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.
such that for all 
is great if the following holds: for any point 
on the side 
, if 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
. Let be the foot of the altitude of triangle issuing from the vertex 
, and let 
be the circumcenter of triangle . Assume that 
. Prove that 
.
Prove that 
When does equality hold?
be a parallelogram such that 
Let 
and 
be the midpoints of and 
respectively. Assuming that 
is a cyclic quadrilateral, find 
be a polynomial with integer coefficients. Define 
and 
for 
.
such that 
, then prove that either 
or 
.
have 
distinct real roots, which sum up to 
. If 
, find the values of 
and their corresponding roots.
be a positive integer greater than . Find all functions 
satisfying:
for all integers 
be a triangle, and let be an arbitrary point on line 
, where is the circumcenter of . Define 
and 
as the orthocenters of triangles 
and 
. Prove that 
passes through a fixed point which is independent of the choice of .
be a triangle inscribed in a circle, where the tangents to the circle at points 
and 
intersect at the point . Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines and 
intersect at point . Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point , and intersects the circumcircle of again at point 
. lies on the circumcircle of 
.
be circumscribed about circle 
, and let 
be the orthocenter of . The circle touches line at . The tangent to the circle 
at meets at 
. Let 
be the midpoint of 
, and let the line 
meet again at 
. The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
.
. Perform an inversion about with radius 
. Let the -median intersect the circle with diameter 
at 
and at . Note that this inversion simply swaps and 
, and and . Morever, inverts to the nine-point circle of . Finally, the line inverts to itself, so these three objects all pass through , a point on , so inverting back yields the desired conclusion.
be the foot of the perpendicular from to . Since the reflection of lies on the circumcircle of , it follows that 
and are congruent. In particular, the reflection of over and lie on . Denote the reflection of over and as and 
, respectively. By power of a point, 
. It follows that 
are concyclic. Since 
, it follows that 
as well.
be the foot of the altitude from to , and 
be the ortic triangle.
and 
are tangent to 
, 
is a harmonic quad. Project it through to line , and goes to , goes to , goes to , and goes to the harmonic conjugate of wrt. (say, 
). But by ceva-menelao, 
and concurr at . Moreover, we can see that is the radical center of , and 
and thus 
is the rad ax of and which implies is in , but since it's also in the median, we have that 
and we're done.
,
and let 
are midpoints of 
,respectively.We know that 
is cyclic or 
, so 
lies on radical axis of the circles 
and 
,hence 
or 
.But we have by 
lies also radical axis 
,so we are done !
,
.
and 
.
is a parallelogram 
and 
.
.
.
is cyclic 
