July 6, 1883.] 



♦ KNOAVLEDGE ♦ 



15 



to take FE equal to Q F ; then clrawing PE to cut AB in C, we 



are certain that C is the required point. In all such cases we should 

 not be equally certain that the proof would be as simple as the 

 analysis, since sometimes the reversal of a process involves pro- 

 perties not so readily seen as their converse theorems. In this case, 

 however, it is obvious (or will at least appear on a moment's inquiry), 

 that the proof is simple. 



For, join C Q (we are going now through the synthetic treatment 

 of the problem, and therefore ignore the prior constructions). 

 Then, because Q F is equal to F E, and C F is common and at right 

 angles to Q E, the triangles C P Q and CPE, are equal in all 

 respects. Therefore, tlie angle Q C F is equal to the angle EOF. 

 But E C F is equal to the vertical angle P C A. Therefore the angle 

 Q C F is equal to the angle P C A. 



It is an excellent practice, when a problem has been solved, to 

 notice results which flo'w from, or are in any way connected with, 

 our treatment of the problem. In Ex. 8, Fig. 13, we notice that 

 the line C Q is equal to the line C E, so that the sum of the lines 

 P C, C Q, is equal to the line P E. It might occur to us to inquire 

 what is the sum of lines drawn from P and Q to any other point, 

 as G, in A B. Join P Q and Q G. Then the fact that C E is equal 

 to C Q reminds us that if we join G E, G E will be equal to G Q. 

 Thus P G and G Q are together equal to P G and G E together. 

 But P G and G B are together greater than P E ; that is, P G and 

 G Q are together greater than P C and C Q together, or P C Q is 

 the shot-test path from P to Q, siihject to the cvnditio7i that a point of 

 the path shall lie on A B. 



Erratum.— On p. 397, 1st col., lines 17 and 18 from bottom, for 

 ' a right-angled triangle " read " an isosceles triangle." 



GEOMETRICAL PROOF OF THE FORMULA FOR DOUBLE 

 ANGLES. 



Let AP B be a semicircle, with centre C. 

 P C N any angle less than 90° (2.r) 



.•.PAN = ii!. 



B PN-90°-PBN = PAN = a 

 PN 



Sin 2a! =- 



Cos 2x = 



AP.AB PB.AB 

 = Cos- .^ — Sin- a;. 



PN_ 2PN _ 2PN 



'■~CN 2CN~CnTcN 



2 PN 



"AN- 



rad -t- rad - N B 

 PN 

 AN 



AN_NB PN 

 AN PNAN 

 2 tan X 



1 — tan- J). 



©ur WA^i^i Column, 



By " Five op Clubs." 



The Hands. B. 



Spades— K, 2. 

 Hearts— Kn, 10, ■i, 2. 

 Clubs— Q, Kn, G, 5. 

 Diamonds — K, 6, 5. 



A. 

 Spades — A, Q, 8, 7. 

 Hearts— Q, 9, 3. 

 Clubs— 9, 8, 7, 3. 

 Diamonds — Q, 3. 



Spades—Kn, 10, 5. 

 Hearts— A, Q, 8, 5. 

 Clubs— 10, 4. 

 Diamonds— A, 10, 8, 4. 



10 ♦] o~o] [♦"♦ 

 ♦ o o o ♦ ♦ 



A. D. C. (Cantab.) 



Spades— %, G, 4, 3. 

 Hearts — 7, 6. 

 Clubs— A, K, i. 

 Diamonds— Kn, 9, 7, 2. 



NOTES ON THE PLAY. 



NoTB. — The card underlined wins th* 

 trick, and card below leade next round. 



1. The score being at love, A 

 would have been justified, we 

 think, in leading a trump, not- 

 withstanding the weakness of his 

 hand outside trumps. Still there 

 are valid reasons for opening his 

 longest suit ; and on the whole, 

 it was, perhaps, the safer course. 

 With A'a cards in Clubs, too, it 

 was not unlikely that the suit 

 would be established at the third 

 round. A knows that the Two is 

 with Z, unless Y or £ is signalling. 



2. Z opens his long suit. I'does 

 not hold the Queen. 



3. i>' should have led his own 

 suit, notwithstanding the major 

 tenace. He knows from liis 

 partner's lead, himself holding 

 Club Ten, that his Hearts are 

 stronger than his partner's Clubs ; 

 and apart from that, the suit is 

 intrinsically so strong that it was 

 liis duty to show it. But there are 

 players who never will lead from 

 a suit headed by Ace, Queen. The 

 Club Two is seen now to be with Z, 

 Eight and Seven with A\ Queen 

 and Knave with Y. 



4. Z (Mr. J. Clay) has been 

 blamed for discontinuing here his 

 own suit. But there were good 

 reasons; although the event turned 

 out unfavourably. He knew that 

 the Diamond Queen was held by 

 either A or B. If by A, then a 

 Club lead from A would give i> a 

 ruff, likely enough to suit him, as 

 Z holds four trumps himself. If B 

 holds the Queen, it is likeU- that, 

 having no Clubs, he holds one or 

 two more Diamonds, and that A 

 holds no more. In this case there 

 is danger of a cross ruff. On the 

 other hand, as A led from a weak 

 suit of Clubs, and B did not lead 

 Hearts, it looked more than likely 

 that both were weaker still in 

 Hearts, and therefore that I' held 

 groat strength in that suit- On 

 the whole, though the question is 

 one of probabilities only, Mr. Clay 

 seems to us to have shown his 

 usual judgment in leading what 

 he thought would be his partner's 

 suit. 



5. Having made his tenace, 7> 

 might ?io)(', at least, have en- 

 lightened his partner as to the con- 

 stitution of his hand. If ho had 

 led Ace of Hearts, Z playing the 

 Six would have showed he had no 

 more. Another lead of Hearts 



