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♦ KNOWLKDGE ♦ 



[October 1, 1886. 



graphic projections of the circles circumscribing these six 

 equal divisions of the sphere. 



A little consideration will show that these circles intersect 

 eacli other on the sphere, and therefore in the stereographic 

 projection at angles of 60^. For the sides A(ib, acIb, ase 

 meet at equal angles; each, therefore, 120°. Hence, a«b 

 and A(7d meet the circle abcd at an angle of 30° (half of 

 180° — 120°). They must in like manner (regarding them 

 as representing sides of the spherical square eabf) meet the 

 circle eabf at the .same angle, 30°. Hence the circles abcd 

 and EABF cut each other at angles of 00°, and the like with 

 all the other circles of the figure. 



Note that the arcs a«b, g)?ih belong to one circle. The 

 reason will be obvious to those familiar with the properties 

 of the stereographic projection. For on the sphere the 

 points A, Ci, lie at the extremities of a diameter, and b, h, 

 at the extremities of another. So that the points abgh 

 necessarily lie on a great circle. Similarly with the arcs 

 b6c, h«e, (tc, ifec. 



Problem VI. To draw iivelve circles cutting each other in 



twentji points at an amjh of 60°. 



The method is shown in fig. 3. The explanation is 



similar to that of Problem V. In fig. 3 the dotted lines 



represent the stereogi-aphic projection of the twelve portions 



Fig. 3. 



of a sphere corresponding to a circumscribing dodeca- 

 hedron — the twelve maps, for instance, of my " Library 

 Star Atlas" and "School Star Atlas," the central point 

 of one face being the centre of projection. The twelve 

 circles are the stereographic projections of the circles 

 circum.scribing these twelve equal divisions of the sphere. 

 (In my atlas the twelve circles are the boundaries, so that 

 the niai)s conveniently overlap.) 



The dotted arcs ah and c'd' form part of the same circle, 

 as do the arcs au, Dd, ic, <tc., kc. For the points «,c' are 

 antipodal, as are t,D' and a,D, kc, kc, kc 



The problems for solution next month are these three : — 

 Problem VII. To mark in a series oj" points on a card, by 

 some simple geometrical co7ist ruction, by joining which 

 with siraif/ht lines a parabola maij be pictured. 

 Problem Vlli. The same for an elliptic. 

 Problem IX. The same for a hyperbola. 

 (To be continued.) 



THE SIXTY-FOUR SIXTY-FIVE PUZZLE. 



T appears that misprints in the description of 

 my plan for improving the construction of the 

 '■61:-G.5 square" puzzle rendered it unintelli- 

 gible to man}'. The printers w-ere not in fault, 

 as this was the firet occasion when any mathe- 

 matiail work has been done by them for 

 Knowledge since the printing passed into 



their hands ; and m}' plan of underlining single letters 



intended to be italic, seems to have be^n misundei'stood in 



some cases as signifying small capitals. 



I will now give a brief but better account of my method 



—which really makes the trick very much better than as 



usualh' pre.sented : — 



In fig. 1 we have the rectangle of 65 squares, 5 by 13. 



The diagonal uh cuts the fifth line fa in li above e, the 



Fig. 1. 



intersection o? f(( and kl ; while it cuts af in g below e, the 

 intersection of fa and kl. In fitting the four pieces B(/a, 

 ad6(/, hah, and Boah, together, this defect may attract 

 attention. If, instead, we divide up the square of Oi squares, 

 a corres])onding defect will arise which a keen eye wiU also 

 raadily detect. 



Let us see what the discrepancy reallj' amounts to. Sup- 

 pose the squares each 1 inch in the side. Then fe is 

 2 inches ; and since./"// : Di : : b/' : bd, we see that fh^=f^ of 

 5 inches. Thus eh = 2 in. — f ;] in. = ^L inch = Y.g. 



To prevent a keenej-ed person from detecting, as he 

 might, thLs 13th of an inch, halve it between the two 

 arrangements. Instead of drawing b6 as in one method 

 (where the rectangle is cut up), or Be and eh as in the other 

 (where the square is cut up), draw a line from b to a point 

 midway between /( and e, and from this same point carry on 

 a line to b. In like manner, draw from h a straight line to 

 a point midway between g and E, and from this same point 

 carry on a line to B. Cut along these lines, by which you 

 cut out a long parallelogram, having b6 as a diagonal, and 

 equal in area to half one of the 05 squares. This is thrown 

 awa}-. Thus practically, in presenting the problem this way, 

 you put together areas equal to 04^ squares, in such ^\•ays 

 that they may be supposed either to make up 04 squares or 

 65 squares : whereas, on the usual plan, you either have to 

 convince folk that 64 .squares are 65, or that 65 squares 

 are 64. 



But on this plan, as on the other, you still fail to have 

 true corners for the squares around e and E. Now this 

 peculiarity of shape is more easily detected than the errors 

 of length and area which aft'ect the drawing. To get rid 

 of it, the lines fa and kl must be altered. INIy plan for this, 

 which makes the puzzle as neaily perfect as it can be, may 

 best be illustrated as follows : — 



Let hem, fig. 2, be an enlarged view of the small triangle 

 which lias he for a side in fig. 1. Let p be the middle 

 point of he, and let pq be part of the cutting line to 6 on 

 the plan described. Then jieq is the triangle whose absence 

 from the .small square having e as its lower left-hand corner 



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