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a+b+c+d = S All-Russian MO 2005 Regional (R4) 11.2

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August 26, 2024
algebra

Problem Statement

It is known that there is a number SS such that if a+b+c+d=S a+b+c+d = S and 1a+1b+1c+1d=S\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d} = S (a,b,c,d(a, b, c, d are different from zero and one)), then 1a−1++1b−1+1c−1+1d−1=S.\frac{1}{a- 1} ++ \frac{1}{b- 1} + \frac{1}{c- 1} + \frac{1}{d -1} = S. Find SS.
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