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.
and![\[
a_n=\frac{a_{n-2}\times a_{n-1}}{2a_{n-2}-a_{n-1}}
\]](http://latex.artofproblemsolving.com/5/0/6/5064223358890cf3a58a6a6b289e2ccba6118861.png)
For each integer 
, where 
, hence find the value of 
+1 w
, which doesn't have right angle, let 
be its circumcenter. Circle 
intersects 
and 
at 
and 
for the second time, respectively. Similarly, circle 
intersects 
and 
at 
and 
, and circle 
intersects 
and 
at 
and 
for the second time, respectively. Let 
and 
be circumcenters of triangle 
and 
, respectively. Prove that 
are collinear.
each time. The 
th term of the sequence is 
. What is the first term of the sequence?
of integers is rootiful if, for any positive integer 
and any 
, all integer roots of the polynomial 
are also in . Find all rootiful sets of integers that contain all numbers of the form 
for positive integers 
and 
.
be an isosceles triangle with 
. Let 
be a point on 
. Let 
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
satisfies the following conditions:
Find the value of 
.
satisfying 
and
for all 
.
is a factor of 
, find the sum of 
be a cyclic quadrilateral with 
being the midpoints of segments 
respectively. Suppose 
is the intersection of diagonals 
of quadrilateral 
Define 
to be the isogonal conjugate point of point in 
Define 
similarly. Suppose 
intersects 
at a point 
Prove that: The Newton-Gauss line of quadrilateral bisects segment 
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
:
Square again and subtract to get 
Thus, 
Roots are 
Negatives are bad because lhs and rhs would have differing signs, and 
fails. Thus, all are bogus except 
. Now observe that for 
, 
, so we must have 
. So, 
, giving 
.
for can be proven by casework.)
:
.
Prove that
Prove that
Prove that
because the LHS is always positive, so the RHS is also always positive. Now we split it into cases:
This gives 
solving gives 
This gives 
so 
again.
