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.
on the board. For each procces, the stutent deletes the number x and replaces it with either 
or 
or 
. Is this possible to make the number 
on the board?
be different real numbers. Prove that![\[\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}=\left\{\begin{array}{ll}
0, & \text { if } n \text { is even; } \\
1, & \text { if } n \text { is odd. }
\end{array}\right.\]](http://latex.artofproblemsolving.com/6/5/2/6525fc6b451ff5a3c83ff2a0541b063a7cf065c3.png)
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
satisfying
for all real numbers 
and 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
that satisfy the condition 
for the natural number 
Here 
and 
are natural numbers. Is there a natural number 
such that each of the numbers 
is the last digit of one of the numbers written by the teacher on the board? Justify your opinion.
we have 
.
. has orthocenter . Let be a point distinct from the vertices on the circumcircle of . Suppose that circle 
meets 
at 
, and circle 
meets at 
. Prove that as moves on the circumcircle, the reflection of across line 
also moves on a fixed circle. there exists an -digit number divisible by 
all of whose digits are odd.
and 
.
?
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and columns and looks at the 
cells in the intersection. Babek wants to have all these cells in a same color, but Azer doesn't. With at least how many colors, Azer can reach his goal?
and 
are constructed in the extirior of a right-angled triangle with 
.Let 
be the midpoint of .Given that 
and 
.Find the length of .
, but I might be wrong with calculations.
are equilateral 
![$[CG] = [CA] = [GA]$](http://latex.artofproblemsolving.com/a/b/4/ab421e7805fb68ff78fa7ef170bc6193534bf750.png)
, ![$[FB] = [AB] = [AF]$](http://latex.artofproblemsolving.com/c/6/f/c6fe82d238ee035deb72c832bd709a8eab1b177f.png)
and 
by SAS ![$[GB] = [FC]$](http://latex.artofproblemsolving.com/4/d/4/4d49ac5d96d4e5de516d1ad4566e59c96f8ad94d.png)
.
are G-, F-medians of 
![$[GM]^2 = \tfrac{1}{4}\left(2[GB]^2 + 2[GC]^2 - [BC]^2\right)$](http://latex.artofproblemsolving.com/e/1/f/e1fb86879f913c0da9bf27f79fb4d5c977e1ac8f.png)
and ![$[FM]^2 = \tfrac{1}{4}\left(2[FB]^2 + 2[FC]^2 - [BC]^2\right)$](http://latex.artofproblemsolving.com/9/b/e/9bed683d4e3a6886c553e24ef14d12ad349dbb38.png)
![$[FM]^2 - [GM]^2 = \tfrac{1}{2}\left([FB]^2 - [GC]^2\right) = \tfrac{1}{2}\left([AB]^2 - [CA]^2\right) = \tfrac{1}{2} [BC]^2 =$](http://latex.artofproblemsolving.com/a/e/4/ae4bfe990229b05c41464d1bba8ed0f5941eeae3.png)
![$[BC]^2 = 144$](http://latex.artofproblemsolving.com/1/d/7/1d729c3d6702a814cca0ff398eb2165df65bd570.png)
![$[BC] = 12$](http://latex.artofproblemsolving.com/f/c/a/fca981613c8213607683ec4f336bfad3402be878.png)
.
and are constructed in the extirior of a right-angled triangle with .Let be the midpoint of .Given that and .Find the length of .
such that
be the midpoint of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
and 
be the midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
, 
and 
.
is right at its vertex 
hence
, 
formula (cosine formula ) gives
, so we have
we get
we get
