Difference between revisions of "1965 IMO Problems/Problem 4"
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Therefore, the solutions are: <math>(1,1,1,1)</math>; <math>(3,-1,-1,-1)</math>; <math>(-1,3,-1,-1)</math>; <math>(-1,-1,3,-1)</math>; <math>(-1,-1,-1,3)</math>. | Therefore, the solutions are: <math>(1,1,1,1)</math>; <math>(3,-1,-1,-1)</math>; <math>(-1,3,-1,-1)</math>; <math>(-1,-1,3,-1)</math>; <math>(-1,-1,-1,3)</math>. | ||
+ | |||
+ | == See Also == | ||
+ | {{IMO box|year=1965|num-b=3|num-a=5}} |
Latest revision as of 12:51, 29 January 2021
Problem
Find all sets of four real numbers ,
,
,
such that the sum of any one and the product of the other three is equal to
.
Solution
Let be the product of the four real numbers.
Then, for we have:
.
Multiplying by yields:
where
.
If , then we have
which is a solution.
So assume that . WLOG, let at least two of
equal
, and
OR
.
Case I:
Then we have:
Which has no non-zero solutions for .
Case II: AND
Then we have:
AND
So, we have as the only non-zero solution, and thus,
and all permutations are solutions.
Case III: AND
Then we have:
AND
Thus, there are no non-zero solutions for in this case.
Therefore, the solutions are: ;
;
;
;
.
See Also
1965 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |