AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
, 
, 
, 
, 
, 
initially contains one coin. The following operations are allowed
, 
, remove one coin from and add two coins to 
;
, 
, remove one coin from and swap the contents (maybe empty) of the boxes 
and 
., , , , become empty, while box contains exactly 
coins.
be a triangle, 
its orthocenter, 
its circumcenter, and 
its circumradius. Let 
be the reflection of the point 
across the line 
, let 
be the reflection of the point 
across the line 
, and let 
be the reflection of the point 
across the line 
. Prove that the points , and are collinear if and only if 
.
be a real number and let 
be arbitrary real numbers in 
. The sequences 
and 
are then defined recursively by
. Prove that 
for all sufficiently large 
.
is a prime number
is inscribed in circle 
. Let 
denote its incenter and 
its -excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and at and respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
be a triangle with circumcircle and incentre . A line 
intersects the lines , 
, and 
at points , , and , respectively, distinct from the points , , , and . The perpendicular bisectors 
, 
, and 
of the segments 
, 
, and 
, respectively determine a triangle 
. Show that the circumcircle of the triangle is tangent to .
be a positive integer. Prove that there are infinitely many primes, each of which divides at least one term of the sequence 
, 
.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
denote the set of single-variable polynomials in with integer coefficients. Find all functions ![$\theta : \mathbb{Z}[x] \to \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/3/2/032a637d4ae21da2c0ddd8dc127723807f11cc5f.png)
(i.e. functions taking polynomials to polynomials)![$p, q \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/7/e/7/7e7eb925258a33e456ef4ff722c1f943dc155d10.png)
, 
;![$p \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/2/6/0267b98c16b400dda1f081877e9b84fe672ce0e7.png)
, 
has an integer root if and only if 
does.
of positive real numbers that satisfy the system:![\[
\begin{aligned}
11bc - 36b - 15c &= abc \\
12ca - 10c - 28a &= abc \\
13ab - 21a - 6b &= abc.
\end{aligned}
\]](http://latex.artofproblemsolving.com/1/7/0/170e1cfaee90f5c37428fa75192ac889261a8f83.png)
that verify the following equation for all 
:
is a function for which the expression 
for all 
is always a perfect square. Prove that 
for all 
.
such that 
.
be angle bisector of 
and 
foot of perpendicular from to . Then by Iran Lemma is fixed point that all of 
passes through it.
be the incenter of . Let 
be the incircle touchpoints with 
, respectively. Let 
. Let 
be the foot from to 
. Let 
be the incenter of 
. Let be the foot from to .
. lies on 
by easy angle chase. By La Hire, it is sufficient to show that lies on the polar of , which is equivalent showing that 
is tangent to 
.
is cyclic due to right angles, we get that
we are done.![[asy]import olympiad;
size(10cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;
pair O,A,B,C,I,X,Y,D,P,Q,R,S,K,Z,L;
O=(0,0);A=dir(140);B=dir(210);C=dir(330);path w=incircle(A,B,C);I=incenter(A,B,C);P=foot(I,B,C);Q=foot(I,A,C);R=foot(I,A,B);
D=0.3A+0.7C;S=incenter(D,B,C);X=foot(S,B,D);Y=foot(S,C,D);K=extension(C,I,R,Q);Z=foot(S,B,C);L=foot(D,K,C);
draw(A--B--C--cycle,deep);draw(w,heavyblue);draw(incircle(D,B,C),heavyblue);draw(D--B,deep);draw(K--C,deep);draw(R--Q,deep);draw(circumcircle(R,I,B),heavyblue);
draw(circumcircle(B,K,X),heavyblue+dashed);draw(X--Z,deep+dashed);draw(L--D,deep);draw(X--S--Z,deep);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$I$",I,dir(I));
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(Q));
dot("$R$",R,dir(R));
dot("$D$",D,dir(D));
dot("$S$",S,dir(S));
dot("$X$",X,dir(X));
dot("$Y$",Y,dir(Y));
dot("$K$",K,dir(K));
dot("$Z$",Z,dir(Z));
dot("$L$",L,dir(180));
[/asy]](http://latex.artofproblemsolving.com/b/9/4/b947bd6019ea3ec44ca477d89bcb155b14633221.png)
and the point of intersection of XY never change.
Prove that
intersects 
and 
at 
,
,
be the common angle bisector of 
.
.
and so we are done.
bisector is also fixed, so by Iran Lemma,