AoPS Academy
with 
and at most one of them being zero, show that![\[\frac{1}{\max(a^2,b)}+\frac{1}{\max(b^2,c)}+\frac{1}{\max(c^2,a)}\geq\frac{27}{4}\]](http://latex.artofproblemsolving.com/9/e/1/9e14a9ef2d1a0a98006bcf716fa33c3d9ea29bd5.png)
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
and 
. Prove that
Equality holds when 
or 
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 
.
with 
. Point 
lies inside the rectangle such that 
is equilateral. Given that 
and the midpoint of are on the same line, compute the length of .
be reals with and 
. Prove that
that 
and 
. Prove that from the lengths 
, 
and 
, you can make a right triangle.
. From the given conditions 
. Choose the point on the bisector of angle 
such that it lies in the halfplane determined by 
not containing 
and 
. Then it is pretty well-known that the triangles 
and 
are equilateral. Thus
and 
. Hence the triangle 
has sidelengths equal to 
and is right-angled.
