Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 6"
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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 09:37, 24 December 2023
Problem
Let the roots of 
be 
.
Find
![\[\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}\]](http://latex.artofproblemsolving.com/c/b/9/cb9a83120ec6b7523b4b0afee9639ff451323c1a.png)
Solution
By Vieta's relation we get, ![\[\alpha+\beta+\gamma=2023=\sum_{cyc}{}\alpha^2=4092529.\]](http://latex.artofproblemsolving.com/8/e/f/8ef195dc2a97e0d7c13042ae0294c61d06e79b74.png)
![\[\alpha\beta+\beta\gamma+\gamma\alpha=0\]](http://latex.artofproblemsolving.com/d/4/7/d47f6ef8e51ad6e6d5a3e93ff75036437ec3bd21.png)
![\[\alpha\beta\gamma=-(2023)^{2023} \implies \sum_{cyc}{}\alpha^3=8279186167-3(2023)^{2023}\]](http://latex.artofproblemsolving.com/7/f/6/7f635119ccc5692b7382e95ba2a6b63e2660ec61.png)
Now we have to find the value of ![\[\frac{\alpha^2+\beta^2}{\alpha+\beta}+\frac{\beta^2+\gamma^2}{\beta+\gamma}+\frac{\gamma^2+\alpha^2}{\gamma+\alpha}=\frac{4092529-\alpha^2}{2023-\alpha}+\frac{4092529-\beta^2}{2023-\beta}+\frac{4092529-\gamma^2}{2023-\gamma}\]](http://latex.artofproblemsolving.com/6/6/d/66dc69517187cef304529032485e9106a54b979a.png)
Therefore we get, ![\[\frac{\sum_{cyc}{}(4092529-\alpha^2)(2023-\beta)(2023-\gamma)}{\prod_{cyc}{}(2023-\alpha)}\]](http://latex.artofproblemsolving.com/9/1/f/91f1c4571521bc9fc8fb581503d93c33f7b4cfed.png)
We get, ![\[\frac{-\alpha\beta\gamma(\sum_{cyc}{}\alpha)+2023\sum_{cyc}{}\alpha^2(\beta+\gamma)-4092529(\sum_{cyc}{}\alpha^2)+4092529(\sum_{cyc}{}\alpha\beta)-8279186167(2(\sum_{cyc}{}\alpha))+50246380847523}{\prod_{cyc}{}(2023-\alpha)}\]](http://latex.artofproblemsolving.com/6/4/8/6488b73f323e8e6644c15119520eb79714ccb317.png)
![\[\frac{4(2023)^{2024}}{2023^{2023}}\implies\boxed{8092.}\]](http://latex.artofproblemsolving.com/b/9/5/b95a183b87299f8fd0a78955fdb93dcca9d27a27.png)
