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

(Created page with "==Problem== Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>p, q, r</math>. Find <cmath>\frac{p^2 + q^2}{p + q} + \frac{q^2 + r^2}{q + r} + \frac{r^2...")
 
(Solution)
 
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==Problem==
 
==Problem==
Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>p, q, r</math>.
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Let the roots of <math>P(x) = x^3 - 2023x^2 + 2023^{2023}</math> be <math>\alpha, \beta, \gamma.</math>.
 
Find
 
Find
<cmath>\frac{p^2 + q^2}{p + q} + \frac{q^2 + r^2}{q + r} + \frac{r^2 + p^2}{r + p}</cmath>
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<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==
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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>

Latest revision as of 10:37, 24 December 2023

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

By Vieta's relation we get, \[\alpha+\beta+\gamma=2023=\sum_{cyc}{}\alpha^2=4092529.\] \[\alpha\beta+\beta\gamma+\gamma\alpha=0\] \[\alpha\beta\gamma=-(2023)^{2023} \implies \sum_{cyc}{}\alpha^3=8279186167-3(2023)^{2023}\] Now we have to find the value of \[\frac{\alpha^2+\beta^2}{\alpha+\beta}+\frac{\beta^2+\gamma^2}{\beta+\gamma}+\frac{\gamma^2+\alpha^2}{\gamma+\alpha}=\frac{4092529-\alpha^2}{2023-\alpha}+\frac{4092529-\beta^2}{2023-\beta}+\frac{4092529-\gamma^2}{2023-\gamma}\] Therefore we get, \[\frac{\sum_{cyc}{}(4092529-\alpha^2)(2023-\beta)(2023-\gamma)}{\prod_{cyc}{}(2023-\alpha)}\] We get, \[\frac{-\alpha\beta\gamma(\sum_{cyc}{}\alpha)+2023\sum_{cyc}{}\alpha^2(\beta+\gamma)-4092529(\sum_{cyc}{}\alpha^2)+4092529(\sum_{cyc}{}\alpha\beta)-8279186167(2(\sum_{cyc}{}\alpha))+50246380847523}{\prod_{cyc}{}(2023-\alpha)}\] \[\frac{4(2023)^{2024}}{2023^{2023}}\implies\boxed{8092.}\]