AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
with diameter 
is given. The point 
lies in the interior of the circle, but not on . The line 
intersects in 
and 
. The tangent to at intersects the line through perpendicular to , at 
. The point 
lies on , and is such that 
is tangent to and 
.
, and are collinear.
has centre 
, and 
is a diameter of . Let be a point of such that 
. Let 
be the midpoint of the arc 
which does not contain 
. The line through parallel to 
meets the line 
at 
. The perpendicular bisector of 
meets at 
and at 
. Prove that is the incentre of the triangle 
lies inside triangle 
such that 
and 
. Point is the midpoint of segment . Point lies on segment with 
. Prove that 
.
be complex numbers satisfying 
. Show that there exist some among the 
complex numbers such that the modulus of the sum of these complex numbers is not less than 
.
be a function such that 
for all pairs 
of positive integers. Prove that there exists a positive integer 
which divides all values of 
.
, points and 
are taken on the legs and , respectively, so that 
and 
. Find the acute angle between line segments 
and 
.
. Find the max 
that 
for all 
.
be real numbers such that ![\[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \]](http://latex.artofproblemsolving.com/8/f/5/8f5d49c9b44e3db3cd85fea7b6521e1dc02d4269.png)
Prove that ![\[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]](http://latex.artofproblemsolving.com/c/0/4/c0438d6eba686d0a11fd4decb4afa9a14aa50bac.png)
be real numbers. Prove that
where the indices are taken modulo 8, i.e., 
, 
, and 
. In which cases does equality hold?
be a circumscribed quadrilateral with incircle 
and no two opposite angles equal. Let be an arbitrary point on the diagonal 
, which is inside . The segments 
and 
intersect at 
and . The tangents to at and intersect at . Prove that lies on the line .
be an acute triangle with 
, midpoint of side , altitude 
(
), and orthocenter 
. A circle passes through points 
and 
, is tangent to line 
, and intersects the circumcircle of triangle at a second point 
. The circumcircle of triangle 
intersects line 
at a second point 
. Prove that the lines 
and 
are perpendicular.
be a convex quadrilateral such that 
. Suppose that a point on the side 
satisfies the equality![\[AE\cdot ED + BE^2=CD\cdot AE.\]](http://latex.artofproblemsolving.com/d/7/5/d75cc3a449a56b43eac574cc44e9932a2e63b33b.png)
.
-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)
and 
.Find max 
.
for all 
be real numbers. Prove that
