MHB Discover Positive Real Solutions in a System of Equations

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The discussion focuses on finding positive real solutions for a system of equations involving variables \( a_1 \) to \( a_{100} \). By applying the AM-GM inequality, it is established that the product of the equations leads to the conclusion that all inequalities must hold as equalities. This results in a relationship where each variable is inversely related to the next, specifically \( a_1 = \frac{1}{a_2} \), \( a_2 = \frac{1}{a_3} \), and so forth. Ultimately, the solutions are determined to be \( a_1 = 2 \) and \( a_2 = \frac{1}{2} \), repeating this pattern throughout the system. The method effectively demonstrates how to derive the solutions systematically.
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Find all positive real solutions of the system below:

$a_1+\dfrac{1}{a_2}=4,\,a_2+\dfrac{1}{a_3}=1,\cdots,a_{99}+\dfrac{1}{a_{100}}=4,\,a_{100}+\dfrac{1}{a_1}=1$
 
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Every term, $a_i$ can be expressed by $a_1$ as follows:
\[a_k = \frac{(k-1)-(\frac{k}{2}-1)a_1}{2k-(k-1)a_1} \; \; \; \; k = 2,4,..,100. \\\\\\ a_j = \frac{2(j-1)-(j-2)a_1}{j-\left ( \frac{j-1}{2} \right )a_1}\; \; \; j =1,3,5,..,99. \\\\\\ a_{100}=\frac{99-49a_1}{200-99a_1}\; \; \; \; \; \;and\;\;\;\;\; a_{100}+\frac{1}{a_1}=1\Rightarrow a_1^2-4a_1+4 = 0\]
The positive (and only) solution is: $a_1 = 2$

So
\[a_k = \frac{1}{2}\; \; \; k = 2,4,..,100 \\\\ a_j = 2\; \; \; \; j = 1,3,..,99\]
 
Thanks for participating, lfdahl! :) I think it really is a great idea to relate $a_{100}$ with $a_1$ again and with the two well defined formulas, how to solve further would then be as clear as daylight. Well done!:)

Another method that I saw online that I would like to share:

By AM-GM:

$a_1+\dfrac{1}{a_2}\ge2\sqrt{\dfrac{a_1}{a_2}},\,\cdots,\,a_{100}+\dfrac{1}{a_1}\ge2\sqrt{\dfrac{a_{100}}{a_1}}$

Multiplying we get

$\left( a_1+\dfrac{1}{a_2} \right)\left( a_2+\dfrac{1}{a_3} \right)\cdots\left( a_{100}+\dfrac{1}{a_1} \right)\ge2^{100}$,

From the system of equations we get

$\left( a_1+\dfrac{1}{a_2} \right)\left( a_2+\dfrac{1}{a_3} \right)\cdots\left( a_{100}+\dfrac{1}{a_1} \right)=2^{100}$,

so all thoseinequalities are equalities, i.e.

$a_1+\dfrac{1}{a_2}=2\sqrt{\dfrac{a_1}{a_2}}$

$\left(\sqrt{a_1}-\dfrac{1}{\sqrt{a_2}} \right)^2=0,\,\,\,\rightarrow a_1=\dfrac{1}{a_2}$

and analogously $a_2=\dfrac{1}{a_3},\,\cdots,\,a_{100}=\dfrac{1}{a_1}$.

Hence, we get $a_1=2,\,a_2=\dfrac{1}{2},\,\cdots\,,a_{99}=2,\,a_{100}=\dfrac{1}{2}$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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