[/quote]
,
be non-negative real numbers such that 
Prove that
,
be the orthocenter of 
.
at 
at 
.
be the midpoint of 
,
meet 
.
at 
.
is tangent to 
such that![\[ f(x)f\left(x + 4f(y)\right) = xf\left(x + 3y\right) + f(x)f(y) \]](http://latex.artofproblemsolving.com/d/b/1/db14b64aa71cb3c8dbf53707f719f5d9a43bd7e4.png)
for any positive real numbers 
.
and 
be coprime integers. Call a set 
balanced if the numbers 
form an arithmetic progression with difference 
,
form an arithmetic progression with difference 
.
and 
,
be an integer. The sequence 
is defined via 
,
and![\[ a_{n+2} = ka_{n+1} + a_n \]](http://latex.artofproblemsolving.com/b/5/e/b5e3e6f9f0b0709113498266e41b372299a2986f.png)
for any positive integer 
.
of positive integers such that 
.
be a circumscribed quadrilateral with incircle 
and no two opposite angles equal. Let 
,
and 
intersect 
and 
.
has exactly 
ordered pairs 
as solutions?
.
be positive reals such that 
.
(Shining Sun, USA)
be an acute triangle with 
,
of side 
)
.
and 
,
,
.
intersects line 
at a second point 
.
and 
are perpendicular.
such that exactly 
of the integers 
are squarefree.
be a non-constant monic polynomial with integer coefficients and let 
be an infinite sequence. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
.
cells of an 
square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell 
,
through a sequence 
of black cells where each consecutive pair 
are adjacent (sharing a common side) for every 
.
denote the maximum possible cost over all initial colorings with exactly 
such that![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](http://latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
Y by feridverdiyev, BEHZOD_UZ, tenplusten, Amir Hossein, mathleticguyyy, Adventure10, Mango247
Let 
be an isosceles triangle with 
, whose incentre is 
. Let 
be a point on the circumcircle of the triangle 
lying inside the triangle . The lines through parallel to 
and 
meet 
at 
and 
, respectively. The line through parallel to meets and at 
and 
, respectively. Prove that the lines 
and 
intersect on the circumcircle of the triangle .
Proposed by Hojoo Lee, Korea
be an isosceles triangle with 
, whose incentre is 
. Let 
be a point on the circumcircle of the triangle 
lying inside the triangle . The lines through parallel to 
and 
meet 
at 
and 
, respectively. The line through parallel to meets and at 
and 
, respectively. Prove that the lines 
and 
intersect on the circumcircle of the triangle .Proposed by Hojoo Lee, Korea
Attachments:
This post has been edited 1 time. Last edited by Amir Hossein, Aug 5, 2013, 6:55 AM
Reason: Note by Darij: This was also Problem 6 of the German pre-TST 2004, written in December 03.
Reason: Note by Darij: This was also Problem 6 of the German pre-TST 2004, written in December 03.
,
.
,
to be tangent to 
is also tangent to 
.
.
,
is to show that 
because 
.
and power 
turns the circle 
into the line 
in the invariant points 
.
.
,
,
.
,
is cyclic. similarly, 
is cyclic.
.
is cyclic. 
.
is also cyclic. 
.
.
.
.
.
where the point 
is the 
with the diameter ![$[II_{c}]$](http://latex.artofproblemsolving.com/4/d/3/4d360c2d7cf645c4c380fd8f59ddcee5827c26fd.png)
; the circumcircles 
, 
of the isosceles trapezoids 
, 
respectively ; the second intersection point 
between the line 
and the circle 
Prove easily that the lines 
From the relation 
obtain 
i.e. the quadrilaterals 
, 
are similarly as 
Thus,
and let 
, 
be two mobile points so that 
and exists a point 
for which 
Construct the points 
for which 
Ascertain the geometrical locus of the intersection 
where the point 
is the 
-
to meet the circumcircle of 
at 
.
intersect 
at 
,
meet 
.
,
is cyclic. Now extend 
,
to meet the circumcircle of 
at 
.
is cyclic; hence, 
.
at 
,
.
,
,
is parallel to 
,
is isosceles, as desired.
denote the intersection of 
and 
.
,
.
so 
is cyclic, and similarly 
is cyclic. The result then follows by direct angle chasing to show that 
,
and 
.
and 
.
and 
are similar. This follows from 
.
.
.
and 
are cyclic. Hence, 
and 
are cyclic too. Thus we have to show that circles 
,
,
have a point in common. Therefore, consider 
.
and 
.
is cyclic too. The conclusion follows.
,
is harmonic. Now 
is tangent to 
(because 
is a paralellogram). So 
harmonic, which yields 
.
,
,
is the center of 
.
.
and it's easy to verify that 
.
.
,
.
.
![[asy]
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draw((1.347581935409076,0.1927304575931286)--(1.660000000000002,3.580000000000004));
draw(circle((2.368300313660502,1.806628929897715), 1.909590083396774));
draw((1.438901086131840,1.182822302271518)--(2.291169545721799,-0.1014028128268362));
draw((2.291169545721799,-0.1014028128268362)--(3.471717322567828,1.994747342753156));
draw((1.438901086131840,1.182822302271518)--(3.471717322567828,1.994747342753156));
draw((2.030481395477170,1.419104784674090)--(2.778764072909347,0.7643574419209349));
draw(circle((2.002030691392678,0.6316259215254196), 0.7879926411260609));
draw(circle((3.105051331405495,0.8207367229440963), 1.229936943302180));
draw((2.030481395477170,1.419104784674090)--(1.939162244754406,0.4290129399956997));
/* dots and labels */
dot((1.660000000000002,3.580000000000004),dotstyle);
label("$C$", (1.718017351869851,3.665804292213880), NE * labelscalefactor);
dot((4.220000000000004,1.340000000000002),dotstyle);
label("$B$", (4.273985312412097,1.416019993611593), NE * labelscalefactor);
dot((1.347581935409076,0.1927304575931286),dotstyle);
label("$A$", (1.398521356802070,0.2711593446187132), NE * labelscalefactor);
dot((2.030481395477170,1.419104784674090),dotstyle);
label("$P$", (2.077450346321104,1.495893992378538), NE * labelscalefactor);
dot((1.939162244754406,0.4290129399956997),dotstyle);
label("$D$", (1.997576347554159,0.5107813409195485), NE * labelscalefactor);
dot((2.778764072909347,0.7643574419209349),dotstyle);
label("$E$", (2.836253334607084,0.8435896691151531), NE * labelscalefactor);
dot((1.438901086131840,1.182822302271518),dotstyle);
label("$F$", (1.491707688696840,1.256271996077703), NE * labelscalefactor);
dot((3.471717322567828,1.994747342753156),dotstyle);
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dot((2.291169545721799,-0.1014028128268362),dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/f/4/1/f411f0672b5772972e351f5d6cda57dce9971a02.png)
so 
and 
.
so 
is cyclic. Similarly, 
is also cyclic. Let 
and 
so ![\[\angle AEX + \angle BDX = 180^{\circ} - \theta \implies \angle DXE = \theta = \angle FXE,\]](http://latex.artofproblemsolving.com/6/9/4/694c56154046571699259a205eb3be9ddfea102c.png)
are collinear. Similarly, 
are also collinear. It hence suffices to show that ![\[\angle AXB = \angle AXP + \angle BXP = \theta + \theta = 180^{\circ} - \angle ACB,\]](http://latex.artofproblemsolving.com/4/4/3/443ac421d8c0f13462a5d31e24a6f939c7f5592a.png)
.
so, by p.o.p. 
,
and the two parallelograms 
are similar, 
,
is cyclic; since 
is an isosceles trapezoid, 
is cyclic as well and 
.
is cyclic and 
.
and 
side by side we get 
,
are collinear.
and 
are isosceles trapezoids, hence they are cyclic. Moreover, since 
,
,
hence 
so that 
are collinear. By symmetry, 
are collinear, so the result is obtaine
where 
.
.
we have 
.
.
and 
then 
in 
So 
.
.
.
which is true since 
.
,
.
is a parallelogram, 
passes through the midpoint 
of 
.
and 
;
.
and 
,
is cyclic, so 
.
,
is cyclic, so 
.
true since 
is cyclic. Thus 
intersect on 
intersect on 
we find 
are concurrent.
and 
are collinear.
where we use parallel lines.
is cyclic
is cyclic
and 
is isosceles cyclic trapezoid
is parallelogram 
and 
is isosceles cyclic triangle
we find 
is cyclic and from 
we have 
is cyclic.
is cyclic.As desired.
,
lying inside the triangle ![[asy]
unitsize(100);
pair A, B, C, M, P, D, E, F, G, K;
A = dir(90); M = dir(270); B = dir(230); C = reflect(A, M) * B;
P = M + abs(M - B) * dir(100);
K = 2 * foot(origin, A, P) - A;
D = extension(P, P + C - A, B, C);
E = extension(P, P + B - A, B, C);
F = extension(P, P + C - B, A, C);
G = extension(P, P + B - C, A, B);
draw(F--G, gray(0.5));
draw(P--D^^P--E, gray(0.5));
draw(A--K, gray(0.5));
draw(F--K^^G--K, gray(0.5) + dashed);
draw(B--P^^C--P, gray(0.5));
draw(A--B--C--cycle);
draw(arc(M, C, B));
draw(circumcircle(B, P, K)^^circumcircle(C, P, K));
draw(circumcircle(B, E, K)^^circumcircle(C, D, K), dotted);
draw(unitcircle);
dot(A^^B^^C^^P^^D^^E^^F^^G^^K);
label("$A$", A, dir(A));
label("$B$", B, dir(200));
label("$C$", C, dir(340));
label("$P$", P, dir(90));
label("$D$", D, dir(290));
label("$E$", E, dir(230));
label("$F$", F, dir(40));
label("$G$", G, dir(140));
label("$K$", K, dir(K));
[/asy]](http://latex.artofproblemsolving.com/a/c/f/acf45e5b89b545cde48d385cc25227a6c2860297.png)
again at 
and 
are tangent to 
.
,
,
,
is tangent to 
.
is tangent to 
.
,
.
bisects 
is a parallelogram, 
also bisects 
followed by reflection in the angle bisector of 
.
):-
be the 
be its circumcircle. Suppose the tangent to 
and 
at 
and 
at 
meet at a point.
.
Also 
lies on the perpendicular bisector of 
,
are collinear. And, 
,
.
.
and 
,
is a parallelogram. Thus, 
is cyclic. Similarly, 
is cyclic. Let 
is cyclic. 
tangent to 
,
and similarly, 
are tangents to 
,
,
and 
are cyclic. So we see that pentagon 
is cyclic too. Thus we have 
. Now note that 
.
.
concur at a point, call it 
,
.
,
,
where 
,
since
is cyclic. Therefore, it suffices to show given 
,
is cyclic. I got stuck here, so if anyone has any ideas, that would be awesome.
,
.
is cyclic.
.
lie on 
and it is well-known that centre of 
Hence, we have 
since 
and 
.
.
to 
.
and 
all concur and since 
and 
,
intersect on the 
