[/quote]
,
be the number of positive integers that divide the integer 
.
of 
,
?
is a square of an integer.
be all positive integers smaller than 
such that 
for all 
is the largest positive integer that divides both 
and 
.
.
,
,
are odd primes and![\[2^ap^b=(p+2)^c-1.\]](http://latex.artofproblemsolving.com/5/8/5/58518a8ebed14836b72af743ff100eae2efba5e2.png)
let 
be the circumcircle of 
and let 
be its radius. Point 
is chosen on 
such taht 
and point 
is the foot of the perpendicular from 
.
mets 
is chosen on line 
such that 
and 
lie on a line in that order. Finally, let 
and 
.
has two real roots, so that the one is the square of the other.
such that: ![\[ x = \sqrt{ x - \frac{1}{x} } + \sqrt{ 1 - \frac{1}{x} } \]](http://latex.artofproblemsolving.com/e/c/f/ecf7fb3c2805290f2cb07b7fdd7c73e1c7a55b99.png)
be a convex pentagon such that 
.
is the circumcenter of triangle 
.
be the circumcenter of triangle 
.
passes through the midpoint of segment 
.
such that for all 
the following holds:![\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]](http://latex.artofproblemsolving.com/a/c/a/aca098172ea7ce5663e9dd0dd10c2f1495630230.png)
.
meet at a point belonging to the diameter 
and 
are the incenter and the 
on the circumcircle, and 
intersects the circumcircle again at 
. For
on the circumcircle of triangle 
intersect 
. Prove
are concyclic.
,
lie on a line in this order and satisfy 
.
,
is acute, and let 
be its orthocentre. Points 
and 
,
and 
and 
and 
and 
be a transversal of 
be intersections of 
with 
be a tangent line of 
is cyclic
Y by Adventure10
Let there be an acute triangle 
, such that 
. Let the perpendicular from 
to 
hit the circumcircle of at 
, and let 
be the midpoint of 
. The tangent to the circumcircle of at hits the perpendicular bisector of at 
, and the circumcircle of 
hits the circumcircle of at 
. Let 
be the foot of the perpendicular from to 
, and 
be the midpoint of 
. Prove that 
are collinear.
, such that 
. Let the perpendicular from 
to 
hit the circumcircle of at 
, and let 
be the midpoint of 
. The tangent to the circumcircle of at hits the perpendicular bisector of at 
, and the circumcircle of 
hits the circumcircle of at 
. Let 
be the foot of the perpendicular from to 
, and 
be the midpoint of 
. Prove that 
are collinear.
.
,
is diameter in 
,
,
.
;
are collinear and 
,
is harmonic, hence 
,
is the midpoint of 
and 
are tangents to 
the circumference of diameter 
is orthogonal to 
since the circumference of diameter 
and 
but 
is 
and 
since 
and let 
is the midpoint of 
are collinear... 
is symmedian of 
,
are collinear.
and we want 
.
(since 
and 
is cyclic.So we have 
which gives the result.
.
is harmonic quadrilateral.
are collinear. Since 
is a right triangle, so 
.
,![\[\frac{\sin{\angle FBD}}{\sin{\angle ABF}}=\frac{\sin{\angle EDF}}{\sin{\angle ADF}}=\frac{FE}{FM}.\qquad (1)\]](http://latex.artofproblemsolving.com/4/6/7/467e9f1cea57dddfe73056709b167f4079056c9f.png)
On the other hand, Since 
is a median of 
and clearly 
,![\[\frac{\sin{\angle NBG}}{\sin{\angle ABN}}=\frac{AB}{BG}=\frac{ED}{DM}. \qquad (2)\]](http://latex.artofproblemsolving.com/5/0/9/5090e9b29ed7683e10844257b72f34958e5eda9d.png)
Compare 
and 
,
,
,
are collinear, as desired.![[asy]
import olympiad; import cse5; size(9cm); defaultpen(fontsize(10pt));
pair O,A,B,C,D,E,G,M,N;
A=dir(60);
B=dir(-170);
C=dir(-10);
O=origin;
D=2*foot(A,O,dir(0))-A;
M=midpoint(A--D);
E=extension(O,M,A,A+(O-A)*dir(90));
G=foot(A,B,D);
N=midpoint(A--G);
pair [] F=IPs(unitcircle,circumcircle(M,D,E));
draw(unitcircle);
draw(A--E);
draw(D--E);
draw(A--B--C--cycle);
draw(A--D);
draw(O--M,dashed);
draw(circumcircle(M,D,E),blue);
draw(B--D);
draw(A--G);
draw(B--F[0],red+dashed);
draw(M--F[0],dashed);
draw(F[0]--E,dashed);
draw(M--E);
draw(D--F[0],dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$D$",D,dir(-100));
dot("$E$",E,dir(E));
dot("$F$",F[0],dir(75));
dot("$G$",G,dir(G));
dot("$M$",M,dir(135));
dot("$N$",N,dir(135));
dot("$O$",O,SW);
[/asy]](http://latex.artofproblemsolving.com/e/9/a/e9a28f01ea494d0138de6d977f274a85c9902241.png)
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= 
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= 
= 
= 
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= 
or 
,
