AoPS Academy
and in 
steps and has to reach again. But in between one of the s steps and can't go . Find probability. For example 
is valid but 
is not valid.
, 
, 
, 
and 
according to the diagram. The edges from points and 
, respectively are pairwise perpendicular. Prove that ![\[[ABCD]^2+[ABFE]^2+[ADHE]^2=[BCGF]^2+[CDHG]^2+[EFGH]^2,\]](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 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
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 is equal to the oriented angle DCB.
lies on the segment 
, then
plus the oriented angle 
equals 
.
plus the oriented angle 
equals 0°, then
and are parallel. plus the oriented angle equals , then and are parallel.
and be points on a plane such that 
Let be a point on that plane such that 
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)
cannot completely cover a square of side length 
, if the three smaller squares do not overlap in their interiors (i.e., they may touch at edges or corners, but no part of one lies over another).
with 
, denote the incenter of the triangle as 
. Extend 
to meet the circumcircle of 
at 
, find the length of 
.
be a triangle, its incenter and 
the touchpoints of the incircle with sides , , and , respectively. Let be the intersection of segment 
with line 
. Prove that 
, and the midpoint of are collinear.
be a triangle, its incenter and the touchpoints of the incircle with sides , , and , respectively. Let be the intersection of segment with line . Prove that , and the midpoint of are collinear.
as 
respectivily. Let 
be a point such that 
is diameter on the incircle. and now let the tangent from to the incircle meet 
at 
. By newton's theorem on 
we have 
concurrent at . Now let 
and 
, then by pascal on the degenerate hexagon 
we have 
colinear and now since is the ortocenter of 
we have 
. Now by brokard on 
(here we take all the polars w.r.t. incircle btw) we have that 
and now by la'hire we have 
and now by la'hire on definitions of 
and by the last fact we got, we have 
and 
and now by brokard on 
and the past facts we got, we have 
colinear and 
colinear. Now note that lies on the 
-symedian on 
and since 
is cyclic we have that is midpoint of 
now let 
be the point of infinity of 
then by projecting:
Thus we are done 
as respectivily. Let be a point such that is diameter on the incircle. and now let the tangent from to the incircle meet at . By newton's theorem on we have concurrent at . Now let and , then by pascal on the degenerate hexagon we have colinear and now since is the ortocenter of we have . Now by brokard on (here we take all the polars w.r.t. incircle btw) we have that and now by la'hire we have and now by la'hire on definitions of and by the last fact we got, we have and and now by brokard on and the past facts we got, we have colinear and colinear. Now note that lies on the -symedian on and since is cyclic we have that is midpoint of now let be the point of infinity of then by projecting:Thus we are done 
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)
