AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, where 
, then maximum value of 
.
be the incentre of acute triangle 
with 
. The incircle 
of is tangent to sides 
, and 
at 
and 
, respectively. The line through 
perpendicular to 
meets at 
. Line 
meets again at 
. The circumcircles of triangle 
and 
meet again at 
.
and 
meet on the line through 
perpendicular to 
.
such that there exists a pair 
of positive integers, such that 
is not divisible by the cube of any prime, and 
+1 w
where 
is a natural number.What is the remainder when is divided by 
![$f : \mathbb{Q}[\sqrt{2}] \to \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/c/a/1/ca172b912bf5bf7178cbbd5745fbc0a6a8b27503.png)
such that for all ![$x, y \in \mathbb{Q}[\sqrt{2}]$](http://latex.artofproblemsolving.com/9/f/e/9fede0b2fd22836372b8afc488ccd8a14db5d0c2.png)
,
where ![$\mathbb{Q}[\sqrt{2}] = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \}$](http://latex.artofproblemsolving.com/f/a/c/fac8bf68761534a139d570bf66b1aa5dfb59e8db.png)
.
and 
are the roots of the cubic equation 
be a circle, and 
are parallel chords and 
is a in its middle point 
and 
. 
is the middle of 
. Prove that 
with orthic triangle 
, let 
, 
, 
be points such that 
. Let 
be the trilinear polar of the polar conjugate of wrt , and define 
and 
cyclically. Let 
be the triangle bounded by , , . and are bilogic.
, a function 
is called interesting if the following two conditions hold:
such that 
., we have 
(where 
denotes the -th iteration of 
).
. for which there is no interesting function
such that Bob can always find the two coins in at most guesses?
of integers satisfying![\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]](http://latex.artofproblemsolving.com/4/8/e/48e7b6f98f2062879377a9a2e507a63eac9269a8.png)
.
.
therefore u can use 1 and -1 alternating coming to the minimum of 0, because there is sequence is periodic with the period 4, and the first 4 
besides 
are 0 and they repeat themselves 250 times. also 0 is the smallest number possible, since 
when is positive. In other words, you can't go from 1 to -1.
. 24 more numbers. Then, you go
Prove that
,
,
and 
and so forth.
.
and this means that 
or 
since we want the minimum. Then 
,
,
,
.