be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
be a parallelogram. Let line 
externally bisect 
and let 
be the line passing through 
which is parallel to line 
.
and 
,
at point 
.
meet line 
at point 
and the internal bisector of 
at point 
.
,
are concyclic.
and 
.
and 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
is the incentre and point 
and 
is the perpendicular bisector of 
and 
be the circles passing through 
and 
respectively such that 
and 
are perpendicular.
weights, each weighing 
and another 
weights with 
each. Write a algorithm that determines the ways in which a scale can be balanced with a weight 
.
.
to meet 
meets the midpoint of 
.
,
.
.
.
as 
.
.
as 
respectively, compute the area of 
given the area of 
.
Y by brendan_cape, centslordm, Didier
Let 
be an altitude of acute scalene triangle 
. Points 
and 
are chosen on sides 
, 
respectively such that 
. The circumcircles of 
and 
intersect for the second time at point 
and intersect ray for the second time at points 
and 
respectively. Prove that the circumcenter of 
lies on the angle bisector of 
.
Proposed by Mykhailo Shtandenko
be an altitude of acute scalene triangle 
. Points 
and 
are chosen on sides 
, 
respectively such that 
. The circumcircles of 
and 
intersect for the second time at point 
and intersect ray for the second time at points 
and 
respectively. Prove that the circumcenter of 
lies on the angle bisector of 
.Proposed by Mykhailo Shtandenko
as 
.
be the midpoint of arc 
be the center of 
.
we have 
,
.
at 
also sends 
and 
.
,
which means 
to 
,
rotation. But we have deduced that 
,
,
as desired.
,
lies 
and 
,
is the midpoint of 
and 
.
.
lies on 
.
,
is the angle bisector of 
and 
