be real numbers such that 
and 
Find the value of 
and 
Find the value of 
?
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，
is a regular polygon with 
diagonals. Randomly select a triangle whose three vertices are the three vertices of polygon 
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 
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 
.
.
then 
which has orthic triangle 
and by Incenter-orthocenter duality we are done.
and 
is cyclic since 
and similarly 
.
and 
implies 
and analogously 
implies that 
