denote the set of all positive integers less than 
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
blocks, each weighing at least 
;
.
with 
you can choose a subset of the blocks whose total weight is at least 
.
such that 
.
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and 
cells in the intersection. Babek wants to have all these 
high-quality questions. This test was created by multiple 130+ scorers, and we have tried our very best to produce the highest quality test possible. We've blended the question types in a good distribution, and maintained a realistic difficulty. (something most AOPS mocks fail at)
,
be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the x-axis. The left edge of 
-
as many lattice points as does 
,
of the lattice points contained in 
is 27 times the fraction of lattice points in 
.
and 
be positive real numbers with 
.
be the maximum possible value of 
for which the system of equations![\[ a^2+y^2=b^2+x^2=(a-x)^2+(b-y)^2\]](http://latex.artofproblemsolving.com/1/3/4/13432a7d6e5c1465e08d513ee64c6c7ce99d75f9.png)
has a solution in 
satisfying 
and 
.
can be expressed as a fraction 
,
and 
are relatively prime positive integers. Find 
.
Y by
Let 
be a fixed equilateral triangle. For every arbitrary line 
that what goes through vertex 
consider the points 
and 
, which represents the foot of the perpendicular taken from the points 
and 
, respectively on the line 
. Determine the locus of the points 
, which form an equilateral triangle 
.
be a fixed equilateral triangle. For every arbitrary line 
that what goes through vertex 
consider the points 
and 
, which represents the foot of the perpendicular taken from the points 
and 
, respectively on the line 
. Determine the locus of the points 
, which form an equilateral triangle 
.This post has been edited 3 times. Last edited by parmenides51, Mar 25, 2021, 7:28 AM
Reason: vertex instead of vertice
Reason: vertex instead of vertice
is midpoint of 
so 
lies on circle and so 
Same way 
and so 
is equilateral.
is symmetric for point 
.
is equilateral too. And obvious, that point 
lies on circle with center 
