160 



♦ KNOAVLEDGE 



[Sept. 7, 1883. 



[Have just found your triscction question cmong unanswered 

 and unopened correspondence. The trisection is perfectly sound, 

 and if the secant could be drawn, by line and circle, to fulfil the 

 conditions named, any angle could \ye trisected by line and circle. 

 Unfortunately it cannot. — E. H. M. S. Would be glad to hear of 

 a good book on the chemical constituents of garden vegetables. — 

 LCTETIA Eediviva. An odd coincidence. But the history of the 

 discovery of Uranus is well-known. Sir W. Herschel himself fully 

 recorded it, and there is nothing in the slightest degree confirming 

 Paris's statement. — E. S. Thanks; the idea is a good one. A 

 series of pictures of the old constellation figures, with 

 stars, names, and explanations of these, would probably 

 interest many. Will try to arrange the matter. — A. Holdsworth. 

 A telescope will not reveal the hull under those conditions. — 

 IXFORMATlox. Pitman's I should say. Other questions belong to a 

 kind we have been obliged to leave unanswered. — H. J. W. JIars 

 has two satellites, both very small, — probably not more than fifteen 

 miles in diameter. There is an essay on them in my " Poetry of 

 Astronomy," entitled " Living in Dread and Terror," these being 

 the Englished names of the two satellites of Mars (Demos and 

 Phobos). — F. R. The "cold spells" are purposely omitted by 

 smoothing off the temperature curves. — Ax Ixquiree. The institu- 

 tion of Infant Baptism is a little outside our line. — T. H. K. The 

 weight lifted is not great, — each raised Jess than 3 stone, each 

 forefinger raised less than li stone. The inspiration, expiration, 

 &c., only help I imagine by directing attention to the proper amount 

 of lifting; so that all act together. But really lifting 40 lb. with 

 the two forefingers is hardly a feat reciuiring special preparation or 

 special explanation when accomplished. — E. T. L. Solution correct. 



MATHEMATICS. 



Peop. — AB, DC are equal chrtrds to the circle AB C T) , meeting 

 inF; and tanr/ents FP, FQ are draicn, and FbOa through the 

 centre 0. Shnxa that chords B D, A C iyitersect at P Q. 



It is clear from symmetry that G, their point of intersection, lies 

 in « F ; and it will suffice to show that a perpendicular chord P G Q 

 cuts the circle in the points P, Q, where tangents from F meet it. 

 Join OB; then in the triangles GB, D GF, angle OGB = opp. 

 angle D GF ; and Z GO B (being angle at centre on B b) = angle 

 at circumference at B C = angle B D F. Hence, the triangles are 

 similar, and 



OG:GB::GD:GF 



.-.OG, GF = GB, GD = sq. on P G 



Wherefore, P G is a right angle, and F P, F Q are tangents. 

 Q.E.D. 



CoR. — Ij'F be n»v external point from which ta7igents FP, FQ are 

 draxcn to a circle, and P Q cut F O at G, then straight lines F B, 

 F D to the extremities of any chord B D G through G are equally 

 inclined to F 0. 



^ur Cbefis Column. 



By Mephisto 



PROBLEM No. 96. 



By J. Beeger Gr.\z. 



Black. 



Whttb. 

 White to play and mate in three moves. 



FRENCH DEFENCE. 



Uerr Bodusilika. 

 P toK3 

 P toQ4 

 KKt to B3 

 P takes P (a) 

 B to K2 

 Castles 

 QKt toQ2 

 K to K sq 

 Kt to B sq (b) 

 P to QKt3 

 B to Kt2 



White. 

 TI.MT lieivcr. 



14. Q takes B 



15. QR to Q sq 

 IG. R to Q3 



17. P to KR-t 



18. P to Ro 



19. P to RG 



20. P to Kt4 



21. P to Kt5 



22. Kt takes B 



23. Q to E4 



24. E to Kt3 



25. Q takes Kt 



Black. 

 Herr Hoduschka. 

 R to B sq 

 P to B3 

 Kt to Kt3 

 Kt to K2 

 K to R sq (c) 

 Kt to B4 

 Kt takes HP 

 B takes P 

 P to KB4 (d) 

 K to Kt sq 

 R to K2 

 P takes Q 



' 26. KttksKP(ch)Resigns(e) 



White 

 Herr Berger. 



1. P to K4 



2. P to y4 



3. QKt to B3 



4. B to Q3 



5. Kt takes P 

 G. KKt to B3 



7. Castles 



8. R to K sq 



9. QB to KB4 



10. B to K5 



11. Q to K2 



12. Kttks Kt (ch)B takes Kt 



13. B to K4 QB takes B 



NOTES BY HERR BERGER. 



(a) P to B4 is the right move here. 



(6) If Kt to R4 tlien'foUows 10. B to K5 in reply. If then Black 

 plays P to KB3, White could answer with KKt to Kt5 ! 



(c) This is a lost move, and strengthens White's attack. 



((J) If Q takes Kt, White will either win a piece or the Q by R to 

 Kt3. 



(e) After the exchange of pieces White will remain with a R 

 ahead, i.e., K to B2. 27. Kt takes Q (ch), R takes Kt. 28. R to 

 Kt7 (ch), K to B sq. 20. R takes R, K takes R. 30. B to B7 

 (ch), and wins. 



SOLUTION. 

 Problem No. 93, by L. P. Rees, p. 128—1. Kt to B3, and mates 

 accordingly. 



ANSWERS TO CORRESPONDENTS. 

 *jf* Please address Chess Editor. 



Berrow.— If in Problem No. 90, 1. Kt to Kt7, P takes Kt, then 

 2. R takes P(ch), K to K3, 3. Kt to Q8 mate. 



H. Seward. — "There is no penalty if a player announces mate and 

 fails to give it. 



Correct solutions received : — Problem 93, Hammond, John Wat- 

 kins ; Problem No. 95, Clarence, M. T. Hooton, H. Seward, Berrow, 

 L. F. Q., End-game, Clarence. 



Contents of No. 96. 



A Naturalist's Tear. Wild Peas. 



By Grant Allen 129 



Pleasant Hours with the Microscope. 



By H.J. Slack 130 



TricVcles in 1833 : Small Wheels u. 



Large Wheels. By John Browning 131 

 The Amateur Electrician. (/«u«.) 132 

 Laws of Brightness. IX. (Illus.) 



By R. A. Proctor 133 



The Chemistry of Cookery. XVU. 



By W. Mattieu WiUiams 135 



Evolution of Human Physiognomy. 



(IHiis.) By E. D. Cope 136 



FAGl 



The Morality of Happiness : Evo- 



lutiou of Conduct. II. By Thos. 



Foster 138 



Pretty Proofs of the Earth's Kotun- 



dity. (Illus). By R. A. Proctor 139 

 Punctuation and Printers. By Sir 



Edmund Beckett 140 



The Face of the Sky Ill 



Correspondence : Luminous Ring — 



Flight of a Vertical Missile, &c. 141 



Our Mathematical Column 143 



Our Whist Column 143 



Our Chess Column 144 



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