AoPS Academy
satisfying![\[f(x+f(y))=x+f(f(y))\]](http://latex.artofproblemsolving.com/8/d/a/8daa6bcb3155768ee1ba4bd22beb53a85a6e3923.png)
and 
, with the additional constraint 
.
be positive odd integers. Prove that 
doesn't divide 
.
: 
such that:
for all x,y E 
denote the number of positive integer divisors of a positive integer 
(for example, 
). Given a polynomial 
with integer coefficients, we define a sequence 
of nonnegative integers by setting![\[a_n =\begin{cases}\gcd(P(n), \tau (P(n)))&\text{if }P(n) > 0\\0 &\text{if }P(n) \leq0\end{cases}\]](http://latex.artofproblemsolving.com/7/5/9/759f4e6c8511d920b858c4234eff7a53f38ac289.png)
for each positive integer . We then say the sequence has limit infinity if every integer occurs in this sequence only finitely many times (possibly not at all). for which the sequence 
, 
, . . . has limit infinity?
, in its exterior we draw the triangles 
, 
, 
so that 
, 
and 
, where 
. Prove that:
(the triangle 
is isosceles in the vertex 
).
we obtain the problem from IMO-1975.
in its exterior
we draw the triangles 
so that 
and
be the equilateral triangle constructed such that 
and 
are on the same side. 
. We have 
from similarity. Also we have 
. So 
. Then 
and 
. Similar calculations for 
. We will have 
and 
. Also from the question we have 
. So 
and 
.
. It is not hard to see that 
, since 
and 
. Using the fact that spiral similarities come in pairs, we discover that 
.![[asy]
size(7cm); pair A = dir(110); pair B = dir(210); pair C = dir(330);
pair R = rotate(15)*(Sin(15)/Sin(150))*(A-B)+B; pair P = rotate(-45)*(Sin(30)/Sin(75))*(C-B)+B; pair Q = rotate(45)*(Sin(30)/Sin(75))*(C-A)+A;
pair T = rotate(60)*(A-B)+B;
draw(T--A--C--cycle, blue); draw(R--A--Q--cycle, red); draw(R--B--T); draw(Q--C--P--B); draw(A--B--C);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$R$", R, dir(110)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$T$", T, dir(T));
/* Source generated by TSQ */ [/asy]](http://latex.artofproblemsolving.com/c/3/a/c3a617fe7e22df5b678ad34ce8bcf4819c47fcbb.png)
![\[ \angle ARQ = \angle ATC \text{ and } RQ = \frac{RA}{TA} \cdot TC. \]](http://latex.artofproblemsolving.com/2/7/0/2707cc014afbbe0daf427120af6c5d8f3e070133.png)
In exactly the same fashion we can show that 
and 
. This implies the result.
on the same side of 
as 
satisfying 
equilateral, and similarly construct 
on the same side of 
as 
satisfying 
equilateral. Then from 
and 
we see 
are right isosceles triangles with right angle at 
and 
so 
and 
similarly oriented. Then by spiral similarity we get 
similarly oriented. But there is a 
rotation at sending to and to 
and this must send 
to 
so we are done.
with the vertices in counterclockwise order.
on the same side of 
as such that triangle 
is equilateral. Then there exists a spiral similarity that rotates 
degrees counterclockwise centered at sending 
to 
, and a one that rotates degrees clockwise centered at sending 
to . These have the same scale factor too, so the result follows. 
and moving 
, we just get that and 
are moving linearly wrt , so it makes sense to spiral similarity them away. Also, I tried the case 
which gave me a really computational angle chasing geo-type vibe, so constructing stuff like this felt right.
concur at a point .
. lies on 
.![\[ 60^\circ = \measuredangle RAB + \measuredangle CAQ = \measuredangle RXB + CXQ = \measuredangle RXQ + \measuredangle CXB = \measuredangle RAQ + \measuredangle CXB \]](http://latex.artofproblemsolving.com/3/c/d/3cdd34dd690cdc61826bc20b3f63ed1eb8bdf264.png)
which implies that 
. ![\[ \measuredangle ARQ + \measuredangle PRB = \measuredangle AXQ + \measuredangle PXB = \measuredangle AXB + \measuredangle PXQ = \measuredangle ACB + \measuredangle PCQ = 60^\circ. \]](http://latex.artofproblemsolving.com/7/c/4/7c4a6008210c5fe56b996500e2c77d9e9f5d638e.png)
so 
. We can similarly angle chase that ![\[ \measuredangle BPR + \measuredangle QPC = \measuredangle BXR + \measuredangle QXC = \measuredangle BXC + \measuredangle QXR = \measuredangle BAC + \measuredangle QAR = 60^\circ \]](http://latex.artofproblemsolving.com/a/2/3/a236de38c78c41dfd4ad317f04b39e54bd7d7919.png)
which implies the result.
is a perfect six point set / perfect hexagon / coharmonic hexagon / super hexagon and that this circle concurrence is a property of said hexagons.
in its exterio
so that 
,
,
.
and
and 
so that 
and 
.
and 
intersect at 
.
and 
,
.
and 
,
is equilateral. Similarly is 
.
if 
and 
are equilateral. Now, 
.
,
,
.
.
,
.
.
.
degrees around 
and 
from this rotation, so 
.
.
is directly similar to 
.
and that 
;
and 
is a right isosceles triangle. If we prove that 
is similar to 
,
,
.
.
,
.
,
.
,
,
.
is equilateral, inducing![\[\triangle BAK \sim \triangle BCP \implies \triangle BKP \sim \triangle BAC.\]](http://latex.artofproblemsolving.com/9/8/0/980376416129753b0bc6c7af351bc206bb7fee51.png)
,![\[\angle RAQ = 60 + \angle B = \angle RKP, \quad QC = \frac{CA}{BC} \cdot PB = PK, \quad AR = KR. \quad \blacksquare\]](http://latex.artofproblemsolving.com/6/8/5/6853bf4d14790e94481c447bcd5f8e7d761471a2.png)
![[asy]
size(250);
pair A, B, C, P, Q, R, K;
A = dir(230);
B = dir(310);
C = dir(110);
P = dir(-45)*.51764*(C-B) + B;
Q = dir(45)*.51764*(C-A) + A;
R = dir(15)*.51764*(A-B) + B;
K = dir(-90)*(A-R) + R;
filldraw(R--A--Q--cycle^^R--K--P--cycle, lightgreen);
draw(A--B--C--cycle^^A--R--B--P--C--Q--A--K--P--R--Q^^R--K--B);
draw(B--K--P--cycle^^B--A--C--cycle, red+linewidth(1.5));
dot("$A$", A, dir(225));
dot("$B$", B, dir(315));
dot("$C$", C, dir(90));
dot("$P$", P, dir(0));
dot("$Q$", Q, dir(180));
dot("$R$", R, dir(270));
dot("$K$", K, dir(135));
[/asy]](http://latex.artofproblemsolving.com/b/f/4/bf484b65b85ed625e20d1cbd1ce7c02d47af997d.png)