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Difference between revisions of "1966 IMO Problems/Problem 5"

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(2 intermediate revisions by the same user not shown)
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<math>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\
 
<math>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\
 
|a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\
 
|a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\
−
|a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3-a_4| x_4 = 1 \\
+
|a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\
 
|a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1</math>
 
|a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1</math>
  
 
where <math>a_1, a_2, a_3, a_4</math> are four different real numbers.
 
where <math>a_1, a_2, a_3, a_4</math> are four different real numbers.
 +
  
 
==Solution==
 
==Solution==
Line 23: Line 24:
  
 
Hence <math>x2 = x3 = 0</math>, and <math>x1 = x4 = 1/(a1 - a4)</math>.
 
Hence <math>x2 = x3 = 0</math>, and <math>x1 = x4 = 1/(a1 - a4)</math>.
 +
 +
 +
==Remarks (added by pf02, September 2024)==
 +
 +
The solution above is in the realm of flawed and incorrect.
 +
It is flawed because you can not claim to have solved a
 +
system of equations by having solved a particular case.
 +
It is correct in solving the particular case, but it is
 +
incorrect in stating that the result obtained is a solution
 +
to the system in general.
 +
 +
Below I will give a complete solution to the problem.  The
 +
first few lines will be a repetition of the "solution" above,
 +
and I will repeat them for the sake of completeness and of a
 +
more tidy writing.
 +
 +
 +
==Solution 2==
 +
 +
There are 24 possibilities when we count the ordering of
 +
<math>a_1, a_2, a_3, a_4</math>, and each ordering gives a different
 +
system of equations.  Let us consider one of them, like
 +
in the "solution" above.
 +
 +
Assume <math>a_1 > a_2 > a_3 > a_4</math>.  In this case, the system is
 +
 +
<math>\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_1 - a_2) x_2 + (a_1 - a_3) x_3 + (a_1 - a_4) x_4 = 1 \\
 +
(a_1 - a_2) x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_2 - a_3) x_3 + (a_2 - a_4) x_4 = 1 \\
 +
(a_1 - a_3) x_1 + (a_2 - a_3) x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_3 - a_4) x_4 = 1 \\
 +
(a_1 - a_4) x_1 + (a_2 - a_4) x_2 + (a_3 - a_4) x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1</math>
 +
 +
Subtract the second equation from the first, and divide by
 +
<math>(a_1 - a_2)</math>.  Also, subtract the fourth equation from the
 +
third, and divide by <math>(a_3 - a_4)</math>.  We obtain
 +
 +
<math>-x_1 + x_2 + x_3 + x_4 = 0 \\
 +
-x_1 - x_2 - x_3 + x_4 = 0</math>
 +
 +
It follows that <math>x_1 - x_4 = 0</math> and <math>x_2 + x_3 = 0</math>.
 +
 +
Subtract the third equation from the second, and divide by
 +
<math>(a_2 - a_3)</math>.  We obtain
 +
 +
<math>-x_1 - x_2 + x_3 + x_4 = 0</math>
 +
 +
Since <math>x_1 - x_4 = 0</math>, it follows that <math>x_2 - x_3 = 0</math>.
 +
Combining with <math>x_2 + x_3 = 0</math>, we get <math>x_2 = x_3 = 0</math>.
 +
Replacing these in the first equation of the system, we
 +
get <math>x_4 = \frac{1}{a_1 - a_4}</math>, so we also have
 +
<math>x_1 = \frac{1}{a_1 - a_4}</math>.
 +
 +
Now we have two ways of proceeding.  We could consider each
 +
of the other 23 cases, and solve it by a similar method.
 +
The task is made easy if we notice that each case is obtained
 +
from the first case by a permutation of indices, so it can be
 +
viewed as a change of notation.  With some care, we can just
 +
write the solution in each case.  For example, in the case
 +
<math>a_2 > a_1 > a_3 > a_4</math>, we will obtain <math>x_1 = x_3 = 0</math>
 +
and <math>x_2 = x_4 = \frac{1}{a_2 - a_4}</math>.
 +
 +
We will proceed differently, but we will use the same idea.
 +
Let <math>m, n, p, q</math> be the indices such that
 +
<math>a_m > a_n > a_p > a_q</math>.  Written in a compact way, our
 +
system becomes
 +
 +
<math>\sum_{\substack{i = 1 \\ i \ne j}}^4 |a_j - a_i| x_i = 1, \ \ \ \ j = 1, 2, 3, 4</math>.
 +
 +
Make the following change of notation:
 +
<math>a_m \rightarrow b_1 \\
 +
a_n \rightarrow b_2 \\
 +
a_p \rightarrow b_3 \\
 +
a_q \rightarrow b_4</math>
 +
 +
and
 +
 +
<math>x_m \rightarrow y_1 \\
 +
x_n \rightarrow y_2 \\
 +
x_p \rightarrow y_3 \\
 +
x_q \rightarrow y_4</math>
 +
 +
In the new notation we have <math>b_1 > b_2 > b_3 > b_4</math> and
 +
the system becomes
 +
 +
<math>\sum_{\substack{k = 1 \\ k \ne l}}^4 |b_l - b_k| y_k = 1, \ \ \ \ l = 1, 2, 3, 4</math>.
 +
 +
This is exactly the system we solved above, just with a
 +
new notation (<math>b, y</math> instead of <math>a, x</math>).  So the solutions
 +
are <math>y_2 = y_3 = 0, y_1 = y_4 = \frac{1}{b_1 - b_4}</math>.
 +
 +
Returning to our original notation, we have
 +
<math>x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}</math>.
 +
 +
In conclusion, here is a compact way of giving the solution
 +
to the system:  let <math>m</math> be the index of the largest of the
 +
<math>a_i</math>'s, and q = the index of the smallest of the <math>a_i</math>'s,
 +
and let <math>n, p</math> be the other two indices.  Then
 +
<math>x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}</math>.
 +
 +
[Solution by pf02, September 2024]
 +
  
 
==See also==
 
==See also==
 
{{IMO box|year=1966|num-b=4|num-a=6}}
 
{{IMO box|year=1966|num-b=4|num-a=6}}
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Latest revision as of 18:12, 10 November 2024

Problem

Solve the system of equations

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1$

where $a_1, a_2, a_3, a_4$ are four different real numbers.


Solution

Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:

\[- x1 + x2 + x3 + x4 = 0.\]

Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:

\[- x1 - x2 - x3 + x4 = 0.\]

Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:

\[- x1 - x2 + x3 + x4 = 0.\]

Hence $x2 = x3 = 0$, and $x1 = x4 = 1/(a1 - a4)$.


Remarks (added by pf02, September 2024)

The solution above is in the realm of flawed and incorrect. It is flawed because you can not claim to have solved a system of equations by having solved a particular case. It is correct in solving the particular case, but it is incorrect in stating that the result obtained is a solution to the system in general.

Below I will give a complete solution to the problem. The first few lines will be a repetition of the "solution" above, and I will repeat them for the sake of completeness and of a more tidy writing.


Solution 2

There are 24 possibilities when we count the ordering of $a_1, a_2, a_3, a_4$, and each ordering gives a different system of equations. Let us consider one of them, like in the "solution" above.

Assume $a_1 > a_2 > a_3 > a_4$. In this case, the system is

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_1 - a_2) x_2 + (a_1 - a_3) x_3 + (a_1 - a_4) x_4 = 1 \\ (a_1 - a_2) x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_2 - a_3) x_3 + (a_2 - a_4) x_4 = 1 \\ (a_1 - a_3) x_1 + (a_2 - a_3) x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a_3 - a_4) x_4 = 1 \\ (a_1 - a_4) x_1 + (a_2 - a_4) x_2 + (a_3 - a_4) x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1$

Subtract the second equation from the first, and divide by $(a_1 - a_2)$. Also, subtract the fourth equation from the third, and divide by $(a_3 - a_4)$. We obtain

$-x_1 + x_2 + x_3 + x_4 = 0 \\ -x_1 - x_2 - x_3 + x_4 = 0$

It follows that $x_1 - x_4 = 0$ and $x_2 + x_3 = 0$.

Subtract the third equation from the second, and divide by $(a_2 - a_3)$. We obtain

$-x_1 - x_2 + x_3 + x_4 = 0$

Since $x_1 - x_4 = 0$, it follows that $x_2 - x_3 = 0$. Combining with $x_2 + x_3 = 0$, we get $x_2 = x_3 = 0$. Replacing these in the first equation of the system, we get $x_4 = \frac{1}{a_1 - a_4}$, so we also have $x_1 = \frac{1}{a_1 - a_4}$.

Now we have two ways of proceeding. We could consider each of the other 23 cases, and solve it by a similar method. The task is made easy if we notice that each case is obtained from the first case by a permutation of indices, so it can be viewed as a change of notation. With some care, we can just write the solution in each case. For example, in the case $a_2 > a_1 > a_3 > a_4$, we will obtain $x_1 = x_3 = 0$ and $x_2 = x_4 = \frac{1}{a_2 - a_4}$.

We will proceed differently, but we will use the same idea. Let $m, n, p, q$ be the indices such that $a_m > a_n > a_p > a_q$. Written in a compact way, our system becomes

$\sum_{\substack{i = 1 \\ i \ne j}}^4 |a_j - a_i| x_i = 1, \ \ \ \ j = 1, 2, 3, 4$.

Make the following change of notation: $a_m \rightarrow b_1 \\ a_n \rightarrow b_2 \\ a_p \rightarrow b_3 \\ a_q \rightarrow b_4$

and

$x_m \rightarrow y_1 \\ x_n \rightarrow y_2 \\ x_p \rightarrow y_3 \\ x_q \rightarrow y_4$

In the new notation we have $b_1 > b_2 > b_3 > b_4$ and the system becomes

$\sum_{\substack{k = 1 \\ k \ne l}}^4 |b_l - b_k| y_k = 1, \ \ \ \ l = 1, 2, 3, 4$.

This is exactly the system we solved above, just with a new notation ($b, y$ instead of $a, x$). So the solutions are $y_2 = y_3 = 0, y_1 = y_4 = \frac{1}{b_1 - b_4}$.

Returning to our original notation, we have $x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}$.

In conclusion, here is a compact way of giving the solution to the system: let $m$ be the index of the largest of the $a_i$'s, and q = the index of the smallest of the $a_i$'s, and let $n, p$ be the other two indices. Then $x_n = x_p = 0, x_m, x_q = \frac{1}{a_m - a_q}$.

[Solution by pf02, September 2024]


See also

1966 IMO (Problems) • Resources
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 6
All IMO Problems and Solutions
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