[/quote]
,
be reals such that 
and 
Find the value of 
.
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
.
and 
be the circumcentre of triangle 
.
and 
,
to tangent 
.
is tangent to two fixed circles.
as the set of all products of an element of 
and 
then 
We call a set 
of positive integers, each with at least two elements and such that the sets 
and 
is good.
,
,
and 
are integers.)
be the midpoint of side 
and 
be the feet of the altitudes from 
and 
intersect at a point 
is perpendicular to line 
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
.
and 
with conditions,
is a prime number.
is a perfect square.
is also perfect square.
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)
such that:
for all 
for all 
is a natural number. A function 
is called \textit
-
with 
,
has a solution in natural numbers 
such that ![\( \frac{y_0}{x_0} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/d/a/6/da67f464a934f890626d188b84940fc82c4f4ff9.png)
,
is a solution. Suppose 
,
is a polynomial with real coefficients, negative leading coefficients, and total degree greater than 2, and for every real number 
,
as ![\( \frac{y}{x} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/6/5/d/65d4154dd5fb612f92f24f98406bc9bcfc713806.png)
.
is not 
such that for all 
be defined by![\[
x_1 = 2,\quad x_{n+1} = x_n + \frac{n}{x_n},\quad \text{for all } n \geq 1.
\]](http://latex.artofproblemsolving.com/f/2/8/f28193d8ea3fa6da8bbdb51f9a7e2e9d8a7d3c9f.png)
Prove that the sequences 
and 
have finite limits, and find those limits.
Y by
Let 
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
.![$[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$](http://latex.artofproblemsolving.com/3/a/0/3a0d813ceb846745ae927892e0a6bbddb6eec0bd.png)
is the incenter of 
.![$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$](http://latex.artofproblemsolving.com/4/5/6/4567bb6a0790e2821b3a2b4ba1592e30649baf97.png)
and 
belong to the 
mixtilinear incircle of 
.
is the midpoint of arc 
and 
belongs to the 
be the feet of perpendicular from 
meet the line 
at points 
.
centered at 
.
and 
.
.
and similarly 
and we have proved the lemma. 
.
at 
.
,![\[CL'^2 = CD\cdot CA = CX^2 = CL^2.\]](http://latex.artofproblemsolving.com/d/c/9/dc9e1ea07a744020713f6dc1137a2caeb0135af7.png)
Also 
lie on same side of 
giving 
lies on 
or 
.
.
is the internal angle bisector of 
.
gives 
and 
m
is the 
and 
at 
and 
respectively. We know (well-known) that 
are collinear and 
.
and we are given 
are collinear. This gives 
and 
which further gives that 
is the 
by 
are tangent to 
is tangent to 
.
is parallel to 
,
.
and 
.![\[\measuredangle XRR' = \measuredangle XPB = \measuredangle R'YX,\]](http://latex.artofproblemsolving.com/e/d/0/ed0b04ceb8f66d55c4dd0f96740056f508bbb1a4.png)
so 
is cyclic. Similarly, 
is cyclic as well.
is acute) inversion centered at 
,
,
.
and 
.
is tangent to 
.
and 
are tangents to 
and 
are also tangents to 
form the 
.
,
by 
.
,
and 
are tangent to 
is harmonic. This means that the tangents to 
,
and 
.
,
.
,
lies on 
(as 
and 
)
are collinear (as 
)
,
(as 
and 
)
Thus, 
,
.
such that 
so 
is tangent to 
and also 
is tangent to 
and hence by PoP we have 
and so 
.
and see that 
and so 
and hence it is the 
-
is 
