be an integer, let 
be a set of 
elements, and let 
,
such that ![\[ A_i \cap A_j \not= \emptyset, A_i \cap A_k \not= \emptyset, A_j \cap A_k \not= \emptyset, \;\textrm{imply}\ \;A_i \cap A_j \cap A_k \not= \emptyset \ .\]](http://latex.artofproblemsolving.com/c/1/1/c113f5cdce00dfec6714c382a8e1bc74f188f025.png)
Show that 
.
such that 
or 
,![\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]](http://latex.artofproblemsolving.com/6/8/e/68e3d73edec43b2cc480832adf23bbbb5f4c7bee.png)
such that![\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]](http://latex.artofproblemsolving.com/2/9/e/29edf6ab8cee2a526e7753a140c5a291e987856a.png)
denote the maximum number of 
-
board such that each 
such that![\[
f(N) \leq cN^2
\]](http://latex.artofproblemsolving.com/7/4/8/748109c3d63203f9a67a167434e10a909f38e83f.png)
for all positive integers 
.
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
.
Let 
.
Let 
.
Let 
such that
Proposed by Ngo Van Trang, Vietnam
s.t. 
and 
be positiv integer numbers.
,
and 
.
.
be a positive integer.Prove that there is a polynomial 
with integer coefficients so that 
,![$$a^{2n+1}+b^{2n+1}+c^{2n+1}=abc[P(a,b)+P(b,c)+P(a,c)]$$](http://latex.artofproblemsolving.com/5/2/0/520813bebb2f297928208f7b7b2eea76d24043ac.png)
.
.
Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr
Let 
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
be a triangle with incenter 
, and let 
be a point on side 
. Points 
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side such that 
and 
are the angle bisectors of angles 
and 
respectively. Let 
be the circle tangent to segment 
, the extension of 
past , and the extension of 
past . Prove that is tangent to the circumcircle of triangle .
and 
,
be midpoint of arc 
on 
,
be 
respectively now let reflection of 
over 
and reflection of 
over 
.
and 
are diameters on 
respectively, now let 
be the 
excenters of 
then notice we have 
which implies 
cyclic and similarily 
is cyclic however since 
we also get that 
is cyclic and similarily 
is cyclic, however it doesn't end here...
so 
on 
however 
are symetric in 
are reflections of 
over 
are colinear, and similarily 
are colinear.
which means 
is cyclic and similarily we have 
cyclic, but also note that 
which means that 
are colinear.
which does in fact show that 
is isosceles and therefore 
,
,
where 
(clearly then 
is A-excenter of 
)
which means that 
is cyclic but by summing arcs we end up realising 
is diameter and in fact now this means 
is 
be the miquelpoint of 
then 
colinear and then Reim's twice gives 
and from double Reim's once again we have that 
and this is excellent news because now we can note that 
which shows that 
are tangent at 
and 
.
and 
be the projections of 
and 
be the projections of 
respectively, and let 
be the tangency point of 
is tangent to 
is a parallelogram, by projecting onto 
.
Simplifying yields 
,
But 
With Thales' Theorem, 
so 
thus getting 
and the conclusion follows.
are in centres, 
meets 
in 
and tha
.
