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.
be a quadrilateral such that the excircles 
and 
of triangles 
and 
touching their sides 
and 
respectively touch the extension of 
at the same point 
. The segment 
meets at point 
, and the line meets at 
and 
. Prove that one of angles 
and 
is right
be the set of all positive integers 
such that 
has a divisor in the range 
. Prove that there are infinitely many elements of of each of the forms 
and no elements of of the form 
and 
, where 
is an integer.
of odd degree with real coefficients such that 
.
. Consider 
real numbers 
and 
. Let be the set of all pairs 
of real numbers for which 
, 
are pairwise distinct. For every such pair sort the corresponding values 
increasingly and let 
be the 
-th term in the list thus sorted. This denes an index permutation of 
. Let 
be the number of all such permutations, as the pairs run through all of . In terms of , determine the largest value may achieve over all possible choices of 
.
be a positive integer. For a set of real numbers, let 
denote the number of increasing arithmetic progressions of length at least two all of whose terms are in . Prove that, if is a set of real numbers, then![\[ f(S) \leq \frac{n^2}{4} + f(\{1,2,\ldots,n\})\]](http://latex.artofproblemsolving.com/c/8/8/c887443e08b9f61e5ce55c3e35fcf2912a6da2a5.png)
, 
is acute, 
, and 
. Let point 
be the intersection of the tangent to the circumcircle of at point 
and the perpendicular bisector of segment 
. Given that 
, find the radius of the circumcircle of 
.
. Yael and Ziad play a game as follows, involving a monic polynomial of degree 
. With Yael going first, they take turns to choose a strictly positive real number as the value of one of the coecients of the polynomial. Once a coefficient is assigned a value, it cannot be chosen again later in the game. So the game rounds, until Ziad assigns the final coefficient. Yael wins if 
for some real
. Otherwise, Ziad wins. Decide who has the winning strategy.
positive integers with least common multiple 
and greatest common divisor 
. It is known that is not a perfect square and is not among the initially written numbers. Two players 
and 
play the following game, taking turns alternatingly, with starting first. In a move the player has to write a number which has not been written so far, by taking two distinct integers 
and 
from the board and write LCM
or LCM/. The player who writes or loses. Who has a winning strategy?
. At a maths camp, there are 
children, each of whom selects tables, with children tables, with children and let 
be a convex polygon in the plane. Let 
be the set of all midpoints 
of segments 
where 
. Assume that all of these midpoints are distinct, i.e. consists of 
elements. Dissect the polygon 
into triangles so that the following hold:
for some pairwise distinct indices 
.
.
, there exists some positive integer 
that makes the following equation have no integer root 
.
are nonnegative because 
. We aim to show that there must exist a less than 
that cannot be equal to the left-hand side of the given equation. We count the number of possible values of each so that 
., assume for the sake of contradiction that 
. When we multiply this by itself 
times we see that 
, which we don't want. to be less than 
. There are possible values for in the range 
. When we take the product of for all such that 
, we get ![\[ \prod\limits_{i=1}^n 2^{2^{n-i}} = 2^{\sum\limits_{i=1}^n 2^{n-i}} = 2^{2^n - 1}\]](http://latex.artofproblemsolving.com/e/4/d/e4d0e5a2f1977c4e12bec6fa560fbe9166f30418.png)
This value is an upper bound for the amount of tuples 
such that the LHS of the given equation is less than . However, this value 
is less than the 
possible values of less than . Thus, by pigeonhole there must exist a less than that cannot be equal to the LHS of the given equation.
is always false. If it turns out this equation can be true for some large