, of which, there are three, whose faces are equal and equilateral triangles ; one, whose faces are equal squares; and one, whose faces are equal and regular pentagons. PROBLEM IV. To construct the five regular solids. The regular Tetrahedron. Each of the angles of an equilateral triangle is one-third of two right angles; a solid angle may therefore be formed by three angles of three equal and equilateral triangles, and the figure formed by the three bases of the triangles is manifestly an equilateral triangle equal in magnitude to each of the three given equilateral triangles. The angles of inclination of every two of the four faces are also equal. The regular Octahedron. Through any point o draw three straight lines perpendicular to each other, take 0A, Oa, OB, Ob, OC, Oc equal to one another, and join the extremities of these lines. The faces ABC, AC, &c. are equilateral triangles equal to one another and eight in number ; also the inclinations of every two contiguous faces are equal. The regular Icosahedron. A solid angle may be formed with five angles, each equal to the angle of an equilateral triangle. At the point A of any equilateral triangle ABC, let a solid angle be formed with it and four other equal and equilateral triangles ABD, ADE, AEF, AFC, each equal to the triangle ABC. Next at the point B, let another solid angle be formed with the triangle ABC and four others BCH, BHK, BKD, BDA, each equal to it. The solid angle at B is equal to the solid angle at A, and the inclinations of every two contiguous faces are equal; also the two solid angles have two faces ABC, ABD common. Next let a third solid angle be formed at C, by placing the two triangles CFG, CGH contiguous to the three CAB, CFA, CHB. The solid angle at C is equal to that at A or B, and the inclinations of the contiguous faces make equal angles. Thus two equal and equilateral triangles are placed contiguous one to another, forming three solid angles at A, B, C, and having every two contiguous faces equally inclined: also the solid angles formed at D, E, F, G, H, K, have alternately three and two angles of the equilateral triangles. In the same manner let another figure equal to this be formed with ten equal and equilateral triangles, each equal to the triangle ABC. If these two figures be connected together, so that the points at which there are two angles of one figure, may coincide with the points which contain three angles of the other, there will be formed at the points D, E, F, G, H, K, six equal solid angles, each contained by five angles of the equilateral triangles, and every two contiguous faces will have the same inclination. Hence a figure of twenty faces is formed each equal to the equilateral triangle ABC, and having the inclinations of every two contiguous faces equal. The regular Hexahedron. Since three right angles may form a solid angle, it is therefore obvious that the solid angle formed by three equal squares, has every two of the faces equally inclined to one another; and with three other squares, each equal to the former, a figure is formed, bounded by six equal squares, and having every two contiguous faces at right angles to one another. The regular Dodecahedron. Since three angles each equal to the angle of a regular pentagon may form a solid angle; let ABCDE be a regular pentagon, and with two others each equal to this, let a solid angle at A be formed; the inclinations of every two contiguous faces will be equal. At the points B, C, D, E successively, let solid angles be formed by pentagons equal to ABCDE. The solid angles at B, C, D, E, are each equal to the solid angle at A, and the inclination of every two contiguous faces is the same. Thus is formed a figure with six equal and regular pentagons, having the inclination of every two contiguous faces equal, and the angles at the linear boundary of the figure alternately consisting of an angle of a pentagon and of two angles of two pentagons equally inclined to each other. * Next, let another figure equal to this be constructed with six pentagons, each equal to the pentagon ABCDE, If these two figures be so placed that the angular points of the plane angles in the linear boundary of one, may coincide with the points at which there are two angles in the other figure; at each of ihese points will be formed ten solid angles, each equal to the angle at A, and having the inclination of every two contiguous faces equal to one another. Hence a regular figure is formed having twelve equal faces, and the inclinations of every two contiguous faces equal to one another. 5. Construct a circle the area of which shall have a given ratio to that of a given circle. 6. Divide a circle into any number of equal parts by means of concentric circles. 7. To divide a circle into any number of equal parts, the perimeters of which shall be equal to the circumference of the circle. 8. Let AB and DC be two diameters of a given circle, at right angles to each other; AEB a circular arc described with radius DB or DA; prove that the area of the lune A.EBC = area of triangle ADB. 9. Two circles touch each other internally, and the area of the lune cut out of the larger is equal to twice the area of the smaller circle. Required the ratio of the diameters of these circles. 10. The diameter of a circle is divided into two parts, upon each of which as diameters circles are described; when the remaining area of the great circle is equal to that of one of these two circles, find the ratio which the parts of the diameter bear to one another. 11. The diameter of a semicircle ADB is divided into two parts in C (so that the length of AC is twice that of BC), and upon them are described the semicircles AEC, CFB. Compare the areas of the circles which are described on each side of the common tangent CD so as to touch it and the two semicircles. 12. The centers of three circles A, B, and C are in the same right line, B and C touch A internally, and each other externally; P, Q, |