(22) From two points A, B (fig. 1), to draw two straight lines to a point P in a given line ON such that AP+BP shall be a minimum. Take the given line as the axis of x, O the origin. Let OP, and let the co-ordinates of A and B be a, b, a1, b1. Then u = AP + BP = {b2 + (x − a)2 } 3 + {b,2 + (a, − x)2 } 3 = minimum. Whence x-a {b2 + (x − a)2 } } ̄ ̄ {b12 + (a, or the angles APM, BPN are equal. (23) To find the point in the straight at which BC subtends the greatest angle; pendicular to AD. line AD (fig. 2), ABC being per If the angle be a maximum its tangent is also a maximum. tan BPC = tan (APC – APB) = (a - b) x is to be a maximum. therefore (24) Through a point M (fig. 3) within the angle BAC draw the line PQ so that the triangle PAQ shall be a mini mum. Draw MN parallel to AQ, and let AN = a, MN = b, AP = x. Then 2a makes PAQ a minimum, and PQ is bisected in M. = (25) Given the length of the arc of a circle, find the angle which it must subtend at the centre in order that the corresponding segment may be a maximum. Let a be the half-length of the arc, & the radius of the a circle; then is the half-angle of the segment. 00 If - tan x = ∞, and u = 0 is a minimum. This last equation might be equally satisfied analytically by a value of between and 00 3 п 2 but such an angle is excluded by the geometry of the problem. A geometrical solution of this problem is given in the Mathematical Collections of Pappus, Book V. Theor. 16. (26) AC (fig. 4) and BD being parallel, it is required to draw from C a line CXY, such that the sum of the triangles ACX and BXY shall be a minimum. If AC = a, AB = b, AX = x, it is easily seen that the area of the triangle ACX is proportional to ax, and that of a (b - x)2 BXY to so that we have Vincent Viviani, Geometrica Divinatio, p. 152. (27) OM, OP (fig. 5) are two arcs of great circles on a sphere, and the arc PM is drawn perpendicular to OM; find when the difference between OP and OM is the greatest. Let POM = a, OP=4, OM=0; then we have 1 + tan20 аф = cos a (1 + tan2 ) d Ꮎ therefore 1 + tan ̊0 = cos a (1 + tan2), or Whence 1 + cos2 a tan2 = cos a (1 + tan2p). tan = (sec a)3, tan 0 = (cos a)3. (28) To determine the dimensions of a cylinder open at the top, which, under the least surface, shall contain a given volume. Let a be the given volume, a the radius of the cylinder, y its base. Then y = a determines the cylinder of least surface. = (29) The content of a cone being given, find its form when the surface is a maximum. be the given content, the radius of the base, (30) To inscribe the greatest cone in a given sphere. Let the radius of the sphere be r, and the distance of the base of the cone from the centre be x. Then = r gives the maximum cone. (31) To find the point in the line joining the centres of two spheres, from which the greatest portion of spherical surface is visible. If r, be the radii of the spheres, a the distance of the centres, and the distance of the required point from the centre of the sphere whose radius is r. (32) A regular hexagonal prism is regularly terminated by a trihedral solid angle formed by planes each passing through two angles of the prism; find the inclination of these planes to the axis of the prism in order that for a given content the total surface may be the least possible. Let ABC abc (fig. 6) be the base of the prism, PQRS one of the faces of the terminating solid angle passing through the angles P, R. Let S be the vertex of the pyramid. Draw SO perpendicular to the upper surface of the prism. Join OM, RP, SQ intersecting each other in N. Then it is easy to see that MN = NO and consequently_SO = QN, and as the triangles POR, PMR are equal, the pyramids PSRO and PMRQ are equal, so that, whatever be the inclination of SQ to OM, the part cut off from the prism is equal to the part included in the pyramid SPR, and the content of the whole therefore remains constant. We have then to determine the angle ONS or OSN so that the total surface shall be a minimum. Let AB, the side of the hexagon, = a, AP, the height of the prism, = b, OSN = 0. ON = NM = 1a, a, and SN The surface of ABPQ = 1a (2b – = a cosec 0, Then and QM = a cot 0. cosec 0. which is to be a minimum. Differentiating we have Hence tan SRN = 24 and tan SRQ = 21. This is the celebrated problem of the form of the cells of bees. Maraldi was the first who measured the angles of the faces of the terminating solid angle, and he found them to be 109°. 28′ and 70°. 32′′ respectively. It occurred to Réaumur that this might be the form, which, for the same solid content, gives the minimum of surface, and he requested Koenig to examine the question mathematically. That geometer confirmed the conjecture;-the result of his calculations agreeing with Maraldi's measurements within 2'. Maclaurin† and L'Huilliert, by different methods, verified the preceding result, excepting that they shewed that the difference of 2' was due to an error in the calculations of Koenig—not to a mistake on the part of the bees. (33) To determine the greatest parabola which can be cut from a given cone. Let ABC (fig. 7) be the cone BC= a, AC-b, BN = x. Then DN = b a x, and EN = (ax − x2)3. 4 3 The area of the parabola is EN. DN, or x (ax-x2), which is to be a maximum. (34) To determine the greatest ellipse which can be cut from a given cone. Let AC (fig. 8) = a, major axis of the ellipse. CD = b, CN= x, BP being the of the ellipse shall be a maximum gives 00= *Mémoires de l'Académie des Sciences, 1712, p. 299. |