board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
steps and has to reach 
is valid but 
is not valid.
,
,
,
and 
according to the diagram. The edges from points 
,![\[[ABCD]^2+[ABFE]^2+[ADHE]^2=[BCGF]^2+[CDHG]^2+[EFGH]^2,\]](http://latex.artofproblemsolving.com/c/f/b/cfb6dc6726af4711bcfe7827b6531471a75270fb.png)
where ![$[XYZW]$](http://latex.artofproblemsolving.com/9/c/2/9c25ef1b5cfb129136334ed811fb699a84f48a13.png)
denotes the area of quadrilateral 
.
and 
be on segment 
and 
.
and 
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
be four distinct points in the plane.
are distinct and parallel, then
lies on the segment 
plus the oriented angle 
e
.
equals 0°, then
a
Let 
Prove that ![$$AP \in \left[\frac{5-\sqrt{5}}{10}, \frac{-1+\sqrt{5}}{2}\right] \cup \left[\frac{5+\sqrt{5}}{10}, \frac{1+\sqrt{5}}{2}\right].$$](http://latex.artofproblemsolving.com/7/8/e/78e428cbc99ce4e9448f5c4cec86f50567123812.png)
be reals such that 
. Prove that
specifies the greatest common divisor of 
and 
.
and 
Prove that
Let 
Prove that
![\[
\frac{(x+y+z)^5 - x^5 - y^5 - z^5}{(x+y)(y+z)(z+x)} = 0
\]](http://latex.artofproblemsolving.com/7/8/f/78f8e6c2570684631dc288144e13568d71e68ec8.png)
cannot completely cover a square of side length 
,
,
.
to meet the circumcircle of 
at 
,
.
Y by
Let 
be an acute-angled triangle. Let 
be the midpoint of the arc 
which contain the point 
. Let 
and 
be the foot(s) of the perpendicular(s) of the point to the lines 
and 
, respectively. Define 
analogously.
a) Prove that the line
cuts in the midpoint.
b) Prove that the lines
and 
are concurrents.
be an acute-angled triangle. Let 
be the midpoint of the arc 
which contain the point 
. Let 
and 
be the foot(s) of the perpendicular(s) of the point to the lines 
and 
, respectively. Define 
analogously.a) Prove that the line
cuts in the midpoint.b) Prove that the lines
and 
are concurrents.This post has been edited 1 time. Last edited by mathisreal, Feb 14, 2022, 9:52 PM
the midpoints of 
.
is perpendicular to 
,
are in the simson line of the point 
is a simson line and 
is a simson line.
,
,
is parallel to the angle bisector of 
.
and 
is parallel to 
.
.
is the simson line of the triangle ABC relative to verex A. So, It suffices to show that the projection of 
,
, C = 
and 
.
but this simplifies to 
,
.
is parallel to the internal bissector going trought A. This is proven checking if the angular coefficients of the lines are equal. The angular coefficient of the line ab is 
- 
Superior part of the fractions
)
Inferior part of the fraction
and do some simplifications.
superior part of the fraction
inferior part of the fraction
. And this let us left with 
.
that is 
are homothetic to the triangle ABC and their respective lines are parallel, the Simson lines meet themselves at the incenter of 
