Subject: Entertaining Problem (~{H$N6NJLb~})

From: [email-masked] (Zhiping You)
Date: 22 Jan 94 16:36:00

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~{!>IOV\NJLbLVB[!?~}

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~{K31cHC4s<R?4?4VP9z9EJ1:r5DJ}Q'>+;*!#~}

>>~{SL<G3uVPTxQ'9}Wfg.#(#G#E#N#G#4#)T-@m#,HtUb2;Kc;}7V#,Tr=bHgOB#~}
>>
>>Let the plane spanned by the axises of the two cylindars be P, the shared
>>
>>part of the two cylindars be V. Consider a ball B with radius 1 and w/ its
>>
>>center on P and an arbitary plane P'//P at a distance d<=1 from P.
>>
>>The intersection of P' w/ V is a square whose sides are   2 x sqrt(1-d^2),
>>
>>and thus has an area of  4*(1-d^2). The intersection of P' w/ B is a circle
>>
>>w/ a radius of sqrt(1-d^2), whose area is hence pi*(1-d^2). Therefore
>>
>>the ratio between the areas of the two intersections is 4:pi, a constant
>>
>>indepentant of d. The ratio between the volumes of V and B is the same
>>
>>constant, according to Zu Geng's principle. So:
>>
>>         vol. of V = 4:pi * vol. of B = 16/3 .
>>
>> 
>>
>>~{Wfg.6+::HK#,0";y_d5B~} Hellenistic ~{J14zHK#,N4V*JkTg#.~}

  ~{6+::#-#-9+T*#2#5#-9+T*#2#2#0!#!!0";y_d5B#-#-#2#0#0#b#c!#~}
>>~{R;Yg~}

~{!>IOIOV\NJLbLVB[!?~}

~{IOIOV\NJLbSI#m#i#n#g!!#x#u#e!!8x3vOjO8=bJM#,P;P;!#~}

>>I understand what he says, let me explain his idea in a more clear way: 
>>
>>1) Look at the chess board:  b-black w-white
>>  
>>   bwbwbwbw 
>>   wbwbwbwb 
>>   bwbwbwbw 
>>   wbwbwbwb 
>>   bwbwbwbw 
>>   wbwbwbwb 
>>   bwbwbwbw 
>>   wbwbwbwb 
>>
>>   where each letter represent a small square (cell), and note that 
>>   all squares on the immediate top, bottom, left or right, if any, 
>>   of any small square, say A, have a different color the square has   
>>
>>2) Now, without loss of generality, say the upper & left most square
>>   and the lower & right most square have been taken, note that they both 
>>   are b(lack).   IF there is such an arrangement of 31 1x2 rectangles
>>   for the remaining 62 squares, of which 30 are black and 32 are white, 
>>   a contradiction comes as follows: 
>>   each 1x2 rectangle put on the chess board must cover exactly one
>>   b(lack) square and one w(hite) square, so 31 rectangles cover
>>   exactly 31 b(lack) squares and 31 w(hite) squares.    
>>
>>3) So, the required arrangement can not be made, or is impossible.
>>-- 
>>Ming XUE ([email-masked])
>>
>>
>>