MISCELLANEOUS EXERCISES ON SOLID GEOMETRY. Pages 428–430. 1. Let ab, cd be the projections of two parl. st. lines AB, CD on any plane XY. Then, because Aa, Cc are both perp. to plane XY, .. Aa is par!. to Cc (xi. 6]. And AB is par!. to CD, .. plane BAа is par!. to plane DCC (xi. 15). But these planes intersect XY in ab and cd respectively; .. ab is parl . to cd [xı. 16]. 2. Draw AE parl to ab. AE will be in plane AabB and will cut Bb in E. Similarly CF, drawn parl. to cd, will cut Dd in F. Because AE and CF are part. respectively to ab and cd which are par!. to one another, .. AE is par!. to CF (xi. 9]. And because the sides of ABE are respectively par!. to the sides of A CDF, .. the _s of A ABE are equal respectively to the į s of A CDF [xı. 10]. .. AB : CD=AE : CF = ab : cd. 3. Let AB, CD be the two given st. lines. Through E any pt. in AB, draw EF par!. to CD: and through H any pt. in CD, draw HG par! to AB. Then the plane containing AB, EF is par!. to the plane containing CD, HG (XI. 15). 4. Let AB, CD be the two given st. lines. As in the last Ex., draw through AB, CD two par. planes. Then it follows from XI. 16 that the projections of AB, CD on any plane perpendicular to the two parl. planes will be parl. 5. In the fig. of p. 421, let AB, CD be the given non-intersecting st. lines, having directions at rt. angles to one another; and let He be the line of constant length. Required the locus of M the middle point of HE. Draw PQ perp. to AB, CD [Ex. 2, p. 421], and let XY be the plane through AB par to CD. Draw HK perp. to the plane XY. Join QK, KE; and let the plane through M par. to XY cut PQ, HK at O, s. Then 0, S are the middle points of PQ, HK (xı. 17]. Join OM, OS, SM; and draw MN perp. to the plane XY, meeting KE at N. Join QN. Then N is the middle point of KE. Now in the rt. angled A HKE, since HE and HK are constant, :. KE is constant. [1. 47.] And in the rt. angled A KQE, since the hyp. KE is constant, and N is its middle point, .. QN = one half of KE = constant. [111. 31.] But OM = QN; .. the locus of M is a 0, of which o is the centre, lying in a plane par. to AB, CD and midway between them. 6. Let o be the angular point. Then from the rt. angled A$AOB, AOC, it follows that BO, OC are less than BA, AC respectively. (1. 18.] But BC?=B0? + OC? (1. 47.] .. BC2 is less than BA? + AC?; 7. Since the opp. faces of a parallelepiped are parallel, .. their common sections with a third plane are parallel (xi. 16). 8. Let da, dB, dc be three edges terminating in d, and let a, b, c, d be the vertices diagonally opposite to A, B, C, d respectively. Join dd, dc. Then, because each of the planes DcBa and DcAb are perp. to the plane dBcA, .. their common section Dc is perp. to the plane dBcĀ [xı. 19). .. Dc is perp. to do which meets it in that plane. .. dD2 = cD? + dc?. Again, because the planes Bdca, BdAc are each perp. to the plane BcDa, :, their common section Bd is perp. to the plane BcDa... Bd is perp. to Bc, which meets in that plane. .. dc* = Bc2 + dB?. .. dD2 = cD+ Bc + dBP = dc? + dA? + dB?, since the faces are parallelograms. 9. Since the edges of a cube are equal, -. (diagonal)2 = three times (edge). [Ex. 8.] 10. See fig. p. 422. In parm. ACA'C', A'A' + C'C? = 2AC2 + 2A'c [Ex. 25, p. 147]; and in parm. BDB'D', B'B? + D'D2 = 2BD? + 2B'D?. .:. A'A? + B'B? + C'C? + D'DP = 2AC2 + 2A'C? + 2BD? + 2B'D?. But in parm. ABCD, AC2 + BD2 = 2AB? + 2BC? ... sum of squares on diagonals of pard = 4AB2 + 4BC2 + 2A'C2 + 2B'D= 4AB2 + 4BC2 + 4A'C? = sum of squares on twelve edges. 11. Let AP be perp. to base BCD of a regular tetrahedron ABCD. Join BP, CP, DP, and produce them to meet the sides of the base in X, Y, Z. + + + Then from the rt. angled As APB, APC, APD, we have PB PC = PD. [Ex. 12, p. 91.] And from the As PBC, PBD, the PBC the PBD. (1. 8.] Lastly from the A8 XBC, XBD, we have XC = XD. [1. 4.] Hence Bx is a median of the base : similarly CY, DZ are medians, and P divides each of them in the ratio 2 : 1. [Ex. 4, p. 105.] 12. Let AP, BQ be perps. from the vertices A, B upon the faces BCD, ACD respectively. Then AQ, BP meet at E, the middle pt. of CD (Ex. 11]. Draw QR par to AP. This will cut BE, because the parls, AP, QR are in the same plane ABE. And because AP is perp. to plane BCD, so also is QR. .. AP : QR = AE : QE = 3: 1. [Ex. ll.] 13. With the fig. of last Ex., AE? = BE” = BC – CE= 3CE. 14. Let ABCD be the given tetrahedron. Bisect AB in E, CD in E', AD in .F, and BC in F'. Then EF, E'F' are both par. to BD, .. EF is parl . to E'F'; and EF', E'F are both par. to AC, :. EF' is par!. to E'F (xi. 9] ... EFE'F' is a parm... FF' bisects EE'. Similarly if G, G' are the middle pts. of AC, BD, GG' also bisects EE'. .. EE', FF', G' intersect one another at the middle pt. of each. 15. In the tetrahedron ABCD let a plane part to AC and BD cut the edges AB, BC, CD, DA in the pts. E, F, G, H respectively. Then, because BD is par!. to the plane EFGH, .. BD is par! to EH, the common section of EFGH with the plane ABD through BD (Ex. 9, p. 418]. Similarly FG is par!. to BD... FG, EH are parł to one another. Similarly EF, HG are parl. to one another. 16. Let E, F be the middle pts. of AB, CD, opp. edges of a regular tetrahedron ABCD. Then the AS CED, AFB being isosceles, EF is perp. to CD and to AB. : . EF is the shortest distance between AB and CD. [Ex. 2, p. 421.] Now EF=CE? – CF? = BC – BE? – CF2 = 2cF”. .:. 4EF2 = 8CF2. But sq. on diagonal of sq. on CD = 2CD2 = 8CF. .. EF = half diagonal of sq. on edge CD. 17. Let AB be at rt. Ls to CD, and AD at rt. s to BC. Draw BL, CM, DN perps. on CD, DB, BC to cut in a the orthocentre of BCD. Then, because CD is at rt. to AB and BL, .. it is at rt. _ s to the plane ABL, and .. at rt. to Aa in this plane. Similarly BC is at rt. _ s to Aa. :. Aa is perp. to plane BCD, and .. perp. to BD in that plane... BD is perp. to Aa and am. :: BD is perp. to plane Aam, and .. perp. to AC in that plane. 18. By last example, the perps. Ad, Cc upon the opp. faces cut those faces in their orthocentres. And the perpå, upon any edge such as BD from the extremities of the opp. edge AC meet BD in the same pt. M. (1) Let Aa, Cc cut in X. Then, since a is on CM, and c is on AM, .. X is the orthocentre of A ACM. .. X is the pt. where MM', the perp. from M upon AC, cuts Aa and Cc. But BD is perp. to plane ACM, .. MM' in this plane is perp. to BD and to AC. .. Aa, Cc and the shortest distance between AC and BD cut in the pt. X. (2) Join BX, and produce it to meet the plane ACD at b. Then, because CD is perp. to AB and Ba, CD is perp. to plane Aba. :: plane ACD is perp. to plane Aba, and similarly it is perp. to plane Coc, .. it is perp. to BX the common section of Aba and cbc. Hence the perpe. from B on ACD and from D on ABC cut at X. .. all the perps are concurrent with one another and with the shortest distance between AC and BD, and therefore with the shortest distances between AB and CD, and between AD and BC. 19. By the last examples, ABP = AMP + BMP and CD2 = CM2 + DMP, and BC2 BM? + CMP and AD2 = AMP + DM?. ::. ABP + CD2 = BC? + AD?. 20. In the tetrahedron ABCD let P, Q, R be the middle pts. of AB, AC, AD; and L, M, N the middle pts. of CD, DB, BC. Join PL, PC, PD. Then 2 (DA? + DBạ) = 4DP2 + 4AP? [Ex. 24, p. 147.] = 40PS + AB? Similarly 2 (CA? + CB”)= 4cP® + AB? :. by addition, DAR + DB+ CA+ CBP = 2DP? + 2CP? + AB? 4PL? + CD? + AB?. (Ex. 24, p. 147.] Adding the three similar equations, ABP + AC? + ADP + DBP + BC? + CD2 = 4PL? + 4QM? + 4RN?. 21. Let the plane ACE, which bisects the — between the planes ACB, ACD cut BD in E. Draw DN perp. to plane ACE, and DP, DQ perp. respectively to AC, CE in that plane. Then NP and NQ are perp. respectively to AC and CE [Ex. 14, p. 418]. :. the 4s NPD, NQD are respectively the inclinations of the plane ACE to the planes ACD and BCD. If now Bn, Bp, Bq are drawn perp. respectively to the plane Ace and to the st. lines AC, CE, then the L npB, nqB are respectively the inclinations of the plane ACE to the planes ACB and BCD. .. L npB= 2 NPD and _ nqB= 2 NQD... by similar A$, BE : DE = Bq: DQ=Bn : DN=Bp : DP=AACB : AACD. 22. If OA, OB, oc are mutually at rt. 2$, and the A ABC is equilateral, it may be proved that OA= OB=OC. Take P any point within the ABC; and draw PL, PM, PN perp. respectively to the planes OBC, OCA, OAB. Through P take a plane obc parl. to OBC, and therefore perp. to OA. Then PM and PN lie in the plane obc. Now PL + PM + PN = PL + ON + Nb = 00 + ob = 00 + A = OA. 23. Let ABCD... be the base, and abcd... the top of the prism. Let two parł. planes cut Aa, Bb, Cc, Dd ..., in H, K, L, M... and h, k, l, m... . Then HK is par. to hk [xı. 16), and Hh is par'. to kk [xl., Def. 14.], :. HK is par!. and equal to hk. H. K. E. 14 |