[/quote]
,
be circumcenter, orthocenter and A-HM point of triangle 
.
intersect 
at 
,
be the triangle formed by 
.
is tangent to 
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)
in only 
steps (Namely, the total number of circles and straight lines drawn is at most 
.
is any point not on (O). 
intersect 
.
be the Lemoine points of triangle 
respectively. Prove that 
are collinear.
be a prime number and 
.![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
Determine 
such that 
Determine 
different ropes. She then performs a experiment that causes 
children, each connected to each other with the same type of rope, to be cured. Two experiments are said to be of the same type, if each of the ropes connecting the children has the same medicine imbued in it. She then unties them and lets them go back home.
be the minimum m such that Myla can perform at least 
For every 
there exists a 
and 
such that for all 
,![\[f(n, k) = a_kn + b_k.\]](http://latex.artofproblemsolving.com/6/6/0/660699c6cec811eee25f727076665a6df3257f57.png)
Find the value of 
for every 
be a prime number. Prove that if 
,
,
are integers (not necessarily positive) satisfying the equations ![\[ a^n + pb = b^n + pc = c^n + pa\]](http://latex.artofproblemsolving.com/9/2/2/922cd81d65164e887b05cc9bc6026998d6a80ce6.png)
then 
.
is an even integer. At the start, a line joining two vertices is chosen arbitrarily and one of its endpoints is chosen as its pivot. Now, Aloo rotates the line around the pivot either clockwise or anti-clockwise until it passes through another vertex of the 
be such that 
and 
.
.
be the foot of altitudes of 
onto the opposite sides, respectively. Consider 
,
about line 
.
cut 
,
be the orthocenter of 
.
are concyclic.
=2$$](http://latex.artofproblemsolving.com/0/c/d/0cd423e22e36e5b16ad19841c5702dd1fefbc354.png)
be the circumcenter and orthocenter of triangle 
be intersections of 
with 
.
be orthocenter and circumcenter of triangle 
.
Y by mathematicsy, Adventure10, Mango247, and 5 other users
Let 
be an integer greater than 
. For a positive integer 
, let 
. Suppose that there exists a 
-element set 
such that
(a) each element of is an -element subset of 
;
(b) each pair of elements of shares at most one common element;
and
(c) each element of is contained in exactly two elements of .
Determine the maximum possible value of in terms of .
be an integer greater than 
. For a positive integer 
, let 
. Suppose that there exists a 
-element set 
such that(a) each element of
is an -element subset of 
;(b) each pair of elements of
shares at most one common element;and
(c) each element of
is contained in exactly two elements of .Determine the maximum possible value of
in terms of .
be the elements of 
.
there is exactly one, unique, pair of 
such that 
(second and third condition), there are possibly some empty intersections, so the number of intersection is not less then the number of elements of 
.
or 
.
is attainable. We construct 
as follows. 
.
then for 
we take an element on the 
-
columns and 
rows. Let each of the 
marked squares, and each column has 
marked squares; and it's easy to see that each pair of elements in 
.
-
,
-
squares, and mark the diagonal downwards again - keep doing this, and some simple computations show that when you finish you just finish filling the table.
.
.
.
elements of 
,
.
edges, so we label the edges with distinct elements of 
.
,
,
.
.
matrix with all entries either 
or 
iff 
.
nice if 
and 
.
.
and 
according to problem condition 
,
ways. Then, according to condition 
,
.
for ![\[\displaystyle A_i = \left\{ \begin{array}{lr} \{x:x=(p-1)(2n-1)+q \cup (n-1)(2n-1)+i \text{ for } 1 \leq p \leq n \text{ and } 1 \leq q \leq 2\} & \text{when}\ \ 1 \leq i \leq n-1 \\ \\ \{x:x=a(2n-1) \cup (n-1)(2n-1)+b \text{ for } 1 \leq a \leq n \text{ and } n \leq b \leq 2n-1\} & \text{when}\ \ i=n \\ \\ \{x:x=(i-n-1)(2n-1)+j \text{ for } 1 \leq j \leq 2n-1\} & \text{when}\ \ n+1 \leq i \leq 2n \end{array} \right.\]](http://latex.artofproblemsolving.com/a/b/f/abfc805380904043b2fa2708581ac481ab0279ec.png)
Then one can easily see that this set 
,
.
:
is indeed attainable and is the desired maximum. 
incidence matrix, where the rows represent the pair of 
and the column 
represents the number 
cell. Note that for each element, there is a pair of set and the intersection of any set is atmost 1. So we get 
,
.
My construction is one that I have not seen on the thread 
of pairs 
where 
.
and 
and using condition (b), we get that 
.
and using (c), we get that 
.
.
of triples ![\[(A_i, A_j, x \in A_i\cap A_j)\]](http://latex.artofproblemsolving.com/1/d/9/1d9b34f0ddb3f73fb4238835c52ec7f68642992e.png)
gives us 
,
.
.
or just 
.
as 
,
built, then the idea is that 
will have the k-th element repeated from each of the 
.
