and in 
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 
.
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
be a function satisfying![\[
f(f(x)) = x^2 - 6x + 6
\quad \text{for all} \quad x \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/8/a/e/8aebed1e260316d675ca7d104d83fabcd8b5717e.png)
Given that![\[
f(i) < f(i+1) \quad \text{for} \quad i = 0, 1, 2, 3, 4, 5,
\]](http://latex.artofproblemsolving.com/9/f/9/9f92574f9fe14e2e30c3e51e194c661fbd4ece07.png)
find the value of![\[
f(0) + f(1) + f(2) + \cdots + f(6).
\]](http://latex.artofproblemsolving.com/4/0/7/4075d7efa78dda4c46a5cf764fb8852e8c0fb09b.png)
be four distinct points in the plane.
and 
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 Adventure10
Let 
be the circumcircle of acute triangle 
, 
-altitude of (
), 
, 
-midpoint of 
, 
-feet of perpendicular from 
to 
.
Proved that point of intersection
and 
lies on circumcircle of triangle 
be the circumcircle of acute triangle 
, 
-altitude of (
), 
, 
-midpoint of 
, 
-feet of perpendicular from 
to 
.Proved that point of intersection
and 
lies on circumcircle of triangle 
This post has been edited 4 times. Last edited by Ovchinnikov Denis, Sep 5, 2010, 9:31 AM
is the infinity point of the line 
is the midpoint of 
Then 
Therefore, 
and the orthogonal projection of 
are collinear 
Lines 
are antiparallel with respect to lines 
Therefore:
intersect circle 
.
.
∽
,
