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
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 | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
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