[/quote]
,
by 
and 
such that 
.
be a random point on the smaller arc 
of the circumcircle of square 
,
be the intersection point of segments 
and 
.
to the circumcircle of the triangle 
are 
, where 
is the center of the square. Prove that points 
,
and 
.
a regular hexagon inscribed in a circle 
.
be a point inside 
its polar reflection wrt 
meet 
.
the polygon formed by the vertices 
using 
,
,
,
,
intersects lines 
and 
at 
and 
,
and 
be distinct points on side 
.
be the circumcircle of 
.
.
be an even integer, and let 
denote the number of (positive) divisors of 
.
such that 
for all 
such that:
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)
with an area of 
,
.
,
.
such that for all 
we have:
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)
Let 
be positive reals such that 
.
pieces of equal weights. Can Vova prevent Pasha's win?
Y by Mango247, Mango247, Mango247
Let 
be a triangle and 
its circumscribed circle. The internal bisector of angle 
meets 
at point 
. Let 
be the circle tangent to 
and . Similarly, let 
be the circle tangent to 
and . Prove that the tangency points of and with 
coincide.
be a triangle and 
its circumscribed circle. The internal bisector of angle 
meets 
at point 
. Let 
be the circle tangent to 
and . Similarly, let 
be the circle tangent to 
and . Prove that the tangency points of and with 
coincide.
fixes 
and so swaps 
to itself, it fixes 
at the point at distance 
.
and 
respectively. Now deffine 
,
and 
,
.
be the midpont of minor arc 
.
and 
,
, but as 
is tangent to 
in 
and 
.
are concyclic. Thus 
, and we're done.
