[/quote]
,
segments 
and 
are parallel. Angle bisectors of 
and 
meet at 
.
and 
meet at 
.
is cyclic
and 
.
be an acute triangle with orthocenter 
and circumcenter 
.
and 
be points on the line segments 
and 
is a parallelogram. Prove that 
.
be the midpoint of side 
.
intersect 
at 
.
is tangent to 
,
.
,
be the orthocenter of it and 
be any point on the side 
.
are on the segments 
,
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
.
and 
and 
are positive integers, we say that 
and 
.
almost prime if the following holds: for any positive integers 
such that 
,
and 
,![$\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$](http://latex.artofproblemsolving.com/b/f/a/bfa17e8bb4cc05a84ed6a757f506d6d0ed2b0081.png)
is an odd number, let 
be a positive integer. If 
then 
.
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
and 
.
?
be positive integers satisfying ![\[\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.\]](http://latex.artofproblemsolving.com/b/8/c/b8c3fe14640a2dcb30b755b2347721da5d080a78.png)
Determine all possible values of 
.
users, some pairs of whom are friends. Whenever user 
is friends with user 
users have 
friends each, and 
Y by
Let 
and 
be monic quadratic trinomials, and let 
and 
be the vertices of the parabolas 
and 
, respectively. Let 
denote the minimum value of the function 
. It is known that the differences 
and 
are equal positive numbers. Find the angle between the line 
and the 
-axis.
and 
be monic quadratic trinomials, and let 
and 
be the vertices of the parabolas 
and 
, respectively. Let 
denote the minimum value of the function 
. It is known that the differences 
and 
are equal positive numbers. Find the angle between the line 
and the 
-axis.This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:25 PM
