Problem
ALG-B1-M06-P019 Two possible ratios
#19
★★★★☆ Level 4 of 5
Let \(x,y>0\) and \(\frac{x^2+y^2}{xy}=\frac{5}{2}\). Find the possible values of \(\frac{x-y}{x+y}\).
Denote \(s=x+y\), \(p=xy\), and express the square of the desired fraction.
Since \(\frac{x^2+y^2}{xy}=\frac{s^2-2p}{p}=\frac{5}{2}\), we have \(\frac{s^2}{p}=\frac{9}{2}\).
Then \(\left(\frac{x-y}{x+y}\right)^2=\frac{s^2-4p}{s^2}=1-\frac{4p}{s^2}=1-\frac{8}{9}=\frac{1}{9}\). Thus the possible values are \(\frac{1}{3}\) and \(-\frac{1}{3}\). Both are attained, for example, by the ratios \(x:y=2:1\) and \(1:2\).
Checking attainability of the values is necessary.