1961 IMO Problems/Problem 1
Problem
(Hungary) Solve the system of equations:
where and are constants. Give the conditions that and must satisfy so that (the solutions of the system) are distinct positive numbers.
Solution
Note that , so the first two equations become
.
We note that , so if equals 0, then must also equal 0. We then have ; . This gives us . Mutiplying both sides by , we have . Since we want to be real, this implies . But can only equal 0 when (which, in this case, implies ). Hence there are no positive solutions when .
When , we divide by to obtain the system of equations
,
which clearly has solution , . In order for these both to be positive, we must have positive and . Now, we have ; , so are the roots of the quadratic . The discriminant for this equation is
.
If the expressions were simultaneously negative, then their sum, , would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when and . But we have already replaced the first inequality with the sharper bound . It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if is positive and . Q.E.D.
Video Solution
https://www.youtube.com/watch?v=_stCjNU0_M4&list=PLa8j0YHOYQQJGzkvK2Sm00zrh0aIQnof8&index=3 - AMBRIGGS
https://youtu.be/e5cuvmW0clk [Video Solution by little fermat]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1961 IMO (Problems) • Resources | ||
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