AoPS Academy
be a triangle with 
. A circle 
is tangent to sides 
and 
, and 
is the midpoint of 
. Points 
and 
lie on sides and , respectively, such that segment 
is tangent to circle at point 
. Let 
and 
be the orthocenters of triangles 
and 
, respectively. Prove that line 
bisects segment 
.
, construct a graph 
on 
, where 
are adjacent if and only if divides 
. Prove that is disconnected for infinitely many
-plane and a tiling of this plane with white and red unit squares in a checkerboard pattern (ensure the vertices of the squares are at lattice points). For each square, draw a regular tetrahedron with the property that each vertex of the square coincides with the midpoint of one of the edges of the tetrahedron. Notice for every square there are exactly two distinct possible ways the tetrahedron could be drawn. On the white squares ensure that the upper edge of the tetrahedron is parallel to the 
-axis and for the red squares ensure that the upper edge of the tetrahedron is parallel to the 
-axis.
rotation about the or axis it is clear that the tetrahedrons fit together seamlessly and their vertices coincide with the midpoints of the edges of adjacent tetrahedron.![[asy]
size(200);
import three;
currentprojection = perspective(3,5,7);
draw(surface((-4,-4,0)--(4,-4,0)--(4,4,0)--(-4,4,0)--cycle), gray+opacity(0.2));
draw((-4,0,0)--(4,0,0), black); // x-axis
draw((0,-4,0)--(0,4,0), black); // y-axis
triple[] tetra1 = {(1,-1,1), (1,3,1), (-1,1,-1), (3,1,-1)};
triple[] tetra2 = {(-1,-3,1), (-1,1,1), (-3,-1,-1), (1,-1,-1)};
triple[] tetra3 = {(-3,1,1), (1,1,1), (-1,-1,-1), (-1,3,-1)};
triple[] tetra4 = {(-1,-1,1), (3,-1,1), (1,-3,-1), (1,1,-1)};
void drawTetra(triple[] T, pen color) {
draw(surface(T[0]--T[1]--T[2]--cycle), color+opacity(0.4));
draw(surface(T[0]--T[1]--T[3]--cycle), color+opacity(0.4));
draw(T[0]--T[2]--T[1]--T[3]--cycle, black+linewidth(0.8));
draw(T[0]--T[1], black+linewidth(0.8));}
drawTetra(tetra2, red);
drawTetra(tetra4, grey);
drawTetra(tetra1, red);
drawTetra(tetra3, grey);
draw(surface((-1,-1,1)--(-1,1,1)--(0,0,0)--cycle), red+opacity(0.4));
draw(surface((-1,-1,1)--(-1,1,1)--(-2,0,0)--cycle), red+opacity(0.4));
draw((-1,-1,1)--(-1,1,1), linewidth(0.8));
draw((2,2,0)--(2,-2,0)--(-2,-2,0)--(-2,2,0)--cycle);
draw((2,0,0)--(-2,0,0));
draw((0,2,0)--(0,-2,0));
draw((0,0,0)--(-1,1,1), linewidth(0.8));
draw((0,0,0)--(-1,-1,1), linewidth(0.8));
draw((-2,0,0)--(-1,1,1), linewidth(0.8));
draw((0,0,-4)--(0,0,4), black, Arrow3);
label("$z$", (0,0,4.5), N);
label("$x$", (4.5,0,0), E);
label("$y$", (0,4.5,0), N);
[/asy]](http://latex.artofproblemsolving.com/8/4/1/84127fa541e6ee5089fdd53383a2ee3fadc4cebe.png)
-plane do this for all planes 
where 
is a non-zero integer (we are "stacking" the sheets of tetrahedron). It is not hard to see that after drawing such tetrahedron, we are left with a collection of octahedron in space. We now go through a separate procedure for partitioning each of these octahedron into tetrahedron.-plane we can simply divide it into four tetrahedron. Otherwise let the octahedron touch the -plane at . Call the octahedron 
. Then let 
be the result of after applying the homothety centered at with factor 
. Then 
can be tiled with a finite number of tetrahedron (specifically 
). Then repeat this process with and so on.-plane intersects each of the original tetrahedron at exactly 
points and non of the tetrahedron formed by portioning the octahedron touch the -plane, we are finished.
![[asy]
import geometry;
size(2cm);
pair A=dir(90); pair B=dir(210); pair C=dir(330);
pair A1=.5(C+B); pair B1=.5(A+C); pair C1=.5(A+B);
pair A2=.5(B1+C1); pair B2=.5(A1+C1); pair C2=.5(A1+B1);
pair A3=.5(B2+C2); pair B3=.5(A2+C2); pair C3=.5(A2+B2);
pair A4=.5(B3+C3); pair B4=.5(A3+C3); pair C4=.5(A3+B3);
filldraw(A--B1--C1--cycle, red+opacity(.5));
filldraw(B--A1--C1--cycle, red+opacity(.5));
filldraw(C--B1--A1--cycle, red+opacity(.5));
filldraw(A1--B2--C2--cycle, red+opacity(.5));
filldraw(B1--A2--C2--cycle, red+opacity(.5));
filldraw(C1--B2--A2--cycle, red+opacity(.5));
filldraw(A2--B3--C3--cycle, red+opacity(.5));
filldraw(B2--A3--C3--cycle, red+opacity(.5));
filldraw(C2--B3--A3--cycle, red+opacity(.5));
filldraw(A3--B4--C4--cycle, red+opacity(.5));
filldraw(B3--A4--C4--cycle, red+opacity(.5));
filldraw(C3--B4--A4--cycle, red+opacity(.5));
dot((0,0));
[/asy]](http://latex.artofproblemsolving.com/9/1/8/9181484fa01d455685df228ad52b299a31e7d185.png)
.
Where 
such that![\[
f(x^4 + 5y^4 + 10z^4) = f(x)^4 + 5f(y)^4 + 10f(z)^4
\]](http://latex.artofproblemsolving.com/3/0/8/308079e75e1acd9a9f896885387b67a7b4020c16.png)
for all 
.
