2023 SSMO Accuracy Round Problems/Problem 6

Problem

Let the roots of $P(x) = x^3 - 2023x^2 + 2023^{2023}$ be $\alpha, \beta, \gamma$ . Find $\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}$

Solution

From Vieta's formulas, we have $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 2023^2 - 2(0) = 2023^2.$

Now, $\frac{\alpha^2 + \beta^2}{\alpha + \beta} = \frac{(\alpha^2 + \beta^2 + \gamma^2) - \gamma^2}{(\alpha + \beta + \gamma) - \gamma} = \frac{2023^2 - \gamma^2}{2023 - \gamma} = 2023 + \gamma.$

Similarly, $\frac{\alpha^2 + \gamma^2}{\alpha + \gamma} = 2023 + \beta \quad \text{and} \quad \frac{\beta^2 + \gamma^2}{\beta + \gamma} = 2023 + \alpha.$

Thus, $\begin{align*} &\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta + \gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha} \\ &= (2023 + \gamma) + (2023 + \alpha) + (2023 + \beta) \\ &= 6069 + (\alpha + \beta + \gamma) = \boxed{8092}. \end{align*}$

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2023 SSMO Accuracy Round Problems/Problem 6

Problem

Solution