[/quote]
,
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 
to the side 
and 
,
,
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 Davi-8191, mathematicsy, jdong2006, Jufri, Adventure10, Mango247
Let 
be an acute triangle and 
is orthocenter. Let 
be the intersection of 
and 
and 
be the intersection of 
and 
. The circumcircle of 
cuts the circumcircle of at 
. Prove that the angle bisectors of 
and 
concur at a point on 
be an acute triangle and 
is orthocenter. Let 
be the intersection of 
and 
and 
be the intersection of 
and 
. The circumcircle of 
cuts the circumcircle of at 
. Prove that the angle bisectors of 
and 
concur at a point on 
of 
of 
is the antipode of 
If 
cuts 
again at 
then 
i.e. 
Since 
are tangents of 
at 
then 
is harmonic 
but 
and 
are similar, due to 
and 
and 
be the points of intersection of the angle bisectors of 
is the Miquel point of quadrilateral 
,
,![\[ \frac{BP_1}{P_1 C} = \frac{FB}{FC} = \frac{EB}{DC} = \frac{\frac{EB}{BC}}{\frac{DC}{BC}} = \frac{\sin \angle ECB}{\sin \angle DBC} = \frac{\sin \angle HCB}{\sin \angle HBC} = \frac{HB}{HC} = \frac{BP_2}{P_2C}, \]](http://latex.artofproblemsolving.com/e/f/8/ef800585553eab7c68649477c8594e2beb25a5c7.png)
,
![\[ -\frac{IC}{IB} = -\frac{DA}{CD}\cdot\frac{EB}{AE} \]](http://latex.artofproblemsolving.com/a/3/c/a3c7ecf2e07355dd762158c6403507fc8e4a5745.png)
![\[ \frac{QB}{QC} = -\frac{DA}{CD}\cdot\frac{EB}{AE} \]](http://latex.artofproblemsolving.com/3/d/c/3dcad594f3c94cb541f10008665db9c99f677754.png)
,
(1)
hence we have ![\[ \angle CFD+ \angle EFC = \angle EFD = \angle BFC = \angle EFC + \angle BFE\]](http://latex.artofproblemsolving.com/9/8/8/988e567a5c1efad3129ae758e27b302326ff36d0.png)
and therefore 
(2)
we have 
and therefore ![\[ \frac{BF}{FC} =\frac{BE}{CD} =\frac{BH}{CH}\]](http://latex.artofproblemsolving.com/7/7/0/770ac2095a5d1c6ebcce561b2877d1307737fc87.png)
and we are done.
are respectively the orthocenter of 
,
are collinear and 
is a parallelogram.
have equal areas; as 
,
;
,
,
.
.
.
are collinear.Now 
and
.
.
.
,
.
.
is simply the circle with diameter 
.
be the midpoint of 
by radical axis, where 
is the circumcenter of 
.
and 
.
or ![\[ (f-b)(\bar f - \bar b)(a+b)(\bar a + \bar b) = (f-c)(\bar f - \bar c)(a+c)(\bar a + \bar c) \]](http://latex.artofproblemsolving.com/1/5/3/15335b41645a046bfc9c5ce5deca2a41537d9ab5.png)
;
and 
.
Now the fact that 
can be rearranged to ![\[ b-1/b = 1/c - c + (2/a-2a) \]](http://latex.artofproblemsolving.com/a/0/5/a05aa1bd0a1f8b11b80ba132d5e2bfb97000a7b8.png)
and also ![\[ b^2+1/b^2 = (c^2+1/c^2) + (4/a^2+4a^2-8) + (4/a-4a)(1/c - c) = (c^2+1/c^2) + (4/a^2+4a^2-8) - (4c-4/c)(1/a) - (4/c-4c)(a) \]](http://latex.artofproblemsolving.com/7/d/e/7de4fc3224af67626eeeb5ec71dea999dcd3dee1.png)
and 
and 
,
.
and 
we have
.
.
.
concur at the harmonic conjugate of 
wrt 
is an harmonic quadrilateral, and clearly the angle bisectors of 
will meet at 
meets the angle bisector of 
so we have 
so we are finished.
be the circumcircle of 
,
be the circumcircle of 
,
be the circumcircle of 
(cyclic because 
)
,
and 
intersect on 
.
be the foot of the 
.
,![\[\left(B,C;F,H'\right)=\left(B,C;J,K\right)=-1\]](http://latex.artofproblemsolving.com/5/5/c/55ca3a892122ba7af04fcdd6fea66b1bcaf2cc98.png)
by projecting from ![\[\left|\frac{BH}{HC}\right|=\left|\frac{BH'}{H'C}\right|=\left|\frac{BF}{FC}\right|.\]](http://latex.artofproblemsolving.com/7/6/c/76c96e1f04f34cf276820104d34407c19f0e003d.png)
By the angle bisector theorem, this implies that the angle bisectors of 
and 
concur at a point on 
so we're done by the Angle Bisector Theorem on triangle 
and 
.
by the angle bisector theorem.
and 
.
is a kite, we have 
,
is harmonic. By Radical Center on 
,
,
,
,
.
by a well known property of concurrent cevians (in this case applied to the altitudes of 
Since 
is well-known to pass through 
and 
are tangents to 
,
is harmonic and so is 
,
,
.
,
,
,
,
concur at 
,
are cyclic, a spiral similarity exists sending 
is harmonic, so 
is harmonic as well and we have 
and 
and 
Let this point of concurrency be 
Let the foot of the altitude from 
We see that 
Taking perspectivity from 
we see that 
where 
implying that 
as desired.
be the reflection of 
be the midpoints of the minor and major arcs 
,
,
and 
,
and vice versa).
.
and 
(we can reflect the diagram otherwise).
.
.
:
,
.
denote the reflections of 
are collinear from 
line configuration. Now note that it just suffices to show 
and as 
.
and now projecting through ![\[-1=(B,C;M,\infty_{BC})\overset{A'}{=}(B,C;F,H'),\]](http://latex.artofproblemsolving.com/f/a/3/fa3bf74ead5115f88187add5c6f0150d12ac0021.png)
and we are done! 
,
to 
.
and 
as required.
passing through 
is cyclic.
.
,
.
is harmonic, as then 
,![\[-1 = (AH \cap BC, AF \cap BC; B, C) \stackrel{A, \gamma}{\longrightarrow} (H', F; B, C),\]](http://latex.artofproblemsolving.com/6/b/6/6b6aab18dc99d1f7636f84297b9e66586df7b705.png)
as desired. 
by angle-bisector. Let 
.
is harmonic.
,
.
be the reflection of the midpoint of major arc BAC over 
.
,
is cyclic. Using radical axis theorem on 
,
,
,
,
because of the angle bisector theorem -- they're supposed to equal 
,
.
means it suffices to prove 
is cyclic -- so 
.
with 
so we're done!
be the reflection of 
concur. We have 
so 
is harmonic, which gives us 
and we are done by Angle Bisector Theorem.
.
.
,
and 
be the midpoints of minor and major arc 
respectively. Let 
and 
.
meet on 
. Since 
.
.
passes through 
.
.
and 
,
and we are done. 
