J0LY 27, 1883.] 



♦ KNOWLEDGE ♦ 



63 



Or, we may produce A B and A C in Fig. 17, and conceive the 

 part of the figure to the right of A E rotated round A E till it 

 falls on the part to the left, and then show the perfect coincidence 

 of the two portions. 



In attacking geometrical deductions we are often compelled to 

 assume in this way the existence of figures which are clearly con- 

 ceivahlc, though we may not know precisely how to construct them, 

 or though it may even be impossible to construct them by any of 

 the ordinary geometrical processes. The following example of a 

 problem in geometrical maxima and minima affords an instance. 



Ex. 11. — A C B, Fig. 19, is part of a circle whose centre is at 0. 

 The points P and Q lie without the circle. Determine under what 

 conditions the sum of the distances F C and Q C will le a minimum. 



— Q 



Fig. 19. 



Here guided by Examples 8, 9, to which the above is supposed to 

 be given as a rider, we are readily led to the inference that P C and 

 Q C should be equally inclined to the tangent at C. Now there is 

 no simple method of determining C so that this relation may hold. 

 But it is clear that there must be some position of C for which it 

 holds. Conceive, then, that P C and Q C are equally inclined to 

 D C E, and let us inquire whether their sum is a minimum. Take 

 any point P in A C, and join P P and Q F. Then we have to sliow 

 that PF and QF are together greater than PC, CQ. Let P F 

 meet D C in G and join G Q. Then P G and G Q are together 

 greater than P C and C Q (Example 9) ; and P F, F Q are clearly 

 greater than P G, G Q (Euclid I. 20). Hence, d fortiori, P P, F Q 

 are together greater than PC,CQ. Therefore the sum of P C 

 and C Q is a minimum. 



Cor. — Join C 0, then the angle P C O is equal to the angle Q C 0, 

 and we may express the relation deduced above thus : — 



The sum of the lines drawn from any jmint without a circle to a 

 point on the circumference will be a m,inimum when the two lines are 

 equally inclined to the radius draivn to the last-named point. 



The subject of geometrical maxima and minima is a wide one, but 

 we shall content ourselves here by adding three in which areas are 

 dealt with. 



Ex. 12. — Two sides of a triangle heivg given it is required to 

 construct the triangle so that its area shall he a maximum. 



Let A B, B C, Fig. 20, be the lengths of the given sides. 



With centre B and radius B C describe the circle C D F E. 

 Then if we draw any radius B D or B E, and join A D or A E, it 

 is clear that the triangle A B D or A B E thus constructed will have 

 sides A B, B D, or A B, B E of the required length, audit is obvious 

 that the area of any triangle thus formed will bo greater or less, 

 according as the distance of its vertex from the line A B C is 

 greater or less. We have not, indeed, any problem in Euclid which 



expressly states this as a truth respecting triangles on the same 

 base, but tho property is clearly involved in the proof of I. 39. 

 Now since tho vertex must lie on the circle C D F E, it is obvious 

 that tho distance of the vertex from ABC can never exceed tho 

 radius of this circle, and can only be equal to the radius when tho 

 side adjacent to A B is at right angles to A B. Draw B F at right 

 angles to A B, and join A F. Then the triangle A B P is the 

 triangle of maximum area under tho given conditions. The proof 

 consists in showing that U G or E H drawn perpendicular to A B C 



is less than B F. This is evident ; for in the right-angled triangle 

 B D G, the angle D B G is less than a right angle, therefore D G is 

 less than B D ; that is, than B F. 



^ur C^eees Column* 



By Mephisto. 



SCORE ON SATURDAY THE 21st inst. OF THE INTER- 

 NATIONAL TOURNAMENT AT NUREMBERG. 



GAME PLAYED IN THE NUREMBERG INTERNATIONAL 

 CHESS TOURNAMENT, JULY 16, 1883. 



Position after White's IGth move of Q to Kt-l :— 



Mason. 

 Slack. 



///m 





Fbitz. 

 French Defence. 



16. Q to Kt4 



17. K to Q2 



18. Q to KtG (ch) 



19. Q to Kt7 (ch) 



20. Q to KtS (ch) 



21. Q takes P (ch) 



22. R takes R 

 White announced mate in four. 



Black. 

 R takes R (ch) 

 P takes P 

 Kto K2 

 K to K sq 

 K toK2 

 K to B sq 

 B to B sq 

 23. R to RS (ch), K to Kt2. 



24. R toR7(ch),KtakesR. 25. Q to B7 (ch), K to R sq. 26. Ktt» 

 KtC mate. 



