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Difference between revisions of "1966 IMO Problems/Problem 5"
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Line 62: Line 62:
 
-x_1 - x_2 - x_3 + x_4 = 0</math>
 
-x_1 - x_2 - x_3 + x_4 = 0</math>
  
−
It follows <math>x_1 - x_4 = 0</math> and <math>x_2 + x_3 = 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
 
Subtract the third equation from the second, and divide by
Line 77: Line 77:
 
Now we have two ways of proceeding.  We could consider each
 
Now we have two ways of proceeding.  We could consider each
 
of the other 23 cases, and solve it by a similar method.
 
of the other 23 cases, and solve it by a similar method.
−
The task is made easy if we notice that each case is just
+
The task is made easy if we notice that each case is obtained
−
a permutation of indices, so it can be viewed as a change
+
from the first case by a permutation of indices, so it can be
−
of notation, and with some care, we can just write the
+
viewed as a change of notation.  With some care, we can just
−
solution in each case.  For example, in the case
+
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>
 
<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>.
 
and <math>x_2 = x_4 = \frac{1}{a_2 - a_4}</math>.

Latest revision as of 18:00, 23 September 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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