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Difference between revisions of "1972 IMO Problems/Problem 4"
 
(2 intermediate revisions by the same user not shown)
Line 51: Line 51:
  
 
<math>(x_1^2 - x_3x_5) \le 0</math> and <math>(x_2^2 - x_3x_5) \ge 0</math>
 
<math>(x_1^2 - x_3x_5) \le 0</math> and <math>(x_2^2 - x_3x_5) \ge 0</math>
 +
<math>\hspace{35pt}</math> (we will call this a "direct" pair of inequalities)
  
−
or  
+
or
  
 
<math>(x_1^2 - x_3x_5) \ge 0</math> and <math>(x_2^2 - x_3x_5) \le 0</math>
 
<math>(x_1^2 - x_3x_5) \ge 0</math> and <math>(x_2^2 - x_3x_5) \le 0</math>
 +
<math>\hspace{35pt}</math> (we will call this an "inverted" pair of inequalities)
  
 
We have a similar conclusion from the other four inequalities.
 
We have a similar conclusion from the other four inequalities.
−
So, we have to look at three cases:
+
So, we have to look at six cases:
  
−
1. all five first factors are <math>\le 0</math>
+
1. all five inequalities imply 5 "direct" pairs of inequalities
  
−
2. four first factors are <math>\le 0</math> and one is <math>\ge 0</math>
+
2. the five inequalities imply 4 "direct" pairs and 1 "inverted" pair
  
−
3. three first factors are <math>\le 0</math> and two are <math>\ge 0</math>
+
3. the five inequalities imply 3 "direct" pairs and 2 "inverted" pairs
  
−
The other three cases are equivalent to these.
+
4. the five inequalities imply 2 "direct" pairs and 3 "inverted" pairs
 +
 
 +
5. the five inequalities imply 1 "direct" pair and 4 "inverted" pairs
 +
 
 +
6. all five inequalities imply 5 "inverted" pairs of inequalities
 +
 
 +
Solving the problem in cases 4, 5, 6 is similar to solving it in
 +
cases 3, 2, 1 respectively; indeed, all we have to do is switch <math>\le</math>
 +
and <math>\ge</math>.  So, we will solve the system of inequalities in cases
 +
1, 2, 3.
  
 
In case 1, we have
 
In case 1, we have
Line 84: Line 95:
 
<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>  
 
<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).
+
which implies <math>x_1 = x_2 = x_3 = x_4 = x_5</math> (remember that <math>x_n</math> are
 +
positive).
  
−
For case 2, let us look at
+
For case 2, let us look at the particular situation when the first
 +
pair of inequalities is "inverted":
  
 
<cmath>
 
<cmath>
Line 100: Line 113:
 
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),
 
we obtain <math>x_1 x_2 \le x_1 x_3 \le x_2 x_4</math>.  From this we deduce
 
we obtain <math>x_1 x_2 \le x_1 x_3 \le x_2 x_4</math>.  From this we deduce
−
<math>x_1 \le x_4</math>.  Combining this with (6), we get <math>x_1 = x_4 = x_5</math>.
+
<math>x_1 \le x_4</math>.  Combining this with (6) again, we get <math>x_1 = x_4 = x_5</math>.
  
 
Now, (2r) and (3l) imply <math>x_4 x_1 \le x_5 x_2</math>, which implies
 
Now, (2r) and (3l) imply <math>x_4 x_1 \le x_5 x_2</math>, which implies
−
<math>x_1 \le x_2</math> (because <math>x_4 = x_5</math>).  It follows that
+
<math>x_1 \le x_2</math> (because <math>x_4 = x_5</math>).  Combining with (6) again, it
−
<math>x_1 = x_2 = x_3 = x_4 = x_5</math>.
+
follows that <math>x_1 = x_2 = x_3 = x_4 = x_5</math>.
 +
 
 +
If the "inverted" pair is the 2nd row, we can make the substitution
 +
<math>x_1 = y_5, x_2 = y_1, x_3 = y_2, x_4 = y_3, x_5 = y_4</math>.  In the new
 +
variables, the 1st row would contain the "inverted" pair, so our
 +
proof applies.  If the "inverted" pair is in the 3rd, 4th or 5th
 +
row, we just repeat this substitution as many times as needed to
 +
reduce the system of inequalities to the one we already solved.
 +
This finishes the proof in case 2.
  
−
For case 3, let us look at
+
For case 3, let us look at the particular situation when the first
 +
two pairs of inequalities are "inverted":
  
 
<cmath>
 
<cmath>
Line 136: Line 158:
 
we get <math>x_1 = x_2 = x_3 = x_4 = x_5</math>.
 
we get <math>x_1 = x_2 = x_3 = x_4 = x_5</math>.
  
−
The only thing left to do is justify why in cases 2 and 3 it is OK
+
Note that by applying the substitution described in case 2 as many
−
to consider the first, and respectively, the first two inequalities
+
times as needed, we can always assume that one of the "inverted"
−
to imply that the first factor <math>\ge 0</math> and the second factor <math>\le 0</math>.
+
pairs is in the 1st row.  Furthermore, if the "inverted" pairs are
 +
in the 1st and 5th rows, we can make a substitution which makes
 +
the "inverted" pairs of inequalities be in the 1st and 2nd rows
 +
(1st becomes 2nd, and 5th becomes 1st).
  
−
There are two possible arguments.  One is that if other inequalities
+
So we have to consider 2 more situations for case 3: the "inverted"
−
would imply the positive/negative signs, the proof would be the same
+
pairs are in the 1st and 3rd rows, and the "inverted" pairs are in
−
except that we would have to change the indexes of <math>x_n</math> appropriately.
+
the 1st and 4th rows.  The proof in each of these situations goes
−
The other argument is that we can just make a substitution
+
along the same lines as the one we just gave, and we will not go
−
<math>x_n = y_{\sigma(n)}</math>, where <math>\sigma</math> is an appropriately chosen
+
into these details, but leave them to the diligent reader.
−
permutation, so that the new inequalities in <math>y_p</math>, would be like
 
−
in the particular cases 1, and respectively 2.  I skip the details
 
−
of this last step.
 
  
 
[Solution by pf02, May 2025]
 
[Solution by pf02, May 2025]

Latest revision as of 18:07, 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 $(x_2^2 - x_3x_5) \ge 0$ $\hspace{35pt}$ (we will call this a "direct" pair of inequalities)

or

$(x_1^2 - x_3x_5) \ge 0$ and $(x_2^2 - x_3x_5) \le 0$ $\hspace{35pt}$ (we will call this an "inverted" pair of inequalities)

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

1. all five inequalities imply 5 "direct" pairs of inequalities

2. the five inequalities imply 4 "direct" pairs and 1 "inverted" pair

3. the five inequalities imply 3 "direct" pairs and 2 "inverted" pairs

4. the five inequalities imply 2 "direct" pairs and 3 "inverted" pairs

5. the five inequalities imply 1 "direct" pair and 4 "inverted" pairs

6. all five inequalities imply 5 "inverted" pairs of inequalities

Solving the problem in cases 4, 5, 6 is similar to solving it in cases 3, 2, 1 respectively; indeed, all we have to do is switch $\le$ and $\ge$. So, we will solve the system of inequalities in cases 1, 2, 3.

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 the particular situation when the first pair of inequalities is "inverted":

\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) again, 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 $x_4 = x_5$). Combining with (6) again, it follows that $x_1 = x_2 = x_3 = x_4 = x_5$.

If the "inverted" pair is the 2nd row, we can make the substitution $x_1 = y_5, x_2 = y_1, x_3 = y_2, x_4 = y_3, x_5 = y_4$. In the new variables, the 1st row would contain the "inverted" pair, so our proof applies. If the "inverted" pair is in the 3rd, 4th or 5th row, we just repeat this substitution as many times as needed to reduce the system of inequalities to the one we already solved. This finishes the proof in case 2.

For case 3, let us look at the particular situation when the first two pairs of inequalities are "inverted":

\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 $x_1 = x_2 = x_3 = x_4 = x_5$.

Note that by applying the substitution described in case 2 as many times as needed, we can always assume that one of the "inverted" pairs is in the 1st row. Furthermore, if the "inverted" pairs are in the 1st and 5th rows, we can make a substitution which makes the "inverted" pairs of inequalities be in the 1st and 2nd rows (1st becomes 2nd, and 5th becomes 1st).

So we have to consider 2 more situations for case 3: the "inverted" pairs are in the 1st and 3rd rows, and the "inverted" pairs are in the 1st and 4th rows. The proof in each of these situations goes along the same lines as the one we just gave, and we will not go into these details, but leave them to the diligent reader.

[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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