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.
let 
be the number whose 
-th digit after the decimal point is the 
-th digit after the decimal point of 
. Show that if is rational then so is 
.
from the set 
of positive integers into itself is defined by the equality ![\[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]](http://latex.artofproblemsolving.com/2/a/5/2a5a20395450d37d0352ab75e9e4875ab7803805.png)
for every two relatively prime 
.
the equation 
has a solution.
such that the equation has a unique solution.
be an even integer and let 
be real numbers satisfying 
.
such that 
, 
, 
, 
be four points such that no three are collinear and is not the orthocenter of 
. Let 
, 
, 
be the orthocenters of 
, 
, 
, respectively. Suppose that the lines 
, 
, 
are pairwise distinct and are concurrent. Show that the four points , , , lie on a circle.
and 
be the circumcenter and orthocenter, respectively, of an acute scalene triangle . The perpendicular bisector of 
intersects 
and 
at 
and 
respectively. Let 
denote the intersection of the circumcircles of triangles 
and 
other than .
and 
analogously by repeating this construction two more times. Prove that , , , and are concyclic.
be a quadrilateral inscribed in a circle 
. Let 
, 
, 
, 
, 
, 
, 
, 
. Prove that 
lies over 
.
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist. be an odd prime and 
be integers so that the integers 
are divisible by . divides each of .
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible. be a triangle (
). 
are the altitudes of the triangle. The bisector of 
intersects 
at 
. Let be a point such that 
and 
. Prove that passes through the midpoint of 
.
is the altitude. A line through the midpoint 
of side 
meets the rays 
and 
at 
and 
respectively such that 
Point 
is the circumcenter of the triangle 
Prove that 
and is the midpoint of 
Let 
pass through 
and 
such that 
Let 
and is the center of the circumcircle of 
Prove that 
and 
by Menelaus' theorem for 
cut by the straight line 
we get 
Thus, 
is the center of the rotation carrying 
into 
and due to 
then 
is the midpoint of the arc 
of 
Obviously, lies on the internal bisector of 
Since 
is the circumdiameter of 
perpendicular to 
it follows that 
is the midpoint of 
Moreover, if 
denote the midpoints of 
then is the intersection of 
with the perpendicular bisector of segment 
since the positive homothety with center 
and factor 
transforms 
into 
into and the perpendicular bisector 
of into the perpendicular bisector of 
Now, it remains to see that quadrilateral 
is an isosceles trapezoid, in which lies on its symmetry axis. Indeed, 
and 
are symmetric about the perpendicular bisector of hence 
parallel to meet line 
at . So 
, so 
. We also have 
, so 
, therefore 
. WLOG assume 
, so 
. Let 
be the projections of onto 
respectively. We get 
, 
, 
, 
. Note that 
, so 
. Therefore 
. Also note that 
. So 
is the midpoint of 
and hence 
, as desired.
.
