1969 IMO Problems/Problem 2
Contents
Problem
Let be real constants, a real variable, and Given that prove that for some integer
Solution
Because the period of is , the period of is also . We can get for . Thus, for some integer
Solution 2 (longer)
By the cosine addition formula, This implies that if , Since the period of is , this means that for any natural number . That implies that every value is a zero of .
Remarks (added by pf02, August 2024)
Both solutions given above are incorrect.
The first solution is hopelessly incorrect. It states that (and relies on it) if has period and then for some integer . This is plainly wrong (think of ). There is an obvious "red flag" as far as solutions go, namely the solution did not use that and .
The second solution starts promising, but then it goes on to (incorrectly) prove the converse of the given problem, namely that if then for any .
Below, I will give a solution to the problem.
Solution
For simplicity of writing, denote . . First, we want to prove that it is not the case that for all . To prove this, we will prove that the maximum value of is at least . This will ensure that .
We do this by induction. The statement is clear for : has a maximum value of . Assume that we have terms in , and the maximum value of is at lease . Now add the term (and remember that ). This additional term has values in , so it can decrease the maximum of by subtracting at most from the previous maximum, which was at least . So, the new maximum is at least .
Now we will the formula .
We get .
If both and then for all . So at least one of these sums is . I will give the proof for the case . The other case is proven similarly.
Using , we get , and similarly for .
If , then and . It follows that both and are odd multiples of , so they differ by for some integer .
If , then we can divide by this quantity, and we get
.
Thinking of the graph of would be enough for many people to conclude that for some integer . If we want to be more formal, we proceed by writing . Some easy computations yield . It follows that , which implies that for some integer .
[Solution by pf02, August 2024]
See Also
1969 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |