x2 =ƒa + go + 2 fg cos y; likewise the triangle TSP, in which TPS cos CSP, will give u2 = z2 + h2 + 2 hz cos CSP. We now want only to find the cosine of the angle CSP, or of the arc FH: and here, in the spherical triangle EFH, we have cos FH cos EH + sin EF sin EH cos E; substituting the values EF COSC-COS & COS Y =cos EF a and sin EH sin y cos E= we obtain cos FH-cos a cos EH + , EH). cos a + sin y +gh cos a. u2 =ƒ2 + g2 Hence, finally, the square of the required diagonal + h2 + 2 fg cos y + 2fh cos + 2 gh cos a. Cor. The solid angle A, is formed by the edges f, g, h, making with each other, two by two, the angles 200° -Y, 2000 6, a; hence it is enough to change the signs of cosy and cos in the expression of SE, in order to have the expression of AM2. By doing the same for the other two diagonals, we shall have the values of their squares as follows: 2fh cos 2 fh cos 2fh cos CP2 =ƒ2 + g2 + h2 + 2 fg cos y - = · 4 ƒ2 + 4g2 + h2 Hence, in every parallelepipedon, the sum of the squares of the four diagonals, is equal to the sum of the squares of the twelve edges. This remarkable theorem is analogous to that concerning the diagonals of a parallelogram (14. III), and might be derived immediately from it. For by means of the parallelograms SCTP, ABMN, we have ST2+ CP2 2 SC2 + 2 SP2 AM2 + BN2 — 2 BM2 + 2 AB2. Add these two equations together, observing that we have SC — BM, and SP2 + AB2 2 SA + 2 SB2; there will result ST2 + AM 2 +BN2+ CP2 4 SA2 + 4 SB2 + 4 SC2. PROBLEM VI. Given the three edges ending at one vertex of a triangular pyramid, and the three angles which these edges form together, to find the solidity of the pyramid. Let SABC be the proposed triangular pyramid, in which are given the edges SA =f, SB=g, SC = h, and the contained angles ASBY, ASC = 6, BSC = «. If on the edges SA, SB, SC, which are given in magnitude and position, the parallelepipedon ST be described, the pyramid, which is the third of the prism, will be the B sixth of this parallelepipedon ST. Hence, calling P the solidity of the pyramid, by Prob. IV. we shall have 2 Pfgh (1 S - COS α sin 2 sin + sin 2 2 2 2 PROBLEM VII. Given the six sides or edges of a triangular pyramid, to find its solidity. Retaining the same denominations as in the foregoing Problem, and, farther, putting BC=ƒ', CA = gʻ, BA = h', we shall have cos y = 2fg substituting these values in the formula already found, and for the sake of abridgment, making g+h? —ƒ12 = F, ƒ2 + h2 fe ƒ2 + g2 — h' ?H, we shall have the solidity P= √ -ƒ2F9-g Gehe H2 + FGH). % = g ? =G, 2 -g' √ (4ƒ3 g2 ha 2 In the application of these formulas, it will be observed, that f', g', h', designate the sides of one face or base, while f, g, h, designate the other three edges which end at the vertex, their arrangement being such, that ƒ is opposite to f', g to g', and h to h'. Scholium. Let A be the sum of the four triangles composing the surface of the pyramid; let r be the radius of the inscribed sphere: it is easy to see that PA× r; for the pyramid may be conceived as divided into four others, having for their common vertex the centre of the sphere, and for their bases the different faces of the pyramid. Hence we have the radius of the inscribed sphere r = 3 P PROBLEM VIII. Given the same things as in Prob. 6., to find the radius of the sphere circumscribed about the pyramid. Let M be the centre of the circle circumscribed about the triangle SAB; MO the perpendicular drawn through the point M to the plane SAB: in like manner, let N be the centre of the circle circumscribed about the triangle SAC; NO the perpendicular raised through the point N to the plane SAC. These two perpendiculars, lying in the same plane MDN perpendicular to SA, must intersect each other in a point O, which will be the centre of the circumscribed sphere; for the point O, as belonging to the perpendicular MO, is equally distant from the three points S, B, A; and this same point, as belonging to the perpendicular NO, is equally distant from the three points S, A, C: hence it is equally distant from the four points S, A, B, C. The point M in the plane SAB may be conceived as determined by means of the quadrilateral SDMH, whose two angles D and H are right, and in which we have SD =}ƒ, SH = 1 g, and ASB=r. Hence (Prob. 3.) we shall have DM = fcos y sin y ; in like man Let us designate the angle MDN, which measures the inclination of the two planes SAB, SAC, by D; in the spherical triangle of which 6,y are the sides D will be the angle opposite the side «; and thus COS a we shall have D = regarded as known. cos y cos sin y sin ; so that the angle D may be This being granted, in the quadrilateral OMDN, in which the two angles M and N are right, the two sides MD, DN, and the contained angle MDN =D being known, we shall have (Prob. 3.) the square of DM2+DN2 —— 2 DM × DN cos D sin 2 D Then in the the diagonal ODo = triangle OSD right-angled at D, we shall have SO2 = OD2+SD2; which is the value of the square of the circumscribed sphere's radius. If it is required to find the expression of the radius SO immediately by means of the data in Prob. 6., substitute the values of DM, DN, and next the values of cos D and of sin D; the result will be 2 [ƒ 2 sina+g sine+he sin2-2 fg (cosy-cost cosα) cosa cos y)-2 gh (cosa cos y cos C -2 fh (cos } NOTE VI. On the shortest distance between two straight lines not situated in the same plane. Let AB, CD be the two straight lines, not in the same plane, whose distance it is required to find. Along AB, extend two planes perpendicular to each other, and meeting CD, the one in C, the other in D; from the points C and D, draw CA and DB perpendicular to AB; in the plane ABD, draw DE parallel and AE perpendicular to BA, thus forming the rectangle ABDE; E in the plane CAE, join CE, and draw N M AI perpendicular to CE; lastly, in the plane CDE, draw IK parallel to DE till it meet CD in K, make AL= IK, and join KL: then, first, the straight line KL will be at once perpendicular to the two given straight lines AB, CD; secondly, this straight line KL will be shorter than any other that can join two points in the lines AB, CD, the line KL, or its equal AI, thus being the shortest distance required. = First. The three straight lines AB, AC, AE being perpendicular to each other by construction, one of them AB is perpendicular to the plane of the other two; hence AB is perpendicular to AI: besides KI is parallel to DE, and DE to AB, hence KI is parallel to AB; and since we made AL KI, the figure AIKL must be a rectangle. This being proved, the angle AIK is right as well as AIC; hence the straight line AI is perpendicular to the plane KIC or CDE; hence its parallel KL is perpendicular to the same plane CDE, and consequently perpendicular to CD. Hence, first, the straight line KL is at once perpendicular to the two straight lines AB, CD. Secondly. Let M be any point in the straight line CD; if through this point, MN is drawn parallel to DE or AB, the distance of the point M from the straight line AB will be equal to AN, the angle BAN being right. Now we have ANAI; hence AI is the shortest distance of the given lines AB, CD. Put the perpendiculars CA a, and DB = AE = b; we shall have CE=√(a+b); and since the area of the triangle ACE is ex pressed equally by ACX AE and by CEX AI, we shall have shortest distance between the given lines. If at the same time we put AB=c, and give the name A to the angle included between the two given lines, that is, to the angle CDE included between the line CD and DE a parallel to AB, the triDE angle CDE, right-angled at E, will give us cos CDE=CD, or C cos A= √(a2+b2+c2); because we have CD2=CE+ED2=a2+ b2+c2. From this also we might infer that sin A = C √(a+b2) √ (a2+b2+c2)' and cot A = √ (a2 + b2)* NOTE VII. ever; On symmetrical Polyedrons. It was for the sake of simplicity that, in Definition 16. VI., we supposed the plane to which the symmetrical polyedrons are referred to be the plane of a face: we might have supposed it to be any plane whatand the Definition in that case would have become more general without causing any change in the demonstration of Prop. 2., by which the mutual relations of the two polyedrons were established. A very correct idea of the nature of those two solids may also be obtained, by considering the one of them as the image of the other formed in a plane mirror, which mirror will occupy the place of the plane we have just been describing. NOTE VIII. On Proposition 25. Book VII. This theorem, which Euler first proved, in the Memoirs of Petersburgh, anno 1758, presents several consequences worthy of being developed. First. Let a be the number of triangles, 6 the number of quadrilaterals, c the number of pentagons, &c., composing the surface of a polyedron; the total number of faces will be a+b+c+d+ &c.; and |