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 acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
. Define 
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
and 
be real numbers. Let 
be the matrix with entries ![\[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\]](http://latex.artofproblemsolving.com/c/0/8/c0856fe7b2b113180659952c4d216a01d8849f64.png)
Suppose that is an 
matrix with entries 
, 
such that the sum of the elements in each row and each column of is equal to the corresponding sum for the matrix . Prove that 
.
. Find 
.
are collinear, and 
are arbitrary points
, let 
be the sum of the squares of the digits of . For example, 
. Determine all integers 
such that 
.
with centroid 
and circumcircle 
centred at 
, the extension of 
meets at 
; lines and 
intersect at ; and lines and 
intersect at . Suppose the circumcentre 
of the triangle 
lies on and 
are collinear. Prove that 
.
, let 
and 
be points on sides and , respectively, such that 
is cyclic. Let lines 
and 
intersect at point , and and 
be the midpoints of 
and 
, respectively. If 
is the foot of the perpendicular from to , and the circumcircles of triangles 
and 
intersect at second point 
different from , prove that 
and are collinear.
, such that 
for all integers 
.
of positive integers, with 
prime, such that 
is a perfect square. inscribed in 
. Two points 
lie on such that 
are isogonal in 
. is the midpoint of 
. 
lies on 
such that 
. 
intersects at . Prove that the line through parallel to 
passes through the Euler center of triangle .
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
be a primitive 
th root of unity, and observe that for 
, 
. Hence 
, and 
are roots of the quadratic equation 
. A quadratic null at 3 different places must be null everywhere, implying 
, so we're done.
![$T\colon\mathbb R[x]\to\mathbb R^5$](http://latex.artofproblemsolving.com/6/8/8/68855b81fbc161a6f3a1e961dad6f2bb9c1bc2b6.png)
be the function given by![\[ \sum a_kx^k\mapsto\left(\sum_{5\mid k}a_k,\sum_{5\mid k-1}a_k,\ldots,\sum_{5\mid k-4}a_k\right). \]](http://latex.artofproblemsolving.com/5/6/1/561dc24cbb1a22a0740d2ca5cbdea919e9b211a7.png)
It is clear that![\[ T(P(x^5)+xQ(x^5)+x^2R(x^5))=(P(1),Q(1),R(1),0,0). \]](http://latex.artofproblemsolving.com/d/1/9/d191251d5402e6fd85d892a7ece504357b262949.png)
It is also easy to see that![\[ T((x^4+x^3+x^2+x+1)S(x))=(S(1),S(1),S(1),S(1),S(1)). \]](http://latex.artofproblemsolving.com/5/6/8/56868c5e37649b8b4a7b689600853baa00032acc.png)
Since 
, these two must be equal, giving![\[ P(1)=Q(1)=R(1)=S(1)=0. \]](http://latex.artofproblemsolving.com/0/d/3/0d3799bfd4e92541002a6055adb144050d877405.png)
Hence, 
. 
divides 
from ,of course assuming that is not .
divides from ,of course assuming that is not .
![\[x^4+x^3+x^2+x+1 \mid x^5-1 \mid \left[P(x^5)+xQ(x^5)+x^2R(x^5)\right]-\left[P(1)+xQ(1)+x^2R(1)\right]\]](http://latex.artofproblemsolving.com/8/1/0/8103bd9135d49b300d5572b67ca69686c97a6da6.png)
Then![\[x^4+x^3+x^2+x+1 \mid P(1)+xQ(1)+x^2R(1)\]](http://latex.artofproblemsolving.com/4/f/d/4fdf2a5d8b5aacdb20c2f15f5f86cb639bc60622.png)
But 
then![\[P(1)+xQ(1)+x^2R(1) \equiv 0 \ \forall \ x\]](http://latex.artofproblemsolving.com/5/8/5/585729b07c902a35838dc3194a7b5b2140aa944d.png)
Hence, 
,
,
.![\[\sum_{i=1}^{n}\frac{1}{1+x_{i}}\cdot \sum_{i=1}^{n}x_{i}\leq \sum_{i=1}^{n}\frac{x_{i}^{k+1}}{1+x_{i}}\cdot \sum_{i=1}^{n}\frac{1}{x_{i}^{k}}\]](http://latex.artofproblemsolving.com/2/6/4/264d110e2a40c7f787c650aa8b8dc30af5d28263.png)
such that![\[
f(x^4 + 5y^4 + 10z^4) = f(x)^4 + 5f(y)^4 + 10f(z)^4
\]](http://latex.artofproblemsolving.com/3/0/8/308079e75e1acd9a9f896885387b67a7b4020c16.png)
for all 
.
be a fifth root of unity not equal to 
.
and 
one at a time for 
and add up the five resulting equations.
and that 
,
while the right hand side is 
,
,
.![\[ x P(x^5) + x^2 Q(x^5) + x^3 R(x^5) = x(x^4 + x^3 + x^2 + x + 1) S(x)\]](http://latex.artofproblemsolving.com/4/b/1/4b10de4e307ba9b817eaa08286dc863ab67c0fa2.png)
for the five roots 
to get![\[ 0 = P(1)\sum_{k = 0}^{4}{\omega^k} + Q(1)\sum_{k = 0}^{4}{\omega^{2k}} + R(1)\sum_{k = 0}^{4}{\omega^{3k}} = 5S(1) = 5P(1).\]](http://latex.artofproblemsolving.com/2/6/1/26187bacf0e34bc169d1f85f2e6a800873613e0c.png)
a primitive fifth root of unity, then the five equation that result from plugging in 
for 
for 
are
.
yields 
.
then from the first two equations we have
.
,
so 
which means 
is a factor of 
as desired.
![\[\begin{vmatrix}1&\omega&\omega^2\\1&\omega^2&\omega^4\\1&\omega^3&\omega\end{vmatrix}=\omega^3+2-2\omega^2-\omega^4\neq0,\]](http://latex.artofproblemsolving.com/c/f/d/cfdae26343895428bcb86efb7bca01910870dddc.png)
one can immediately deduce that 
.
,
to the original equation, we get 
.
gives 
.
denote the given assertion, and let 
Adding up all of these yields 
.
which is more than enough. Solving the system, we easily get 
and the fact that if x-1 is a factor of P(x) suggests P(1)=0 gives suggestions to find ways with powers of 5 and 1. With experience, you will find motivation to consider the roots of unity. Now summing over 
with x=a fifth root of unity not equal to 1, and 
,
(in some order, 1, x, …, x^4 each appear once in each root of unity) 
,
,
,
