AoPS Academy
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)
such that for all 
the followings hold:
be a triangle with circumcentre 
. The points 
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
. (By interior we mean, for example, that the point 
lies on the line and is between 
and 
on that line.)
be the circumcentre of triangle 
. Prove that the lines 
and are perpendicular.
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 
.
with orthocenter 
. On its circumcenter, choose an arbitrary point 
(other than 
) and let 
be the mid-point of 
. Now, we find three points 
on the line 
, respectively, such that 
. Show that 
are colinear.
with orthocenter . On its circumcenter, choose an arbitrary point (other than ) and let be the mid-point of . Now, we find three points on the line , respectively, such that . Show that are colinear.
be the circumrectangular hyperbola passing through . Suppose 
is the orthocenter of 
, and define 
analogously. Then 
. Also, one can easily see that 
, et al. are parallelograms. In particular, 
, and so 
(with similar results for 
). Then Pascal on 
gives that line passes through 
. But, due to the parallelogram condition, is the midpoint of segments 
and 
also. Thus, passes through . Similarly, 
, and so 
are collinear.
be . We only need to prove that 
. Denote the midpoint of 
and the antipode of by 
and 
respectively. Then it's obvious that 
, 
is a parallelogram. Let the projection of 
to 
be 
respectively. By 
and is the midpoint of 
, we can see 
. However, 
, it means that 
, and now it's trivial to see that 
are both lying on the circle with a diameter 
. So . Q.E.D.
as a unit circle in the complex plane centered at 
. lies on iff 
, hence 
. 
iff 
. Substituting 
from the previous equation, we obtain 
. Similarly 
and 
. Also, 
. 
and 
are cyclical. Now we evaluate the determinant
, 
, 
are collinear and , , and are collinear, which suffices to prove the statement of the problem. 
be . We only need to prove that . Denote the midpoint of and the antipode of by and respectively. Then it's obvious that , is a parallelogram. Let the projection of to be respectively. By and is the midpoint of , we can see . However, , it means that , and now it's trivial to see that are both lying on the circle with a diameter . So . Q.E.D.![[asy]
size(150);
pair A=dir(130),B=dir(-160),C=dir(-20),H=A+B+C,P=dir(-93),M=1/2*(H+P),L=1/2*(B+C);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H$",H,dir(-120));
dot("$L$",L,dir(L));
dot("$M$",M,dir(M));
dot("$P$",P,dir(P));
dot(extension(A,P,L,M));
draw(unitcircle,red);
draw(A--B--C--A,blue);
draw(H--P,brown);
draw(L--M^^A--P,green);
markscalefactor=0.01;
draw(rightanglemark(P,extension(A,P,L,M),L));
markscalefactor=0.008;
add(pathticks(P--M,1));
add(pathticks(M--H,1));
[/asy]](http://latex.artofproblemsolving.com/9/8/8/988a0255beaecf42b595d08b6f7ade574df2c582.png)
. Like, this can of course be proven using complex numbers, but how to prove it synthetically ?
be . We only need to prove that . Denote the midpoint of and the antipode of by and respectively. Then it's obvious that , is a parallelogram. Let the projection of to be respectively. By and is the midpoint of , we can see . However, , it means that , and now it's trivial to see that are both lying on the circle with a diameter . So . Q.E.D.. Like, this can of course be proven using complex numbers, but how to prove it synthetically ?
with ratio 
and use euler circle
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)
and others analogously. So we need to show collinearity of 
and 
.
and then its trivial.
be the circumcenter of 
be the second intersection of 
with 
and 
be the midpoint of 
respectively, then 
and 
so 
hence we conclude that 
and 
