IMO Shortlist 1987 problem 1
Dodao/la:
arhiva2. travnja 2012. Let f be a function that satisfies the following conditions:
![(i)](/media/m/2/b/9/2b900b6092ba26af1415020bbd427e84.png)
If
![x > y](/media/m/e/b/2/eb2d9b827b93932d1bd216752a5dc6c0.png)
and
![f(y) - y \geq v \geq f(x) - x](/media/m/3/b/8/3b86e8a6ec55a1247b3cd5d1f855917a.png)
, then
![f(z) = v + z](/media/m/b/b/f/bbf9af8af1aaa0e40871c96e8d0a94c1.png)
, for some number
![z](/media/m/d/2/4/d241a79f1fdd0ce9a8f3f91570ba5d62.png)
between
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
and
![y](/media/m/c/c/0/cc082a07a517ebbe9b72fd580832a939.png)
.
![(ii)](/media/m/a/6/5/a65ca45870cb4c928be31b9a8a6fc8e5.png)
The equation
![f(x) = 0](/media/m/1/a/0/1a0332a8813a1110e758c328354c4121.png)
has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions;
![f(0) = 1](/media/m/2/f/8/2f8d7e5312aeea192bbf7dd6a5556409.png)
.
![f(1987) \leq 1988](/media/m/f/e/2/fe225bf784a0204f1b781aba8b15672e.png)
.
![f(x)f(y) = f(xf(y) + yf(x) - xy)](/media/m/0/e/6/0e6671817a719fbd2dd8a6eb9f79a032.png)
.
Find
![f(1987)](/media/m/1/f/c/1fc28dbe8f06946006333c812a206ee9.png)
.
Proposed by Australia.
%V0
Let f be a function that satisfies the following conditions:
$(i)$ If $x > y$ and $f(y) - y \geq v \geq f(x) - x$, then $f(z) = v + z$, for some number $z$ between $x$ and $y$.
$(ii)$ The equation $f(x) = 0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions;
$(iii)$ $f(0) = 1$.
$(iv)$ $f(1987) \leq 1988$.
$(v)$ $f(x)f(y) = f(xf(y) + yf(x) - xy)$.
Find $f(1987)$.
Proposed by Australia.
Izvor: Međunarodna matematička olimpijada, shortlist 1987