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 towersfreak2006. The original thread for this problem can be found here: [1]
Solution 2
Taking gives , or equivalently . Taking for yieldsSince , we must have , which forcesNow 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 |