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 a trapezoid. By 
we denote the angle bisector of angle 
. Let 
and 
. Prove that 
is cyclic.
such that
Proposed by Ngo Van Trang, Vietnam
write all its divisors in increasing order: 
. such that 
.
such that 
.
.
be an inscribed quadrilateral. Circles of diameters 
and 
intersect at points 
and 
, and circles of diameters 
and 
intersect at points 
and 
. The circles of diameters 
and 
intersect in two points 
and 
. Prove that the lines 
and 
are concurrent.
be distinct real numbers such that 
Find the value of 
be distinct real numbers such that 
Find the value of 
be distinct real numbers such that 
Find the value of 
for which there exists a sequence 
of positive integers satisfying the following conditions:[list]
for all integers 
for all integers 
the sequence 
is a permutation of
[/list]
.
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be non-negative real numbers such that 
Prove that
be positive real numbers such that 
and 
, prove that
moves.
. Then suppose 
and 
correspond to the same state (some distinct values of 
and 
must satisfy this by PHP). Therefore, applying the algorithm 
times to a state returns that state back to itself. This is true no matter what state we are currently at, since every piece is unique on the cube. This means that after algorithms at the beginning, we will return to the solved state.
moves.
and then 
then trivially the puzzle gets resolved after applying this finite sequence of legal moves.
and 
Prove that
Let 
Prove that
