[/quote]
,
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
,
an arbitrary point on side 
meets the circumcircle of 
again at 
,
meets it again at 
.
,
and 
are concurrent.
be non-negative real numbers. Prove that
be a triangle such that 
.
lie on segments 
,
and 
are the angle bisectors of triangle 
.
.
satisfy 
prove in equality.
and circumcenter 
.
with center 
intersects the segment 
,
are all different and lie on line 
be the points of intersection of 
be the second point of intersection of the circumcircle of triangle 
and the segment 
be the second point of intersection of the circumcircle of triangle 
and the segment 
and 
are different and intersect at the point 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
,
be the incenters of triangles 
,
,
,
,
meet at a point.
satisfying
Prove
cups labeled 
,
-
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and 
,
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
,![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
Y by Adventure10, Mango247
The incircle of a triangle 
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
touches the sides 
, 
, 
at the points 
, 
, 
respectively. Let the line 
intersect this incircle of triangle at a point 
(apart from ). Assume that this point is the midpoint of the segment , this means, 
. Let the line 
meet the incircle of triangle at a point 
(apart from ), and let the line 
meet the incircle of triangle at a point 
(apart from ). Show that 
.
Power of the vertex A to the incircle (I) is 
Since 
we get 
BX is the B-median of the triangle 
Thus
In exactly the same way, we can show that 
As a result, EFYZ is a cyclic trapezoid with bases 
i.e., isosceles and with equal diagonals, hence 
,
.
,
.
,
should be parallel to 
should be parallal to 
where 
and 
.
,
,
and 
.![\[AX AD=AE AE\]](http://latex.artofproblemsolving.com/9/3/c/93caff3f2af3b6860646f8be2043925780283396.png)
![\[AD = (s-a)\sqrt{2}\]](http://latex.artofproblemsolving.com/5/a/9/5a909b8347358201ac5d614a89b494251e7d318e.png)
From triangle ADC we get![\[\frac{\sin \varphi}{\sin \gamma}= \frac{CD}{AD}\]](http://latex.artofproblemsolving.com/4/1/7/417e7cbf461aad0d1c90986ba0144a5ce4bf6055.png)
![\[\sin \varphi= \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/3/3/7/337d63702d62161786276a2f0a5dc02530d7d8ed.png)
![\[\sin \theta = \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\]](http://latex.artofproblemsolving.com/7/6/5/7657af71cb7cc74df44169e0bf4a0ef530ade9f4.png)
In triangle AXB: ![\[\frac{\sin \theta}{\sin \xi}= \frac{BX}{AX}\]](http://latex.artofproblemsolving.com/7/b/b/7bbd38c28039a075bc6d8a9f5792db2442327b7a.png)
From triangle BFY: ![\[\frac{\sin \mu}{\sin \xi}= \frac{BY}{FY}\]](http://latex.artofproblemsolving.com/c/7/f/c7f11ff3938970fd8dedb1111b79dae3f6b676d5.png)
We have: ![\[BX BY = BD^{2}\]](http://latex.artofproblemsolving.com/a/0/7/a0794d8e643604053100d0467a00a01ee871b6da.png)
![\[\sin \xi = \sqrt{2}\sin \mu \sin \frac{\beta}{2}\]](http://latex.artofproblemsolving.com/a/2/5/a25e513203a495ff3ed1360f16ea0865f784c43c.png)
Also we have ![\[\sin \nu = \sqrt{2}\sin \eta \sin \frac{\gamma}{2}\]](http://latex.artofproblemsolving.com/6/7/0/6701c3c33033da33415e5decaa72b50d09748f6a.png)
From Ceva' s theorem we have![\[\frac{\sin \varphi}{\sin\theta}\frac{\sin \xi}{\sin \nu}=\frac{\sin(\beta-\xi)}{\sin(\gamma-\nu)}= \frac{CX}{BX}\]](http://latex.artofproblemsolving.com/8/6/9/869d298f4e7f0d9b79dc9b9c7d96ce98552e0abf.png)
but I don't know how to
and 
then using Pascal's theorem on hexagon 
we obtain 
and 
are collinear, and because 
,
.
,
making 
an iscoceles trapezium. So the diagonals are congruent hence 
.
and that 
is the 
s
,![\[
\angle BFY=\angle FXY=\angle MXD=\angle FAD
\]](http://latex.artofproblemsolving.com/8/0/f/80fd56fe2f42e13cf49eea31a510441dce9a6792.png)
meaning that 
