AoPS Academy
be the set of all positive integers. Determine all subsets 
of 
such that
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 
.
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 .
and straight line 
are given at the plane. Construct using a compass and a ruler the straightline which is parallel to and bisects the area of triangle 
is projected by a parallel projection into an isosceles right triangle 
with the angle 
One focal point of the normal hyperbola with the asymptotes A'B', A'C' is the midpoint F of B'C', the other one a reflection F' of F in A'. One hyperbola vertex is on the ray A'F at 
, the other is a reflection V' of V in A'. Circle (A') centered at A' and with radius A'V = A'V' is the hyperbola pedal circle. A normal to the line 
(obtained from by the used parallel projection) through the focus F meets the pedal circle (A') at points P, Q. Parallels to through P, Q are hyperbola tangents. One is tangent to the other hyperbola branch and it does not cut the triangle at all. If the other one touches the hyperbola branch passing through the triangle on the hyperbolic segment MN, where M, N are midpoints of B'- and C'- medians (i.e., the constructed parallel to intersects the segments A'B', A'C'), it divides in half the triangle area.
) the line passing through this vertice and parallel to 
contains points within the triangle and divides it into two parts. Let 
be the point of intersection of this line and 
, we assume that 
(if 
everything is clear), also let 
be the line we need to construct.
and 
the points of intersection with 
and respectively. Apparently, 
and 
. Multiplying these two equations, we obtain 
, whence we can easily construct the length of 
and the line therefore.
, whence we can easily construct the length of and the line therefore.
at some way, but how would you place it parallel with 
, can you write out your solution complete, so we can see if it is simpler?
at some way, but how would you place it parallel with , can you write out your solution complete, so we can see if it is simpler? or how to draw a parallel line? The latter is trivial when the line and one point is given. As to expressing , we know it equals 
. Since we know how to construct a proportional segment if three others are given, and how to construct the geometrical mean, we rewrite the needed expression as 
and after a chain of transformations obtain a segment equivalent to .
and similar at an easy way (I don't know, but seems it will have an easy way, but how?) and how do you finish a length 
if 
are already constructed?
and similar at an easy way (I don't know, but seems it will have an easy way, but how?) and how do you finish a length if are already constructed?
and 
, we can take another segment 
, the point 
on it such that 
, 
. Then by using as the diameter to draw a circle containing 
and 
and intersecting it with a straight line through perpendicular to , we'll get two segments 
units long.
and 
.
