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 
.
be a triangle, and , , points on the segments , 
, and respectively such that
Show that if the centres of the circumscribed circles of the triangles 
and coincide, then is an equilateral triangle.
as the origin, we have that there is a real number 
such that
, we have that 
. But
and we can find a similar formula for 
and 
. Equating these and remembering that 
where 
is the circumradius of , we find that
which implies that
which implies that is equilateral.
be the common circumcentre of and , and let 
and 
. Stewart's Theorem in 
give us that
where 
is the circumradius of . We thus have that 
, and a similar calculation gives us that if 
and 
, then 
where 
is the circumradius of . It follows that 
, and hence that 
.
and will always coincide? Then if we say that if this is so, then having the circumcentres of and coincide means that the circumcentre must coincide with its centroid, hence both triangles are equilateral?
to the circuncircle of ABC is 
and 
. The power of point of is 
and 
. As 
, the product of the lengths is the same. Together with the stated condition of ratios, gives us that 
and 
.
be 
and let the circumcenter of be . Clearly, 
. Now, let 
and 
. By Stewart’s theorem on 
, we have that:
By the same reasoning,
and
Since 
, 
, and is equilateral.
, 
and 
be points on , and respectively such that 
; 
, 
and 
. (Reflections over midpoints of the sides of the triangle )
.
These give 
.
. Let be the circumcenter of and . By the definition of the midpoint of 
is the same point as the midpoint of , hence the perpendicular bisector of is the same as the perpendicular bisector of since the points are all collinear. Therefore, the perpendicular bisector of passes through as well as the perpendicular bisector of since it is the circumcenter of , hence, is also the circumcenter of 
. Similarly, is also the circumcenter of 
and 
. Therefore, the perpendicular bisectors of , 
and 
all pass through . This implies 
and are concyclic points. Now, 
and 
, we have 
is a parallelogram, therefore, 
, and 
from 
. Also 
is a cyclic quadrilateral, therefore, 
, hence, 
. Also, and 
, we have 
is a parallelogram, therefore, 
and 
from 
. Also 
is a cyclic quadrilateral, therefore, 
, hence, 
. Therefore, 
, hence the triangle is equilateral.
be the common circumcentre of and , and let and . Stewart's Theorem in give us thatwhere is the circumradius of . We thus have that , and a similar calculation gives us that if and , then where is the circumradius of . It follows that , and hence that .
Prove that
Let 
Prove that
.
