3. The regular octaedron. Cons. Upon AB construct the square ABCD. At its centre 0 erect EOF to the plane ABCD, so that OE OF = OA. Join the pts. E and F to all the vertices of the square. Then E-ABCD-F is a regular octaedron. A E *· Proof. The lines from E and F to A, B, C, D are all equal. And since the rt. As AOE, AOB are equal, ... AE AB. (496) (104) ... the twelve edges of the octaedron are all equal, and the faces are eight equal equilateral As. Since the diagonals BD and EF are equal and bisect each other at rt. /s, (Cons.) ... BEDF is a square = ABCD. And since AO is to BD and EF, .. AO is to the plane of BEDF. (500) pyramid A-BEDF = pyramid E-ABCD. (633) .. polyedral A = polyedral E. Similarly, it can be shown that any other two polyedral Zs are equal. ... E-ABCD-F is a regular octaedron. 4. The regular dodecaedron. Cons. Upon AB construct the regular pentagon ABCDE, and join to its sides the sides of five other equal pentagons, so inclined to the plane of ABCDE as to form five triedral Zs at A, B, C, D, E. There is then formed a convex surface FGHIK, etc., composed of six equal regular pentagons. Construct a second convex surface fghik, etc., equal to the first. Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular dodecaedron. Proof. Because the faces of the triedrals in FK are equal respectively to the faces of the triedral s in fk, (Cons.) .. the triedral s of FK and fk are equal, each to each. (56%) Now suppose the convexity of FK to be up, and the convexity of fk to be down. Put the two surfaces together so that the pt. O and the side OP shall coincide with the pt. n and the side no, respectively. Then two consecutive faces of one surface will unite with a single face of the other. Thus the three faces 1,5, 7, will enclose a triedral, since the diedral contained by 1 and 5 is the diedral of the equal triedral formed at E. The vertex N will coincide with m, and a like triedral will be formed at that pt., and so of all the others. .. the triedrals are all equal, and the solid is a regular dodecaedron. 5. The regular icosaedron. Cons. Upon AB construct a regular pentagon ABCDE. At its centre erect a to its plane, and in this take a pt. S so that SA = AB. Join SA, SB, SC, SD, SE. Then S-ABCDE is a regular pentagonal pyramid (496), and each of its faces is an equilateral A. Complete the polyedrals at A and B by adding to each three equilateral s each equal to SAB, and making the diedrals around A and B equal. There is then formed a convex surface CDEF, etc., composed of ten equal equilateral as. Construct a second convex surface cdef, etc., equal to the first. Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular icosaedron. Proof. Suppose the convexity of DG to be up, and the convexity of dg to be down. Put the two surfaces together so that the pt. D, where two faces meet, falls upon the pt. c, where three faces meet. Then two consecutive faces of one surface will unite Iwith three consecutive faces of the other. Thus, the two faces 1 and 9 will unite with the three faces 10, 11, 12, forming a polyedral of five faces equal to S, without in any way changing the form of either surface, since the three diedrals contained by 1 and 9, 10 and 11, 11 and 12 are those which belong to such a polyedral . The vertex C will fall at h, and a like polyedral will be formed at that pt., and so of all the others. .. the polyedrals are all equal, and the solid is a regular icosaedron. Q. E. F. 654. SCH. Models of the regular polyedrons may be easily constructed as follows: Draw the following diagrams on cardboard, and cut them out. Then cut half way through the board in the dividing lines, and bring the edges together so as to form the respective polyedrons. TETRAEDRON HEXAEDRON OCTAEDRON DODECAEDRON ICOSAEDRON GENERAL PROPERTIES OF POLYEDRONS. EULER'S THEOREM.† Proposition. 27. Theorem.* 655. In any polyedron the number of edges increased by 2 is equal to the whole number of vertices and faces.‡ Hyp. Let E, F, and V denote the number of edges, faces, and vertices respectively of the polyedron S-ABCD. E+2=V+ F. To prove Proof. Beginning with one. face ABCD, we have EV. If we annex to this a second face SAB, by applying one of its edges, as S B AB, to the corresponding edge of the first face, we form a surface having one edge AB, and two vertices A and B common to both faces; therefore, the whole number of edges is now one more than the whole number of vertices. ... for 2 faces ABCD and SAB, E = V + 1. If we annex a third face SBC, adjacent to both ABCD and SAB, we form a new surface having two edges SB and BC, and three vertices S, B, C in common with the preceding surface; therefore the increase in the number of edges is again one more than the increase in the number of vertices. or .. for 3 faces, E · V + 2. In like manner, for 4 faces, E = V + 3. And so on until every face but one has been annexed. .. for (F1) faces, E = V + F -2. But annexing the last face adds no edges nor vertices. .. for F faces, E = V + F E+2=V+F. 2, + The proof of this theorem was first published by Euler (1752). Q.E.D. |