100公尺長隊伍 - 拼圖
By Yedda
at 2011-05-10T11:03
at 2011-05-10T11:03
Table of Contents
※ 引述《uorol (豆腐喵的勒)》之銘言:
: 雖然已經有解答了,
: 不過我想請問有沒有其他比較...平易近人的解法(?)
: 假設隊伍行進速度為 x, 小兵前進速度為 y
: 式子 A. [小兵抵達隊伍頂端, 耗時t1]
: x * t1 = y*t1 + 100
: 式子 B. [小兵折返, 耗時t2]
: x * t2 = 100 - y * t2
: 式子 C. [隊伍總經過距離為100]
: y(t1+t2) = 100
: 但感覺還少了一個式子 Q_Q
感謝squirrel1085跟cutecpu兩位版友的答案 <(_ _)>
由A得到 => t1 = 100/(x-y)
B得到 => t2 = 100/(x+y)
帶入C => 100/(x-y) + 100/(x+y) = 100/y
(配方) => x^2 - 2xy + y^2 = 2y^2
=> x = (1+2^0.5)*y
所求即為 x(t1+t2) = 100(1+2^0.5) #
--
: 雖然已經有解答了,
: 不過我想請問有沒有其他比較...平易近人的解法(?)
: 假設隊伍行進速度為 x, 小兵前進速度為 y
: 式子 A. [小兵抵達隊伍頂端, 耗時t1]
: x * t1 = y*t1 + 100
: 式子 B. [小兵折返, 耗時t2]
: x * t2 = 100 - y * t2
: 式子 C. [隊伍總經過距離為100]
: y(t1+t2) = 100
: 但感覺還少了一個式子 Q_Q
感謝squirrel1085跟cutecpu兩位版友的答案 <(_ _)>
由A得到 => t1 = 100/(x-y)
B得到 => t2 = 100/(x+y)
帶入C => 100/(x-y) + 100/(x+y) = 100/y
(配方) => x^2 - 2xy + y^2 = 2y^2
=> x = (1+2^0.5)*y
所求即為 x(t1+t2) = 100(1+2^0.5) #
--
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拼圖
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By Kyle
at 2011-05-11T01:26
at 2011-05-11T01:26
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at 2011-05-14T20:29
at 2011-05-14T20:29
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at 2011-05-16T10:18
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at 2011-05-18T17:47
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