[/quote]
,
such that for all 
,
.
such that, for all integers 
that satisfy 
,![\[f(a)^2+f(b)^2+f(c)^2=2f(a)f(b)+2f(b)f(c)+2f(c)f(a).\]](http://latex.artofproblemsolving.com/a/9/e/a9edf1a3207d29facc970b6ee288d4fabdd43a4b.png)
denotes the set of integers.)
be a 
board. Some of its cells are colored black in such a way that every 
board of 
black cells. Find the maximum amount of black cells that the board may have.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
,
be the 
-
and 
are placed on line 
such that 
is between 
,
and 
,
be the 
,
be the reflection of the orthocenter of triangle 
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
,
,
be the incenter, 
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
Y by Armanb, Adventure10, Mango247, and 1 other user
Let 
be a triangle, and let the tangent to the circumcircle of the triangle at 
meet the line 
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. Prove that the points , and are collinear.
Valentin Vornicu
be a triangle, and let the tangent to the circumcircle of the triangle at 
meet the line 
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. The perpendicular to at 
meets the perpendicular bisector of 
at 
. Prove that the points , and are collinear.Valentin Vornicu
This post has been edited 2 times. Last edited by Valentin Vornicu, Oct 24, 2005, 12:54 AM
(I don't know what Thales' theorem is, and I didn't use the intersection of EF and BC) - my solution doesn't quite reach half a page.
Let be the points 
and 
tangent to the circumcircle 
of 
at points 
respectively.
where 
is the circumcenter of 
and 
and 
intersect the pencils 
respectively, we have that 
and 
we conclude that the line segments 
are concurrent at one point, here the point 
and the proof is completed.
are points on the mid-perpendiculars of the side-segments 
respectively, such that 
and in this general case also hold, the proofs presented by 
,
the circumradius.
,
,
,
are collinear.
,
and 
,
,
meet the line 
at 
meets the perpendicular bisector of 
at 
.
meets the perpendicular bisector of 
at 
.
.
be the orthic triangle. Note 
so if 
then 
where 
is the orthocentre of 
,
.
to be the antipodes of 
respectively. Apply pascals theorem on hexagons 
and 
,
are collinear and 
are collinear. Its suffice to show 
are collinear.Consider the circles through 
and 
,
and have diametres 
.
,
.
about 
and 
such that 
.
be the anti-complementary triangle of 
goes to 
,
so 
but trivially 
are collinear.
through 
,
.
meet 
is a harmonic quadrilateral, so the tangents meet at 
and the conclusion follows (Co-axial circles).
be tangent to the circle,with 
and 
where 
the perpendicular bisector of 
.
.
.
thus 
.
gives 
.
is cyclic.This gives 
.
.
.
be the orthogonal projections of 
respectively.
is cyclic.
.
.
.
be the orthogonal projection of 
upon 
(attachment on the top)
.
we get that the quadrilaterals 
are cyclic.
.
is the Simson line 
of 
and 
thus 
which proves the collineartiy.
and factor 
,
and 
are collinear .
![\[\frac{FC}{EB}=\frac{CD}{BD}=\frac{CB+BD}{BD}=1+\frac{CB}{BD}\]](http://latex.artofproblemsolving.com/5/4/b/54b0d714e25f665d773bcaae3cf56024d8018b7a.png)
Let's say that 
,
.
And again By Sine Law in triangle 
we have that ![\[\boxed{BD=\frac{2R\sin^2\alpha}{\sin{\beta-\alpha}}}\]](http://latex.artofproblemsolving.com/2/7/9/279729571251058a1cac252c2d07cbca2f018672.png)
And we know 
it remain to find 
and 
.
and 
we have that ![\[\boxed{EB=\frac{R\cdot \sin\alpha}{\sin \beta} \ , \ \ CF=\frac{R\cdot\sin\beta}{\sin\alpha}}\]](http://latex.artofproblemsolving.com/3/c/4/3c45e2fdef6621231d61e6b5446244da42e9019b.png)
And we must prove that ![\[\frac{FC}{EB}=1+\frac{CB}{BD}\to \frac{CB}{BD}=\frac{FC}{EB}-1\longrightarrow \frac{\sin(\alpha+\beta)\cdot\sin(\beta-\alpha)}{\sin^2\alpha}=\frac{\sin^2\beta-sin^2\alpha}{sin^2\alpha}\longrightarrow \sin(\alpha+\beta)\cdot\sin(\beta-\alpha)=\sin^2\beta-sin^2\alpha \]](http://latex.artofproblemsolving.com/1/6/c/16c48ef67268b94395b9bb68900f3a40478f0603.png)
And The lat part is 
Let's mark 
with 
and 
with 
is center of the circle)
we have 
(where 
is radius of the circle) 
we get 
we have 
so the points 
are collinear.
are tangent to the circle 
with diameter 
and 
is the radical axis of 
.
and 
be the circles centered and 
and passing through 
respectively. Both 
be their exsimilicenter. i.e. the intersection of 
.
be the second intersection of 
.
because of the homothety at 
Hence 
is tangent to 
at 
.
,
![$AD\cap \odot (E,EB) = [A,L]$](http://latex.artofproblemsolving.com/9/8/b/98b809132e509b6b6b977ca585cf8be239cd4ebd.png)
.
.
.
.
from bc being a chord 
from OEꞱAB you get 
from BEꞱCB you get 
combining them you get 
you get F by this way aswell because it's symmetric. Now we just have to prove collinearity we just need do calculate d-e and d-f for d-e you get 
for d-f you get the same but just switch b with c so plugging everything in you just need to prove that 
which isn't hard to prove 
