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 an integer greater than 
. For a positive integer 
, let 
. Suppose that there exists a 
-element set 
such that is an -element subset of 
; shares at most one common element; is contained in exactly two elements of . in terms of .
be an odd prime number. Suppose 
and 
are polynomials with integer coefficients such that 
, there is no nonconstant polynomial dividing both and , and![\[
1 + \cfrac{x}{1 + \cfrac{2x}{1 + \cfrac{\ddots}{1 +
(p-1)x}}}=\frac{P(x)}{Q(x)}.
\]](http://latex.artofproblemsolving.com/b/c/a/bca9034e2affa1d80b167451d77434fbcd8bd88a.png)
Show that all coefficients of except for the constant coefficient are divisible by , and all coefficients of are not divisible by .
other knights where friendship is mutual. Prove that King Arthur can place his knights around a round table in such a way that every knight is friends with the 
people adjacent to him and between them there are at least 
friendships of knights that are not adjacent to each other.
be 
integers which satisfy![\[ a_{n+5}+a_n>a_{n+2}+a_{n+3}\]](http://latex.artofproblemsolving.com/8/8/b/88bbb2f304525a3261851571d34863bf4d4a1e51.png)
for all integers 
. Find the minimum possible value of the difference between the maximum value and the minimum value among .
be a cyclic quadrilateral inscribed in circle 
whose diagonals meet at 
. Lines 
and 
meet at 
. Segment 
intersects at 
. Lines 
and meet at 
, and lines 
and meet at 
. Prove that 
and 
concur with the tangent to at .
, point 
lies on side and point 
lies on side 
. Let and be points on segments 
and 
, respectively, such that 
is parallel to . Let 
be a point on line 
, such that lies strictly between and , and 
. Similarly, let 
be the point on line 
, such that lies strictly between and , and 
.
, and are concyclic.
be the circumcircle of a triangle 
. A circle passing through points 
and 
meets the sides 
and 
at 
and 
, respectively. The lines 
and 
meet again at 
and 
, respectively. The tangent lines of at and meet the line 
at 
and 
, respectively. Prove that the lines 
and 
meet at .
is the orthocenter of an acute triangle , and let be the midpoint of . Suppose 
meets and at 
respectively. 
meets 
at , and the line through perpendicular to meets 
at . Prove that 
are collinear.
prime number, there exists only one positive number that makes the equation 
a perfect square of a positive integer
be a real-valued function defined on the set of real numbers that satisfies![\[f(x + y) \leq yf(x) + f(f(x))\]](http://latex.artofproblemsolving.com/c/9/4/c94075d26e09712f5393fb66ecbfdc2573c239b6.png)
and 
. Prove that 
for all 
.
, 
. The internal angle bisector of 
and the external angle bisector of intersect the ray at points 
and , respectively. Given that 
, prove that 
.
is not 
, then there is some number not represented, say 
. But then whenever 
, we also have 
. So the number of times 
is represented in the multiset 
is an even number. But it should be exactly times, because for each 
there is exactly one 
with by the second condition. This is a contradiction with the fact that is odd, so must be in 
for all , and so 
.
array. Given that every row in the array has distinct numbers and that the 
row is exactly same as the column, you are basically asked to prove that the left diagonal is filled with all the distinct numbers only..
.
denote the set of all the tuples 
such that 
. Now note that for each 
, there is exactly one 
such that 
. Thus the set has at exactly elements. is in , 
is also in . Now if 
, then we get a pair of two tuples that are present in . But the cardinality of is odd. Thus there must be some such that 
is also in .
for each . Now note that a tuple cannot be present in two sets. This is because otherwise, 
and 
which implies 
, contradiction. So each has at least one such that this tuple is not present in any other 
. many 
tuples and there is at least one distinct present in each of the sets of . Thus the equality holds and there is exactly one in each . in each and we are done, because this forces 
.
Prove that
Let 
Prove that
.
where 
be a function defined on 
taking values in 
for all 
for all 
had done the right thing. I have also this solution same logic with him, Hence nothing is wrong in his solution
There is no such 
