AoPS Academy
be a cyclic quadrilateral, and let 
, 
, 
, and 
be the midpoints of 
, 
, 
, and 
respectively. Let 
, 
, 
and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.
with real coefficients is the average of two monic polynomials of degree with real roots.
of positive integers, for which![$$(a, b) + 3[a, b] = a^3 - b^3$$](http://latex.artofproblemsolving.com/b/8/3/b83f52a0152dc39b6d966880a1d6ca824a0f3786.png)
denotes the greatest common divisor of , and ![$[a, b]$](http://latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
denotes the least common multiple 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.
