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Difference between revisions of "1972 IMO Problems/Problem 4"
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Just like in case 1, this implies
 
Just like in case 1, this implies
  
−
<math>x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2 \hspace{30pt} (6)</math>.
+
<math>x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2 \hspace{30pt} (6)</math>
  
 
(4r) and (5l) imply <math>x_1 x_3 \le x_2 x_4</math>.  Combining this with (6),
 
(4r) and (5l) imply <math>x_1 x_3 \le x_2 x_4</math>.  Combining this with (6),
Line 139: Line 139:
 
to consider the first, and respectively, the first two inequalities
 
to consider the first, and respectively, the first two inequalities
 
to imply that the first factor <math>\ge 0</math> and the second factor <math>\le 0</math>.
 
to imply that the first factor <math>\ge 0</math> and the second factor <math>\le 0</math>.
 +
(For example, in case 2, we would have to consider the possibility
 +
that the second inequality implies
 +
<math>(x_2^2 - x_4x_1) \ge 0</math> and <math>(x_3^2 - x_4x_1) \le 0</math>, while all
 +
the other inequalities give <math>\le 0</math> and <math>\ge 0</math>.)
  
−
There are two possible arguments.  One is that if other inequalities
+
There are two possible arguments.  One is that in the other situations
−
would imply the positive/negative signs, the proof would be the same
+
the proof would be the same except that we would have to change the
−
except that we would have to change the indexes of <math>x_n</math> appropriately.
+
indexes of <math>x_n</math> appropriately. The other argument is that we can
−
The other argument is that we can just make a substitution
+
just make a substitution <math>x_n = y_{\sigma(n)}</math>, where <math>\sigma</math> is an
−
<math>x_n = y_{\sigma(n)}</math>, where <math>\sigma</math> is an appropriately chosen
+
appropriately chosen permutation, so that the factors in the new
−
permutation, so that the new inequalities in <math>y_p</math>, would be like
+
inequalities in <math>y_p</math> would be like the factors in the particular
−
in the particular cases 1, and respectively 2.  I skip the details
+
situations of cases 1, and respectively 2 that we looked at.  I skip
−
of this last step.
+
the details of this last step.
  
 
[Solution by pf02, May 2025]
 
[Solution by pf02, May 2025]

Revision as of 02:44, 5 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 case 1, 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 case 2, 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*}

Just like in case 1, this implies

$x_2^2 \le x_3^2 \le x_4^2 \le x_5^2 \le x_1^2 \hspace{30pt} (6)$

(4r) and (5l) imply $x_1 x_3 \le x_2 x_4$. Combining this with (6), we obtain $x_1 x_2 \le x_1 x_3 \le x_2 x_4$. From this we deduce $x_1 \le x_4$. Combining this with (6), we get $x_1 = x_4 = x_5$.

Now, (2r) and (3l) imply $x_4 x_1 \le x_5 x_2$, which implies $x_1 \le x_2$ (because $x_4 = x_5$). It follows that $x_1 = x_2 = x_3 = x_4 = x_5$.

For case 3, 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) \ge 0 \hspace{20pt} (2l) \hspace{40pt} (x_3^2 - x_4x_1) \le 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*}

Just like in case 1, this implies

$x_3^2 \le x_2^2 \le x_1^2 \hspace{55pt} (7)$

$x_3^2 \le x_4^2 \le x_5^2 \le x_1^2 \hspace{30pt} (8)$

Now, (3r) and (4l) imply $x_5 x_2 \le x_1 x_3$, and (4r) and (5l) imply $x_1 x_3 \le x_2 x_4$. This implies $x_5 x_2 \le x_2 x_4$, so $x_5 \le x_4$. Using (8), it follows that $x_4 = x_5$.

Now, (2l) and (1r) imply $x_4 x_1 \le x_3 x_5$. Simplifying with $x_4 = x_5$, we get $x_1 \le x_3$. Combining this with (7) and (8), we get $x_1 = x_2 = x_3 = x_4 = x_5$.

The only thing left to do is justify why in cases 2 and 3 it is OK to consider the first, and respectively, the first two inequalities to imply that the first factor $\ge 0$ and the second factor $\le 0$. (For example, in case 2, we would have to consider the possibility that the second inequality implies $(x_2^2 - x_4x_1) \ge 0$ and $(x_3^2 - x_4x_1) \le 0$, while all the other inequalities give $\le 0$ and $\ge 0$.)

There are two possible arguments. One is that in the other situations the proof would be the same except that we would have to change the indexes of $x_n$ appropriately. The other argument is that we can just make a substitution $x_n = y_{\sigma(n)}$, where $\sigma$ is an appropriately chosen permutation, so that the factors in the new inequalities in $y_p$ would be like the factors in the particular situations of cases 1, and respectively 2 that we looked at. I skip the details of this last step.

[Solution by pf02, May 2025]


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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