[/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 foot of the altitude from 
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, El_Ectric, Adventure10, Rounak_iitr, GeoKing, Tastymooncake2, ItsBesi, and 8 other users
Consider five points 
, 
, 
, 
and 
such that 
is a parallelogram and 
is a cyclic quadrilateral. Let 
be a line passing through . Suppose that intersects the interior of the segment 
at 
and intersects line 
at 
. Suppose also that 
. Prove that is the bisector of angle 
.
Author: Charles Leytem, Luxembourg
, 
, 
, 
and 
such that 
is a parallelogram and 
is a cyclic quadrilateral. Let 
be a line passing through . Suppose that intersects the interior of the segment 
at 
and intersects line 
at 
. Suppose also that 
. Prove that is the bisector of angle 
.Author: Charles Leytem, Luxembourg
,
and 
be the midpoints of 
,
and 
respectively. Since 
also passes through 
,
;
.
.
be the foot of perpendicular from 
,
and 
(angles in the same segment), 
is similar to 
.
.
and the Simson line 
are related by a homothety, so they are parallel. Therefore the angle between 
is equal to the angle between 
and 
.
at 
such that 
.
be the midpoints of 
resp
similarly 
but 
But 
(
we get 
is midline of 
triangle 
,
.
)
,
.
is isosceles.
all lie on the circle with center 
.
be the intersection between this circle and the circumcircle 
of 
is a trapezoid and 
.
,
and by LAL 
.
be the second intersection of 
and 
,
be the second intersection of 
with 
.
.
is cyclic.
bisects 
,
and 
,
and 
are congruent, hence 
.
is cyclic and 
and 
implies that 
is the incenter of triangle 
.
and we are done.
is the bisector, then the center of 
lies on 
.
is the arc midpoint of 
.
be the point on 
such that 
is an isosceles trapezoid, which by symmetry gives 
.
.
hence 
.
hence 
bisects 
.
,
.
(where the second equality comes from 
)
is the bisector of 
,
.
,
.
,
intersecting 
at 
and 
at 
,
of 
was on 
,
strictly increases.
,
of 
(from 
)
lies on the opposite side of 
,
than 
,
,
,
.
,
.
,
,
,
collinear. Hence they are collinear on 
,
then 
lies on 
.
.
.
.
,
.
lie on 
or 
is the angle bisector of 
by AA we have 
,
.
is the image of 
lies on segment 
within 
is homothetic to 
and therefore passes through the center of 
.
so it's trivial that 
and hence 
as desired.
.
,
is the point at which the altitude at 
.
,
is thus both the midpoint and foot from 
,
bisects 
,
be midpoints of 
,
,
and 
.
.
are midpoints of 
,
,
respectively, 
.
,
and 
so 
.
which implies 
.![\[\angle FAD=\angle CGF=\angle CFG=\angle BAF\]](http://latex.artofproblemsolving.com/8/8/8/8885e00857ad1f95b9bab27feac9725e6ea0c805.png)
as desired.
and 
be the midpoints of 
and 
.
combined with 
and 
.![\[\frac{\frac{1}{2}CF\cdot\tan\angle EFC}{\frac{1}{2}CG\cdot\tan\angle ECG} = \frac{EH_{d}}{EH_{b}} = \frac{DH_{d}}{BH_{b}} = \frac{DF + FH_{d}}{BC + CH_{b}} = \frac{DF + \frac{1}{2}CF}{BC + \frac{1}{2}CG} = \frac{CF}{CG}\cdot \frac{\frac{DF}{CF}+\frac{1}{2}}{\frac{AD}{CG}+\frac{1}{2}} = \frac{CF}{CG}\]](http://latex.artofproblemsolving.com/6/c/b/6cbcad288fdbf1a23da26ecf7ebca778d4b9d0d4.png)
where the last equality follows from 
and the angles are acute, hence 
.
,
and therefore 
,
as desired.
as desired. 
be the midpoint of 
and 
,
,
be a triangle, and let 
and 
and 
such that 
Let 
be on 
such that 
is a parallelogram. Then regardless of the position of points 
will pass through a fixed point other than 
as the reference triangle. Let 
and 
Also let 
and 
where 
and therefore 
A point 
on this circle must satisfy an equation of the form![\[
a^2 yz + b^2 zx + c^2 xy = ux + vy + wz,
\]](http://latex.artofproblemsolving.com/f/9/a/f9a51fd2e9e6eb58bec42bfe9d1a0f17b6a1a802.png)
where 
and 
Plugging in 
gives 
so![\[
a^2 yz + b^2 zx + c^2 xy = ux + wz.
\]](http://latex.artofproblemsolving.com/f/6/c/f6c01a738186b2f98830810bdc59158ac3c6400f.png)
Plugging in 
so 
Similarly, plugging in 
gives 
so 
![\[
\boxed{a^2 yz + b^2 zx + c^2 xy = c^2 \lambda x + a^2 (1-\lambda) z}.
\]](http://latex.artofproblemsolving.com/b/a/4/ba46b80115b78ccfaf9be00f4e59a84564db3759.png)
If we let 
and 
be two distinct real values of 
then the radical axis of the two corresponding circles is given by![\[
(c^2 \lambda_1 - c^2 \lambda_2) x + (a^2 (1 - \lambda_1) - a^2 (1-\lambda_2))z = 0.
\]](http://latex.artofproblemsolving.com/b/8/a/b8a5fb88b1fc47afdd3580883e5c4a538e4e6eb8.png)
Dividing both sides by 
gives 
which is clearly fixed. This concludes our (lengthy) proof of the claim.
Since 
it has no relevance to 
or 
so 
we discover that 
is tangent to 
Hence 
However, by sum of angles in a triangle, 
so 
Since 
and 
we get 
and 
Thus 
and we conclude.
and 
call triangle 
'
.
and 
,we also know 
.
.
which is indeed true.
denote the midpoints of 
.
.
.
is in fact the perpendicular bisector of segment ![\[\measuredangle BAD = \measuredangle DCB = \measuredangle DEB = 2\measuredangle DBE = 2 \measuredangle FCE = \measuredangle FEC = 2 \measuredangle FGC = 2\measuredangle FAD \]](http://latex.artofproblemsolving.com/f/9/c/f9c2a10f5b0a8388a64c60660f86a71f9e866284.png)
which indeed implies that 
b
,
,
respectively, and they are collinear by simson line
is parallel to 
so 
since 
is cyclic, and 
also since 
and we can prove 
and 
,
centered at 
