336 



♦ KNOWLEDGE ♦ 



[September 1, 1886. 



the evidence above adduced with regard to the Orion 

 nebula and the nebula in Sagittarius, and further con- 

 sidering the evidence given by the Andromeda nebula 

 recently, and the Argo nebula long since, should retain the 

 opinion that the nebuiie are external star-clouds only seen 

 by accident in o[itical connection with much nearer stars 

 belonging to our galaxy, there c;in be nothing to prevent him 

 from rejoicing in such an idai. Certainly no evidence could 

 have any influence in that direction. He could not be con- 

 vinced though " one rose from the dead," seeing that the 

 orbs of heaven would have spoken in vain for him. Inapti- 

 tude such as this for appreciating evidence might not pre- 

 vent him from being a fairly good astronomer of the 

 surveying sort. But the philosophy of astronomy must ever 

 remain a sealed book to him. 



THREE PUZZLES. 



(See last Number.) 



I. To arrange nineteen trees, so that tlieif inai/ form nine 

 rows, five in each. 



ARK out the ground as follows : — Let 

 A, D, B, F, 0, E, fig. 1, be the angles of a 

 regular hexagon. Join ab, bc, ca, de, ef, 

 ED, and also dc, af, eb, intersecting in one 

 point, o. Then if a tree be set at each of 

 the points where the nine lines thus 

 drawn intersect, viz., at a, d, u, f, c, e, at 

 fi, H, K, L, M, X, at o, and at p, Q, R, s, T, 

 I', there will be five trees on each line, or nine rows of five 

 trees in each. 



The hexagon need not be regular, nor need the figure be 

 in any way symmetrical. But the farmer M'ould naturally 

 prefer to have his ground neatly plotted out. 



This puzzle may be extended as follows : — Let c, A, D, b, E 

 be the angles of a regular figure, having '2n sides; draw 

 AK and CD, de cutting, and cutting ae in f and g. Then let 

 straight lines be drawn from c, A, d, b, e to o, the centre of 

 the circumscrilnng circle (as in fig. 1). This being done 

 fiom ever}' such angle, as at D, A, B, c, E, iSrc, it is evident 



D 



c 



Fig. 1. 



O 



Fig. 2. 



that on every such side as AB, Df, CE, &c., there be will be 

 five intersections, and on every line through the centre./(»tc 

 (for the lines to the centre will give straight lines through 

 the centre), so that the centre itself will make a fifth point 

 precisely as in fig. 1). 



Now let us see how many rows of five trees will be 

 formed if a ti-ee is put at each intersection, besides a tree at 

 o ; and how many trees will be retpiired : 



(1) There are 2n sides, and u lines through the centre; 

 that is, there are 3ti rows in all. 



(2) Counting the trees in triplets (as thus in fig. 1 — 

 D, H, G ; A, T, N ; (fee), we see that there are 2n triplets ; or 

 adding the central tree, the total number of trees is found 

 to be 6n + 1. 



Thus the general problem is : 



Given Qn + 1 trees it is required to arrange them in 3n 

 rows, .5 in each row. (But n must not be less than 3.) 



The solution for the general case is indicated in fig. 2. 



For the general problem regular figures are not actually 

 required but make a better show than irregular ones, which 

 also cannot be drawn casually. We know that straight 

 lines drawn from the angles of a regular figure of an even 

 number of sides inscribed in a circle will pass severally 

 through the opposite angles. But lines connecting opposite 

 angles (as in fig. 1) of a polygon with an even number of 

 sides will not intersect in a point unless the polygon is 

 either legular, or has its angles set on lines drawn through 

 one point. 



Tbe best regular arrangements are obtained from figures 

 inscribed either in a circle or an ellipse. It is only thus 

 that this particular puzzle is associated with the ellipse. 



II. To divide a rectangle (or, if preferred, a parallelogram) 

 into two parts lohich shall fit so as to form anot/ier 

 rectangle. 

 We prefer taking the general problem here, showing (he 

 limits within which .solution is possible. 

 Let ABCD (fig. 3) bj the paralltlogram. 



—^6 



c " 



Fig. H. 



Divide AB into any number of equal parts — .say, four — at 

 the points efg; and divide AD into equil parts, one more 

 in number — or in this case five — at the points hkhn. By 

 parallels through these points to the sides ab, ad, divide 

 ABCD into equal parallelograms, as shown in the figure. 

 Let ABCD be then divided into parts along the zigzag line 

 hopqrstc, and let the portion on the right of this line be 

 shifted to the position a6s, as shown in the figure. Then 

 will these two portions form a parallelogi-am, b6. 



This puzzle is worth examining geometrically. 



It is obvious that the longest parallelogram which can be 

 formed out of ad is obtained by dividing AC into two halves 

 by a line parallel to its longest sides. The parallelogram 

 be, thus formed, will have its angle at e, where /e is 

 twice BC. 



Similarly the longest parallelogram in direction ba will 

 be BF, where vm =: 2ba and bc = 2mf. 



The next longest either way will have angles at g and h, 

 two-thirds of ab and bc from bc and ab respectively, and 

 one-lialf iwrihev than d from ba and bc respectively. 



The next will have angles at k and l, threefourths of ab 

 and Br from bc and ab respectively, and one-third further 

 than D from ab and bc respectively. 



