[/quote]
,
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
such that for any integer 
there exists an integer 
for which 
is divisible by 
are distinct prime numbers which satisfy
for natural number 
.
and 
of the circumscribed circle of 
.
meets the circumcircle 
.
:
be reals such that 
and 
Prove that
Where 
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
such that 
that satisfy the equation
.
and 
,
and 
be the incenter of triangles 
and 
,
.
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
Y by rmtf1111, Math_tricks, miguell, Adventure10, Mango247
Quadrilateral 
is inscribed in circle 
with center 
. Point 
is the incenter of triangle 
, and point 
is the incenter of the triangle 
. Line 
intersects segments 
at points 
and, respectively 
. The perpendicular from 
to line 
intersects the perpendicular from 
to line 
at point 
. The perpendicular from 
to line 
intersects the perpendicular from to line 
at point 
. Prove that 
are colinear.
is inscribed in circle 
with center 
. Point 
is the incenter of triangle 
, and point 
is the incenter of the triangle 
. Line 
intersects segments 
at points 
and, respectively 
. The perpendicular from 
to line 
intersects the perpendicular from 
to line 
at point 
. The perpendicular from 
to line 
intersects the perpendicular from to line 
at point 
. Prove that 
are colinear.This post has been edited 1 time. Last edited by XxProblemDestroyer1337xX, Mar 8, 2019, 4:47 PM
is our complete quadrilateral with Miquel point 
and 
the intersection of diagonals. Drop the incircles and only use that 
.
and let 
be the point at infinity on the external bisector of 
.
is projective because 
.
and 
are projective. Note that 
is projective, because it is a rotation by 
about 
and the bisector of 
is projective, analogously 
is projective, thus 
is projective. It is enough to check the assertion of the problem for three positions of 
be a point on 
,
,
is cyclic by Miquel, following that 
,
and 
in this case. Analogously we do the case 
.
,
is the point at infinity on 
are parallel, we have that 
.
the circle that touches 
where 
is the intersection of the diagonals. By one consequence Sawayama-Thebault for 
and 
passes through 
and 
are those points and 
without 
,
.
lies "outside"). So we only need to prove that 
are concyclic, so 
,
are also concyclic. But the same Sawayama for 
are concyclic where 
since both of them lie on the same arc 
internally, 
are concyclic;
lies on 
a line 
cuts 
at 
resp
cuts 
again at 
are concurrent
are reciprocal pair, so 
are concurrent
so 
are concurrent at 
are concurrent at 
moves is 
cuts 
at 
resp
cuts 
then 
are concurrent on 
be the midpoint of arc 
of 
are concurrent at 
to obtain that the circle centered at 
is simply the circle tangent to 
and 
)
be the point where the circle centered at 
Let 
be the midpoints of arcs 
not containing 
we know that 
are collinear, where 
e
It would suffice to show that 
which would imply that 
and hence the result. However, this is easy when applying the Japanese Rectangle Theorem and also applying Fact 5 in 
:
at 
; 
:
at 
By Sawayama we get 
whence 
are the centers of 
so by the lemma from 
hence 
are collinear.
is tangent to 
and to 
in 
and 
is tangent to 
.
are concurrent and 
are also concurrent. Thus 
and the problem is clear.
are colinear.
