112 



♦ KNOWLEDGE ♦ 



[Aug. 17, 1883. 



tkrongh P lie on the circle E H P." Bnt it is clear that points on 

 the arc L E II bisect chords through P ; and also that every point 

 ■on this arc bisects some chord throngh P. 



A readiness in determining the loci corresponding to different con- 

 ditions will often be found serviceable to the student engaged in 

 solving problems of different classes. 



Suppose, for instance, that the following problem is set : — 

 Ex. 19. — Let A, B, C (Fig. 29), he three given points, D a given 

 straight line. It is refjuired lo find a point which shall be equidistant 

 Jrom the points A and B, and at a distance from equal to the 

 line B. 



Fig. 29. 



In order that the distance of the point from C may be equal to 

 the line D, it is clearly necessary that the point should lie some- 

 where on the circumference of the circle described with centre C, 

 and radius equal to D. Let G E P be this circle. 



Xext, we inquire whether there is any locus containing all points 

 equidistant from A and B. We join A B and bisect in H, giving 

 one point, H, clearly belonging to such a locus. Xert, either by 

 applying tentative methods, as in the above instances, or by the 

 consideration of a few obvious facts, we find that the indefinite" line, 

 K H G E, drawn through H at right angles to A B, contains all 

 points equidistant from A and B. The line K H G E does not 

 necessarily intersect the circle F G E. If it intersects that circle 

 in two points, G and E, it is clear that each of these points satisfies 

 the required conditions. For C G is equal to D (const.), and G A 

 is equal to G B (Euc. I., 4. See also Ex. 1.) Also, C E is equal to 

 ■D and EAtoEB. IfKHGE touch the circle there is only one 

 point satisfying the given conditions. And clearly, if K H G"E do 

 not meet the circle, there is no point satisfying the given con- 

 ditions. For if there were such a point, it would be at a distance 

 D from C ; and, therefore, would lie on the circle F G E. Also it 

 would be equidistant from the points A and B, and therefore would 

 lie on K H E. In other words, the circle F G E a'ould have a point 

 in common with the line K H G E, which we have supposed not to 

 be the case. 



&ur €heS9 Column. 



By Mephisto. 



PROBLEM Xo. 92. 

 By J. C. 

 Black. 



GAME PLATED RECENTLY BETWEEN MEPHISTO AND 

 ANOTHER STRONG PLATER (SCOTCH GAMBIT). 



White. Black. i WMte. Black. 



Mepbisto. Amateur. I Mephisto. Amateur. 



1. PtoK4 P to K4 |lO. KttoK4 B to Kt3 (/) 



2. KttoKB3 KttoQBS i 11. P to B4 Kt to Kto 



3. P to Q4 P takes P ' 12. KttoB6(ch)(<7) P tks Kt (h) 



4. Kt takes P Kt to B3 (a) 13. B tks P (ch) K takes B (t) 



5. Kt takes Kt KtP takes Kt 14. Q to R5 (ch) K to Kt sq 



6. B to Q3 (b) B to B4 (c) ] 15. Q to Kt4 (ch) K to R2 



7. P to Ko Q to K2 ; 16. Q to E4 (ch) K to Kt sq 



8. Castles Kt to Q4 i 17. P takes P Resigns (j) 



9. Kt to Q2 (d) Castles (e) 



NOTES. 



(a) This defence was adopted by Zuckertort in his match against 

 Blackburne. It leads to a difficult game for Black. 



(6) Steinitz is of opinion that the P should at once advance to 

 K5. 



(c) P to Q4 ought to be played here ; it is the principal move in 

 Black's defence, for if White plays P to Ko the Kt can conveniently 

 retire to Q2. 



(d) Here P to QE3 first might have been preferable. 



(e) Castling is always dangerous when all the pieces are placed 

 on the Q side, and especially when the adversary's B is posted on 

 Q3. Black ought to have played Kt to Bo instead, which would 

 have enabled him, if necessary, to play Kt to K3, but his position 

 was a difficult one to handle. 



(/) If P to B4, then 11. P takes P en pas, Kt takes P. 12. B 

 to Kto. 



(g) This combination forces the game. 



(h) Obviously Black could not reply with K to R sq., on account 

 of 13. Q to R5, P to KE3. 14. QB takes P. 



(i) Here again Black has no resource but to accept his fate ; if 

 K to R sq., then 14. B to B5 threatens mate by Q to E5, or, if 

 K to Kt2, then Q to Kt4 (ch) brings about the same position. 



(j) There was no feasible way of preventing the threatened 

 mate on Kt" ; an interesting variation arises from the following 

 play: B takes P (ch). 18. K takes B best (R takes B would be 

 bad' on account of Q to K8 (ch), and the R could not cover; 

 Q takes B would be good, but it would render winning a little 

 more difficult,) Q to B4 (ch). 19. K to Kt3, Q to Q3 (ch). 20. R 

 to B4, Q to Q6 (ch). 21. K to B2, Q to R2, best. 22. Q to 

 Kt3 (ch) and wins the Queen. 



Contents of Xo. 93. 



PAGE 



VesaviuB and Ischia. By E. A. 

 Proctor SI 



The Birth and Growth of Jlyth. 

 XIII. Bv Edward Clodd S! 



Pretty Proofs of the Earth's Eoti.n. 

 ditf. (Illua.) By B, A. Proctor SI 



The Moon in a Three-Inch Tele- 

 scope {Illus.) By F.H.A.S S6 



Weather Forecasts, and How to 

 Make Them. Bv John Browninj -,7 



The Fisheries Exhibition. IT. 

 [Illitt.) Bj John Ernest Ady ... b7 



PAGE 



Sea Anemones : The White Carna- 

 tion. (Illua.) BtT. Kimber 89 



A Steamer goes Safely Past Nia- 

 gara Whirlpool 90 



Principles oi Dress Reform. By 

 E.M. King 91 



Chemistrr of the Cereals. III. By 

 William Jago, F.C.S 93 



Editorial Gossip 94 



Corresponde: 



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