be the set of all positive integers. Find all functions 
such that 
for all 
.
be real numbers . Prove that
![$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$](http://latex.artofproblemsolving.com/9/9/9/9997397eb38cd4bceec2ff8c85951c78e2797d0f.png)
and orthocenter 
.
intersect 
respectively. Lines through 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
Y by Adventure10, Mango247, Rounak_iitr
Squares 
and 
are constructed on the sides of acute-angled triangle 
, outside of the triangle. Line 
intersects line segment 
at 
, while line 
intersects line segment 
at 
. Point 
, lying inside triangle , is an intersection of the circumcircles of triangles 
and 
. Point 
is the midpoint of 
. Prove that angle 
.
and 
are constructed on the sides of acute-angled triangle 
, outside of the triangle. Line 
intersects line segment 
at 
, while line 
intersects line segment 
at 
. Point 
, lying inside triangle , is an intersection of the circumcircles of triangles 
and 
. Point 
is the midpoint of 
. Prove that angle 
.
which is clearly the 2nd intersection of 
and 
Since 
then 
is cyclic 
is on radical axis 
of 
is the reflection of 
about 
then 
by SAS, because 
and 
Therefore, 
but since 
then 
and 
are then the altitude and circumdiameter of 
issuing from 
which in turn are isogonals WRT 
are isogonals WRT 
is cyclic, and so is 
.
,
,
and 
is the radical axis of the two circles. As 
and 
we infer that 
symmedian of 
and we are done (any point of symmedian has distances to the adjacent sides proportional to the lengths of the sides).
intersect 
at 
.
,
,
.
is the radical axis of the given two circles. Since 
,
,
is a symmedian. Thus, we are done.
.
and 
are cyclic. From 
we deduce that 
,
,
,
,
-
.
about ![\[ \{ \begin{gathered} L \to A \\ N \to N' \\ T \to T'. \end{gathered}\]](http://latex.artofproblemsolving.com/7/c/0/7c01f9f0d9c6ff882f5bfb5a40714e4d98c6baa1.png)
From midpoints, 
,
,
,
.
,
.
are isogonal: i.e. the circumcenter 
of 
is a diameter of 
,
and 
,
,
to intersect 
at 
and 
are cyclic. Indeed, for ![\[ \angle{MQK} = 180 - \angle{NCL} = \angle{ACB} = 90 - \angle{YCB} = \angle{CYB} = 180 - \angle{MYL} \]](http://latex.artofproblemsolving.com/5/6/0/560c4d56f005435c320b401de1e3ea9bc8df8729.png)
, and similar for 
and 
intersect on the 
.
be its isogonal conjugate in 
.
so 
is tangent to 
.
is also tangent to 
is a median; hence 
.
we obtain 
so 
just as we desired. 
![[asy]
size(9cm);
pair A=(0,0),B=(10,0),C=(2,7),K,L,M,Nn,X,Y,P,Ss,Q,O;
K=C*(0,1);L=C+(A-C)*(0,-1);M=B+(C-B)*(0,-1);Nn=C+(B-C)*(0,1);
X=extension(C,Nn,A,K);Y=extension(C,L,B,M);
pair[] x=intersectionpoints(circumcircle(K,X,Nn),circumcircle(L,Y,M));
P=x[1];Q=x[0];
O=(C+Q)/2;Ss=(A+B)/2;
D(MP("A",A,S)--MP("S",Ss,S)--MP("B",B,S)--MP("C",C,N)--A);
draw(circumcircle(K,X,Nn)^^circumcircle(L,Y,M)^^circumcircle(L,C,Nn),green);
D(A--MP("X",X,S)--MP("K",K,S)--MP("L",L,W)--C--MP("N",Nn,E)--MP("M",M,SE)--MP("Y",Y,S)--B,magenta);D(MP("Q",Q,N)--MP("O",O,NE)--MP("P",P,S),heavycyan+dashed);
draw(X--C--Y^^L--Q--Nn,magenta);
draw((C+L)/2--O--foot(O,C,A)--C,red+dotted);
dot(A^^B^^C^^K^^L^^M^^Nn^^X^^Y^^P^^Q^^O^^Ss);[/asy]](http://latex.artofproblemsolving.com/3/5/8/3583fd95654c494933de1afbf0b5a019e537b775.png)
.
is cyclic (since 
)
and 
,
and 
are cyclic too.
and 
,
Therefore 
.
is a diameter of 
,
is 
from the red dotted rectangle above. Similarly, 
Therefore from a well-known property of symmedians, 
So 
is actually the 
