Aug. 3, 1883.] 



♦ KNOWLEDGE ♦ 



79 



arc led, therefore, at once to the consideration that oui- triangle 

 will have its greatest area when the angles at A and B are equal. 



To see whether this is the case, we construct a new figure (Fig. 

 22), in which we omit all unnecessary parts of the former figure, 

 and draw A KF, so that when the triangle A K B is completed, the 

 angle KAB shall be equal to the angle K B A. We then draw 

 K L M parallel to A B, knowing that it is on the distance of this 

 parallel from A B that the area of the triangle A K B depends. We 

 take A E pretty near to A F (seeing that the triangle has obviously 

 nearhj a maximum area when the angles at A and B are equal, so 

 that any great departure from equality makes the triangle consider- 

 ably smaller). Let AE intersect K L JI in L. Then, if we can 

 show that B H, drawn as before, falls between B A and B L, our 

 surmise will have been proved to be correct. Xow the angle U B E, 

 by our construction, is equal to the angle H E B, therefore we 

 must show that the angle L B E is less than the angle L E B, or 

 L E less than L B (Enc. I., 19) ; therefore, adding A L, we have to 

 show that A E (or C) is less than A L, LB together. This is the 

 problem dealt with in Example 5, and thus the rest of the work 

 corresponds with the work in that example. We find that A L and 

 L B are together greater than A E, so that H does fall below L ; 

 and the triangle A K B is greater than the triangle A H B. Our 

 surmise is, therefore, shown to be correct, and the problem is 

 solved. 



It will be noticed that a problem in maxima and minima loses a 

 large part of its difficulty when, as is usually the case, we are 

 merely asked to prove that such and such relations supply a maxi- 

 mum or a minimum. In the case of Example 13, indeed, inspection 

 supplied a tolerably obvious solution, but this seldom happens. 

 Presented in the usual form, the above problem would run. 



Of all triangles on a given hase, and havi7ig a given perimeter, the 

 isosceles triangle is the greatest. 



Thus given, the problem reduces immediately to the case of 

 Example 5. 



Example 13 fitly introduces the following, which belongs to a 

 class often found perplexing : — 



Example 14. — Of all triangles having a given perimeter, the equi- 

 lateral triangle is the greatest. 



The difficulty in a problem of this sort resides in the fact that we 

 have three elements to consider, all of which admit of being 

 changed. In Example 13 we only had two sides to consider, and when 

 a length had been selected for one, the other was determined at the 

 same time. In Example 11 we have three sides, and must assign 

 lengths to two before the final condition of the triangle is deter- 

 mined. This would bo found to afford no assistance towards tho 

 solution of the problem. The way to proceed is to assign a length 

 to one side, pro%nsionally, and then to consider what relation must 

 hold between tho two remaining sides, whose sum ia now assigned. 



Fig. 23. 



Fig. 24. 



in order that the triangle may be as largo as possible. This we 

 have learned already from Example 13. Those two sides must be 

 equal. Hence, whatever side we suppose assigned, the remaining 

 two must bo equal, to make the area of the triangle a maximum. 



Therefore, obviously, the triangle must be equilateral. The proof 

 of this would ran as follows : — 



Let ABC (Fig. 23) be the triangle having the greatest possible 

 area with a given perimeter. Then ABC must be the greatest 

 possible triangle on a given base B C and with the sum of the 

 remaining sides equal to the sum of B A and AC. Hence B A is 

 equal to A C. But also, A B C is the greatest triangle on the base 

 AB with the given perimeter; hence, as before, AC is equal to 

 B C. Therefore A B, B C, and C A are all equal. 



As another instance of the application of this important method, 

 we give the following : — ■ 



Examplel5.—AB C (Ftj. 24) is an acute-angled triangle. It is 

 required to determine the position of a point P within the triangle, 

 such that the .s«ra of the distances PA,PB, PC, shall be a minimum.^ 



Assume P to be the required point. Then PA, P B, and P C 

 together have a minimum value. Therefore, also, P A and P B have 

 the least sum they can have so long as the length of P C remains 

 unchanged : so that if we draw the arc D P E with radius C P and 

 centre C, A P and P B are together less than the sum of any two 

 lines which can be drawn from A and B to meet on the arc D P E. 

 Hence (Example 11, Cor.), A P and P B are equally incUned to C P. 

 Similarly AP and PC are equally inclined to B P. Hence the 

 angles A P B, B P C, and C P A are all equal, and each, therefore, 

 is one-third part of four right angles. 



©m Cbtss Column. 



By Mephisto. 



SCORE OF THE NUREMBERG INTERNATIONAL TOURNA- 

 MENT, ON SATURDAY THE 28th JULY, 1883. 



As will bo seen from the score table published, this great tourna- 

 ment is all but concluded, Winawer being at the head of the list, 

 and Blackb\irne and Jlason following close behind. To speak of 

 Winawer's play as being absolute best would be doing injustice to 

 Blackburue, tliau whom none in this Tournament have played finer 

 games — a fact acknowledged by all other competitors. Winawer, 

 however, is fond of originality. He generally avoids the known 

 openings by playing P to Q3. As second player he also deyelopes 

 groat resources in difficulties. As a notable instance of this may 

 bo cited Iiis game with Hrnby, one of the most complicated games 

 iu the Tournament. At the adjournment the position stood very 

 unfavourably for Winawer, in spite of his being a rook to the good, 

 and it was the general impression that he would lose. Nevertheless, 

 he scored a comparatively easy victory. 



Blackbuine, who defeated Winawer, has played several games in 

 which, for powerless ]ilay in one given position, he has played in 

 matchless style ; but he has erred in some instances in judgment of 

 position, which cost him the first prize, notably in his game with 

 Mason, when he repeatedly avoided the draw in a position slightly 

 in his favour, and finally lost. This is a quality in play which re- 

 flects credit upon him to the disadvantage of his score. The fact 



