are collinear, and 
are arbitrary points
be an acute triangle and 
be a point on the parallel through 
to 
such that 
.
and 
and through 
similarly. If 
are positioned around the sides of 
.
and 
are chosen on segments 
and 
such that 
.
is the midpoint of the arc 
and 
is the midpoint of the arc 
,
.
of positive integers such that ![\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]](http://latex.artofproblemsolving.com/e/c/7/ec778edf97100f3b19350d8a817317ebf1f391bd.png)
.
lie on 
are isogonal in 
.
.
lies on 
such that 
.
intersects 
passes through the Euler center of triangle 
be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
integers 
,
,
at 
is the in-centre of 
.
and 
intersect at 
,
and 
intersect at 
lie on a circle.
be the number of positive divisors of 
.
be the number of positive divisors of 
when divided by 
.
.
kids in a class. Is it true that for all positive integers 
it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
with 
.
meets 
in 
and 
intersect in 
and 
are parallel.
Y by
Let 
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
.![$[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$](http://latex.artofproblemsolving.com/3/a/0/3a0d813ceb846745ae927892e0a6bbddb6eec0bd.png)
is the incenter of 
.![$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$](http://latex.artofproblemsolving.com/4/5/6/4567bb6a0790e2821b3a2b4ba1592e30649baf97.png)
and 
belong to the 
mixtilinear incircle of 
.
is the midpoint of arc 
and 
belongs to the 
be the feet of perpendicular from 
meet the line 
.
centered at 
.
and 
.
.
and similarly 
and we have proved the lemma. 
.
at 
.
,![\[CL'^2 = CD\cdot CA = CX^2 = CL^2.\]](http://latex.artofproblemsolving.com/d/c/9/dc9e1ea07a744020713f6dc1137a2caeb0135af7.png)
Also 
lie on same side of 
lies on 
or 
.
.
is the internal angle bisector of 
.
gives 
and 
m
is the 
and 
at 
and 
respectively. We know (well-known) that 
are collinear and 
.
and we are given 
are collinear. This gives 
and 
which further gives that 
is the 
by 
.
are tangent to 
is tangent to 
.
is parallel to 
,
.
and 
.![\[\measuredangle XRR' = \measuredangle XPB = \measuredangle R'YX,\]](http://latex.artofproblemsolving.com/e/d/0/ed0b04ceb8f66d55c4dd0f96740056f508bbb1a4.png)
so 
is cyclic. Similarly, 
is cyclic as well.
,
,
.
and 
.
is tangent to 
.
and 
are tangents to 
and 
are also tangents to 
form the 
.
,
by 
.
,
and 
are tangent to 
is harmonic. This means that the tangents to 
,
and 
.
,
.
,
lies on 
(as 
and 
)
are collinear (as 
)
,
(as 
and 
)
be the tangent to 
Thus, 
,
.
such that 
so 
is tangent to 
and also 
is tangent to 
and hence by PoP we have 
and so 
.
and see that 
and so 
and hence it is the 
is 
