AoPS Academy
are positiv real number, such that 
prove inequality 
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 
, solve the given equation in 
., the equation has at most 
real solutions.
be a parallelogram. Let line 
externally bisect 
and let 
be the line passing through 
which is parallel to line 
. Suppose that meets line 
at point 
and at point 
, and that meets the internal bisector of 
at point 
. Further let circle 
meet line 
at point 
and the internal bisector of meet circle 
at point 
., , 
, and 
are concyclic.
and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
is the incentre and point is defined such that 
and 
is the perpendicular bisector of 
let 
and 
be the circles passing through 
and 
respectively such that 
is tangent to and define 
to be a point such that it lies on both the circles and prove 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 on the left pan, and display the number of possible solutions. (The weights can be placed on both pans and the program starts with the numbers 
. What will be displayed after three successive runs: 5,2,5,1,4 | 5,2,5,1,11 | 5,2,5,1,20?
with orthocenter 
. Extend 
to meet at . The angle bisector of 
meets the midpoint of 
. If 
, compute the area of 
.
with side length of 
. The point 
is selected on the diagonal such that 
. Denote the circumcenters of 
as 
. Find the length of 
such that 
. Denote the midpoints of 
as 
respectively, compute the area of 
given the area of is 
.
? (The triangles may not overlap and may not exceed the margins of the square.)
can be tiled with 
congruent right-angled triangles, whose sides are of lengths 
, 
, and 
.![\[ \frac{b^2 \sqrt{3}}{2} \]](http://latex.artofproblemsolving.com/0/5/3/0534d2d3c3c9a4255ad911f638395ac8bbdbad27.png)
![\[
S_{\text{sq}} = k \cdot \frac{b^2 \sqrt{3}}{2}. \tag{1}
\]](http://latex.artofproblemsolving.com/c/d/8/cd85edeb4f81a5d97aff694c7b08606b151e0b98.png)
, is obtained by the contribution of an integer number of lengths , , and , hence![\[
a = m b \sqrt{3} + n b,
\]](http://latex.artofproblemsolving.com/3/d/1/3d1848972199602a32d9a6a8903e6e91b521a842.png)
where 
, and at least one of the numbers 
and 
is different from zero.![\[
S_{\text{sq}} = a^2 = \left(m b \sqrt{3} + n b\right)^2 = b^2 \left(3m^2 + n^2 + 2 \sqrt{3} m n\right). \tag{2}
\]](http://latex.artofproblemsolving.com/0/7/b/07b4a34ff9d1c9cbace32383f66c8ed22b6d40f6.png)
![\[
3m^2 + n^2 + 2 \sqrt{3} m n = k \cdot \frac{\sqrt{3}}{2},
\]](http://latex.artofproblemsolving.com/f/d/8/fd8cfc1180c70e32577631eab0b11511a5d1e5b3.png)
i.e.,![\[
6m^2 + 2n^2 = (k - 4mn) \sqrt{3}. \tag{3}
\]](http://latex.artofproblemsolving.com/c/4/d/c4dcb8e4abb426ce0ca323c336f6e62915df1b95.png)
and from equality (3), it follows that 
.![\[
\sqrt{3} = \frac{6m^2 + 2n^2}{k - 4mn},
\]](http://latex.artofproblemsolving.com/f/c/4/fc47449a7d0ec56473243683c71d56946a491565.png)
which contradicts the fact that 
is irrational, because 
is a rational number.
