for which the inequality ![\[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]](http://latex.artofproblemsolving.com/4/d/b/4db42ebabc7309bba9629a9a7125be43abf113f0.png)
holds for all real numbers 
,
,
,
.
and 
are given. Let 
be the midpoint of the segment joining their centers, 
,
be arbitrary points on 
.
.
,
be the circumcenter and the incenter of an acute-angled scalene triangle 
;
,
,
be the touching points of its excircle with the side 
and the extensions of 
,
respectively. Prove that if the orthocenter of the triangle 
lies on the circumcircle of 
.
be a point inside a quadrilateral 
such that 
.
,
,
are isogonally conjugated to 
,
,
,
concur.
can be written in the form![\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]](http://latex.artofproblemsolving.com/9/e/9/9e96c07ea3b0750af8bed344df6e8a86fcf0d772.png)
where 
and 
are positive integers. Find the value of 
.
be an acute triangle with 
,
.
be the altitude from vertex 
.
and 
be the feet of the perpendiculars from 
to the lines 
and 
,
intersect the circle 
and 
such that 
.
be the midpoint of segment 
intersect line 
at point 
.
.
is parallel to line 
.
be the orthocenter of triangle 
,
be the midpoint of segment 
is the center (HBC). Let point 
lie on line 
such that ray 
is the angle bisector of 
.
intersect at a point lying on circle 
(i.e., the square on 
has 
,
,
to the centers of the opposite squares are concurrent.
be positive integers. Maximize 
if 
.
,
,
.
where 
is a natural number.What is the remainder when 
Y by Adventure10, Mango247
We are given a ruler with two marks at a distance 1. With its help we can do all possible constructions as with a ruler with no measurements, including one more: If there is a line 
and point 
on , then we can construct points 
for which 
. By using this ruler, construct a perpendicular from a given point to a given line.
and point 
on , then we can construct points 
for which 
. By using this ruler, construct a perpendicular from a given point to a given line.This post has been edited 1 time. Last edited by Pinko, Sep 19, 2019, 8:04 AM
