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 |