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.
with 
. Let 
be the midpoint of side 
, and let 
and 
be points on segments 
and 
, respectively, such that 
. Let 
be the circumcircle of 
, and 
the circumcircle of 
. The common tangent 
to and , which lies closer to point 
, touches and at points 
and 
, respectively. Let the line 
intersect again at 
, and the line 
intersect again at 
. Prove that the circumcircle of triangle 
is tangent to both and .
, 
, 
be positive real numbers. Prove that![\[ \dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.
\]](http://latex.artofproblemsolving.com/0/5/e/05e90cc0c9d60f8943608b690f75d20bcd4f6220.png)
, where 
is a positive integer.
with exactly 
elements such that the sum of its elements is a prime number.
![\[{x^3}{y^2}\left( {2y - x} \right) = {x^2}{y^4} - 36\]](http://latex.artofproblemsolving.com/d/5/6/d5668b2a486f5264e1cf2faeb363ac69bf91e3f0.png)
satisfies the following condition.
, ![$\sum_{k=1}^{n}\frac{1}{2}\left(1 - (-1)^{\left[\frac{n}{k}\right]}\right)a_k=1$](http://latex.artofproblemsolving.com/3/1/0/3109e3f024dc59c2cdad60efa270494d6095c330.png)
holds.
, compute 
.
such that 
for all 
.
. 
amount of people stand on coordinates 
where 
. Every person got a water cup and two people are considered to be neighbour if the distance between them is 
. At the first minute, the person standing on coordinates 
got litres of water, and the other 
people's water cup is empty. Every minute, two neighbouring people are chosen that does not have the same amount of water in their water cups, and they equalize the amount of water in their water cups. will not have more than 
litres of water.
be positive real numbers such that 
. Prove that 
has centroid 
. A line parallel to 
passing through intersects the circumcircle of at a point 
. Let lines 
and intersect at 
. Suppose a point 
is chosen on such that the tangent of the circumcircle of 
at , the tangent of the circumcircle of at 
and concur. Prove that 
. is called beautiful if, for every integer 
, the base-
representation of contains the consecutive digits 
, 
, , 
(in this order, from left to right). Determine whether the set of all beautiful integers is finite.
such that for all positive integers 
,![\[
f(mn) = \mathrm{lcm}(m, n) \cdot \gcd(f(m), f(n)),
\]](http://latex.artofproblemsolving.com/6/0/7/6071692e2716901e88cc1cfc19ed60388c10701d.png)
where 
and 
denote the least common multiple and the greatest common divisor of 
and 
, respectively.
.
, so solutions come from the second one.
.
.
or from 
has nonnegative real roots (
. However, that equation in 
has no real roots.
does have two real roots. This becomes clear when you note that at 
the polynomial is negative, but for very positive or very negative 
, the polynomial is positive.
and 
easily.
and easily.
. We manipulate the polynomial as follows:
, and 
, and thus the answer is 
, as claimed.
The solutions to 
are 
so our answer is 
.
. By Difference of Cubes, 
, so our expression is equivalent to 
. It is clear that 
cannot have real roots since for all reals 
, so we are looking for the roots of 
. The root in the desired form is 
.
If 
then 
or 
where w is a cube root of unity. to be real w=1 but contradiction since that would make 10x^2-1 in the denominator 0. Hence instead it's just solution to first factor which is 
so solution is 
such that 
coprime, 
and 
.
has exactly two real roots, one of which is 
where 
and 
are integers, 
and 
Find 
.
,
.
.
.
.
.
.
and rearranging, we get
,
and 
and subtracting 
terms must cancel in order to get this polynomial equal to 
into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying 
,
and we get
.
since we have to solve a quadratic with a negative discriminant, then multiply by 
.
.
(using 
)
,
)
for a final answer of 
Thus, the two real roots are 
Hence, the desired answer is 
has no real roots since it's 
