be a quadrilateral that has both an incircle and a circumcircle, with radii 
and 
respectively. If the diagonals 
are perpendicular, find the area of 
be a tower of number fields. Let 
be a prime of 
let 
and let 
be the residue class degree of 
in 
that satisfy the following equation for all real numbers 
:![\[
f\big(f(x - y)\big) = f(x) - f(y).
\]](http://latex.artofproblemsolving.com/c/3/c/c3c423f0ceb2fcb91064c5de218ec88f69c95e6d.png)
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)
Y by samrocksnature, Mango247, Mango247, Mango247
Let 
be a triangle with 
altitude. The point 
is chosen on the segment as follows such that 
. Let 
be the center of the circle circumscribed around by triangle 
and 
be the midpoints of 
and AB respectively. Lines 
and 
intersect at the point 
, a perpendicular to 
passing through point intersects lines and 
at points 
and 
respectively. Prove that 
.
be a triangle with 
altitude. The point 
is chosen on the segment as follows such that 
. Let 
be the center of the circle circumscribed around by triangle 
and 
be the midpoints of 
and AB respectively. Lines 
and 
intersect at the point 
, a perpendicular to 
passing through point intersects lines and 
at points 
and 
respectively. Prove that 
.
meet 
at 
,
be the orthocenter of 
.
,
and 
are concyclic. Therefore by spiral similarity, we have![\[\triangle OZX\sim\triangle ONA\sim\triangle EHB.\]](http://latex.artofproblemsolving.com/6/7/b/67bb04b0f30ffc0356f234aa2a2ad575f40da640.png)
Now it suffices to show that 
.
be the centroid of 
be the point on segment 
such that 
.
.![\[\frac{DK}{KZ}=2=\frac{HG'}{G'Z},\]](http://latex.artofproblemsolving.com/c/1/f/c1fdbf6802148e734804306de1634404b1b4e8fa.png)
so 
as well, and hence 
are collinear. Finally, since 
,
as desired.![[asy]
defaultpen(fontsize(10pt));
size(12cm);
pair A = dir(120);
pair B = dir(222);
pair C = dir(318);
pair M = midpoint(A--C);
pair N = midpoint(A--B);
pair H = foot(A,B,C);
pair K = (1/3)*A+(2/3)*H;
pair O = circumcenter(A,B,C);
pair Z = extension(M,N,O,K);
pair D = extension(B,C,O,K);
pair G = extension(B,M,C,N);
pair G1 = (1/3)*H+(2/3)*Z;
pair E = orthocenter(A,B,C);
pair X = 2*foot(circumcenter(O,M,Z),A,C)-M;
pair P = midpoint(B--C);
draw(A--B--C--cycle);
draw(A--H);
draw(N--Z);
draw(Z--D);
draw(O--E);
draw(Z--H);
draw(D--B);
draw(K--G1, dashed);
draw(A--P, dotted);
draw(circumcircle(A,B,C));
draw(circumcircle(A,M,N));
draw(circumcircle(O,M,Z));
draw(K--Z--X--cycle, fuchsia+1);
draw(K--H--B--cycle, fuchsia+1);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$H$",H,S);
dot("$E$",E,W);
dot("$D$",D,W);
dot("$M$",M,dir(M));
dot("$N$",N,dir(N));
dot("$K$",K,dir(135));
dot("$O$",O,dir(135));
dot("$G$",G,(0,1));
dot("$G'$",G1,(0,1));
dot("$Z$",Z,dir(45));
dot("$X$",X,(1,0));
dot("$P$",P,dir(270));
[/asy]](http://latex.artofproblemsolving.com/7/9/5/7954b877e2885f71438667e89a65b5619b3f7639.png)
