the point O (38.): O is evidently the centre of a sphere, passing through the points A, B, C, D, and circumscribing the pyramid. Therefore, &c. PROP. 48. Prob. 13. To inscribe a sphere in a given tetrahedron or triangular pyramid A B C D. Bisect the dihedral angles at the edges BC, CD, DB of any the same face, (46.) and let the bisecting planes meet one another in the point 0: from O draw OP, OQ, OR, OS, perpendicular to the faces B CD, ABC, ABD, ACD, respectively (36): the sphere, which is described from the centre O, with the radius O P shall pass through the points Q, R, and be inscribed in the tetrahedron. D R B S For, the point O being in the plane which bisects the dihedral angle at BC, is equidistant (46. Cor.) from the planes ABC, BCD, that is, O Q is equal to OP; and, for the like reason, O'R and OS are each of them equal to OP; therefore, the four distances, O P, OQ, OR, OS, are equal to one another: and the sphere, which is described from the centre O, with the radius OP, is inscribed in the tetrahedron A B C D. Therefore, &c. Cor. If the faces and altitude AL of the tetrahedron be given, the length of the radius OP may be found. For, since the pyramid, which has the altitude A L and the base B C D, is equal to the sum of four pyramids having the common altitude OP, and for their bases the four faces of the first pyramid; or, which is the same thing, to a pyramid, which has the altitude OP, and a base equal to the sum of those faces (30.); it may easily be shown (32. Cor. 2. and II. 10. Cor.) that OP is to A L as the base BCD to the sum of the four faces. Scholium. In these problems, 12. and 13., it is as-. sumed that, three planes being given, a point may be found which is common to the three, or that, any three planes being drawn, there is some point through which each of them passes, and which may be called their point of intersection. It is evident, however, that if the common section of two of the planes be parallel to the third plane (in which case, also, by Prop. 10, the common sections are all parallel to one another), there can be no such point; for, since all the points which are common to the two first planes are found in their common section, if this common section does not cut the third plane (that is, in other words, if there is no point common to this common section and the third plane), there can be no point which is common to the three planes. When, on the contrary, the common section of two of the planes is not parallel to the third, it may be produced to cut that plane, and the point in which it cuts it is the point which is common to the three planes. When a point is to be determined, therefore, by the intersection of three planes, in order to be satisfied with regard to the meeting of the planes, it is only necessary to consider the common section of any two of them, and examine whether it is or is not parallel to the third plane. Thus, in Problem 13, the question occurs, whether OC is, or not, parallel to the plane which is made to pass through DB; and it is easily perceived that it is not parallel; for, if the plane DBA be supposed to be turned round DB, in order to arrive at the position in which it will be parallel to O ̊C, before it arrives at that position it must evidently become parallel to A C, and DBO is necessarily more distant from the position of parallelism than DAB when parallel to AC, for it passes between the face ADB of the pyramid, and the base CDB. In the second solution of Prob. 12, the three planes, which are supposed to meet in O, are at right angles to the edges, and therefore the common section of any two of them, as, for instance, the two which are at right angles to A B, A C, is perpendicular to the face ABC of the pyramid (18. and 18. Cor. 2.), for which reason any plane which is parallel to it must be perpendicular to the same face, which is evidently not the case with the plane which is at right angles to the edge AD. PROP. 49. Prob. 14.' Three plane angles AOB, BOC, COA, which form a solid angle O, being given; to find by a plane construction the angle contained by two of the planes, viz. A O B and B OC. At the point B, in the plane BOC, and upon either side of the angle BOC, make the angles BOD, COE equal to the angles BOA, COA, each to each P C E (I. 47.): in OD take any point D, and make O E equal to OD; from the points D, E, draw D B. EC perpendicular (I. 45.) to AB, AC respectively, and let them be produced to meet one another in P: from P (I. 44.) draw Q P perpendicular to P B, and from the centre B, with the radius B D, describe the circle D Qd cutting PQ in Q, and join BQ; the angle QBP shall measure the angle of inclination contained by the two planes A OB, BO C. 157 struction becomes impossible, when the third angle COE is greater than the sum or less than the difference of the other two. For, in order that it may be applied, it is necessary that BP be less than BD. Now, if D F be drawn perpendicular to O C, and produced to meet the circle described from the centre O with the radius OD in G, it may easily be shown that the angle FOG is equal to the angle FOD (III. 3. and I. 7.); and, therefore, the angle COG to COD, that is, to the sum of BOC and BOD. And in like manner, if df be drawn perpendicular to O C, and produced to meet the circumference of the same circle in g, the angle C Og will be equal to the angle CO d, that is, (because B Od, by III. 3. and I. 7., is equal to BOD) to the difference of BOC and B OD. But if COE be greater than C O G, or less than C Og, in either case the perpendicular E C, which is drawn to O C from a point E in the circumference of the circle D C g, will not cut Dd between the points D and d; and, therefore, BP will be greater than BD. The limits of possibility are, therefore, those above stated. The inferences from this are evidently the same with 19. and 19. Cor. Cor. 1. The same construction, slightly modified, may be used to find the third angle COA from the two AO B, BO C, and their angle of inclination. For, if PBQ be made equal to the angle of inclination, and if B`Q be taken equal to BD, the point P will be determined by drawing Q P perpendicular to BP; and thence the angle COE or COA, by drawing PC pendicular to OC, and producing it to permeet the circumference DGg in E. Take OA equal to OD or OE, and join AB, AC, AP. Then, because the triangles OBA, OBD have the two sides OB, OA of the one equal to the two sides OB, OD of the other, each to each, and the included angles equal to one another, B A is equal to BD (I. 4.), and the angle OBA to the angle OBD, that is to a right angle. And, because OB is at right angles both to B A and BP, it is at right angles to the plane ABP (3.), and therefore the plane B O C, which passes through O B, is at right angles to the same plane ABP (18), or, which is the same thing, the plane ABP is at right angles to the plane O B C. And in the same manner it may be shown that the plane A C P is likewise at right angles to the plane OBC. Therefore the straight line A P, which is the common section of the planes A B P, ACP, is at right angles to the plane OBC (18. Cor. 2.), and the angle APB is a right angle. And, because in the rightangled triangles QBP, AB P, the hypotenuse QB is equal to the hypotenuse A B, and the side B P common to both the triangles, the angle QBP is equal to the angle ABP (I. 13.). But ABP measures the inclination of the planes AOB, BO C, because AB and BP are each of them perpendicular to the common section OB (17. Schol.) Therefore, the angle QBP measures the same inclination. Therefore, &c. Cor. 2 (Euc. xi. A.) It follows, from the solution of this problem, that if two solid angles be contained each of them by three plane angles, and if the plane angles which contain one of them be equal to those which contain the other, each to each, the dihedral angles of the one will likewise be equal to the corresponding dihedral angles of the other, each to each. Cor. 3. And in like manner it follows, from Cor. 1, that if two solid angles be contained each of them by three plane angles, and if two of the plane angles which contain one of them be equal to two of those which contain the other, each to each, and equally inclined to one another, the third angle of the one shall be equal to the third angle of the other, and its plane inclined at equal angles to It will be found that the above con- the adjacent planes. 1. Let the solid angle of the polyhedron in question be contained by three angles of equilateral triangles. Upon the given edge AB, describe (I. 42.) an equilateral triangle ABC: take the centre O (III. 26.) of the triangle ABC, and from O draw OD perpendicular to Α B the plane A B C (37.): in OD take the point D such that AD may be equal to AB, and join DA, DB, DC. Then, because OB and O C are each of them equal to OA, DB and D C are each of them equal to D A or AB (8.), and, therefore, the faces DA B, DAC, DBC of the solid D A B C are equilateral triangles: and because its solid angles are each of them contained by three angles of equilateral triangles, they are equal to one another (49. Cor. 2.). Therefore, D A B C is a regular solid; and since it has four faces, it is a regular tetrahedron. 2. Let the solid angle be contained by four angles of equilateral triangles. Upon the given edge A B describe square A B C D (I. 52.); take the a centre O (III. 26.), D/ and from O draw A E to E A, (I. 4.) or AB. Therefore, the eight faces of the solid E ABCDF are equilateral triangles. Again, each of the quadrilaterals EAFC, EBFD is a square, because its diagonals are equal, and (I. 22. Cor. 2.) bisect one another at right angles: therefore, the planes of any two adjoining faces are inclined to one another at the same angle as the planes of any other two, (49. Cor. 2.) viz., the angle which is contained by two angles of equilateral triangles forming with the angle of a square a solid angle. Therefore, any two of the solid angles, as E and F, may be made to coincide, and are equal to one another. Therefore, E ABCDF is a regular solid; and, since it has eight faces, it is a regular octahedron. 3. Let the solid angle be contained by five angles of equilateral triangles. Upon the given edge A B describe a regular pentagon ABCD E(III. 63. Cor.): take the centre O (III. 26.), and from O draw OF perpendicular to the plane ABCDE (37.): take OF such that FA may be equal to A B (which may be done by joining OA, and taking B (I. 59.) OF such that the square of OF may be equal to the difference of the squares of O A, A B), and join FA, FB, FC, FD, FE. Then, because the point O is equally distant from the points A, B, C, D, E, FB, FC, FD, FE are equal each of them to FA or A B (8.), and therefore FAB, FBC, FCD, FDE, FEA are equilateral triangles. Join BD; then because the triangles B FD, B C D have the three sides of the one equal to the three sides of the other, each to each, the angle BFD is equal to the angle BCD (I. 7.), that is, to the angle of a regular pentagon; and the same is evidently true of the plane angle formed by any two of the principal edges of the OE (37.) perpendi- * That is, take O D2 equal to A B2-A 02 (1. 59.) pyramid FABCDE, which are not adjacent to one another. Complete the regular pentagon BFDGH, and join CG, CH. Then, because the point C is equally distant from the three points B, F, D, it lies in that perpendicular to the plane BFD, which passes through the centre of the circle circumscribing the triangle B FD (37. Cor.), that is, through the centre of the pen. tagon BFDGH; therefore CG, CH are each of them equal to CB (8.) or AB, and C DG, CG H, CHB are equilateral triangles. Join CE, and let it cut BD in the point Z. Then, because the diagonal BD is common to the two pentagons ABCDE, HB FDG, and that another diagonal CE of the former cuts it in Z, it may easily be shown that the diagonal FG of the latter cuts it in the same point Z. Therefore EF, FC, and CG lie in one plane; and because the angles EFC, FCG, as shown in the case of the pyramid first constructed, are equal to the angles of a regular pentagon, let the regular pentagon EF C G K be completed, and join DK. Then, as above, it may be shown that DK is equal to DE or AB, and that DEK, DKG are equilateral triangles. Complete in like manner the pentagon AFDK L, and join LE; and in like manner the pentagon BFE L M, and join MA: then, as before, E L is equal to EA or AB, and EAL, ELK are equilateral triangles; and AM is equal to AB, and ABM, AML are equilateral triangles; and because FA, AM, FC, CH may be shown in the same way as above to be in one plane, and that they make with one another angles of a regular pentagon, the line MH will complete the regular pentagon FAMH C, and B M H will be an equilateral triangle. Again, because G K is parallel to the diagonal C E of the pentagon F CGKE, and K L to the diagonal D A of the pentagon FDKLA, and so on, the lines GK, K L, LM, M H, H G lie all in the same plane parallel to the plane ABCDE (15.); and the angle contained by every adjacent two is equal to the angle of a pentagon, because KGH is equal to E Z B (15.), that is, to E A B (I. 22), and so on. Therefore G K L M H is a regular pentagon. From the centre P of this Because the diagonals of a regular pentagon divide one another always in the same (viz. the medial) ratio. In fact, the diagonals BD, EC of the pentagon ABCDE being parallel to the sides AE. AB respectively (III. 26. 111. 12. Cor. 1. and III. 18.), it follows (1.22.) that EZ is equal to AB or DC; and the triangles ECD, DCZ (I. 6.) being isosceles and similar, EC is to CD (or EZ) as CD (or EZ) to CZ (II. 31.). pentagon draw (37.) a perpendicular PN' and take P N such (I. 59.) that N G may be equal to G K, and join N G, NK, NL, NM, NH: then N G K, N K L, NLM, NMH, NHG are equilateral triangles. Therefore all the faces of the solid FN are equilateral triangles. And it is evident that any two adjoining faces on the same side of the plane GKLMH are inclined to one another at thesame angle as any other two (49. Cor. 2.), viz., the angle which is contained by two angles of equilateral triangles forming a solid angle with the angle of a pentagon. Also, that any two upon opposite sides of that plane, as A LM, NLM, are inclined to one another at the same angle, may be shown by comparing it with the angle contained by any two G D C, FD C, upon opposite sides of the plane ABCDE: for, since a solid angle is formed at C by the three plane angles B CD, BC F, DC F, of which the first is an angle of a regular pentagon, and the two others angles of equilateral triangles, and in the same manner a solid angle at L by the three plane angles KLM, K LN, M LN, of which the first is an angle of a regular pentagon, and the two others angles of equilateral triangles, the dihedral angle, contained at the edge CD by the planes B CD, FCD, is equal to the dihedral angle contained at the edge L M by the planes K L M, NLM (49. Cor. 2.): and again, since a solid angle is formed at C by the three plane angles BCD, BCG, DC G, of which the two first are angles of regular pentagons, and the other an angle of an equilateral triangle, and in the same manner a solid angle at L by the three angles K L M, KLA, M LÅ, of which the two first are angles of regular pentagons, and the other an angle of an equilateral triangle, the dihedral angle contained at the edge CD by the planes BCD, GCD, is equal to the dihedral angle contained at the edge L M by the planes KLM, ALM (49. Cor. 2.); therefore, the whole dihedral angle contained at the edge CD by the planes FCD, GC D, is equal to the whole dihedral angle contained at the edge L M by the planes N L M, A L M (I. ax. 2.). Therefore, all the plane and likewise all the dihedral angles of the solid FN are equal to one another; and, consequently, any two of its solid angles, as at A and K, may be made to coincide, and are equal to one another. Therefore, FN is a regular solid (def. 9.); and, since it has twenty faces (viz. five, forming the pyramid which has the ver tex F and base ABCD; five more, forming the pyramid which has the vertex N and the base GKLM H; and ten more between the two bases ;) it is a regular icosahedron. F 4. Let the solid angle be contained by three angles of squares, that is, by three right angles. Upon the given edge A B describe a square ABCD: from A draw AE at right angles to the plane E ABCD (37.), and therefore at right angles both to AB and AD; take A E equal to AB, and complete the parallelopiped A D B EC. Then it is evident that the six faces of the parallelopiped EC are squares (I. def. 20. and IV. 22.); and because each of its solid angles is contained by three right angles, any two of them may be made to coincide, and are equal to one another. Therefore EC is a regular solid, and since it has six faces, is a regular hexahedron. This figure is the same as the cube (def. 13.). 5. Let the solid angle be contained by three angles of pentagons. Upon the given edge AB describe a pentagon ABCDE: find the angle, I, at which two angles of regular pentagons must be mutually inclined, in order that with a third angle, which is likewise an angle of a regular pentagon, they may contain a solid angle (49.); and through the sides A B, BC, &c. of the pentagon, already described, draw planes, each making with it that angle of inclination, and intersecting one another in the lines AF, BG, CH, DK, FL: then the angles K BCHNG, &c.: then, since the angles BGM, BG N are angles of pentagons, and so inclined, that with a third angle of a pentagon they may form a solid angle, the third angle M G N is an angle of a pentagon; and for the like reason the third angles at H, K, L, F are likewise angles of pentagons. Now, it is evident from the figure thus far constructed, that, if at adjacent angles FAB, GBA of a pentagon, other angles FAE, GBC of pentagons be placed at the inclination I, the edges A E, A B, BC will lie in one plane.* If, therefore, the pentagon GMRSN be completed, not only will H N S be an angle of a pentagon, for the reason before stated, but, for the reason just mentioned, OH, HN, NS will lie in one plane, because at the adjacent angles CH N, GNH, other angles CHO, GNS of pentagons are placed at the inclination I: and for the like reasons FMR is an angle of a pentagon, and QF, FM, MR lie in one plane. In like manner, if the pentagon HNSTO be completed, PK, KO, OT will be in one plane; and if KOTUP lie completed, QL, LP, PU will lie in one plane; and if LPUVQ be completed, M F, FQ, QV will lie in one plane, viz., the plane QFM, or QFMR, and therefore M R being joined will complete the pentagon FQVR M. Lastly, also, by the same rule, the lines RS, ST, TU, UV, VR lie in one plane, and make with one another angles of pentagons, and therefore RSTUV is a pentagon completing the solid A T. And because AT has all its faces regular pentagons, and all its solid angles (49. Cor. 2.) equal (for each of them is contained by three angles of pentagons), it is a regular solid; and, since the faces are twelve in number, it is a regular dodecahedron. And, in every case the regular solid is described with the given edge A B. Therefore, &c. As it is possible that the dotted or occult lines, which have been necessarily introduced in the foregoing constructions, may prevent the reader's acquiring from them a clear notion of the solids intended, we have here added shaded representations of the five regular solids, each in two different positions, in which only so much of the convex surface is exhibited as would present itself to the eye if they were opaque bodies. The same will evidently be the case if FAB, GBA are any other equal angles, FAE, GBC any other two likewise equal to one another, and the common inclination I any whatever. |