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 $p^2+q^2+r^2 = (p+q+r)^2-2(pq+pr+qr) = 2023^2-2(0) = 2023^2.$

Now, $\frac{p^2+q^2}{p+q} = \frac{(p^2+q^2+r^2)-r^2}{(p+q+r)-r} = \frac{2023^2-r^2}{2023-r} = 2023+r.$

Similarly, $\frac{p^2+r^2}{p+r} = 2023+q$ and $\frac{r^2+q^2}{r+q} = 2023+p.$

Thus, $\begin{align*} &\frac{p^2 + q^2}{p + q} + \frac{q^2 + r^2}{q + r} + \frac{r^2 + p^2}{r + p} = (2023+p)+(2023+q)+(2023+r) \\ &= 6069+(p+q+r) = \boxed{8092}. \end{align*}$

Page

Toolbox

Search

2023 SSMO Accuracy Round Problems/Problem 6

Problem

Solution