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.
are different at 
. If 
are also different at , we call the ordered 
-tuple 
lucky. For which positive integers , one can find a lucky -tuple?
, 
be the orthocenter of it and 
be any point on the side 
. The points 
are on the segments 
, respectively, such that the points 
and 
are cyclic. The segments 
and 
intersect at 
is a point on 
such that 
is tangent to the circumcircle of triangle 
at 
and 
intersect at 
. Prove that the points 
and lie on the same line.
. Prove that
denote a triangular grid of side length consisting of 
pegs. Charles the Otter wishes to place some rubber bands along the pegs of such that every edge of the grid is covered by exactly one rubber band (and no rubber band traverses an edge twice). He considers two placements to be different if the sets of edges covered by the rubber bands are different or if any rubber band traverses its edges in a different order. The ordering of which bands are over and under does not matter.
different ways to cover 
using exactly two rubber bands; the full list is shown below, with one rubber band in orange and the other in blue.
denote the total number of ways to cover 
with any number of rubber bands. Compute the remainder when is divided by 
.
be a triangle, and let 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
. Let 
be the second intersection of the circumcircles of triangles 
and 
. Prove that the lines 
, 
, and 
-median of 
are concurrent.
be an integer and let 
be real numbers such that 
. Prove that
be a positive integer which is not a perfect square, and consider the equation ![\[k = \frac{x^2-a}{x^2-y^2}.\]](http://latex.artofproblemsolving.com/c/2/2/c223ccbc272d07c19632c5b8883571023e47a395.png)
Let be the set of positive integers 
for which the equation admits a solution in 
with 
, and let 
be the set of positive integers for which the equation admits a solution in with 
. Show that 
.
red cards, 
white cards, and blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
and 
be two finite subsets of the half-open interval 
such that 
and 
for no 
and no 
. Prove that the set 
has at least 
elements., 
, 
be fixed positive integers. There are 
ducks sitting in a ducks ducks picking paper, and ducks picking scissors., , and , the maximum number of moves which could take
such that the sequence 
defined by ![\[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \]](http://latex.artofproblemsolving.com/2/1/b/21b2a1f93b11a94b84e7b55f4b4f679aa20e36c6.png)
contains at least one integer term.
be distinct positive integers. Prove that
Further, determine when equality holds.
does not have a real root. what is the maximum number of coefficients to be 
?(14 points)
needs have no real roots and 
, it means we need have 
for all real 
. Since we then need 
, it means the number of negative coefficients can be at most 
(out of the total 
coefficients). negative coefficients. Take 
. But 
, while 
. But a polynomial 
has only one real root (root 
), so polynomial 
has no real roots, hence is positively defined. Therefore all terms in the sum above are non-negative, and may be zero only at values 
and/or . However, 
, so 
has no real roots.
has only one real root (root ), so polynomial 
has no real roots.
does not have a real root. what is the maximum number of coefficients to be ?(14 points)
, we see that there doesn't may more than 
, each term is positive.
and 
and 
and add it, we find 
:
and , add them and we find 
