2023 SSMO Accuracy Round Problems/Problem 6
Problem
Let the roots of 
be 
.
Find
![\[\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}\]](http://latex.artofproblemsolving.com/c/b/9/cb9a83120ec6b7523b4b0afee9639ff451323c1a.png)
Solution
From Vieta's formulas, we have
![\[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 2023^2 - 2(0) = 2023^2.\]](http://latex.artofproblemsolving.com/c/c/7/cc75ed8d97ff97aa0b47908efde16fe4e3fa625c.png)
Now,
![\[\frac{\alpha^2 + \beta^2}{\alpha + \beta} = \frac{(\alpha^2 + \beta^2 + \gamma^2) - \gamma^2}{(\alpha + \beta + \gamma) - \gamma} = \frac{2023^2 - \gamma^2}{2023 - \gamma} = 2023 + \gamma.\]](http://latex.artofproblemsolving.com/a/1/6/a16bb2a21eff0d46ed2eb4f7bc0aab78ba11e229.png)
Similarly,
![\[\frac{\alpha^2 + \gamma^2}{\alpha + \gamma} = 2023 + \beta \quad \text{and} \quad \frac{\beta^2 + \gamma^2}{\beta + \gamma} = 2023 + \alpha.\]](http://latex.artofproblemsolving.com/c/a/6/ca6499f9c45846e0c368551d6470e2acde24eda4.png)
Thus,
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