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2006 Cyprus MO/Lyceum/Problem 16

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Problem

If x_1,x_2 are the roots of the equation x^2-2kx+2m=0, then \frac{1}{x_1},\frac{1}{x_2} are the roots of the equation

\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1...

Solution

By Vieta’s, x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m.

The equation with roots x = \frac{1}{x_1}, \frac{1}{x_2} is 0 = \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{x_2}\right) = x^2 - \left(\frac{1}{x_1} + \frac{1}{x_2}\right) + \frac{1}{x_1x_2} = x^2 - \left(\frac{x_1 + x_2}{x_1x_2}\right) + \frac{1.... Substituting from above, we get x^2 - \frac{k}{m}x + \frac{1}{2m}= 0 \Longrightarrow \mathrm{(B)}.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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