Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem26

11&12 Video Solutions 2011 problem26

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Video Summary

The problem involves finding the smallest possible value of \( a^2 + b^2 \) where \( a \) is the sum of three rational numbers \( x, y, \) and \( z \), and \( b \) is the sum of their reciprocals, with both \( a \) and \( b \) being integers. Through exploration, setting \( x = y = \frac{2}{3}k \) and \( z = -\frac{1}{3}k \) leads to \( a = k \) and \( b = 0 \). Choosing \( k = 1 \) gives \( a^2 + b^2 = 1 \), confirmed as the smallest possible value, after verifying that \( a = 0 \) is not feasible. Therefore, the answer is 1.

Keywords

smallest value

rational numbers

sum reciprocals

integer constraints

optimization problem