108 



♦ KNOVSTLEDGE 



[March 1, 1887. 



sellers for a small price. To hold it, make a wire clip 

 about six inches long, and stick one end iu the cork. 



The figures below arc copied from Vogelsang and repre- 

 sent the sulphur globules and crystals. 



SOLUTIONS OF PUZZLES. 



■ UZZLE XIX. To find the square sections of a 

 tetrahedron ; and to fill a square hole with a 

 tetrahedral plug. 



First, let ABCD, fig. 1 , be a tetrahedron ; E, F, 

 II, K, L, M, the bisections of its six equal edges. 

 Then lfkh is a square ; for lf and hk, being 

 each parallel to BC, are parallel to each other, 

 as are hl and kf. Also hl, lf, fk, and kh are all equal, 

 each being half an edge of the tetrahedron. Therefore hf is 

 a rhombus. But by symmetry the angle KFL=the angle 

 FLii ; and hl being parallel- to kf, these angles are together 



Fig. 1. 



equal to two right angles ; each therefore is a right angle. 

 Therefore hlkf is a square. So also mj'EH and elmk are 

 squares. 



Note. — If the student does not feel the force of the argu- 

 ment fi'om .symmetry, he may for his own satisfaction prove 

 LFKH a square, as follows : join ae, de, cutting lf, hk in I 

 and k respectively. Then be being perpendicular both to 

 ae and DE, is perpendicular to aed ; so therefore are hk and 

 LF perpendicular to aef, for they are both parallel to BC. 

 Also lZ=h^-, being the halves of equals. Hence lh is 

 parallel to Ik, and the angles Ihii and kuh are right 

 angles. 



Secondli/, to fill the square hole abdc, fig. 2, with a tetra- 

 liedral phig. The figure shows the tetrahedron of fig. 1 

 fitted into the liole abdc ; abc, bcd of fig. 2 being the faces 



so lettered in fig. 1 , and abd, dac of fig. 2 being the other 

 faces so lettered in fig. 1, supposed to be seen througli the 

 tetrahedron. The bi-sections of ab, bd, dc, ca are similarly 

 lettered in both figures, while m and e of tig. 1 are brought 

 into coincidence at o in fig. 2. 



Puzzle XX. To find the regular hexagonal S'-ctions of a 

 cube ; and to fill a hexagonal hole vjith a cubical plug. 



First, let ag, fig. 3, be a cube ; k, l, m, n the bi-aections of 

 the edges ab, bc, ae, and eh. It is clear that if kl be pro- 

 duced each way, it will meet da and dc produced in points 

 p, Q, such that Ar=AK=LC=CQ. So mx produced will pass 

 through p, and meet dh produced in R, such that hr^iin'=: 

 .MA=:AP. And obviously rq will bisect hg, gc in s and t. 

 Hence the plane of the equilateral triangle pqr gives the 

 section required. For, joining mk, ns, lt, and noting that 

 the points k, l, t, s, n, and .m are the trisections of lh;; sides 



;>Q 



Fig. 



of an equilateral triangle, we see that lkmnst is a regular 

 hexagon, from the well-known obvious properties of equi- 

 lateral triangles. 



And obviously there are four such hexagonal sections, 

 which may be obtained by completing the square inscribed 

 in ABCD, of which kl is a side and the construction as 

 above. 



Secondly, to fill the hexagonal hole abcghd with a cubical 

 plug. The figure shows the cube of fig. 3 fitted into the 

 hole abcghd ; fa, fc, fh on the near side being the edges so 

 lettered in fig. 3, and fb, fg, fd, supposed to be seen through 

 the cube, are also the same as the edges similarly lettered in 

 fig. 3. 



Puzzle XXI. To find the regular hexagonal sections of 

 an octahedron, and to fill a hexoAjonal hole with an octa- 

 hedral plug. 



First, let abcdef, fig. 5, be an octahedron ; g, h, the 

 bisection of the opposite edges av, bc ; kl, nm respectively 



E 



Fig. 



parallel to ab and dc, and bisecting the sides ae, eb, fd, fc. 

 Then kl, cm, and n.m are parallel and in the same plane 



