[/quote]
,
is valid while 
is not)
of positive integers such that 
is an integer.
are positive integers such that for every natural number 
the set of prime factors of 
are same. prove that 
.
Prove that
Where 
square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always two congruent rectangles.
such that![\[f(x(x + f(y))) = (x + y)f(x),\]](http://latex.artofproblemsolving.com/b/0/6/b0668edacd78de23f7611f47a7972a2dd148a4a0.png)
for all 
.
.
such that ![\[ \forall x, y \in \mathbb{R}^+ \ , \ \ f(x+f(xy))=f(x)+xf(y).\]](http://latex.artofproblemsolving.com/f/2/7/f27575aa4092155d285af0eccebe3865043011bc.png)
be a triangle with incentre 
and circumcircle 
.
and 
be the second intersection points of 
and 
,
meets 
at a point 
,
at a point 
.
be the intersection point of the line through 
and the line through 
.
and 
meet at a point 
.
and 
are either parallel or concurrent.
,
and 
is divisible by 
is not divisible by 
.
,
,
,
.
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For 
represents 
.
,
Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only., ItsBesi, cubres
Two circles 
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
Hence KA=KB.
Hence 
cuts the tangent length segment AB at its midpoint C. Since 
,
are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii 
of the circles 
are isosceles and the angles 
are equal. Since the lines 
are parallel and 
,
are equal and similarly, the angles 
are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals 
are perpendicular to each other. Consequently, the lines 
are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle 
is isosceles with EP = EQ.
bisects 
.
is the radical axis of 
,
.
and 
be the projections of 
,
.
,
(as 
.
and 
.
be the point such that 
and 
.
.
bisects 
.
and 
is concyclic.
and it is known that 
.
is radical axis for 
circles. also we know that radical axis passes through the midpoints of comment tangents. Thus 
lies on the same line. 
is the midpoint of 
.
[*] the last equation follows from the fact that 
is cyclic.
,
,
and since 
,
.
.
,
.
and 
and 
and therefore proving 
is equivalent to proving 
since 
.
because 
.
and 
and therefore 
as desired. 
and 
are isosceles and 
so 
.
is the midpoint of 
meets 
are not at all collinear.
is harmonic!
.
is cyclic, so 
.
.
,
by ASA congruency. Then 
is a kite, so 
,
.
,
,
.
,
are collinear, and 
are collinear, a homothety centered at 
takes 
is isosceles, and thus, 
as the point where they intersect has equivalent powers to the respective circles. From this we discover that 
due to the homothety at 
It then suffices to show 
.
![\[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]](http://latex.artofproblemsolving.com/6/4/e/64ea7480f2012c4f9052db0044dc365941c75e1e.png)
Similarly by symmetry, we find that 
Then by SAS, we conclude that 
as was to be shown. 
is a kite, so that 
as 
.
and 
so 
and 
.
.
As 
Taking the homothety centered at 
Then 
and similarly 
so by ASA congruence 
,
,
,
,
are reflections about 
=
from PoP we know 
and 
which means 
and since 
is a trapezoid, 
and we're done. label 
we know from 
that 
and 
being tangent that 
and 
the above facts indicate the similarity of 
so 
=> 
then clearly 
Therefore 
by the parallel lines.
Therefore 
and 
Similarly 
and 
Therefore, 
are the circumcenters of 
respectively so by Thales' Theorem it follows that 
Hence 
and we are done. QED
,
Thus, 
and 
and 
hence 
is a kite 
implying 
.
,
implies 
.
and 
.
![\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]](http://latex.artofproblemsolving.com/3/3/c/33cb77c57e505ab008a08d9152c6c0025237e8b4.png)
![\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\]](http://latex.artofproblemsolving.com/1/7/b/17b761977cdf9496d808500e94dc78cebace60fd.png)
, implying that 
,
are isosceles. Therefore,![\[AE = AM = AC\]](http://latex.artofproblemsolving.com/0/5/9/059eab3cda6c66376c2362600d95b1f8aa3473db.png)
so 
,
.
.
,![\[LA = LB\]](http://latex.artofproblemsolving.com/4/a/1/4a15f4404bb42649f5bc6cf5f3df3284f9e64144.png)
and since 
,
,
,
.
,
,
bisects 
,
,
and similarly, 
.
.
,
is the perpendicular bisector of 
.
.
and 
,
.
and 
are isosceles. Furthermore, 
.
,
.
and similarly 
so 
and we're done
.
be the center of 
is the side bisector of 
,
.
and similarly 
,
(because they also share 
and analogously 
,
,
and similarly 
,
,
![[asy]unitsize(1.5cm);
import olympiad;
real root3 = sqrt(3);
pair O1 = (-root3, -1);
pair O2 = (root3, -1);
real radius = 2;
pair A = (-root3, 1);
pair B = (root3, 1);
pair M = (0, 0);
pair N = (0, -2);
pair C = (-2 * root3, 0);
pair D = (2 * root3, 0);
pair E = extension(A, C, B, D);
pair K = extension(A, B, M, N);
pair P = extension(A, N, C, D);
pair Q = extension(B, N, C, D);
draw(circle(O1, radius), heavycyan + dashed + linewidth(1.4));
draw(circle(O2, radius), heavymagenta + dashed + linewidth(1.4));
draw(A--B, heavygreen + linewidth(1.5));
draw(C--D, deepmagenta + linewidth(1.4));
draw(A--C, orange + linewidth(1.1));
draw(B--D, orange + linewidth(1.1));
draw(A--N, heavyblue + dashed + linewidth(1.1));
draw(B--N, heavyred + dashed + linewidth(1.1));
draw(E--N, black + dotted + linewidth(1.2));
draw(E--A, red + linewidth(2.5));
draw(E--B, blue + linewidth(2.5));
draw(E--P, cyan + linewidth(2.5));
draw(E--Q, magenta + linewidth(2.5));
dot(E, deepgreen);
dot(K, purple);
dot(P, brown);
dot(Q, orange);
dot(M, blue);
dot(N, red);
label("$G_1$", O1, SW, fontsize(9));
label("$G_2$", O2, SE, fontsize(9));
label("$A$", A, NW);
label("$B$", B, NE);
label("$M$", M, NE);
label("$N$", N, S);
label("$C$", C, W);
label("$D$", D, E);
label("$E$", E, NE);
label("$P$", P, SW);
label("$Q$", Q, S);
label("$K$", K, dir(45));
size(10cm);
pair min = (-2.5 * root3, -3.5);
pair max = (2.5 * root3, 3.5);
clip(box(min, max));
[/asy]](http://latex.artofproblemsolving.com/7/4/0/740fd20749ff72e5436121d9a8aa542c708ea487.png)
Let point 
Since 
lie on the same line, we can immediately see that this implies (from homothety) that 
.
.
All that remains to show now is that
(from 
)
which had to be proven. Therefore, we are done.
and 
and 
then by PoP, 
is the midpoint of 
by playing with Thales theorem on 
and 
