[/quote]
,
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
=
be an altitude of that triangle and points 
and 
be the midpoints of 
is the circumcenter of the triangle 
,
and 
,
)
be the anticomplement of 
be the 
-
is parallel to the real asymptote of the circular pivotal isocubic with pivot 
and 
.
such that 
.
is a power of a prime number, but for arbitrary 
and, for each 
,
be the unique integer in the range 
for which 
is divisible by 
.
the obtained sequence is 
.
eventually becomes constant.
lie on sides 
,
.
meet at orthocenter 
.
such that 
and 
.
and 
intersect at point 
.
.
.
.
over 
and 
and 
lies on 
.
such that 
prove that the triangle must have a right angle.
such that for any real numbers 
and 
,![\[
f\left(x^2+2y f(x)\right) + (f(y))^2 \leq (f(x+y))^2
\]](http://latex.artofproblemsolving.com/8/4/3/8437449cd3f8f0da24d17a60b55dcb563e6b5655.png)
defined by 
and![\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/9/9/4/994aa754cc1288ce4f28a95a0276e64282fb5f66.png)
a) Prove that there exist infinitely many positive integers 
such that 
.![\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]](http://latex.artofproblemsolving.com/f/f/1/ff174e7431cbfbc17c650d109651241286756a1a.png)
is a cyclic quadrilateral. Let 
and 
.
and 
the orthocenters of triangles 
and 
,
,
be a triangle. A circle passing through 
and 
intersects the sides 
and 
again at 
and 
respectively. Prove that 
,
and 
are concurrent, where 
and 
are the orthocentres of triangles 
respectively.
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 
and 
are the circumcenter and the symmedian point of 
bisects segment 
and 
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.
