[/quote]
,
be a positive integer and let 
be odd primes. Prove that the positive integers 
can be colored in 
colors, such that for any 
of the same color, 
is not divisible by 
and 
.
be its circumcenter and 
its incenter.
be the isogonal conjugate of 
be the Gergonne point of the triangle 
is parallel with line 
.
be a polynomial with real coefficients such that 
for all 
.
for some real constant 
and 
.
= 2∠AMP.
.
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
,
such that 
and 
are both divisible by 
over the positive integers as follows:
where 
such that 
.
such that 
and 
is a perfect square.
or 
.
and 
and 
operated by two different companies, then there must exist flight route 
operated by the third company different from 
,
and 
denote the number of flight routes operated by companies 
.
such that![\[f(x^{n+1}+y^{n+1})=x^nf(x)+y^nf(y),\forall x,y>1.\]](http://latex.artofproblemsolving.com/a/4/7/a47d53806f2148cc3339335fb990b33641085896.png)
be a fixed integer. Find the least constant ![\[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C
\left(\sum_{i}x_{i} \right)^4\]](http://latex.artofproblemsolving.com/d/d/e/dde2d1dd0c412d8ba640f1d4c044e2fa2dcccc82.png)
(the sum on the left consists of 
summands). For this constant 
,
be the least prime that does not divide 
,
to be the product of all primes less than 
For 
,
.
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
.
.
such that 
.
be a cyclic quadrilateral. Let 
and 
and 
meet at 
and 
.
denote the circumcircle of triangle 
the midpoint of 
.
,
.
and 
similarly.
Y by Adventure10
Let 
be the circumcircle of acute triangle 
. Two tangents of from 
and 
intersect at 
, 
and 
intersect at 
. Point 
, 
are on 
and 
such that 
and 
.
(1) Prove that
are concyclic.
(2) Denote
the centre of the circle passing through . 
, 
are difined similarly. Prove that 
, 
, 
are concurrent.
be the circumcircle of acute triangle 
. Two tangents of from 
and 
intersect at 
, 
and 
intersect at 
. Point 
, 
are on 
and 
such that 
and 
.(1) Prove that
are concyclic.(2) Denote
the centre of the circle passing through . 
, 
are difined similarly. Prove that 
, 
, 
are concurrent.
Let O be the triangle circumcenter and K the symmedian point.
wich means that the points B, C, E, F are concyclic.
of the A-symmedian 
meet the rays (AB, (AC at B', C'. The triangles 
are centrally similar with the similarity center A and D is the symmedian point of the triangle 
It immediately follows that the circumcircle 
of the quadrilateral BCEF is the 1st Lemoine circle of the triangle 
centered at the midpoint X' of the segment DO', where O' is the circumcenter of this triangle. Therefore, 
also at its midpoint X, the center of the 1st Lemoine circle of the original triangle. Simiarly, 
cut KO at its midpoint X, hence all three are concurrent at X.
), but it is much neater than I had originally expected it to be, which is the reason I decided to mention it anyway.
and 
.
is the symmedian, 
(as the symmedian is the reflection of the median over the angle bisector).
etc. (we will use these in part 
as well):
and concyclicity follows.
be 
.
.
and 
we get, using equations 
and 
above,
and 
![$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2}AB. AA_1 sin \angle BAA_1}{\frac{1}{2}AC.AA_1 sin \angle CAA_1} = \frac{c}{b}.\frac{sin \angle BAA_1}{sin \angle CAA_1} ...(6)$](http://latex.artofproblemsolving.com/b/0/1/b0163de7913c73eaceb55852b1f3f5f123d0ac8f.png)
![$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2} AB. BA_1 sin \angle ABA_1}{\frac{1}{2} AC.CA_1 sin \angle ACA_1} = \frac{c}{b}.\frac{sin(90-\alpha)}{sin(90-\alpha-B+C)} = \frac{c}{b}.\frac{cos \alpha}{cos (\alpha-B+C)} ...(7)$](http://latex.artofproblemsolving.com/7/1/2/71299edb5b2bfe3bf3ad60d4ccece1c97fc07d6b.png)
and 
together imply
we get,
(on 
duh 
) and 
in relations 
and 
to finish the proof by the trigonometric form of Ceva's theorem.
as follows. Let the line passing through the midpoint of 
which is antiparallel to 
intersect 
cyclic. Since 
-
the midpoint of 
)
is a parallelogram, so 
.
and 
.
.
.
gives us 
by symmetry. Therefore the corresponding 
is the same for all 3 vertices.
to 
.
.
is a parallelogram by definition, 
bisects 
.
s
,
where 
are the circumcenter and the symmedian point of 
bisects segment 
and 
denote the midpoints of segments 
be the intersection of lines 
and 
.
denote the intersection of the 
.
is cyclic due to the right angles. Let 
be the intersection of the ![\[-1=(EF;M\infty)\overset{A}{=}(BC;DY)\]](http://latex.artofproblemsolving.com/f/c/a/fca819a3d4f7d865596b0655bd3f535392a56586.png)
Thus,![\[DY \cdot DM_a = DB \cdot DC \]](http://latex.artofproblemsolving.com/3/8/d/38ddab8e5cb79318d5c744c9ddef70f618630706.png)
Further, from the midpoint theorem it follows that 
.![\[DM \cdot DK_a = \frac{DA\cdot DK_a}{2} = \frac{DB\cdot DC}{2} = \frac{DY \cdot DM_a}{2} = DX \cdot DM_a\]](http://latex.artofproblemsolving.com/8/b/7/8b7fcf4b245037394ec631b721c45921e6d5d044.png)
which implies that 
is also cyclic. Putting these observations together proves the claim.
and 
are parallel.
.![\[PA_1 \cdot PM_a = PK_a \cdot PM\]](http://latex.artofproblemsolving.com/b/4/e/b4e88483ac85c29908ca1526acc2a7968cb516fb.png)
Also,![\[PM_a \cdot PO = PB^2\]](http://latex.artofproblemsolving.com/4/4/4/4441caaea44d29a39ba47b7c3b6d8131f20f0282.png)
This then implies,![\[\frac{PA_1}{PO} = \frac{PK_a \cdot PM}{PB^2} = \frac{PM}{PA}\]](http://latex.artofproblemsolving.com/e/d/8/ed807ae81a2cf375282ded308ad1c4348323547e.png)
Now, let 
denote the intersection of the 
s![\[-1=(AB;CK_a)\overset{C}{=}(AD;PK)\]](http://latex.artofproblemsolving.com/7/c/d/7cdaeae6ef54a133b0756937a5d32e3927340006.png)
Thus,![\[PD \cdot PA = PK \cdot PM\]](http://latex.artofproblemsolving.com/0/9/9/0998011c586cc303f1eba11ae0bdc43eec652082.png)
Thus,![\[\frac{PA_1}{PO} = \frac{PM}{PA}=\frac{PD}{PK}\]](http://latex.artofproblemsolving.com/5/4/7/547d60031d7decc4910020af52f6cfcee2338415.png)
which implies that 
as claimed.
we have,![\[(OK;X\infty)\overset{A_1}{=}(PK;AD)=-1\]](http://latex.artofproblemsolving.com/2/7/3/27306f8bf59d168678246233bea30b9993d636bb.png)
which implies that 
and 
.
and 
are cyclic. So by PoP we have![\[FD \cdot DF' = BD \cdot DC = ED \cdot DE'\]](http://latex.artofproblemsolving.com/d/2/f/d2f6f4257e90a5d735d713174908dc3fe69ef938.png)
so the points 
are cyclic.
passes through the midpoint of 
,![[asy]
import geometry;
import olympiad;
size(9cm);
filldraw(unitcircle, purple+white+white, blue);
pair A = dir(110); pair B = dir(225); pair C = dir(315);
pair O = (0,0);
pair M = B / 2+ C / 2;
pair P = extension(B, B+rotate(90)*(B-O),O,M);
pair D = extension(A,P,B,C);
pair E = intersectionpoint(parallel(D,line(A,B)),line(A,C));
pair Ep = extension(E,D,B,P);
pair F = intersectionpoint(parallel(D,line(A,C)),line(A,B));
pair Fp = extension(F,D,C,P);
circle BFEC = circle(B,F,E);
pair A_1 = circumcenter(B,F,E);
pair N = B / 2 + Ep / 2;
pair Q = intersectionpoint(parallel(B,line(N,D)), line(A,P));
draw(A -- B -- C -- cycle);
draw(line(P, false, B));
draw(line(P, false, C));
draw(E -- Ep);
draw(F -- Fp);
draw(O -- P);
draw(Q -- O, darkblue+1);
draw(D -- A_1, darkblue+1);
draw(B -- Q, darkblue+1);
draw(N -- D, darkblue+1);
draw(A -- P);
draw(circumcircle(B,F,E), red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$D$", D, dir(D));
dot("$Q$", Q, NW);
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$E'$", Ep, dir(Ep));
dot("$F'$", Fp, dir(Fp));
dot("$O$", O, NW);
dot("$A_1$", A_1, SE);
dot("$N$", N, dir(N));
[/asy]](http://latex.artofproblemsolving.com/4/1/4/41494d502488bc45959cd86b86b462b6fe29a4b7.png)
.
and 
.![\[\frac{PA_1}{PO} = \frac{PM}{PB} = \frac{PD}{PQ}\]](http://latex.artofproblemsolving.com/c/3/d/c3db297e3c7d53bffe0a5ae2fe7a2447971d5b0f.png)
which implies that 
.![\[-1 = (A,D;Q,P) \overset{A_1}{=} (AA_1 \cap QO, QO_{\infty}; Q, O)\]](http://latex.artofproblemsolving.com/5/f/b/5fbad499f45205692bf2b43866940f8f9c081b71.png)
implies that 
is the midpoint of 
as needed.
Let 
denote the Lemoine point(intersection of symmedians).
be the intersections of 
with 
so 
so 
are collinear.
are collinear.
and 
Thus, the 
Hence, 
to 
Projection through 
finishes.
