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.
be a positive integer larger than 
. Build an infinite set 
of subsets of 
having the following properties: distinct sets of have exactly one common element;
distinct sets of have void intersection.
that satisfy the equation 
?
and 
Find the value of 
such that 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of again at 
is the reflection of 
over . Suppose points 
lie on such that 
are parallel to 
. Prove that: The intersection of 
lie on the circumcircle of 
![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a sequence of positive real numbers such that for every 
, we have:![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
, the following holds:![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and and 
are on different sides of 
, 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of w.r.t 
. Prove that 
and are collinear.
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points to the circumcircle of 
intersects at point 
Prove that: 
are parallel., 
, , 
be four points in the plane, with and on the same side of the line , such that 
and 
. Find the ratio![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles and are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles and at the point are perpendicular.)
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer . Prove that 
for every 
.
be a cyclic quadrilateral, and 
the midpoints of sides 
respectively. Prove that the perpendiculars from 
to 
, from 
to 
, from 
to 
, and from 
to 
are concurrent.
)
)
be a cyclic quadrilateral, and the midpoints of sides respectively. Prove that the perpendiculars from to , from to , from to , and from to are concurrent.
is a parallelogram, we get 
, moreover 
then 
is a parallelogram
. Similarly, 
to 
and 
to meet at 
If 
is the circumcenter of 
then 
and 
implies that 
is a paralelogram 
is the reflection of about the midpoint 
of 
But is also the midpoint of 
Hence, perpendiculars from 
to 
and to also pass through
be a cyclic quadrilateral, and the midpoints of sides respectively. Prove that the perpendiculars from to , from to , from to , and from to are concurrent.
and the point of concurrency be 
. Since we can drop perpendiculars from to , at their midpoints and respectively, 
is a parallelogram. Similarly, 
is also a parallelogram.
as the vertices of . The point of concurrency is nothing but the vector sum of the midpoints of opposite sides.
which is the same point as 
. Thus, we arrive at the same point for each pair of opposite sides.
-gon, being even, if I'm not mistaken.
, 
, 
, 
the projections of , , , onto , , , , respectively. Now, It is easy to see that is parallelogram. Let 
be the intersection of its diagonals, 
be the circumcenter of . maps into , maps into , hence, the line 
is maped in to the line 
. But KM is perpendicular to , therefore, is also perpendicular to .
, 
, 
, respectively, is perpendicular to , , 
.
. I am so sorry, I have thought that these diagnols of this ABCD are perpendicular to each other 
. Sorry for my big mistake.