PROBLEM. 579. The side of a regular polyedron being given, to find the radius of the inscribed and that of the circumscribed sphere. Solution. It must first be shown, that every regular polyedron is capable of being inscribed in a sphere, and of being circum scribed about it. Fig. 292. Let AB (fig. 292) be the side common to two adjacent faces; Cand E the centres of those faces; CD, ED, the perpendiculars let fall from these centres upon the common side AB, and therefore terminating in D the middle point of that side. The two perpendiculars CD, DE, make with each other an angle which is known, being the inclination of two adjacent faces, and determinable by the last problem. Now, if in the plane CDE, at right angles to AB, two indefinite lines CO and OE be drawn perpendicular to CD and ED, and meeting each other in O; this point o will be the centre of the inscribed and of the cir. cumscribed sphere, the radius of the first being OC, that of the second 0A. For, since the perpendiculars CD, DE, are equal, and the hypothenuse DO is common, the right-angled triangle CDO must (56) be equal to the right-angled triangle ODE, and the perpendicular OC to OE. But, AB being perpendicular to the plane CDE, the plane ABC (349) is perpendicular to CDE, or CDE to ABC; likewise CO, in the plane CDE is perpendicular to CD, the common intersection of the planes CDE, ABC; hence (351) CO is perpendicular to the plane ABC. For the same reason, EO is perpendicular to the plane ABE; hence the two straight lines CO, EO, drawn perpendicular to the planes of two adjacent faces, through the centres of those faces, will meet in the same point O, and be equal to each other. Now, suppose that ABC and ABE represent any other two adjacent faces; the perpendicular CD will still continue of the same magnitude; and also the angle CDO, the half of CDE; consequently the rightangled triangle CDO, and its side CO will be equal in all the faces of the polyedron; hence, if from the point O as a centre with the radius OC, a sphere be described, it will touch all the faces of the polyedron at their centres, the planes ABC, ABE, &c., being each perpendicular to a radius at its extremity; therefore the Fig. 293. sphere will be inscribed in the polyedron, or the polyedron circumscribed about the sphere. Again, join OA, OB; since CA = CB, the two oblique lines 0.A, OB, being equally remote from the perpendicular, will be equal ; so also will any other two lines drawn from the centre O to the extremities of any one side; hence all those lines will be equal to each other; and, if from the point O as a centre, with the radius OA, a spherical surface be described, it will pass through the vertices of all the solid angles of the polyedron; hence the sphere will be circumscribed about the polyedron, or the polyedron inscribed in the sphere. This being settled, the solution of the problem presents no farther difficulty, and may be effected thus : One face of the polyedron being given, describe that face; and let CD (fig. 293) be a perpendicular from its centre upon one of its sides. Find, by the last problem, the inclination of two adjacent faces of the polyedron, and make the angle CDE equal to this inclination; take DE=CD; draw CO and EO perpendicular to CD and ED, respectively; these two perpendiculars will meet in a point 0; and CO will be the radius of the sphere inscribed in the polyedron. On the prolongation of DC, take CA equal to a radius of the circle, which circumscribes a face of the polyedron ; AO will be the radius of the sphere circumscribed about this same polyedron. For, the right-angled triangles CDO, CAO, in the present diagram, are equal to the triangles of the same name in the preceding diagram; and thus, while CD and CA are the radii of the inscribed and the circumscribed circles belonging to any one face of the polyedron, OC and OA are the radii of the inscribed and the circumscribed spheres which belong to the polyedron itself. 580. Scholium. From the foregoing propositions, several consequences may be deduced. 1. Any regular polyedron may be divided into as many regular pyramids as the polyedron has faces; the common vertex of these pyramids will be the centre of the polyedron; and at the same time, that of the inscribed and of the circumscribed sphere. 11. The solidity of a regular polyedron is equal to its surface multiplied by a third part of the radius of the inscribed sphere. 11. Two regular polyedrons of the same name are two similar solids, and their homologous dimensions are proportional; hence the radii of the inscribed or of the circumscribed spheres are to each other as the sides of the polyedrons. iv. If a regular polyedron is inscribed in a sphere, the planes drawn from the centre, along the different edges, will divide the surface of the sphere into as many spherical polygons, as the polyedron has faces all equal and similar among themselves. Improved Demonstration of the Theorem for the Solidity of the Triangular Pyramid. BY M. QUERET OF ST. MALO: THEOREM. 568. Two triangular pyramids, having equivalent bases and equal altitudes, are equivalent, or equal in solidity. Let S-ABC, s-abc (fig. 294) he two triangular pyramids of Fig. 294. which the two bases ABC, a b c, supposed to be situated in the same plane, are equivalent, the altitude TA being the same in both. If they are not equivalent, let s-a b c, be the smaller; and suppose A a to be the altitude of a prism, which having ABC for its base, is equal to their difference. Divide the altitude AT into equal parts A x, xy, yz, &c., each less than A a, and let k be one of these parts; through the points of division suppose planes parallel to the plane of the bases ; the corresponding sections formed by these planes in the two pyramids will be respectively equivalent by art. 409, namely, DEF, to d e f, GHI, to g h , &c. This being granted, upon the triangles ABC, DEF, GHI, &c., taken as bases, construct exterior prisms having for edges the parts AD, DG, GK, &c., of the side SA; in like manner, on the bases de f, ghi, klm, &c., in the second pyramid, construct interior prisms having for edges the corresponding parts of sa. It is plain that the sum of all the exterior prisms of the pyramid S-ABC will be greater than this pyramid; and also that the sum of all the interior prisms of the little pyramid s-abc will be less than this. Hence the difference between the sum of all the exterior prisms and the sum of all the interior ones, must be greater than the difference between the two pyramids themselves. Now, beginning with the bases ABC, a b c, the second exterior prism DEFG is equivalent to the first interior prism defa, because they have the same altitude k, and their bases DEF, def, are equivalent; for like reasons, the third exterior prism GHIK and the second interior prism ghid are equivalent; the fourth exterior and the third interior; and so on, to the last in each series. Hence all the exterior prisms of the pyramids S-ABC, excepting the first prism DABC, have equivalent corresponding ones in the interior prisms of the pyramid s-abc; hence the prism DABC is the difference between the sum of all the exterior prisms of the pyramid S-ABC; and the sum of all the interior prisms of the pyramid s-abc. But the difference between these two sets of prisms has already been proved to be greater than that of the two pyramids; which latter difference we supposed to be equal to the prism a ADC; hence the prism DABC must be greater than the prism a ABC. But in reality it is less; for they have the same base ABC, and the altitude A x of the first is less than A a the altitude of the second. Consequently the supposed inequality between the two pyramids cannot exist; therefore the two pyramids S-ABC, s-abc, having equal altitudes and equivalent bases, are themselves equivalent. THEOREM. 569. Every triangular pyramid is a third part of the triangular prism having the same base and same altitude. Fig. 216. Demonstration. Let F-ABC (fig. 216) be a triangular pyramid, ABCDEF a triangular prism of the same base and the same altitude; the pyramid will be equal to a third of the prism. Cut off the pyramid F-ABC from the prism, by a section made along the plane FAC; there will remain the solid FACDE, which may be considered as a quadrangular pyramid, whose vertex is F, and whose base is the parallelogram ACDE. Draw the diagonal CE; and extend the plane FCE, which will cut the quadrangular pyramid into two triangular ones F-ACE F-CDE. These two triangular pyramids have for their common altitude the perpendicular let fall from F on the plane ACDE; they have equal bases, the triangles ACE, CDE, being halves of the same parallelogram; hence (568) the two pyramids F-ACE, F-CDE, are equivalent. But the pyramid F-CDE |