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Difference between revisions of "1972 IMO Problems/Problem 4"
Line 9: Line 9:
 
\end{align*}</cmath>
 
\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.
 +
  
 
==Solution==
 
==Solution==
Line 14: Line 15:
 
Add the five equations together to get
 
Add the five equations together to get
  
−
<math>(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</math>
+
<math>(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) +</math>
  
−
Expanding and combining, we get
+
<math>(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</math>
  
−
<math>(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</math>
+
Expanding, multiplying by <math>2</math>, and re-combining terms, we get
 +
 
 +
<math>(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 +</math>
 +
 
 +
<math>(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</math>
  
 
Every term is <math>\geq 0</math>, so every term must <math>= 0</math>.
 
Every term is <math>\geq 0</math>, so every term must <math>= 0</math>.
  
−
From the first term, we can deduce that <math>x_2 = x_4</math>. From the second term, <math>x_3 = x_5</math>. From the third term, <math>x_4 = x_1</math>. From the fourth term, <math>x_5 = x_2</math>.
+
From the first term, we can deduce that <math>x_2 = x_4</math>.
 +
From the second term, <math>x_3 = x_5</math>.
 +
 
 +
From the third term, <math>x_4 = x_1</math>. From the fourth term,
 +
<math>x_5 = x_2</math>.
  
 
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.
  
 
Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln724.html]
 
Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln724.html]
 +
  
 
== See Also == {{IMO box|year=1972|num-b=3|num-a=5}}
 
== See Also == {{IMO box|year=1972|num-b=3|num-a=5}}

Revision as of 13:35, 29 April 2025

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, multiplying by $2$, and re-combining terms, 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
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