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.
towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be “one way” only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns.
questions.
be any point on side 
of 
.Let 
and 
be points on 
and 
such that 
and 
respectively. Prove that the locus of circumcenter of 
is a line.
and 
pebbles, respectively. In the first round Barry chooses a positive integer 
and Andrew takes away 
pebbles from one of the two heaps (if is bigger than the number of pebbles in the heap, he takes away the complete heap). In the second round, the roles are reversed: Andrew chooses a positive integer and Barry takes away the pebbles from one of the two heaps. This goes on, in each round the two players are reversing the roles. The player that takes the last pebble loses the game.
be a triangle with 
. The point 
lies inside the triangle in such a way that 
. Let 
be the midpoint of . Prove that 
.
, is the midpoint of side , is the midpoint of side 
, is the midpoint of side .The line(different from line ) that is tangent to the inscribed circle of triangle and passing through point intersect line 
at 
.Define 
similarly.Prove that 
are collinear.
grid with 
rows and 
columns, and each square of the grid is filled with a non-negative integer. Let be the average of all the numbers in the grid. Prove that if 
the there exist two identical rows (meaning same numbers in the same order).
such that p is a prime, 
and 
is a divisor of 
.
such that![\[ \frac {2^n + 1}{n^2}
\]](http://latex.artofproblemsolving.com/9/2/2/922444e7669c326efcad5fde3ea00f8c2d9fa5d8.png)
is an integer.
on a circle and points 
and on segment , such that 
Find the ratio 
, where is a point on arc , such that 
is maximized. let 
and 
be the altitudes. Let the points 
and 
fulfil the following requirements:
is the circumcentre of triangle .
and 
are equal to the circumradius of triangle .
has the same direction as the oriented segment 
. Similarly, 
has the same direction as 
, and 
has the same direction as . is the incentre of triangle .
be a triangle with circumcenter 
. Let be a point on the perpendicular bisector of (see figure) and 
, 
be the intersections of the perpendicular bisectors of and , respectively, with 
and 
. Prove that the midpoints of 
and and the point are collinear. such that 
is a divisor of 
.
, let and 
denote the midpoints of 
and 
. Let be the projection of onto and let be the reflection of over the midpoint of 
. Assume lies in the interior of quadrilateral . Prove that 
.
as the unit circle.
. Also, 
.
is a parallelogram thus, 
and we get, 
. Now we have, 
and 
.
then we want 
. Now,
as desired. Thus 
. (
) 
![$$\left(a^{3}+b^{3}+c^{3}-3abc\right)^{2}-\left[a^{3}+b^{3}+c^{3}+3abc-ab(a+b)-bc(b+c)-ca(c+a)\right]^{2}\ge 2k\cdot(a-b)^{2}(b-c)^{2}(c-a)^{2}$$](http://latex.artofproblemsolving.com/d/2/4/d24b46e65a587fbd977467b6e4e31838a0eb57aa.png)
forall 
and 
and 
respectively. Without loss of generality, let 
and 
lie on the unit circle.
Therefore, it follows that the midpoint of 
Then, since we obviously have 
it follows that
Now, in order to show that 
it suffices to show that 
Using all previous results and the fact that 
we see that
Then, taking the conjugate of this, we get 
concluding the proof. 