1986 IMO Problems/Problem 5
Problem
Find all (if any) functions taking the non-negative reals onto the non-negative reals, such that
(a) for all non-negative
,
;
(b) ;
(c) for every
.
Solution 1
For all , there exists a nonnegative real
such that
and
. Hence,
for all
.
For , take
.
Since
for
, then
.
Thus,
for all
.
Suppose so that
for some
and
.
Then
. The left hand side is equivalent to
, but the right hand side is nonzero, since
and
for all
.
Hence, for
.
For
,\begin{eqnarray*}f(xf(y))f(y) &=& f\left((x)\left(\frac{2}{2-y}\right)\right)\left(\frac{2}{2-y}\right)\\ &=&\left(\frac{2}{2-\frac{2x}{2-y}}\right)\left(\frac{2}{2-y}\right)\\ &=&\frac{4}{4-2y-2x}\\ &=&\frac{2}{2-y-x}\\ f(x+y)&=&\frac{2}{2-y-x}.\end{eqnarray*}
Indeed,
, so the desired function is
This solution was posted and copyrighted by Shen Kislay Kai. The original thread for this problem can be found here: [1]
Solution 2
Taking gives
, or equivalently
. Taking
for
yields
Since
, we must have
, which forces
Now taking
, for
gives,
Thus,
Hence, this forces
for
, giving a final answer of
This solution was posted and copyrighted by winnertakeover. The original thread for this problem can be found here: [2]
Solution 3
Let be the assertion. Note that
Since is defined for non-negative reals, we have
So,
Now,
Here, we are working with Since we may not have
due to condition (iii), we must have
From our first solution, it is imperative that So,
Now consider . We get,
Similar to the previous step, we have
Re arranging this, we get
Since satisfies both
and
we must have
Of course, we plug both solution back to see if they satisfy the three conditions. Sure enough, they do!
This solution was posted and copyrighted by proshi. The original thread for this problem can be found here: [3]
See Also
1986 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |