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.
is acute triangle. 
is orthocenter, 
is the midpoint of 
, 
is the midpoint of smaller arc . Point 
is on 
such that, 
is perpendicular to 
. Prove that: 
is paralelogram(Synthetic sol needed).
, 
.
, 
be positive integers. Julia and Florian play a game on a 
board. Julia cells. such that Florian has a strategy to always deduce the entire tiling.
has circumcenter 
. 
is the foot from 
to , and 
is apoint on 
. The feet from to 
are 
, respectively, and the foot from to 
is 
. 
meets 
again at 
. 
meets again at 
. If 
is a point on satisfying 
, prove that 
lie on acircle and 
is tangent to it.
be a point inside triangle . Let 
be lines passint trough 
. We denote 
. Circumcircles of 
and 
intersect at 
. Prove that
be an isosceles triangle with 
. Let 
and points over 
and 
, respectively, such that 
. Bisector of 
intersects line at 
. Show that 
passes through the midpoint of 
.
be an equilateral triangle of side length 
. Point is chosen on side such that the length of 
is 
. Likewise points 
and 
are chosen on sides 
and . with 
. Line segments are drawn from points 
to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled 
. Prove that 
is an equilateral triangle with side length 
.
, 
, 
and prime numbers 
and 
, such that:
+1 w
be a cyclic quadrilateral with 
. Let be the midpoint of the arc 
not containing . Suppose there is a point inside such that 
and 
.
, and are concurrent.
on ![$[1,n]$](http://latex.artofproblemsolving.com/2/2/4/2240bb122bdadbe086c0a6291dcd5a3eb82ad5cf.png)
as a word with 
in some order, hence 
is a permutation on 5 symbols. We write 
for the wellness of namely the amount of pairs 
that appear in the same order in the word. Here 
.
for the number of 
with wellness . Hence we want to determine 
. hence 
. The symbol k adds 
to the wellness hence now we count the permutations on ![$[1,7]\setminus \{k\}$](http://latex.artofproblemsolving.com/2/a/3/2a3fb0361b0937dbfba193cae8ae422eb3328c31.png)
with wellness . Since we do not consider the symbol anymore, we may as well count the number of permutations on ![$[1,6]$](http://latex.artofproblemsolving.com/a/b/0/ab036e7da17bee704b3b7ee0ca5ea79f424465e5.png)
with wellness . This equals 
.
.. If the first symbol is 
then it contributes 
to the wellness hence 
. We find 
. The lower bound may be 
, in that case we start counting at 
. This reduces the problem to finding the numbers 
. Etc.
and then work your way upwards.
of 
by a unique 
-tuple of integers 
such that 
for all valid , where each 
represents the number of pairs of integers 
satisfying 
. Thus, we seek the number of -tuples such that for all valid and 
.
in the expansion of 
. But notice that 
is a root of 
. Hence, the sum of the coefficients of the terms of with degree divisible by is 
. Observe that the coefficients of 
and 
are trivially each 
, and the coefficients of 
and 
are equal due to symmetry. Thus, the coefficient of is simply 
.
s.t 
satisfying 
and 
.