AoPS Academy
types of candy, numerated type 1, type 2, ... type n. Initially she takes 
candy pieces and places them in a row on a table. Then, she chooses one of the following operations (if available) and executes it:
She eats a candy of type 
, and in its position in the row she places one candy type 
followed by one candy type 
(we consider type 
to be type 1, and type 0 to be type ).
She chooses two consecutive candies which are the same type, and eats them. for which Celeste can leave the table empty for any value of 
and any configuration of candies on the table.
be a positive integer. Given is a subset 
of 
with 
elements. Prove that there exist three elements 
from such that 
.
be a convex hexagon with 
and 
.
which is equidistant from sides 
and 
.
and 
are the centers of mass of 
and 
, show that 
.
such that if we choose 5 if them then all of them doesn't belong to a circle and 
be cyclic . So Now we can conclude we can choose 5 of them such that problem conditions be wrong .(why?) So there is no 7 point like this . And you can see if we haven't atleast a circle which contain 
of them we have contradiction.
and 
. Assume for the contrary there is two points 
, none of them lies on 
.
. Then four of these form cyclic quad and it should include the points . WLOG, let's say 
is cyclic.
. Then 
cyclic follows from 
and is not on the .
, we get the the contradiction. So such does not exist.
. One can check that this number is not an integer.
points on a plane, no three of which are collinear. Among any 
of these points, at least 
lie on the same circle. Is it necessarily true that at least 
of the points lie on the same circle?
points.
points do not lie on the same circle. Let the four cyclic points be 
and the non-cyclic points be 
Then we consider the case where 
are cyclic. From here the cases with 
and 
are also satisfied.
Then we consider the case where 
are cyclic. From here the case with 
is also satisfied.
We can't have all three 
points and a 
point be cyclic because we already have two intersections of the circumcircle of 
and 
are cyclic. (Note that this case is considering one of the 
)
We can verify that for any selection of four points, the circumcircle of three is already defined, and the fourth point lies on a different circle (also already defined). Moreover, the two intersections of the circles on which the fourth point could lie are already defined by a different point, either 
or 
are cyclic. Then we consider 
and since two circles have at most two intersections, we need to have two different 
are cyclic. Similarly for 
we consider 
cyclic (since we're considering that we only consider one intersection of the circumcircle of the three 
namely, 
)
We can again verify (similarly to the previous case) that it is impossible to have any four of these points concyclic based on how we defined them in this case.
necessary pairs of points 
For every pair of 
there are at most two corresponding pairs of 
which intersect, and since for each pair of 
points do not lie on the same circle. Let the four cyclic points be 
and the non-cyclic points be 
Notice that if we consider both 
are concyclic. Now consider 
WLOG assume 
Since the second intersections of the circumcircles of 
and 
are already defined, and 
cannot pass through either of them due to this as well, we have a contradiction.
be a circle with 4 points say 
.
are points we on 
and 
are concyclic.
4 of these must be concyclic. If 
be concyclic. Then consider 
.
.
,
and 
are concyclic. However, we have only 3 points 
are concyclic on one circle and the other two are concyclic on another circle. And 
implying that we have only 
into two sets of two elements. Clearly, as 
,
not on 
.
and 
are concyclic. Similarly, 
are cyclic. This cannot be possible as 
do not lie on the circumcircle of triangle 
.
,