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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 6"

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==Solution==
 
==Solution==
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By Vieta's relation we get, <cmath>\alpha+\beta+\gamma=2023=\sum_{cyc}{}\alpha^2=4092529.</cmath> <cmath>\alpha\beta+\beta\gamma+\gamma\alpha=0</cmath> <cmath>\alpha\beta\gamma=-(2023)^{2023} \implies \sum_{cyc}{}\alpha^3=8279186167-3(2023)^{2023}</cmath> Now we have to find the value of <cmath>\frac{\alpha^2+\beta^2}{\alpha+\beta}+\frac{\beta^2+\gamma^2}{\beta+\gamma}+\frac{\gamma^2+\alpha^2}{\gamma+\alpha}=\frac{4092529-\alpha^2}{2023-\alpha}+\frac{4092529-\beta^2}{2023-\beta}+\frac{4092529-\gamma^2}{2023-\gamma}</cmath> Therefore we get, <cmath>\frac{\sum_{cyc}{}(4092529-\alpha^2)(2023-\beta)(2023-\gamma)}{\prod_{cyc}{}(2023-\alpha)}</cmath> We get, <cmath>\frac{-\alpha\beta\gamma(\sum_{cyc}{}\alpha)+2023\sum_{cyc}{}\alpha^2(\beta+\gamma)-4092529(\sum_{cyc}{}\alpha^2)+4092529(\sum_{cyc}{}\alpha\beta)-8279186167(2(\sum_{cyc}{}\alpha))+50246380847523}{\prod_{cyc}{}(2023-\alpha)}</cmath> <cmath>\frac{4(2023)^{2024}}{2023^{2023}}\implies\boxed{8092.}</cmath>
+
From Vieta's formulas, we have <cmath>p^2+q^2+r^2 = (p+q+r)^2-2(pq+pr+qr) = 2023^2-2(0) = 2023^2.</cmath>
 +
 
 +
Now, <cmath>\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.</cmath>
 +
 
 +
Similarly, <math>\frac{p^2+r^2}{p+r} = 2023+q</math> and <math>\frac{r^2+q^2}{r+q} = 2023+p.</math>
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 +
Thus,
 +
<cmath>\begin{align*}
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&\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*}</cmath>

Revision as of 21:04, 9 September 2025

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*}

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