July 20, 1883.] 



♦ KNOWLEDGE ♦ 



47 



quickly rejects any points depending on the equi-division of the line. 

 For instance, he cannot suppose that C E is necessarily a fourth 

 part of D E ; for since there is nothing to prevent P from being at 

 the same distance as Q from AB, it is clearly not absolutely neces- 

 sary that C E should be at all unequal to C D. The inequality 



Fig. 15. 



depends on the inequality of P D and Q E, and may naturally be 

 supposed to vary with the extent of the latter inequality. Our 

 student can hardly fail, we think, to light on the supposition that C 

 ought to be so taken that the angles PCD and Q C E should be 

 equal. He would try this, drawing now a new figure, as follows : — 

 Draw a line A B (Fig. 16), and from a point, C, in it draw C P 

 and C Q, inclined at equal angles to AB, and unequal. Take 

 another point, Cj, and join PCj, Cj Q. Produce PC, PCj, to 

 L and H, making C L, Ci H, equal respectively to C Q, Cj Q. Then 

 we have to prove that P Cj H is greater than P li. 



.loin 11 1j, tlicn we should have to prove the angle P L II greate'' 

 than the angle P H L. This does not seem very easy. 



But next we notice that the angle LC E is equal to the vertical 

 angle P C A (Euc. I. 15) and therefrom (/ii/p.) to Q C E. Also C L 

 is equal to C Q and the triangle L C Q (hero we draw in Q E L) is 

 isosceles, C E being the bisector of the angle contained by the 

 equal sides. Hence, C E is at right angles to Q L and bisects Q L 

 in E. It> is a very obvious consideration, at this point, that if we 

 join Ci L wo shall have Q C, h an isosceles triangle, C, Q being 

 clearly equal to C, L (Euc. I. 4). Hence, PC, and C, L together 

 are equal to PC, and C, Q together. But PC,, C, L together are 

 greater than PL (Euc. I. 20). Hence PC, and C, (,) ar(3 together 

 greater than PC, CQ together. We find, then, that our surmise 

 is correct, for what we have proved for P C,, C, Q can be jiroved 

 equally well wherever C, may be taken. Thus the problem is 

 solved. It is not necessary to give the synthetical statement of our 

 solution, since this has already been given in the scholium to 

 Example 8. 



It may bo argued that such tentative processes as we began with, 

 herr, iiic uMi mill lumatics. To this it is to be answered — first, lliat tho 

 art i>r "ih ' ill wi'll is an important aid to the matluMnatician ; and 

 seconilly, Ihiil \vi> deal with our gviesses by means of mathonuitical 

 I'easoniiig, and tlius gain all tho benefit available from niathomatical 

 processes. But further, there are no laws for applying simple 

 geometry — that is geometry resembling Euclid's— to deductions, 

 and therefore in many cases we have no choice but to make use of 

 jciitativo methods. 



<j^ur 21896 (St Column. 



By "Five of Clubs." 



B. 



Spades — 0, 4. 

 Hearts — A, 6, 

 Clubs— 10, 5. 

 Diamonds — K,9,5,3 



Spades — K, 10, G. 

 Hearts — K, 9. 

 Clubs— A, Q, 0, 7, 4, 3 

 Diamonds — A, Q. 



Spades — A, Kn, 7, 5. 

 Hearts— Kn, 10, 8, 7- 

 Clubs— K, 6, 2. 

 Diamonds — Kn, 10. 



Z. 



Spades— Q, 8, 3, 2. 

 Hearts— Q, 4. 

 Clubs- Kn, 8. 

 Diamonds -8, 7, 6, 4, 2. 



A T B Z 



•M,\ ♦ ♦ ♦ ♦ ♦ 



B * 



'Mm * ♦ ♦ ♦ * 



♦ ' ♦ ♦ » , ♦ 



* ♦ o * ♦ «> ♦ 



♦ ♦ o ♦ ♦ ♦ ♦ 



0^0 





NOTES OS THE PLAT. 



tho 



1. Neither 4, 3, nor 2 appearing, 

 someone is signalling unless A 

 holds all three. 



2. A leads the ante-penultimate, 

 showing he held originally six in 

 the suit. Y and Z hold no more, 

 and B has signalled. Both points 

 should be clear to all ; the second, 

 with j4's indication of six in suit, 

 showing that neither i'nor Z holds 

 the Queen. Hence affairs look 

 well for A and B, A having estab- 

 lished his suit, of which B holds a 

 small card to return with, and B 

 having either one honour five 

 trumps, or two honours four 

 trumps. (B is scarcely justified 

 in signalling.) 



3 and 4. B had four trumps, two 

 honours. A returned his highest 

 (he cannot hold the knave), and 

 has but one more at most. B" 

 knows that Z holds the Queen and 

 one more, but A may hold the 8, 

 in which case B can clear out 

 trumps, and if A has a re-entering 

 card the game is won. But as it 

 is obvious that if B leads trumps 

 he will not probably be able to re- 

 turn A'a Clubs, he should have 

 done this at once, forcing Z effec- 

 tually. Therefore 



5. B's lead of trumps here is very 

 bad. Z of course wins with the 

 eight, and has full command ; but 



G. Z does not driiw B's last 

 trump, for he sees that if he does, 

 any re-entering card either in A's 

 hand or B's, would enable A to 

 make his long Clubs. Z therefore 

 takes the only chance of saving 

 and winning the game, leading tho 

 penultimate card of his weak long 

 suit. 



7. r deems it better to return 

 his partner's suit than to lead his 

 own, seeing the value his Heart 

 Ace has, first as a re-entering card, 

 and second in keeping the control 

 of the Heart suit for the nonce. 

 If the King is with the enemy, 

 leading the Ace will free it and 

 give them tho game. Though 

 having four Diamonds originally, 

 Y rightly returns the best of two 

 left. 



9, 10, &e. The rest of the gam© 

 plays itself. Y Z make the odd 

 trick and the game, though A B 

 held twoby honours, seven trumps, 

 three aces, throe kings, two queens, 

 tliree knaves, three tens, and two 

 nines. 



