[/quote]
,
such that 
.
.
which satisfy the condition![\[ f(m + n) \geq f(m) + f(f(n)) - 1
\]](http://latex.artofproblemsolving.com/4/b/d/4bddce124cf0da82cb8bb6c6ae2325239354f27c.png)
Find all possible values of 
.
be a positive integer and 
be a positive integer coprime to 
.
,
,
Find, in terms of 
,
for which there exists an index 
such that 
is divisible by 
.
table are initially white. Alice and Bob play a game. First Alice paints 
rows and 
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
,
as the product of the digits of 
.
be a function satisfying the following conditions:
,
and 
,
;
is surjective.
.
and 
.
and 
respectively. The point 
such that 
is equilateral. Let 
and 
.
.
.
,
.
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
where 
for some non-negative 
such that
for every 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
Y by Adventure10, PikaPika999
Suppose that 
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
,
-
and 
are collinear; therefore, the perpendicularity of 
and 
is equivalent with the fact that 
.
(the power of 
and so 
.
.
and 
(since the circumcircle of 
is the nine-point center of 
cuts the circumcircle 
of 
and 
is a circle with center 
cuts 
is cyclic and the angle 
is right. Since 
is a diameter of 
and 
goes through 
)
is perpendicular to 
,we are done.
.
and 
have 
are concurrent, then 
is cyclic. So 
is 
to 
as 
is the diameter of 
.
;
w.r.t. 
we get (already proven 
are collinear): 
,
is cyclic, i.e. 
.
,
,
are collinear, and we are done.
is the midpoint of the arc 
,
is the 
intersects 
is the incenter of 
and 
,
and ![\[\angle B_{1}BI\\=\angle IBC+\angle B_{1}BC
\\=\angle IBC+\angle B_{1}AC
\\=\angle IBC+(\angle R-\frac{1}{2}\angle AB_{1}C)
\\=\angle R+(\angle IBC-\frac{1}{2}\angle ABC)
\\=\angle R\cdot \cdot \cdot (6)\]](http://latex.artofproblemsolving.com/6/7/7/677420e109fa1a2b2bed326bd115276ddb6cd335.png)
and 
,
and 
are collinea
,
be a point such that 
and 
,
are concyclic
and 
and 
are concyclic
, 
and 
are concyclic
and 
,
and 
are collinear
(
is a radical ceneter)
,
,
and 
,
.
.
be the line through 
perpendicular to 
.
.
are collinear on the external angle bisector of 
are collinear by the definition. Now consider the circles 
.
while 
since 
and 
since 
is a diameter, where 
is the 
.
is 
,
is 
,
is 
,
are collinear, showing 
.
,
,
has coordinates 
.
(as you don't believe check here 
be the 
excenter, By Brokard's Theorem on Quadrilateral 
and 
inversion, 
swap, so we're done because then we must have
which means that 
.
.
.
is cyclic.
,
.
is cyclic too. As 
Its consequence leads to the fact that 
.
)
the intersection of line 
with the circumcircle of 
.![\[B_2B \cdot B_2B_1 = B_2C \cdot B_2A = B_2P \cdot B_2I_B.\]](http://latex.artofproblemsolving.com/9/6/4/96421bc720fe61b02e3919130ca77c79e47d729b.png)
Thus 
are concyclic. Now by the incenter/excenter lemma, we have that 
.![\[\angle I_BPI = 90 = \angle I_BBB_1 = \angle I_BPB_1.\]](http://latex.artofproblemsolving.com/a/2/7/a27b81ccfacba67e9030325fdeee756a43c97e95.png)
Therefore 
are collinear, and 
,
.![\[BI\cdot BI_B = BB_1\cdot BB_2.\]](http://latex.artofproblemsolving.com/e/1/f/e1f805bf6fd4471a5ebaeb9de8300a2c7d69bce5.png)
Let 
be the intersections of line 
with 
.
,
.
and 
,
.![\[BB_1\cdot BB_2 = BD\cdot BM = BI\cdot BI_B\]](http://latex.artofproblemsolving.com/3/5/7/3579b2a835b31fdcea59a6196ca4462e6417e23f.png)
as desired.
.![\[\measuredangle IAI_b=\measuredangle ICI_b=90^\circ\]](http://latex.artofproblemsolving.com/d/4/b/d4b74190fff0dbce1ca7c2ecc0aefacb690144b9.png)
and thus, 
.
,
also lies on 
.
.![\[B_2I\cdot B_2B_3' = B_2B \cdot B_2B_1 = B_2A\cdot B_2C\]](http://latex.artofproblemsolving.com/d/4/0/d400d91248e047569cc5dd80d05fe0750e552886.png)
Thus, 
must also lie on 
and indeed, 
as required.
![[asy]
import olympiad; size(400);
draw(unitcircle);
pair B = dir(130), A = dir(200), C = dir(-20), M = midpoint(A--C), O = origin, B1 = intersectionpoint(O--(O+3*(O-M)),unitcircle);
pair I = incenter(A,B,C), L = intersectionpoint(B--(I+3*(I-B)),unitcircle), Ib = L + L-I;
pair exB = rotate(90,B) * I, B2 = extension(exB, B, A, C);
dot("$A$",A); dot("$B$",B); dot("$C$",C); dot("$I$",I); dot("$O$",O); dot("$M$",M); dot("$L$",L); dot("$I_B$",Ib); dot("$B_1$",B1); dot("$B_2$",B2);
draw(C--B2--B--C^^A--B--Ib^^Ib--B1--L^^B2--extension(B2,I,B1,Ib));
draw(B--B1);
// draw(incircle(A,B,C);
[/asy]](http://latex.artofproblemsolving.com/7/e/4/7e4b00a850d18aa54a1429fef14ca203eb05877b.png)
(since 
is diameter), all 
,
(from right triangle 
)
(from cyclic quad 
)
so from similar triangles 
.
,
so 
.
wrt 
and 
be 
are concyclic by incenter/excenter lemma we have that 
and because 
,
are concyclic.Then it succifies to show that 
are collinear which is trivial by the radical axis concurrence lemma on 
,
are concurrent at 
and 
.
lie on the same line, and 
