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.
has a right angle at 
. The centre of the circumcircle is called 
, and the base point of the normal from to 
is called 
. The point 
lies on 
with 
. The angle bisector of 
meets 
in 
. The lines 
and 
intersect in 
. Show that the lines 
and 
are parallel.
let 
denote the number of (positive) divisors of 
. Find all functions 
with the following properties: [list][*] 
for all 
.
divides 
for all 
, 
.[/list]
of positive integers such that ![\[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]](http://latex.artofproblemsolving.com/e/c/7/ec778edf97100f3b19350d8a817317ebf1f391bd.png)
be a triangle with circumcenter and orthocenter 
. 
the altitudes of 
. A point 
lies on line 
such that 
. A point 
lies on the circumcircle of such that 
are concurrent. 
meets 
at 
. Prove that 
are concyclic.
be a triangle. Distinct points , , lie on sides 
, , and 
, respectively, such that quadrilaterals 
and 
are cyclic. Line 
meets the circumcircle of again at 
. Let denote the reflection of across . Show that lies on the circumcircle of 
. be a triangle with 
, and its shortest side. Let be the orthocenter of . Let 
be the circle with center 
and radius 
. Let be the second point where the line 
meets . Let be the second point where meets the circumcircle of the triangle 
. Let be the intersection point of the lines 
and 
.
is tangent to the circumcircle of the triangle 
.
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 
.
be an acute triangle with 
the feet of the altitudes lying on 
respectively. One of the intersection points of the line and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
is tangent to the circle 
at . and cut the circle at and respectively. 
is the in-centre of , and is on the line 
. 
and intersect at , while 
and 
intersect at . Prove that the points 
lie on a circle.
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 an acute triangle, its orthocenter and the center of its nine point circle. Let be a point on the parallel through to such that 
and and are on different sides of and a point on the parallel through to such that 
and and are on different sides of . If 
and 
are the reflections of over and respectively, and are the intersections of 
and 
respectively with the circumcircle of , prove that the intersection of lines 
and 
lies on 
.
, is the angle bisector of 
. The circumcircle of intersects 
in 
respectively and the line through parallel to cuts 
in 
respectively. Prove that point 
lie on a circle.
be the inscribed circle of a triangle 
. Let 
and be the tangency points of and the sides 
and , respectively, and let 
and intersect at 
and , respectively, such that 
and are all distinct. The tangent line of at 
intersects at , the tangent line of at 
intersects at 
, and the tangent line of at M intersects at 
. Prove that 
and are collinear.
, 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, 
be a fifth root of unity not equal to 
.
and 
one at a time for 
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 
.
is a root of 
,
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) 
,
,
,
