[/quote]
,
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 
and 
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
,
,
,
and 
.
to meet the circumcircle of triangle 
at 
.
.
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 
.
such that![\[2^{n!}+1\mid 2^{m!}+19\]](http://latex.artofproblemsolving.com/8/7/e/87ede5616cffcb33aa187f5d8a13827470f741b4.png)
,
and a positive integer 
is given.
and for each 
![\[f_n(k)=\frac{a_n+\dots+a_{k+1}}{n-k}-\frac{a_k+\dots+a_1}{k}\]](http://latex.artofproblemsolving.com/0/4/1/041f80cf63f17c0f62c166021af474a5d305eb87.png)
Then for every integer 
the condition![\[a_{n+1}=\max_{0\leq k<n}(f_n(k))\]](http://latex.artofproblemsolving.com/b/1/f/b1f5aff787de8fc9c113cf8441a6b2900a19f24f.png)
is satisfied. Prove that this sequence must be eventually constant.
be a positive integer. For every positive integer 
the sequence 
is defined, where 
are integers. Among these 
.
and 
and 
respectively such that 
.
and 
meet at 
.
and 
is the orthocenter of 
.
and 
are circles with diameter 
and 
,
.
with 
with all the unit squares are initially white coloured. Every move, we can turn the color (from white to black or otherwise) from the 5 unit squares that form this T-pentomino which can be rotated or reflexed (see the image below). Determine all natural numbers 
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 
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 
to 
.
and 
,
which means that 
is cyclic. Similarly, we get that 
is also cyclic. Thus, 
This means that 
as diameter, proving Part (i).
,
,
in 
,
bisects 
,
.
bisects 
,
.
.
,
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 
.
