APC. For a like reason, the surface FQECPB, and the surface DQE APB; hence we have that is, = DQF+FQE-DQE APC+CPB- APB, = DEF = ABC; therefore, the two symmetrical triangles ABC, DEF are equal in surface. Scholium. The poles P and Q might lie within the triangles ABC, DEF; in which case, it would be requisite to add the three triangles DQF, FQE, DQE together, in order to make up the triangle DEF; and, in like manner, to add the three triangles APC, CPB, APB together, in order to make up the triangle ABC. In all other respects, the demonstration and the result would still be the same. THEOREM XVIII. If two great circles intersect each other on the surface of hemisphere, the sum of the opposite triangles thus formed will be equivalent to the lune whose angle is equal to the angle formed by the circles. Let the circumference AOB, COD intersect on the hemisphere OACBD; then will the opposite triangles AOC, BOD be equal to the lune whose angle is BOD. B D N A For, producing the arcs OB, OD in the other hemisphere, till they meet in N, the arc OBN will be a semi-circumference, and AOB one also; and taking OB from each, we shall have BN=AO. For a like reason, we have DN=CO, and BD = AC. Hence the two triangles AOC, BDN have their three sides respectively equal; besides, they are so placed as to be symmetrical; hence (T. XVII.) they are equal in surface, and the sum of the triangles AOC, BOD is equal to the lune OBNDO. whose angle is BOD. THEOREM XIX. The surface of any spherical triangle is measured by the excess of the sum of its three angles above two right angles. G F A E H B I Let ABC be the proposed triangle: produce its sides till they meet the great circle DEFGHI, drawn anywhere without the triangle. By the last proposition, the two triangles ADE, AGH are together equal to the lune whose angle is A, and which is measured (T. XVI., C. II.) by 2 A; hence we have ADE+AGH = 2 A; and, for a like reason, BGF+ BID=2 B, and CIH+CFE = 2 C. But the sum of those six triangles exceeds the hemisphere by twice the triangle ABC, and the hemisphere is represented by 4; therefore twice the triangle ABC is equal to 2 A + 2 B + 2 C − 4, and consequently once ABC=A+B+C-2; hence, every spherical triangle is measured by the sum of all its angles minus two right angles. Cor. I. However many right angles there be contained in this measure, just so many tri-rectangular triangles, or eighths of the sphere, which (T. XVI., C. II.) are the unit of surface, will the proposed triangle contain. If the angles, for example, are each equal to of a right angle, the three angles will amount to 4 right angles, and the proposed triangle will be represented by 4 — 2, or 2; therefore it will be equal to two tri-rectangular triangles, or to the fourth part of the whole surface of the sphere. Cor. II. The spherical triangle ABC is equal to the lune A+B+C whose angle is 1. Likewise the spherical pyramid 2 which has ABC for its base, is equal to the spherical ungula Scholium. While the spherical triangle ABC is compared with the tri-rectangular triangle, the spherical pyramid, which has ABC for its base, is compared with the tri-rectangular pyramid, and the same ratio is found to subsist between them. The polyedral angle at the vertex of the pyramid is, in like manner, compared with the polyedral angle at the vertex of the tri-rect angular pyramid. These comparisons are founded on the coincidence of the corresponding parts. If the bases of the pyramids coincide, the pyramids themselves will evidently coincide, and likewise the polyedral angles at their vertices. From this, the following consequences are deduced: First. Two triangular spherical pyramids are to each other as their bases; and since a polygonal pyramid may always be divided into a certain number of triangular ones, it follows that any two spherical pyramids are to each other as the polygons which form their bases. Secondly. The polyedral angles at the vertices of those pyramids are also as their bases; hence, for comparing any two polyedral angles, we have merely to place their vertices at the centres of two equal spheres, and the polyedral angles will be to each other as the spherical polygons intercepted between their planes or faces. The vertical angle of the tri-rectangular pyramid is formed by three planes at right angles to each other: this angle, which may be called a right polyedral angle, will serve as a very natural unit of measure for all other polyedral angles; and if so, the same number that exhibits the area of a spherical polygon, will exhibit the measure of the corresponding polyedral angle. If the area of the polygon is, for example; in other words, if the polygon is of the tri-rectangular polygon, then the corresponding polyedral angle will also be of the right polyedral angle. 4 THEOREM XX. The surface of a spherical polygon is measured by the sum of all its angles, MINUS the product of two right angles by the number of sides in the polygon MINUS two. E C D B From one of the vertices A, let diagonals AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many triangles, minus two, as it has sides. But the surface of each triangle is measured by the sum of all its angles minus two right angles, and the sum of the angles in all the triangles is evidently the same as that of all the angles in the polygon; hence the sur A face of the polygon is equal to the sun of all its angles, diminished by twice as many right angles as it has sides minus two. Scholium. Let s be the sum of all the angles in a spherical polygon, and n the number of its sides; the right angle being taken for unity, the surface of the polygon will be measured by s — 2 (n − 2), or s− 2 n +4. REGULAR POLYEDRONS. Since the faces of any regular polyedron are all equal and regular polygons (B. VI., D. XII.), it follows that all its polyedral angles must be equal. There can be only five kinds of regular polyedrons. For, we already know that three faces at least are necessary to form a polyedral angle; and moreover that the sum of all the plane angles which form the polyedral angle is less than 4 right angles. It is sufficient then to consider all the cases in which the angles of equal and regular polygons, taken 3 at a time, 4 at a time, 5 at a time, etc., do not equal or exceed 4 right angles. Now, of a right 1. The angle of an equilateral triangle being angle, we may construct a polyedral angle by using three, four, or five equilateral triangles for faces. But we cannot use six or a greater number, since of a right angle × 6 = 4 right angles. 2. The angle of a square being 1 right angle, we may form a polyedral angle by using three squares for the faces. But we could not use four or a greater number, since 1 right angle x 4 = 4 right angles. 3. The angle of a regular pentagon being, a polyedral angle may be formed with three regular pentagons, which gives x333. 3 Since the angle of a regular hexagon is, and × 3 = 4, it follows, that a polyedral angle cannot be formed with regular hexagons for faces. And of course, regular polygons of a greater number of sides cannot be used in forming a polyedral angle. Thus, the only possible regular polyedrons are those of which each polyedral angle is formed by three, four, or five equilateral triangles, by three squares, and by three regular pentagons, making in all five polyedrons. When three equilateral triangles are used in forming each polyedral angle, the whole number of faces of the regular polyedron will be four, and it is called a Tetraedron. When four equilateral triangles unite in forming each polyedral angle, the polyedron will have eight faces, and it is called an Octaedron. When five equilateral triangles are used in forming a polyedral angle, the polyedron will have twenty faces, and it is then called an Icosaedron. When three squares are used in forming each polyedral angle of a regular polyedron, there will be six faces, and the polyedron is called a Hexagon, or Cube (B. VI., D. V.). When three regular pentagons are used for each polyedral angle, the number of faces will be twelve, and the solid is called a Dodecaedron. These five regular solids may obviously be inscribed in a sphere, as well as circumscribed about a sphere; or a sphere may be inscribed in each, and circumscribed about each. The inscribed and circumscribed spheres have a common centre, which may be regarded as the centre of the polyedron. If we suppose planes to pass through the centre of a regular polyedron and each of its edges, they will divide the polyedron into as many equal pyramids as the polyedron has faces, since each face of the polyedron will thus become a base of a pyramid, whose altitude will be the radius of the inscribed sphere; and since each pyramid is measured by the area of its base multiplied by one third of its altitude, it follows that the volume of any regular polyedron has for its measure its surface multiplied by one third the radius of its inscribed sphere. Were we to regard the sphere as a regular polyedron of an infinite number of infinitely small and equal faces, we should, from the above, at once infer that the volume of a sphere has for its measure its surface multiplied by one third its radius. Whether we regard the sphere as a regular polygon or not, it is essentially a regular body. Hence, there are then six regular bodies: the Tetraedron, Hexaedron, Octaedron, Dodecaedron, Icosaedron, and the Sphere. |