AoPS Academy
such that for any two infinite sequences of positive integers 
and 
, such that 
for all 
, there always exists a real number 
such that 
for all 
.
+1 w
be a positive integer. There are circles drawn on a chalkboard such that any two circles intersect at 
distinct points and no 
circles pass through the same point. Turbo the snail slides along the circles in the following manner, leaving snail goo behind. Initially he moves on one of the circles in clockwise direction. He keeps sliding along until he reaches an intersection with another circle. Then, he continues his journey on this new circle and also changes the direction he is moving in. We define a snail orbit to be the covering of the whole surface of a circle with turbo's goo, and specifically only a single layer of it. Prove that for every odd there exists at least one configuration of circles with a single snail orbit, and find all such that there is exactly one of the aforementioned configuration type.
and 
be two isogonal conjugate with respect to triangle 
. Let 
and 
be two points lying on the circle 
such that 
and 
are perpendicular and parallel to bisector of 
, respectively. Prove that 
and 
bisect two arcs 
containing 
and not containing , respectively, of 
.
such that there exists an integer 
satisfying
, with 
is drawn externally on side 
of 
. Point is taken on 
. Find 
if 
.
for all 
.
for all 
.
, , 
, 
, 
, only. Note that 
, and 
, so that 
.
satisfys the condition.
Note that 
rectangle can be cut to 
rectangle if 
. Partition the grid to four of the 
rectangle, as following.![[asy]
draw((0, 0) -- (45, 0));
draw((45, 0) -- (45, 45));
draw((45, 45) -- (0, 45));
draw((0, 45) -- (0, 0));
draw((0, 23) -- (23, 23));
draw((22, 0) -- (22, 23));
draw((23, 45) -- (23, 22));
draw((22, 22) -- (45, 22));
fill((22, 22) -- (22, 23) -- (23, 23) -- (23, 22) -- cycle, black);
[/asy]](http://latex.artofproblemsolving.com/c/d/a/cdaf5762b57ff4a04c17001a7b3f73a0cde7b496.png)
or 
, each of the rectangle can be cut to 
, or rectangle, so we are done.
. The idea is to color each of the square, either black or white. Moreover, let (, 
) denotes the square with the th row, th column. Then color the grid such that the square is black if and only if is odd and 
, or even and 
. The example with 
replaced to 
are shown below.![[asy]
unitsize(1cm);
for(int i = 0 ; i <= 5 ; i = i + 1) draw((0, i) -- (5, i));
for(int i = 0 ; i <= 5 ; i = i + 1) draw((i, 0) -- (i, 5));
for(int i = 0 ; i <= 4 ; i = i + 2) fill((i, 5) -- (i + 1, 5) -- (i + 1, 3) -- (i, 3) -- cycle, black);
for(int i = 1 ; i <= 3 ; i = i + 2) fill((i, 3) -- (i + 1, 3) -- (i + 1, 1) -- (i, 1) -- cycle, black);
for(int i = 0 ; i <= 4 ; i = i + 2) fill((i, 0) -- (i + 1, 0) -- (i + 1, 1) -- (i, 1) -- cycle, black);
[/asy]](http://latex.artofproblemsolving.com/9/3/e/93e8fdfd44e05ba88096780859aafcc7aace2184.png)
black squares and same for white. but since the centre (, ) is colored white, the number of the black square is ![\[
23\cdot 23 + 22\cdot 22 = 1013.
\]](http://latex.artofproblemsolving.com/c/f/5/cf58b758901a6796c8c4facd6641d3502ad05ad4.png)
So the number of white square is 
, contradiction.
and 
.
Equality holds when 
or 
or 
grid has had the central unit square removed. For which positive integers 
and 