Difference between revisions of "1972 IMO Problems/Problem 4"

(Problem 4)
 
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==Problem 4==
 
 
 
Find all solutions <math>(x_1, x_2, x_3, x_4, x_5)</math> of the system of inequalities
 
Find all solutions <math>(x_1, x_2, x_3, x_4, x_5)</math> of the system of inequalities
<cmath>(x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \\
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<cmath>
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\begin{align*}
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(x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \\
 
(x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \\
 
(x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \\
 
(x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \\
 
(x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \\
 
(x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \\
 
(x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \\
(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0</cmath>
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(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0
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\end{align*}</cmath>
 
where <math>x_1, x_2, x_3, x_4, x_5</math> are positive real numbers.
 
where <math>x_1, x_2, x_3, x_4, x_5</math> are positive real numbers.
  
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Therefore, <math>x_1 = x_4 = x_2 = x_5 = x_3</math> is the only solution.
 
Therefore, <math>x_1 = x_4 = x_2 = x_5 = x_3</math> is the only solution.
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Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln724.html]
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== See Also == {{IMO box|year=1972|num-b=3|num-a=5}}

Latest revision as of 14:39, 29 January 2021

Find all solutions $(x_1, x_2, x_3, x_4, x_5)$ of the system of inequalities \begin{align*} (x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \\ (x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \\ (x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \\ (x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \\ (x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0 \end{align*} where $x_1, x_2, x_3, x_4, x_5$ are positive real numbers.

Solution

Add the five equations together to get

$(x_1^2 - x_3 x_5)(x_2^2 - x_3 x_5) + (x_2^2 - x_4 x_1)(x_3^2 - x_4 x_1) + (x_3^2 - x_5 x_2)(x_4^2 - x_5 x_2) + (x_4^2 - x_1 x_3)(x_5^2 - x_1 x_3) + (x_5^2 - x_2 x_4)(x_1^2 - x_2 x_4) \leq 0$

Expanding and combining, we get

$(x_1 x_2 - x_1 x_4)^2 + (x_2 x_3 - x_2 x_5)^2 + (x_3 x_4 - x_3 x_1)^2 + (x_4 x_5 - x_4 x_2)^2 + (x_5 x_1 - x_5 x_3)^2 + (x_1 x_3 - x_1 x_5)^2 + (x_2 x_4 - x_2 x_1)^2 + (x_3 x_5 - x_3 x_2)^2 + (x_4 x_1 - x_4 x_3)^2 + (x_5 x_2 - x_5 x_4)^2 \leq 0$

Every term is $\geq 0$, so every term must $= 0$.

From the first term, we can deduce that $x_2 = x_4$. From the second term, $x_3 = x_5$. From the third term, $x_4 = x_1$. From the fourth term, $x_5 = x_2$.

Therefore, $x_1 = x_4 = x_2 = x_5 = x_3$ is the only solution.

Borrowed from [1]

See Also

1972 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions