AoPS Academy
be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
. Given this information, determine all possible values of 
.
..
which is close to 1/2, so C.
which is close to 1/2, so C.
, 
, and 
. Each of those inequalities, combined with the cube, graphs a pyramid with volume 1/6. Thus the volume of all of them is 1/2, out of the total volume of 1. So the answer is 1/2.
, consider two cases:
, then
, then
is the same. Thus the answer is 
.
..
..
..
, with 
, then multiply by 
.
is the same even though we have multiplied by , which is greater than 
, just note that the range of values for 
remains the same!
..
..
.., with , then multiply by . is the same even though we have multiplied by , which is greater than , just note that the range of values for remains the same!
. Then when we scale each number by they increase and 
, 
, 
. Since 
and 
they are both still in the range. So are you saying that it is obvious that all numbers in the domain can be created with 
and 
? For some reason that doesn't seem obvious. So now we try finding the probability of 
and thats how you WLOG?
, 
and 
where ![$x, y, z \in [0, 1]$](http://latex.artofproblemsolving.com/1/e/1/1e1fb5d92c0538c2ff164558642c22b672eb77a9.png)
.
to 
.
.
is just the tetrahedron formed by 
.
.
.
.
in the end.
on the 
plane. It looks like this.![[asy]
import graph;
size(3.9539168343137123cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
real xmin = -0.9249032795535634, xmax = 3.029013554760149, ymin = -0.4308791743516649, ymax = 1.778813028007358; /* image dimensions */
pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666);
pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.);
fill((0.,1.)--(0.5020527889324907,0.5020527889324906)--(1.,1.)--cycle, red);
fill((0.,1.)--(0.5020527889324907,0.5020527889324906)--(0.,0.)--cycle, blue);
draw((1.,1.)--(0.,0.));
draw((0.,1.)--(1.,1.));
draw((0.,1.)--(1.,0.));
draw((0.,1.)--(0.5020527889324907,0.5020527889324906), red);
draw((0.5020527889324907,0.5020527889324906)--(1.,1.), red);
draw((1.,1.)--(0.,1.), red);
draw((0.,1.)--(0.,0.));
draw((0.,1.)--(0.5020527889324907,0.5020527889324906), blue);
draw((0.5020527889324907,0.5020527889324906)--(0.,0.), blue);
draw((0.,0.)--(0.,1.), blue);
/* dots and labels */
dot((0.,0.),uuuuuu);
dot((1.,0.),xdxdff);
dot((0.,1.));
dot((1.,0.),uuuuuu);
dot((1.,1.),uuuuuu);
dot((0.5020527889324907,0.5020527889324906),xdxdff);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/1/d/3/1d33de61bc2b195dc0b8c87310472e5ef82832e6.png)
in terms of 
and 
is ![$[y, \min{(x+y, 1)}]$](http://latex.artofproblemsolving.com/2/f/b/2fb8aed2f58610c6f4e8bd7c0c922e278f6f97c2.png)
.
,
.
,
.
,
,![\[ \int_{0.5}^{1} \int_{1-y}^{y} 1-y dx dy \]](http://latex.artofproblemsolving.com/f/2/f/f2f2866e6b0a7559ea0e4fa17bd3ea18574cd618.png)
This is equal to ![\[ \int_{0.5}^{1} (1-y)(2y-1) dy = -\int_{0.5}^{1} 1-3y+2y^2 dy \]](http://latex.artofproblemsolving.com/5/e/2/5e2b783b91ff15f8bd60179d1c2c7edd3f5d1b76.png)
This is now equal to 
,
.![\[ \int_{0}^{0.5} \int_{x}^{1-x} x dy dx \]](http://latex.artofproblemsolving.com/6/2/3/623c6b9e82625d2e0b7d0515d99954038ae66e2a.png)
, or ![\[ \int_{0}^{0.5} (1-2x)(x) dx \]](http://latex.artofproblemsolving.com/e/5/5/e55d9e138b8302172e8f4bd949ad9d651c3203f1.png)
, which evaluates to 
,
,
has a region of 
to be in, so this is 
.
,
,
,
.