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.
be an acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
. Define 
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be an acute triangle with 
, its orthocenter and 
it's circumcenter. Let be a point on the parallel through to such that 
and and are on different sides of . Denote by 
the intersection of the circumcircle of and 
, where 
is the reflection of over , 
the midpoint of 
, the intersection of 
and the parallel through to 
, the intersection of 
and the perpendicular through to 
and 
a point on such that 
, where 
is the midpoint of 
. Prove that 
lie on a line.
is inscribed in a circle 
. Inscribed in a triangle circle touchs the sides in a point . 
— the circle inscribed in a segment 
circle of , and passing through a point . Let points and 
— the centers of circles and an extra inscribed circle (touching side ) respectively. Prove, that lines 
and 
are parallel.
be cyclic quadrilateral. Let and 
intersect at , and let and 
intersect at 
. Let and are points on and such that 
. Let and be the intersections of 
with the diagonals of . Prove that circumcircles of triangles 
and 
are tangent at a fixed point.
be an acute triangle. Points 
, and lie on a line in this order and satisfy 
. Let and be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let be its orthocentre. Points and lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
with diameter 
is given. The point lies in the interior of the circle, but not on . The line 
intersects in and . The tangent to at intersects the line through perpendicular to , at . The point 
lies on , and is such that 
is tangent to and 
.
, and are collinear.
colours (so that adjacent vertices have different colours).
edges can be removed from the graph so that it remains connected.
be the incenter and the circumcenter of triangle , and 
be the circumcenters of triangle 
. Let 
be the midpoints of segments 
. Prove that the circumcenter of triangle 
, , is the midpoint of segment 
.
Each entry of an 
matrix equals either 
or 
Let 
denote the sum of eight products of entries in each row. Also, let 
denote the sum of eight products of entries in each column. Find the maximum possible value of 
In other words, find![$$ {\rm max}\ \left[ \sum_{i=1}^8\ \prod_{j=1}^8\ a_{ij} -
\sum_{j=1}^8\ \prod_{i=1}^8\ a_{ij} \right]
$$](http://latex.artofproblemsolving.com/2/d/e/2de7fd40366073445d289356d0f4e66b542cb60c.png)
Same question, but for a 
matrix.
Show that there exists a constant 
such that for all positive integer 
![\[\sum_{k\le n^{\varepsilon}}(n\text{ mod } k)>cn^{2\varepsilon}.\]](http://latex.artofproblemsolving.com/f/e/9/fe933e886e28c2ecf1fa880d0e650ac7a2998aaa.png)
Proposed by Cheng Jiang
is written on the board. A move consists of adding either , 
or 
to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is 
, then in a move a player can't choose the number , but he can choose either or ). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
of positive integers is called central if for every positive integer 
, the arithmetic mean of the first 
terms of the sequence is equal to .
of positive integers such that for every central sequence 
there are infinitely many positive integers with 
.
using the Menelaus theorem to the triangle 
: 
then we take that 
using the Menelaus theorem to the triangle 
: 
then we take that 
and 
. Then clearly, 
and 
. The point is given by
Now 
has equation 
, so the point 
. Since 
and 
are collinear, we have
parallel to 
and apply thales theorem.
using the Menelaus theorem to the triangle : then we take that using the Menelaus theorem to the triangle : then we take that 
Prove that
are on segments 
and 
.
such that 
.
.
