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 the centers of the circles circumscribing triangles ADE, BDF, CEF. Prove that the orthocenter of triangle 
belongs to d.
for which the following statement holds: there exists at least one triple 
of integers such that
and all triples of real numbers, satisfying the equations, are such that 
are integers.
be a triangle, 
its circumcircle and 
the center of . Let 
be a point on the segment 
. We denote by 
the second intersection point of the circumcircles of triangles 
and 
. Prove that the line 
and the tangent to at point 
intersect on the circumcircle of triangle 
.
, 
, be positive real numbers such that 
. Prove that:![\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]](http://latex.artofproblemsolving.com/d/1/8/d18152c5635ee2f2059a860bf2fa362fe525ebb3.png)
satisfying the following two conditions:
such that, for all indices 
, 
.
and for any index , the number![\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]](http://latex.artofproblemsolving.com/5/4/a/54ac17063104439114f3c7b07de8fd27450ea1eb.png)
is a multiple of . be a circle with center . Let 
and 
be two fixed points on the circle and let be a variable point on . We denote by 
the intersection point of lines 
and 
, assuming 
. Let 
be the circumcircle of triangle 
. Let 
be the second intersection point of with . The tangent to at intersects at 
. The line 
intersects at 
. The perpendicular bisector of segment 
intersects line 
at 
, and line 
intersects at . We denote by 
the second intersection point of the circumcircle of triangle 
with . Prove that, as point varies, points and remain fixed.
into two regions, two parallel planes divide it into three regions, and two intersecting planes divide space into four regions. Consider the six planes which the faces of the cube 
lie on, and the four planes that the tetrahedron 
lie on. How many regions do these ten planes split the space into?
be a sequence of positive integers satisfying the following property: for all positive integers 
, for all distinct integers 
and for all distinct integers 
,![\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]](http://latex.artofproblemsolving.com/d/1/6/d1632ff7e5e55319cdbf56f45dca9ed84343aa47.png)
Prove that there exist two integers 
and such that 
for all 
.
on the board. Charlotte has two operations available: the GCD operation and the LCM operation. and written on the board, erasing them, and writing the integer 
. and written on the board, erasing them, and writing the integer 
. is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is . Find all winning integers among 
and, for each of them, determine the minimum number of GCD operations Charlotte must use. denotes the greatest common divisor of and , while the number denotes the least common multiple of and .
be an integer. We consider a square grid of size 
divided into 
unit squares. The grid is called balanced if:
, 
or 
.
.
, the smallest integer 
such that any balanced grid always contains an 
square whose absolute sum of the 
cells is less than or equal to 
.
of distinct positive real numbers is called radiant if it satisfies the following property: if and are two distinct elements of , then 
is also an element of .
?
or 
.
.
be a cyclic quadrilateral, and 
the midpoints of sides 
respectively. Prove that the perpendiculars from 
to 
, from 
to 
, from 
to 
, and from 
to 
are concurrent.
)
)
be a cyclic quadrilateral, and the midpoints of sides respectively. Prove that the perpendiculars from to , from to , from to , and from to are concurrent.
is a parallelogram, we get 
, moreover 
then 
is a parallelogram
. Similarly, 
to 
and to 
meet at 
If 
is the circumcenter of 
then 
and 
implies that 
is a paralelogram 
is the reflection of about the midpoint 
of 
But is also the midpoint of 
Hence, perpendiculars from to 
and to also pass through
be a cyclic quadrilateral, and the midpoints of sides respectively. Prove that the perpendiculars from to , from to , from to , and from to are concurrent.
and the point of concurrency be 
. Since we can drop perpendiculars from to , at their midpoints and respectively, 
is a parallelogram. Similarly, 
is also a parallelogram.
as the vertices of . The point of concurrency is nothing but the vector sum of the midpoints of opposite sides.
which is the same point as 
. Thus, we arrive at the same point for each pair of opposite sides.
-gon, being even, if I'm not mistaken.
, 
, 
, 
the projections of , , , onto , , , , respectively. Now, It is easy to see that is parallelogram. Let 
be the intersection of its diagonals, 
be the circumcenter of . maps into , maps into , hence, the line 
is maped in to the line 
. But KM is perpendicular to , therefore, is also perpendicular to .
, 
, 
, respectively, is perpendicular to , , 
.
such that 
.
. I am so sorry, I have thought that these diagnols of this ABCD are perpendicular to each other 
. Sorry for my big mistake.