AoPS Academy
be a cubic polynomial given by 
, where 
are integers and 
. Suppose that 
for infinitely many pairs 
of integers with 
. Prove that the equation 
has an integer root.
be real numbers such that 
. Determine the minimum value of 
and determine all values of 
such that the minimum value is achived.
cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer 
such that no matter how Starways establishes its flights, the cities can always be partitioned into groups so that from any city it is not possible to reach another city in the same group by using at most 
flights.
distinct circles in a plane, with radius 
.
circles, which form a set 
, which satisfy the following., they intersect with each other or, they don't intersect with each other.
, so that MBD is tricolored and right-angled.
, a green point 
and a blue point . Consider the circle 
with diameter 
. Note 
. Let 
meet again at 
respectively. If 
is red then 
contradicts the hypothesis. Since is not blue, it must be green. Now if any point on line 
is red then 
contradicts the hypothesis; if it is blue then 
contradicts the hypothesis. Thus, the entire line is green. Similarly line 
is red. However 
intersect at the orthocenter 
of 
. Thus, is both green and red; and we don't buy that logic. Contradiction! lies on line . Consider lines through 
perpendicular to line 
. Then it is clear that all points on each of these lines have the same color. Now it is easy to find a tri-colored right triangle.
be a triangle with three different colored vertices. Say that is colored red, is colored yellow, and green. WLOG say that the orthocenter of this triangle, , is colored green. Let the perpendicular foot to the sides of the triangle 
be 
, 
, and 
. First, notice that should be green. Then, should be red and should be yellow! But then we have a contradiction in the right triangle 
.
be the intersection of the circle with diameter AC and the line perpendicular to AC at point B(one of them). can't be in b because of 
can't be in a because of 
can't be in c because of 
contradiction be the circle with diameter AB. CB is not tangent to or is different-colored triangle. Let 
.
is not in a because of the lemma is not in b because of 
is not in c because of 
contradiction.
-red,-blue and -yellow.
, ,, are not collinear.
,if is on that circle then we get a contradiction with ....(1) is not on the circle with diameter is that we always want to create perpendicular angles,and we know that each point in that arc will gives us 
,
and 
with the circle be 
and respectively.If is yellow than we get contradiction with ...(1) is red than the triangle 
will get us a contradiction with ...(1) is blue,,so if any point on line 
is Differen color of these 2 points we get a contradiction with ...(1)
is colored all red so can we blue(which means that 
GIVES CONTRADICTION) can be red (which gives that 
GIVES CONTRADICTION)-- are collinear
and points ,, on it,
, 
and 
lines perpendicular to and passind through , , respectively, we see that the colors of these lines should be the same as their point passing throught,otherwise we get a contradiction.
be a red point and be a yellow point and let 
be the point perpendicular to at . only contains red points
then 
. If there is some yellow point on then let us choose a blue point, not on , and call it (such a blue point exists since there are no blue points on 
). Choose some 
such that 
. If is red then we can choose a yellow point 
which gives 
. If is yellow then 
. Thus, contains no blue or yellow points, proving the claim.
(Defined similarly) only contains yellow points. Thus, we have a pair of parallel lines, one all red and the other all yellow. If we take to be a blue point we can choose a sufficiently far away 
such that the circle with radius 
intersects , say... at 
. This gives 
. Contradiction.
such that ....
such that ....
, and be the 3 colours. We must have a column where, w.l.o.g, appears infinitely. In this column, choose 3 points which are coloured , points , and 
. On the 's row, no point can be coloured , or else we would have an triangle, and therefore there should be infinite points coloured, w.l.o.g, on this row. On 's row, every point that lies in a column where, over the 's row the point is coloured , must be painted or else we would have a triangle. Now, take the intersection from a column which there's a on the 's row and a on the 's row, and 's row. This point can't be coloured with any of the 3 colours and we get a contradiction. 
axis contains points 
with colors 
. with color 
is on the axis, then for an arbitrary point not sharing a column with or , one of triangles 
works where is the projection of 
onto the axis. Otherwise, select an arbitrary with color ; 's projection onto the axis is one of colors , hence one of the triangles 
works.
. Also, sorry for the horrible write-up, I couldn't come up with a better one 
All of the vertical lines in the grid are of same colour. In this case we choose any point (say ) on one of the vertical lines and draw 
diagonal lines through that point, each one of which will intersect each vertical lines at exactly one lattice points and of these vertical lines must have different colours than , giving us the desired triangle.
There exist at least one vertical line with two distinct colours. Say are two points of different colours on a vertical line. Now if we have a point on the perpendiculars through which are of different colours than then we are done, if not, then pick a point of different colour than and drop a perpendicular on the line containing at . Then one of triangle 
or 
must have vertex of different colours.
There exist at least one vertical line with 
different colours. In this case, say those 3 points are 
. We draw the perpendicular through each of . If any of the perpendiculars have a point of different colour then we are done. Suppose not. Then the perpendicular lines must be all monochromatic. In fact every perpendicular on the vertical line must be monochromatic, which makes it similar to case 1 (this time we have horizontal monochromatic lines instead of vertical ones) hence we draw the diagonals through a point as given in case 1, and we will get our desired triangle.
and 
, in the same row or column such that is red and is green. Now, consider a blue point 
The foot from to 
, which we call 
, must also be a blue point, since otherwise we would have a right triangle with different colors.
be the line through perpendicular to line 
, and define 
and similarly. By our earlier logic, for any color , all points on 
must be of color .
be some point on . Let 
and 
. Then, consider 
The slope of 
is 
, and the slope of 
is 
, so this is a right triangle, and its vertices are all different colors, so we are done.
. Let 
be the perpendicular line to through the red point and 
be the perpendicular line to through the green point. is red, and every point on is green.. Then, fix some blue point , and consider the projection of onto , say . This point must be red. But taking the right triangle formed by , , and any other point 
, that point must also be red. Hence every point on is red, and a similar result follows from . in the plane, and draw two lines 
through that are perpendicular to each other. Then 
intersects at some red point , and 
intersects at some green point 
. It follows 
satisfies the desired condition.
with the same or 
coordinate (meaning collinear but the line is parallel to the or axis). WLOG, assume that this line is vertical. Then clearly every point on each of the horizontal lines passing through these points have the same color as it. But then we can easily find a right triangle, a contradiction. or axis. Then there are two points colored differently with the same or coordinate, and a third point colored that does not lie on the line connecting the two previous points. WLOG, let the first two points be colored 
and have the same coordinate, while the point with color is somewhere not on their line. Then just consider the foot of the perpendicular from this point to the line. This clearly cannot be colored any of 
so we have a contradiction. QED
in the same order and have the colour 
.
be the perpendicular line to 
.
,
as 
and 
.
and 
.
be distances 
and 
(WLOG, assume, 
)
.
and the construction is obvious when 
,
.
.
,
.
is all one color, WLOG let it be red. Then, the line 
for any integer 
,
is all red since any 
lies on 
,
and let the coordinate of the blue point be 
.
.
and 
.
is red, since if 
would be right and its vertices would all be pairwise different colors, and if 
would be right and its vertices would all be pairwise different colors. Similarly, we can show that 
is blue. Then, it follows that 
,
contains three points 
of pairwise distinct colors. Now, this means that the whole line 
is a right triangle with vertices all pairwise different colors, and if 
is a right triangle with all vertices pairwise different colors. Similarly, the whole line 
is blue and the whole line 
and 
,
is 
and the slope of 
is 
and these multiply to 
,
,
is right. However, 
is blue since it lies on 
is green since it lies on 
the line passing through 
such that 
and 
such that 
.
,
and 
be the projections of 
,
