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.
board, each square is colored either white or black, with all the squares on the border of the board being black. Additionally, no 
square within the board has all four squares of the same color. Prove that the board contains a square colored like a chessboard.
, where 
, then the number of integers in the range of 
is __
is inscribed in circle 
. Let 
denote its incenter and 
its 
-excenter. Let 
denote the midpoint of arc 
. Line 
meets a second time at 
. The perpendicular to 
at meets sides 
and 
at 
and 
respectively. The circumcircle of 
meets 
a second time at 
, and the circumcircle of 
meets 
a second time at 
. Prove that 
passes through the antipodal to on .
. Prove that if the leading coefficient of the polynomial equals then the greatest common divisor of the other coefficients also equals .
be a triangle with circumcenter 
. Point 
is the intersection of the parallel line from to with the perpendicular line to from 
. Let 
be the point where the external bisector of 
intersects with . Let 
be the projection of onto 
. Prove that the lines 
have a common point.
dwarves, whose weigths are 
, who want to cross a river. They have a small boat which can lift at most 
each time without sinking. For each journey of the boat a non-empty subset of the dwarves to be taken to the other side is chosen and one of these dwarves is chosen as the 
for that journey. Since return journeys are counter-current, no dwarf is able to do the rower for more than one return journey. Is it possible for all the dwarves to reach the other side of the river?
with circumcenter and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line such that 
. Let and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
. Consider the sets 
of 
points 
for 
, which partition the vertices of 
-gon (indices are taken modulo ). If there exists a set such that all elements are vertices of the 
-gon, then both 
and 
are clearly sides of the -gon ("clearly" because the convex hull of the -gon will be a subset of the convex hull of the -gon). This is a contradiction since 
. So from each set a maximum of 
vertices can be vertices of the -gon. This gives an upper bound for of 
. This is achievable by considering the 
-gon formed by the segments 
and 
.
a maximum of vertices can be vertices of the -gon. This gives an upper bound for of 
., not 
. Luckily, you lose only one mark...
-gon such that 
.First consider all of its vertices.Let be the number of sides that have a side that is paralel to them.At first 
.Now we take 
edges that share a vertice and we see that there is a pair of paralel sides which also contains a common vertice.We remove that vertice and consider our 
-gon.We see that these edges can no longer have any paralel sides and we see that the new edge won't have paralel sides neither.Also number of edges is now less for edge which is a total of edges that don't have paralel sides so now 
.By now taking the remaining edges with the same trait we can repeat this.This way we will remove ![$[\frac{n}{4}]$](http://latex.artofproblemsolving.com/8/9/2/89210620a0e728259d9979e90a62914ff06d41fe.png)
vertices.The proof that this will get remove minimum nubmer of vertices: Suposse there exists a lesser number of vertices that we can remove.Start removing the vertices of the original one to obtain this one.We see that in every step can't go down by more than so it is impossible to remove less vertices.
we easyily obtain the answer is 
one-arcs. We now need to cover 
arcs, and each side must cover at least two arcs, so we have at most 
sides. This is easily attainable; have two-arcs covering the upper semicircle and have one-arcs covering the bottom semicircle.
-gon is inscribed in a circle. Find the maximal such that we can choose vertices from given and construct a convex -gon without parallel sides.
We claim that 
First assume 
to be the number of chords with strictly 
points in between the endpoints on the circle. Now the non-parallel condition implies 
Note that the total number of points 
To see this, for a chord having 
point in between, consider the left 
points out of the total 
( in between, and end points). Then each vertex on the circle is counted exactly once.
a contradiction.
the construction is easy; if the points are 
then draw the chords 
. For a construction, label the vertices 
in clockwise order, and take vertices 
. Edges 
and 
are parallel iff 
. Note that all edges where 
have 
unique, as with all edges where 
, which covers all edges. Furthermore, since 
is odd in the first case and even in the second, there are no overlaps "between" groups either, as desired., where 
is clockwise along the polygon from 
, to be 
, and define the pseudoperimeter of a polygon formed from vertices of our -gon in the obvious way. Suppose that we have 
edges of odd pseudolength and 
edges of even pseudolength, so we wish to maximize 
. Obviously, the pseudoperimeter of the polygon formed is always , so 
. Furthermore, the pseudolength of an edge has the same parity as , so every odd-pseudolength edge must occupy an odd residue class modulo . Since there are 
of these, we have 
. Hence
as desired. 
gon. We claim that the maximum number of vertices we can pick is 
. Label the vertices 
in the clockwise direction. It is known that edges and are parallel if and only if .
.
has elements. Now, note that the first 
sides (in the first 'half' of our gon) are non-parallel among themselves quite clearly. Similarly the last sides are also non-parallel among themselves. Also one of the first sides and one of the last sides cannot be parallel. This is because the last sides are of the form 
and the first sides are of the form 
, so![\[a+(a+1) = 2a+1 \equiv 1 \not \equiv 0 \equiv 2c +2 = c + (c+2) \pmod{2}\]](http://latex.artofproblemsolving.com/b/9/9/b99db934ed0c2db0aa045d7948e658ee33f2e903.png)
and since 
this implies that they are also distinct 
so the parallel condition is not satisfied.
pair the vertices with its neighbors. For example, we pair 
, 
, 
, 
. For any pair let its antipodal pair 
denote the pair joining the two vertices diametrically opposite to the points in pair . Note that if a pair 
it must be a side of our polygon as there are no other vertices between the two vertices in . Thus, all four vertices in and cannot be in since 
quite clearly. Thus, out of these 4 points at most 3 can be in our polygon, which summing across all pairs implies that we can have at most vertices in and we are done.
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
satisfying 
