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be an integer. We consider a square grid of size 
divided into 
unit squares. The grid is called balanced if:
, 
or 
.
.
, the smallest integer 
such that any balanced grid always contains an 
square whose absolute sum of the 
cells is less than or equal to 
.
is tangent to sides 
and 
at 
and 
, respectively. The bisectors of the angles at 
and 
intersect 
at points 
and 
, respectively. Let 
be the incentre of 
. Prove that 
.
be a positive integer. Each cell of an 
table contains an integer. Suppose that the following conditions are satisfied: modulo . modulo .
be the product of the numbers in the 
row, and 
be the product of the number in the 
column. Prove that the sums 
and 
are congruent modulo 
.
be a triangle with incentre 
, and let 
be the circumcircle of triangle 
. Let 
be a point in the interior of segment such that 
. Suppose that the angle bisector of 
intersects at points 
and 
such that and lie on the same side of , and that the angle bisector of 
intersects at points 
and 
such that and lie on the same side of .
.
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
. Suppose that 
we have to prove that 
. Indeed 
but 
hence result 
result 
and result
, 
, 
and we must solve this system in 
be defined by:
We need to prove:
First, check that composing 
with itself times gives 
.
: 
For 
: 
By induction: Assume 
for some 
. Then:
Suppose 
satisfies 
for some . We show .
: 
Since 
, repeated composition forces 
. Thus, 
.
: Now 
. Let’s write it as:
Compose 
twice: Compute 
:
For , we need:
Solve the equations: From 
and 
, we get 
. Then:
Thus, 
. is:
but hence result result and result , , and we must solve this system in 
then 
with 
and 
just work
should be the only solution to 
not 