AoPS Academy
be a set of Natural numbers from 
to 
.
(
) distinct elements from this set.
. If 
and 
, show that the area of 
is not less than the area of 
, and the pairwise comparison of the sums of elements of all its subsets (with the empty set defined as having a sum of 0), which amounts to 
inequalities, these given comparisons satisfy the following three constraints:
and 
, if the sums of elements of and are greater than those of subsets 
and 
respectively, then the sum of elements of the union 
is greater than that of 
.
that satisfies all these conditions?
then prove \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2}\cdot\frac{abc}{ab+bc+ca}\ge \frac{5}{3}.$$
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?
that satisfy 
with orthocenter 
. Extend 
to meet at 
. The angle bisector of 
meets the midpoint of 
. If 
, compute the area of 
.
be a triangle. A circle intersects side at points 
and 
, side 
at points 
and , and side 
at points 
and 
. The points 
lie on the circle in that order. Suppose that 
and 
. Prove that 
.
be a triangle. A circle intersects side at points and , side at points and , and side at points and . The points lie on the circle in that order. Suppose that and . Prove that .
be the circle, with center 
and let 
Then 
and 
so 
Thus 
is the circumcenter, so because 
it follows that 
are equidistant from the midpoint of 
so 
