2003 Pan African MO Problems/Problem 6
Problem
Find all functions such that: for .
Solution
By letting , we have , and by letting , we have . Substitution and simplification results in Therefore, if , where is a real number, then . Thus, the function has rotational symmetry about .
Additionally, multiplying both sides by results in , and rearranging results in . The equation seems to resemble a rearranged slope formula, and the rotational symmetry seems to hint that is a linear function.
To prove that must be a linear function, we need to prove that the slope is the same from to all the other points of the function. By letting , we have . Additionally, by letting , we have . By rearranging the prior equation, we have
The slope from points and is . By substitution,
The slope from point to is the same as the slope from point to any other point on the function, so must be a linear function.
Let . Using the function on the original equation results in
Thus, can be any linear function, so , where are real numbers.
See Also
2003 Pan African MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All Pan African MO Problems and Solutions |