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Difference between revisions of "1972 IMO Problems/Problem 4"
Line 7: Line 7:
 
(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
 
(x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0
−
\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.
  
Line 39: Line 40:
 
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]
  
 +
 +
==Solution 2==
 +
 +
This solution is longer and less elegant than the previous one,
 +
but it is more direct, and based on a different idea, so it is
 +
worth mentioning.
 +
 +
Looking at the first inequality as an example, we can see that
 +
we have either
 +
 +
<math>(x_1^2 - x_3x_5) \le 0</math> and <math>(x_2^2 - x_3x_5) \ge 0</math>
 +
 +
or
 +
 +
<math>(x_1^2 - x_3x_5) \ge 0</math> and <math>(x_2^2 - x_3x_5) \le 0</math>
 +
 +
We have a similar conclusion from the other four inequalities.
 +
So, we have to look at three cases:
 +
 +
1. all five first factors are <math>\le 0</math>
 +
 +
2. four first factors are <math>\le 0</math> and one is <math>\ge 0</math>
 +
 +
3. three first factors are <math>\le 0</math> and two are <math>\ge 0</math>
 +
 +
The other three cases are equivalent to these.
 +
 +
In the first case, we have
 +
 +
<cmath>
 +
\begin{align*}
 +
(x_1^2 - x_3x_5) \le 0 \hspace{20pt} (1l) \hspace{40pt} (x_2^2 - x_3x_5) \ge 0 \hspace{20pt} (1r) \\
 +
(x_2^2 - x_4x_1) \le 0 \hspace{20pt} (2l) \hspace{40pt} (x_3^2 - x_4x_1) \ge 0 \hspace{20pt} (2r) \\
 +
(x_3^2 - x_5x_2) \le 0 \hspace{20pt} (3l) \hspace{40pt} (x_4^2 - x_5x_2) \ge 0 \hspace{20pt} (3r) \\
 +
(x_4^2 - x_1x_3) \le 0 \hspace{20pt} (4l) \hspace{40pt} (x_5^2 - x_1x_3) \ge 0 \hspace{20pt} (4r) \\
 +
(x_5^2 - x_2x_4) \le 0 \hspace{20pt} (5l) \hspace{40pt} (x_1^2 - x_2x_4) \ge 0 \hspace{20pt} (5r)
 +
\end{align*}
 +
</cmath>
 +
 +
(1l) and (1r) imply <math>x_1^2 \le x_2^2</math>, and we have four similar inequalities
 +
from the other four rows.  Putting them together, we obtain
 +
 +
<math>x_1^2 \le x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2</math>
 +
 +
which implies <math>x_1 = x_2 = x_3 = x_4 = x_5</math> (remember that <math>x_n</math> are positive).
 +
 +
For the second case, let us look at
 +
 +
<cmath>
 +
\begin{align*}
 +
(x_1^2 - x_3x_5) \ge 0 \hspace{20pt} (1l) \hspace{40pt} (x_2^2 - x_3x_5) \le 0 \hspace{20pt} (1r) \\
 +
(x_2^2 - x_4x_1) \le 0 \hspace{20pt} (2l) \hspace{40pt} (x_3^2 - x_4x_1) \ge 0 \hspace{20pt} (2r) \\
 +
(x_3^2 - x_5x_2) \le 0 \hspace{20pt} (3l) \hspace{40pt} (x_4^2 - x_5x_2) \ge 0 \hspace{20pt} (3r) \\
 +
(x_4^2 - x_1x_3) \le 0 \hspace{20pt} (4l) \hspace{40pt} (x_5^2 - x_1x_3) \ge 0 \hspace{20pt} (4r) \\
 +
(x_5^2 - x_2x_4) \le 0 \hspace{20pt} (5l) \hspace{40pt} (x_1^2 - x_2x_4) \ge 0 \hspace{20pt} (5r)
 +
\end{align*}
 +
</cmath>
 +
 +
This implies <math>x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2</math>.
 +
 +
 +
 +
 +
 +
 +
 +
TO BE CONTINUED.
  
 
== 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 20:56, 4 May 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 inequalities 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]


Solution 2

This solution is longer and less elegant than the previous one, but it is more direct, and based on a different idea, so it is worth mentioning.

Looking at the first inequality as an example, we can see that we have either

$(x_1^2 - x_3x_5) \le 0$ and $(x_2^2 - x_3x_5) \ge 0$

or

$(x_1^2 - x_3x_5) \ge 0$ and $(x_2^2 - x_3x_5) \le 0$

We have a similar conclusion from the other four inequalities. So, we have to look at three cases:

1. all five first factors are $\le 0$

2. four first factors are $\le 0$ and one is $\ge 0$

3. three first factors are $\le 0$ and two are $\ge 0$

The other three cases are equivalent to these.

In the first case, we have

\begin{align*} (x_1^2 - x_3x_5) \le 0 \hspace{20pt} (1l) \hspace{40pt} (x_2^2 - x_3x_5) \ge 0 \hspace{20pt} (1r) \\ (x_2^2 - x_4x_1) \le 0 \hspace{20pt} (2l) \hspace{40pt} (x_3^2 - x_4x_1) \ge 0 \hspace{20pt} (2r) \\ (x_3^2 - x_5x_2) \le 0 \hspace{20pt} (3l) \hspace{40pt} (x_4^2 - x_5x_2) \ge 0 \hspace{20pt} (3r) \\ (x_4^2 - x_1x_3) \le 0 \hspace{20pt} (4l) \hspace{40pt} (x_5^2 - x_1x_3) \ge 0 \hspace{20pt} (4r) \\ (x_5^2 - x_2x_4) \le 0 \hspace{20pt} (5l) \hspace{40pt} (x_1^2 - x_2x_4) \ge 0 \hspace{20pt} (5r) \end{align*}

(1l) and (1r) imply $x_1^2 \le x_2^2$, and we have four similar inequalities from the other four rows. Putting them together, we obtain

$x_1^2 \le x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2$

which implies $x_1 = x_2 = x_3 = x_4 = x_5$ (remember that $x_n$ are positive).

For the second case, let us look at

\begin{align*} (x_1^2 - x_3x_5) \ge 0 \hspace{20pt} (1l) \hspace{40pt} (x_2^2 - x_3x_5) \le 0 \hspace{20pt} (1r) \\ (x_2^2 - x_4x_1) \le 0 \hspace{20pt} (2l) \hspace{40pt} (x_3^2 - x_4x_1) \ge 0 \hspace{20pt} (2r) \\ (x_3^2 - x_5x_2) \le 0 \hspace{20pt} (3l) \hspace{40pt} (x_4^2 - x_5x_2) \ge 0 \hspace{20pt} (3r) \\ (x_4^2 - x_1x_3) \le 0 \hspace{20pt} (4l) \hspace{40pt} (x_5^2 - x_1x_3) \ge 0 \hspace{20pt} (4r) \\ (x_5^2 - x_2x_4) \le 0 \hspace{20pt} (5l) \hspace{40pt} (x_1^2 - x_2x_4) \ge 0 \hspace{20pt} (5r) \end{align*}

This implies $x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2$.




TO BE CONTINUED.

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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