April 1, 1887.] 



♦ KNOWLEDGE ♦ 



131 



Into the space emkl fit a tetrahedron of which ek and lm 

 are opposite edges, and in like manner fill in the spaces 

 KMFN. KSGO, and KOHL. Into the space lmno fit a pjTaniid 



apex downwards, and on its base lm\o add another pyramid. 

 We thus get the pyramid abcd having equilateral triangular 

 faces AKB, BKC, CKD, and dka. 



Cor. 1. It is obvious that we can in like manner set nine 

 pyramids to fit a square base; and fit in between them 12 

 tetrahedrons ; and then fit in 4 pyramids apex down- 

 wards, whose i bases \W1I give a surface such as abcd, 

 fig. 1. On this we build up as before a pyramid in which 

 are 6 pyi'amids and i teti'ahedrons. \Ye thus build up 

 19 pyramids and 16 tetrahedrons into a pyramid on a square 

 base and with equilatei-al triangular faces. 



Cor. 2. So can we build up 16 pyramids apex upwards 

 + 9 apex downwards +9 apex upwards +i apex down- 

 wards + 4 apex upwards -f 1 apex downwards -|- 1 apex 

 upwards, into a square-based equilateral-faced pyramid, by 

 fitting in 24 tetrahedrons in lowest layer, 12 in the 2nd, 

 and 4 Ln the 3rd — making 44 pyramids and 40 tetrahedrons 

 in all. 



Cor. 3. And manifestly, if a square base has a side 

 containing an edge of a pyramid or tetrahedron, n times the 

 number of pyramids and tetrahedrons required to build up 

 a pyramid on that base will be as follows : — 



Xo. of pyramids = w--|-2[(n-l)--|-(7i-2)--|- ... +9 



+ 4-hl] 



= n'' + '!LZ}{2n-l)n 



= 3(2n-^+l) 



(Observe that if « is not divisible by 3, 2n- + l is neces- 

 sarily divisible by 3 ; for in that case n is of one of the 

 forms 3r±l, and .•.2/(--t- 1 is of one of the forms 18'/-±12r 

 + 3.) 

 No. of tetrahedrons = 4 [(1-1-2 -f 3. . . +«) + (1 -|-2-|-3-|- 

 . .. +n-l)^- .. . +6j-£-|-l)] 

 = i[{\+2 + 3+ ... +,i-l) + (H. 

 2-1-3+ .. . -fw-2)-|- .. . -I-IO-F 

 6-1-3 + 1] 



= |(»l-l)«(7l+l) 



- -3 W'-- 



1) 



(Obviously if n is not divisible by 3, «'- — 1 

 divisible by 3, its two factors being n + 1 and n ■ 



must be 

 1, one of 

 which mast be a multiple of 3 when n is of one of the 

 forms 3?'±1.) 



When /( Ls very large, we see that the numbers of the 

 tetrahedrons and pyramids approach a ratio of equality. 

 The number of pyramids, however, always exceeds the 

 number of teti-ahedrous by n. Thus, if a = 1,000, or there 

 are a million pyi-amids Ln the base layer, the entire 

 pyi-amid, which will require --^ (999,999), or 666,666,000 



tetrahedrons, will require '-^' (2,000,001), or 666,667,000 

 pyramids — i.e., 1,000 more pyi-amids than tetrahedrons. 



Plzzle XXVI. To build up six octahedrons and eiy/it 

 tetrahedrons having eqiud square faces, into a single 

 octahedron. 



It is obvious that if aekh, ebfk, kfco, and kgdh, fig. 1, 

 represent octaliedrons projecting below as well as above the 

 plane abcd, a fifth octahedron LSixo, with four tetrahedron.? 

 in the previous case, will complete the semi-octahedron 

 abode above the plane abcd, whUe a sixth octahedron and 

 four tetrahedrons heloir this plane, will complete the other 

 semi-octahedron, or build up the entire octahedron. 



for. 1. Clearly we can buildup 9 octahedrons + 4 octa- 

 hedi'ons above +4 below the first nine^l octahedron above 

 + 1 below, with 1 2 tetrahedrons above +12 below the 

 nine octahedi'ons +4 aliove +4 below, to make a single 

 octahedron, using in all 19 octahedrons and 32 tetrahedrons. 



Cor. 2. Similarly can we buikl up a single octahedron of 

 41 octiihedrons and 80 tetrahedrons, simply dealing with the 

 case as in Cor. 2 on the previous puzzle, but using as many 

 tetrahedrons below as above the square base of the upper 

 semi-octahedron, and octahedrons instead of semi-octahedral 

 pyramids throughout. 



Cor. 3. And manifestl}-, if the edge of the containing 

 octahedrons is n times as long as any of the equal edges of 

 the component octahedrons and tetrahedrons, the numbers 

 of these solid figures will be as follows : — 



No. of tetrahedrons=i (2?i'' + 1), 



No. of octahedrons=-— -(w- — 1), 

 o 



the number of octahedrons being always less than twice the 

 number of tetrahedi'ons by 2« (or the number of octa- 

 1 hedrons always n greater than half the number of tetra- 

 hedrons). But when n is very great, the number of 

 tetrahedrons becomes very nearly twice as great as the 

 number of oct;ihedrons. 

 I Note. I propose to leave till next month the inquiry into 

 j the way in which, when space is filled in with octahedrons 

 I and tetrahedrons, or with semi-octahedral pyramids and 

 ' tetrahedrons, the component solids can be shifted in slic&s 

 [ or in strips. One of the puzzles given this month will be 

 found to bear on these points. 



PrzzLE XXVII. To find a semi-regular solid of such a 

 form that ant/ number of such solids, of equal size, can be 

 Imilt up into a single solid without interstices ; as a solid 

 I may be built up of equal cubes, but within the solid thus 

 I built up no shifting or sliding must be possible (as hi the 

 Cfise of a cube-built solid) vAthout inteiTuption of con- 

 tinuifi/. 



I propose to leave the discussion and demonstration of 

 this puzzle to next month, giving for the present only the 

 solution :— 



The solid required is a semi-regular dodecahedron. It 

 has twelve faces, each a rhombus or diamond, so that all its 



F16. 2. 



edges are equal. The solid angles of the figure are made up 

 of three of the larger, or four of the smaller angles of the 



