Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>\alpha, \beta, \gamma | + | Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>\alpha, \beta, \gamma</math>. |
Find | Find | ||
<cmath>\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}</cmath> | <cmath>\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}</cmath> | ||
==Solution== | ==Solution== | ||
− | + | From Vieta's formulas, we have | |
+ | <cmath>\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 2023^2 - 2(0) = 2023^2.</cmath> | ||
+ | |||
+ | Now, | ||
+ | <cmath>\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.</cmath> | ||
+ | |||
+ | Similarly, | ||
+ | <cmath>\frac{\alpha^2 + \gamma^2}{\alpha + \gamma} = 2023 + \beta \quad \text{and} \quad \frac{\beta^2 + \gamma^2}{\beta + \gamma} = 2023 + \alpha.</cmath> | ||
+ | |||
+ | Thus, | ||
+ | <cmath>\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*}</cmath> |
Latest revision as of 21:05, 9 September 2025
Problem
Let the roots of be
.
Find
Solution
From Vieta's formulas, we have
Now,
Similarly,
Thus,