Aug. 31, 1883. 



• KNOWLEDGE ♦ 



143 



the same order, though beginning with a different digit of the set ; 

 in the case of 8000000 and 9000000 we have 999,999 x 8 + If , and 

 999,999 X 9 + If .— R. P.] 



LETTERS RECEIVED, AND SHORT ANSWERS. 

 W. Bavley. — Thanks for cuttings ; but of course the toads were 

 never vomited. — J. Clayton. Handbook of stars out of print. — 

 Jas. Paxman. Illness following i-ailway accident drove the matter 

 from my thoughts. — J. Hakkison, sen. Slany thauks for the weather 

 charts. — A Reader. You will have seen ere this that I quite agree 

 with your objections to cholera-prescribing here, as in Bealth, and 

 other such organs. — Ignoramus. Stephenson said to a perpetual 

 motionist, " Carry yourself round the room by your own waist-band, 

 and I will consider your plan." That is about what it comes- to. — 

 W. W. W. Regret, but no space. — Ja.s. Lucking. Do not know 

 where model of first pair of spectacles can be seen. Am also 

 unable to say where a photograph of Adam and Eve is preserved. 

 — A. H. Swinton. — Your determination of sun-spot maxima 

 from years of great natural phenomena seems to me as funny as 

 determining comets — otherwise unknown — from deaths of kings 

 and rulers. Prove the connection between volcanoes and sun-spots 

 and wo will see about it. — H. H. Thanks for kind wishes. — D. N. 

 Certainly not new. A handred such relations could be written 

 down in a few hours. — E. C. Castlebar. Earth's shape seems to 

 trouble you. When and where have I said I admired the person 

 you mention ? But reading the sickly works of the other writer 

 would have had no influence one way or the other. — E. H. Stutter. 

 Many thanks for your kind letter.— Senex. Not unusual. Showing 

 want of symmetry, not that penumbra was not lower than photo- 

 sphere. — H. W. Jones. No space or time for explanations of so 

 much text-book matter. 



<^ur iHatDfinatiral Column, 



GEOMETRICAL PROBLEMS. 



By Richard A. Proctor. 



PART XIV. 



A READER of these mathematical notes sends for solution, by 

 use only of EucUd's first three books, the following problem : — 

 Prob. — A B C is a right-angled triangle. From any point D in 

 the hypothenuse B C a straight line is drawn at right angles to B C, 

 meeting C A at E and B A produced at F : show that the square in 

 D E is equal to the difference of the rectangles B D, D C and A E, 

 K C ; and that the square on D F is equal to the sum of the rectangles 

 B D, D C omd A i\ P B. 



Wo notice at onco, that the rectangle A E, E C mentioned in first 

 part of the problem is equal to the rectangle D E, E P, since the 

 angles at D and A are right angles so that a circle will pass through 

 tlio points C, D, A, P. Also wo notice at onco that the rectangle 

 A P, P B mentioned in the second part of the problem is equal to 

 the rectangle E F, FD, since a circle will pass through the points 

 A, B, D, K. This leads us at once to think that wo may lind the 

 solution of our problem, by using those theorems of the second book 

 of Kuclid in which a squ£Lre on D E, a part of a line, is shown to bo 

 the difference of two rectangles, and a square on D F, the whole 



of a divided line, is shown to be the sum of two rectangles. But in 

 each case we must bring in the rectangle related to the line D E F 

 which we have seen to be equal to a rectangle referred to in the 

 puzzle. 



Thus, first, we ask how the rectangle D E, E F is related to the 

 square on D E. We know that 



DE' = rect. DE . DP-rect. DE . EF. 

 If then we can show that the rect. D E. D F is equal to the rect- 

 angle CD, D B, what is required, so far as first part of our 

 problem is concerned, is done. Now as we are dealing with the 

 third book, this naturally leads to the idea that if a circle is carried 

 around the points C B F, we may be able to show that F D which 

 intersects C B in D, will when produced cut this circle in G, such 

 that D G = DE. Foi v^e want to show that the rect. CD, DB = 

 rect. D F, D E ; and we know that the rect. CD, D B = rect. F D, 

 D G. But this leads us directly to the solution of this part of the 

 problem. For if we suppose the circle through C, B, F, to cut F D 

 produced in G, we have /;CGF=ZCBF in same segment = 

 Z C E D (.since each is the complement of Z B C A) : hence the 

 triangles C GD and O E D are equal in all respects, and D E = D G. 

 The other part of the problem is found to depend on the same 

 construction : for 



DF= = DF.EF + DF.DE = BF.AF -i- DF.DG 

 = BF.AF-i-CD.DB 

 We may put our solution into the following form : — 

 Produce FD to G making D G = D E and join GC. Then since 

 the angles at D are right angles 



ZCGD= i!;DEC = compt. ofZDCE= /CBA. 

 Hence a circle will pass through the points C, 6, B, F ; also 

 obviously a circle will pass through the points C, D, A, F ; and 

 another circle through the points D, E, A, B. Hence 



rect. C D, D B = rect. G D, D F = rect. D F, D E 



rect. C E, E A = rect. E F, D E 



and rect. A P, F B = rect. D P, F E 



.-.rect. C D, D B -rect. C E, E A = re';t. D F, D E -rect. EF,D E = DE» 



and rect. CD,DB + rect. A F, F B = rect. D F, D E -h rect. DP, FE = DF' 



Q. E. D. 



<!Pur 2iaabist Column, 



By " Five of Clubs." 



TAKING PARTNER'S TRICK. 

 1* T Y partner leads a Queen original lead, I hold Ace, King, and 

 i.TjL "Two. Not a possible trick in any other suit, save Trumps. 

 Love all. To his lead, I play the King, lead tli» Two of Trumps, 

 having Two, Three, Knave, and Quepn. My |iarrtu>r wins with the 

 King, leads Ace, follows with the Four, leaving me with the 

 thirteenth trump and the lead ; I play Ace of his suit, then the 

 small one. Have I made the most of my hand ? Did I lose any- 

 thing by taking his Queen with my King ? Are there any circum- 

 stances under which 1 can lose by playing King on my partner's 

 Queen, holding Ace, King, and one small one? Supposing my 

 partner knows the conventionalities of modern Whist, I expect 

 him to play from his long suit. No card in my hand or played by 

 second player tells mo whether his lead is from weakness or 

 strength; therefore I put on my King in order to get out 

 of his way. If his lead is from two I assume ho lias 

 not four trumps, and wishes to make a trump on that suit. 

 If his lead is from Queen, Knave, and a small one, or Queen, 

 Knave, and anything, I contend I lose nothing by playing my King. 

 If he hold Queen and one other and I let it go, he plays the small 

 one I put on my King. If I then play my Ace his discard is no 

 use to him, or me eiilior, and instead of playing to win the game I 

 am forced to feel my way to the end. He has exposed his hand to 

 the entire party, and leaves me to get out of the mess the best way 

 I can. If he holds a long suit, of which he leads the Queen, I 

 contend that to play the small card would be wrong. The play 

 succeeded. His suit proved to be a long one. Queen, Knave, ten, 

 and others. He afterwards told me I knew nothing about the game, 

 and ncv^r to do so again, although it " came off." " He had 

 played Whist forty years." I being a young player, gave up the 

 argument, believing I was right. He also advised me never to take 

 my ]iartniu-'s trick ; but having brought off DeschapeUes' coup five 

 times last winter, I voted my old friend a " duffer." 



Hoping I have not exhausted your patience, and that I may bo 

 iufoiinod in your next, I am, dear Sir, yours, Ac. Mrrp. 



[If you had played your small card, the evil consequences which 

 would "abnosl inevitablj- have followed would have emphatically 

 justified the name you assume — without having made out any case 

 for its use. Your partner only thinl;s he has played llViisf forty 

 years. — Five of Clubs.] 



