[/quote]
,
,
for all positive reals 
.
in the 
-
,
such that the minimum value of 
is positive.
such that![\[(a^2 + b^2 + c^2)^2 = 3(a^3b + b^3c + c^3a) + 1. \]](http://latex.artofproblemsolving.com/9/e/4/9e400fd807a9535ede983b75dcc307d8b75a35b6.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
,![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
,
is fantabulous, then all the elements are fantabulous. Does it follow that the number 
is fantabulous?
and 
.
that satisfies:![\[
\sum_{i=1}^{n} a_{\left\lfloor \frac{n}{i} \right\rfloor} = n^k
\]](http://latex.artofproblemsolving.com/b/b/e/bbe7fe81e366fe390be5d17193d1581cc40dbece.png)
be a positive integer. Prove that for all integers 
,![\[
\frac{c^{a_n} - c^{a_{n-1}}}{n} \in \mathbb{Z}
\]](http://latex.artofproblemsolving.com/0/4/4/04464c1a350e3f9a929ab51534018498680068fc.png)
such that![\[
f(f(x)+y) = x+f(f(y))
\]](http://latex.artofproblemsolving.com/9/d/7/9d780faa5901646f7035e88400a9e3d8023f04ff.png)
for all 
.
be a function from the Euclidean plane to the real numbers such that 
for any acute triangle 
with circumcenter 
,
and orthocenter 
.
is constant.
passing through 3 points 
and 
.
vary in the range of 
,
is defined by 
and 
Show that there exist infinitely many prime number that divide at least one number in this sequences
be the orthogonal projection of 
on 
and 
are the midpoints of 
and 
of the circumcircle of triangle 
.
and 
are tangent if and only if 
Y by anantmudgal09, Adventure10, Rounak_iitr
3. A quadrilateral 
has the incircle 
and is such that the semi-lines 
and 
intersect at point 
and the semi-lines 
and 
intersect at point 
. The lines 
and 
intersect at point 
. Let 
be the point of closest from line . Prove that the line 
passes through the incenter of triangle 
.
has the incircle 
and is such that the semi-lines 
and 
intersect at point 
and the semi-lines 
and 
intersect at point 
. The lines 
and 
intersect at point 
. Let 
be the point of closest from line . Prove that the line 
passes through the incenter of triangle 
.
be the points of tangency with 
,
,
the foot of the altitude from 
.
)
be the intersection of 
.
to intersect 
.
are perpendicular to 
respectively, so 
,
are concyclic. So 
.
.
lies on 
,
,
bisects 
.
intersect 
,
be the intersection of 
are harmonic. Let 
be the second intersections of 
with 
are concyclic, 
,
are concyclic. 
,
,
.
,
are harmonic. Thus 
,
.
.
,
be the inverse of 
.
concur.
lie on 
respectively. Suppose 
is tangent to 
at point 
as reference. Let 
.
denote the sides, 
the inradius, 
the 
-
the length of 
the length 
,
the length of 
the angle 
,
Also, we have 
hence these cross-ratios are equal. Clearly, we obtain 
concur. 
.
lies on 
are collinear; proving the conclusion. 
and 
be the incircles of 
and 
resp tangent to 
Since 
but 
and 
Since 
and 
then by Monge & d'Alembert theorem it follows that the exsimilicenter of 
is on 
But as their radii 
and 
are parallel with the same direction, then 
are collinear.
and let 
be the antipode of 
Since 
WRT 
then 
i.e 
Since 
then 
are collinear. From 
we get 
and from the complete 
we get 
are collinear.
be the incircle of 
.
be the centers of 
,
and 
be the exsimilicenter of 
,
,
,
.
.
passes through 
And 
Hence 
are collinear.
,
meet 
,
,
sends 
are collinear. Let 
are collinear. Suppose the incenters of 
are 
.
,
,
.
which completes the proof.
be the contact points of 
,
,
,
the incenter of 
.
,
is cyclic. Since 
and 
,
.
But,
Therefore,
The proof is complete.
,
,
and 
the incenter of 
lies on 
and 
,
.
such that 
,
are collinear.
.
of 
of 
.
(both lie inside the segment 
.
,
and 
.
is the exmilicenter of 
.
,
.
,
are collinear, then 
are collinear.
and 
,
so 
,
are collinear, as desired.
be the tangency points on sides 
respectively and let 
.
be the intersection of lines 
and 
.
shows that 
are collinear where 
.
are collinear, hence 
and 
gives 
concurrent. Hence, 
and by the Miquel's point theorem on quadrilateral 
,
(which is the line joining the center and the miquel point of 
are collinear 
triangles 
and 
are homothetic, which is iff 
But one may notice that 
since triangles 
and 
are similar. Hence, our problem is iff 
.
the inradius of 
.
,
.
,
applying inversion distance formula (recall that 
when inverting w.r.t 
(2). Finally, 
and 
.
, which miraculously simplifies to 
,
is not collinear with 
,
?
and and define 
Then animate 
and define 
with respect to line 
with the reflection of line 
with respect to line 
Then define the other points acordingly, i.e. 
then points 
be the incircle of 
be the point of 
Then point 
and, furthermore, the map 
from line 
to line 
is linear. 
the homothety centered at 
So that point 
Then it follows that the map from 
from line 
centered at the point of infinity of the pencil of lines perpendicular to line 
be the incircle of 
Further, let 
Equivalently, 
and 
and 
at 
and 
respectively. From the fact that the two tangents drawed from a point to a circle are equal it follows that
similarly we have 
and since 
and 
(
be the hyperbola with focus 
from 
are tangent at 
So it follows from Monge's Theorem that 
and 
so that 
Then we are done if we show that the map 
from line 
is linear, since the map 
from 
by a projection at 
is linear. Then recall that 
Then 
Now one can assign an coordinate system to line 
for constants 
and as in the problem statement 
Then by Claim 1 we have that the map 
from 
is linear. Also, since point 
from 
is also linear. So to finish the problem it is sufficient to verify that 
for only three choices of 
So taking 
the limit as 
and 
,
.
.
.
.
.
be the harmonic conjugate of 
.![\[-1=(K,L;I,\infty_{KL})\stackrel{T}{=}(K,R';R,N)\stackrel{\infty_{OT}}{=}(I,I_C;R\infty_{OT}\cap II_C,O)\]](http://latex.artofproblemsolving.com/c/9/0/c90a222ca52e8d5bb810a44a792c11b60058d186.png)
thus 
.
with incenter 
.
be its harmonic conjugate with respect to 
.
.
are harmonic conjugates with respect to 
is constant. In particular let ![\[x=MU^2=MG\cdot ME=y\cdot MF\cdot ME\]](http://latex.artofproblemsolving.com/6/8/2/6822dfd7fccb4d74a41f63c1230c4b34a34cb037.png)
is constant, thus 
is constant. It suffices to prove the question for a single choice of 
works fine, and we are done. 
so I figure it is fine.
,
and 
and finding tangent lengths. Then, ![\[2PS=CP+PQ-CQ=PQ+(CP-CQ)=PQ+(AP-AQ)\]](http://latex.artofproblemsolving.com/c/d/d/cdd43d315608b76cd60cba973282527c8277ec55.png)
as desired. 
.
or 
collinear.
.
are collinear.
are collinear. Thus, combined with above, 
are collinear.
,
.
concur. But 
pass through 
does as desired.
respectively. 
at 
respectively. Let 
be the incircle of 
at 
.
.
and 
are collinear.
.
be the line tangent to 
.
.
are collinear.
be the point 
.
,
.
,
are collinear.
and 
swap with each other. Collinearity of 
is equavilent to 
.![\[\measuredangle YEF=\frac{\measuredangle YJF}{2}=\frac{\measuredangle YJC-\measuredangle FJC}{2}=\frac{\measuredangle SIC-\measuredangle LIC}{2}=\frac{\measuredangle SIL}{2}=\measuredangle SML\]](http://latex.artofproblemsolving.com/3/d/5/3d5d9f08d4b5df3dff85b81deeb82ad0c4fea35d.png)
![\[\measuredangle EFY=\frac{180-\measuredangle ZJE}{2}=\frac{180-\measuredangle CJE+\measuredangle CJZ}{2}=\frac{180-\measuredangle CIM+\measuredangle CIT}{2}=\frac{180-\measuredangle TIM}{2}=\measuredangle MLS\]](http://latex.artofproblemsolving.com/7/b/a/7ba098981536439f92e6c2f6da5a2f689c28eeba.png)
Which yields 
.
are collinear.
.
.
hence 
which is folowed by 
.![\[\frac{JZ}{JY}=\frac{IX}{IS}=\frac{XT}{TH}\]](http://latex.artofproblemsolving.com/4/e/f/4efea1c73b0e4c63e335ba60eaf1f8983d4a9f87.png)
Thus, by homothety centered at 
to 
we get that 
the incircle of 
(and similarly for quadrilaterals). Let 
.
,
so the 
is also tangent to 
is true, and by homothety at 
is true. Using Monge's theorem on the 3 incircles, we get that the exsimilicenter of 
and 
.
since 
.
as well, projecting which through 
.
,
unless 
in which case the problem is trivially true.
is a diameter of the incircle 
,
is the incircle, 
.
aligns with the length for triangle 
sending 
and 
sending 
(alternatively Monge) and the common internal tangent 
is collinear by the incircle diameter lemma.
,
are collinear.
of 
and 
,
,
as a result.
passes through the incenter 
