[/quote]
,
people at a party. Prove that there are two people such that, of the remaining 
people, there are at least 
of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that 
denotes the greatest integer less than or equal to 
.
be a polygon formed by the edges of an infinite chessboard, which does not intersect itself. Let the numbers 
represent the number of unit squares that have exactly 
edges on the boundary of 
such that the inequality 
holds for each polygon constructed with these conditions.
such that, for all 
,
is defined as, 
and for all 
,
then 
.
.
,
is also an arithmetic sequence, show that the sequence 
,
are coprime positive integers. Find all 
such that:
such that![\[f(f(f(\frac{x+y}{2}))+x+y)=f(x)+f(y)+f(\frac{x+y}{2})\]](http://latex.artofproblemsolving.com/6/2/4/624afa78b68f810aaaecf0e1e2251047beeddc36.png)
for all rational numbers 
.
in term of a function 
such that 
vertices, each edge has one of the colors 
,
,
.
,
groups such that any two vertices connected by an edge of color 
are in different groups, find the minimum possible value of 
.
be given pairwise coprime positive integers where 
.
be positive integers. We call 
to be a grandson of 
for all functions 
,
for all non-negative real numbers 
.
and 
be sequences of real numbers for which 
and
for all positive integers 
for which 
for all 
)
and 
be two concentric circles. Let 
be any two equilateral triangles inscribed in 
are any two points on ![\[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]](http://latex.artofproblemsolving.com/2/a/7/2a73de753f0332329b743d494272d3fb85f60ef9.png)
(meaning 
takes pairs of positive real numbers to positive real numbers) such that for any real numbers 
,
holds; and
then 
.
Prove that
Y by FreakLM10, MatBoy-123, Adventure10, Rounak_iitr
Let 
be an acute-angled triangle with 
. Let 
be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line 
at 
respectively. Further assume that both 
and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
be an acute-angled triangle with 
. Let 
be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line 
at 
respectively. Further assume that both 
and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
being the orthic triangle of 
and more angle chase in right-angled cyclic quadrilaterals
This was stupidly way too easy for RMO.
and 
.
.
that maps 
and 
to 
.
and 
,
which means that 
is cyclic. Similarly, we get that 
is also cyclic. Thus, 
This means that 
as diameter, proving Part (i).
also lies on 
,
,
in 
,
bisects 
,
.
bisects 
,
.
be the orthocenter of 
.
,
is its orthic triangle, and so by a well known property, we get that 
,
.Done 
and 
are concylic.
.
thus 
.
.
is cycli
.
is cycli
is cyclic, as desired.
.
.
is cyclic 
.
is cycli
.
bisects 
bisects 
.
,
is cyclic. Thus, since 
,
as well. Now let 
be the midpoint of 
.
is right, we have 
,
,
are collinear.
and so, 
is cyclic. Also, as 
are cyclic with 
by fact 5 we get 
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as 
has its orthocenter as 
and 
.
a homothety sending 
,
and we complete our first part.
(
is cyclic). Hence 
bisects 
and by Iran Lemma we have 
.
externally bisects 
i.e. internally bisects 
also bisects 
internally bisects 
in projective terms.
And since 
we get our desired conclusion
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, so we get 
, also we have 
, hence 
are concyclic points, hence 
are concyclic points , hence 
hence we have points 
concyclic 
and we also quadrilateral 
to be cyclic quadrilateral hence we have 
and similarly 
, by using the fact that quadrilateral 
is also cyclic , it is not hard to observe that 
hence we have 
.
are concyclic, So 
and 
.
,
.
,
(say). 
(say).
,
,
.
,
.
and 
are tangents to the incircle of 
.
(vertically opposite). Thus, by angle chasing in 
,
is cyclic. Part (i) proved.
,
is cyclic.
is cyclic 
.
is cyclic 
.
is cyclic and 
.
