Math Book Cambodia: Vietnamese Olympiad 2022: Find the value of `1/(a^2023)+1/(b^2023)+1/(c^2023)`

Friday, September 9, 2022

Vietnamese Olympiad 2022: Find the value of $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)$

If $a + b + c = 2022$ $a+b+c=2022$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2022}$ $1/a+1/b+1/c=1/2022$ Find the value of $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)$

Solution

We had: $a + b + c = 2022$ $a+b+c=2022$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2022}$ $1/a+1/b+1/c=1/2022$

Then, $\frac{1}{a + b + c} = \frac{1}{2022}$ $1/(a+b+c)=1/2022$ So that, $\frac{1}{a + b + c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ $1/(a+b+c)=1/a+1/b+1/c$

$\to \frac{1}{a + b + c} - \frac{1}{a} = \frac{1}{b} + \frac{1}{c}$ $\rightarrow1/(a+b+c)-1/a=1/b+1/c$

$\leftrightarrow \frac{a - a - b - c}{a (a + b + c)} = \frac{b + c}{b c}$ $\leftrightarrow(a-a-b-c)/(a(a+b+c))=(b+c)/(bc)$

$\leftrightarrow - \frac{b + c}{a (a + b + c)} = \frac{b + c}{b c}$ $leftrightarrow-(b+c)/(a(a+b+c))=(b+c)/(bc)$

$\leftrightarrow \frac{b + c}{a (a + b + c)} + \frac{b + c}{b c}$ $leftrightarrow(b+c)/(a(a+b+c))+(b+c)/(bc)$

$\leftrightarrow (b + c) (\frac{1}{a (a + b + c)} + \frac{1}{b c}) = 0$ $leftrightarrow(b+c)(1/(a(a+b+c))+1/(bc))=0$

$\leftrightarrow (b + c) (\frac{a (a + b + c) + b c}{a b c (a + b + c)}) = 0$ $leftrightarrow(b+c)((a(a+b+c)+bc)/(abc(a+b+c)))=0$

Therefore, $(b + c) (b c + a^{2} + a b + a c) = 0$ $(b+c)(bc+a^2+ab+ac)=0$

$\leftrightarrow (b + c) [c (b + a) + a (a + b)] = 0$ $leftrightarrow(b+c)[c(b+a)+a(a+b)]=0$

$\leftrightarrow (a + b) (b + c) (a + c) = 0$ $leftrightarrow(a+b)(b+c)(a+c)=0$

There are 3 cases of above equatoin:

+ Case 01: $a + b = 0$ $a+b=0$ Then $a = - b$ $a=-b$ It is easy to see that $\frac{1}{a^{2023}} = - \frac{1}{b^{2023}}$ $1/a^2023=-1/b^2023$

Also from this $a + b + c = 2022$ $a+b+c=2022$ We can see that $c = 2022$ $c=2022$

Then, $\frac{1}{c^{2023}} = \frac{1}{2022^{2023}}$ $1/c^2023=1/2022^2023$

Hence, $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}} = \frac{1}{2022^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)=1/2022^2023$

+ For other 2 cases $b + c = 0$ $b+c=0$ and $a + c = 0$ $a+c=0$ We do the same and we will get the answer is $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}} = \frac{1}{2022^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)=1/2022^2023$

Solution by: Thin Sokkean

You can download PDF file here:

Math Book Cambodia

Pages

Friday, September 9, 2022

Vietnamese Olympiad 2022: Find the value of $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)$

No comments:

Post a Comment

Popular

Pages

Friday, September 9, 2022

Vietnamese Olympiad 2022: Find the value of 1a2023+1b2023+1c20231a2023+1b2023+1c20231/(a^2023)+1/(b^2023)+1/(c^2023)

No comments:

Post a Comment

Vietnamese Olympiad 2022: Find the value of $\frac{1}{a^{2023}} + \frac{1}{b^{2023}} + \frac{1}{c^{2023}}$ $1/(a^2023)+1/(b^2023)+1/(c^2023)$