1966 IMO Problems/Problem 5
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[hide]Problem
Solve the system of equations
where 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:
Subtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:
Hence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:
Hence , and .
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 , and each ordering gives a different system of equations. Let us consider one of them, like in the "solution" above.
Assume . In this case, the system is
Subtract the second equation from the first, and divide by . Also, subtract the fourth equation from the third, and divide by . We obtain
It follows that and .
Subtract the third equation from the second, and divide by . We obtain
Since , it follows that . Combining with , we get . Replacing these in the first equation of the system, we get , so we also have .
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 , we will obtain and .
We will proceed differently, but we will use the same idea. Let be the indices such that . Written in a compact way, our system becomes
.
Make the following change of notation:
and
In the new notation we have and the system becomes
.
This is exactly the system we solved above, just with a new notation ( instead of ). So the solutions are .
Returning to our original notation, we have .
In conclusion, here is a compact way of giving the solution to the system: let be the index of the largest of the 's, and q = the index of the smallest of the 's, and let be the other two indices. Then .
[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 |