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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 6"
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>\alpha, \beta, \gamma.</math>.
+
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>p^2+q^2+r^2 = (p+q+r)^2-2(pq+pr+qr) = 2023^2-2(0) = 2023^2.</cmath>
+
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{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>  
+
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, <math>\frac{p^2+r^2}{p+r} = 2023+q</math> and <math>\frac{r^2+q^2}{r+q} = 2023+p.</math>
+
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,
 
Thus,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
&\frac{p^2 + q^2}{p + q} + \frac{q^2 + r^2}{q + r} + \frac{r^2 + p^2}{r + p}
+
&\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta + \gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha} \\
−
= (2023+p)+(2023+q)+(2023+r) \\ &= 6069+(p+q+r) = \boxed{8092}.
+
&= (2023 + \gamma) + (2023 + \alpha) + (2023 + \beta) \\
 +
&= 6069 + (\alpha + \beta + \gamma) = \boxed{8092}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Latest revision as of 21:05, 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 \[\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = 2023^2 - 2(0) = 2023^2.\]

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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