# PUZ­ZLES

The Guardian Weekly - - Culture - By Maslanka

1 Down at The Last Chance

Saloon the Hin­doo Jar Broth­ers have placed a num­ber of white balls and a num­ber of black balls– fewer than 20 in to­tal– into an urn. Gullible Gus with­draws a ball and finds it is black. He re­places it. Sher­iff Ein­stein then with­draws a ball and finds it is white. The chances that two such with­drawals should re­sult in a mis­match (a black ball fol­lowed by a white; or a white ball fol­lowed by a black) are 42%. How many balls must you with­draw to be sure you have at least one of each colour?

2 In this logo de­sign a right-an­gled tri­an­gle fits snugly into a semi­cir­cle in the ori­en­ta­tion shown. What should the height and the width of the tri­an­gle be if its area is to oc­cupy the great­est frac­tion of the area of the semi­cir­cle? Try to spot the an­swer with­out go­ing too deeply into the work­ing!

Word­pool

Find the cor­rect def­i­ni­tion: SEA PUSS a) cap­tain’s pet b) sea lion c) pi­rate’s catamite d) dan­ger­ous cur­rent

Find a word which fol­lows the first word in the clue and pre­cedes the sec­ond in each case mak­ing a fresh word or phrase. a) fire band b) trip cut­ters c) tele­graph cat d) fenc­ing city e) sweets scotch f) end ache

1 The chances of a black fol­lowed by a white are the same as of a white fol­lowed by a black: b/(b + w) X w/ (b + w) ; so we are given that 2bw/ (b + w) 2 = 42/100; this boils down to 200bw = 42(b 2 + w 2 + 2bw); that is: 58bw = 21b 2 +21w 2. Di­vide both sides by w 2 to get: 21B 2 -58B + 21 = 0. A re­as­sur­ingly sym­met­ri­cal quadratic with B = b/w. Solv­ing us­ing the for­mula (or by in­spired fac­tor­ing) we find re­as­sur­ingly that B = 3/7 or 7/3. Since b + w < 20, b + w = 10, and we have ei­ther 3 black and 7 white or 7 black and 3 white. Ei­ther way, we need to with­draw 7 + 1 = 8 to be sure of one of each. 2 Feel­ing lazy we com­plete the rec­tan­gle. When the rec­tan­gle is a max­i­mum our tri­an­gle will be max­i­mal, too. Then we get lazier and com­plete the cir­cle and ask our­selves what sort of rec­tan­gle fills most of the cir­cle: the an­swer is a square. So the tri­an­gle must have w = 2h. Such ex­er­cises of “see­ing it” are in­valu­able; but they can be backed up by other means.

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