AoPS Academy
be a convex quadrilateral and let 
be the point of intersection of 
and 
. Suppose that 
. Prove that the internal angle bisectors of 
, 
and 
meet at a common point.
be a positive integer and 
be a positive integer coprime to 
. Let 
, and for 
, define
Find, in terms of and 
, the greatest positive integer 
for which there exists an index 
such that 
is divisible by 
.
+1 w
be an infinite sequence of positive integers, and let 
be a positive integer. Suppose that, for each 
, 
is equal to the number of times 
appears in the list 
.
and 
is eventually periodic.
is eventually periodic if there exist positive integers 
and 
such that 
for all 
.)
and a 
grid are given. We write down one integer to each of the square of the grid so that each of integer 
to 
are written exactly once in the grid, and no two adjacent square have sum prime.
+1 w
round any of its endpoints. Is it possible that after several rotations the needle returns to initial position with the endpoints interchanged?
, where 
are linearly independent over 
, this is a unique representation.
is invariant for both endpoints. They start at different parity. So they can't interchange positions.
![${\mathbb Z}[\omega]$](http://latex.artofproblemsolving.com/d/1/2/d12fd2e831d0da64de29e065253c03b71f18bad6.png)
where 
. Draw the needle as a directed segment from 
to in the plane.
. Rotations around the other endpoint correspond to ![\[ z \mapsto z + \omega^k - \omega^{k-1} \]](http://latex.artofproblemsolving.com/2/e/0/2e04e71d6294a14fa622b3807200a6ed85095f58.png)
for some choice of 
.. To prove this we only need show the following claim, which proves the relevant invariant.
in .![${\mathbb Z}[\omega]/(\omega-1)$](http://latex.artofproblemsolving.com/3/a/0/3a0dc4b76ee55649dcdacc1f56d8a38433a9862d.png)
is not trivial. Write ![${\mathbb Z}[\omega] = {\mathbb Z}[T] / (T^4+1)$](http://latex.artofproblemsolving.com/5/5/2/5528f519ce9305f3ab167cca0e09a578155aae11.png)
, then ![\[ {\mathbb Z}[\omega]/(\omega-1) \cong {\mathbb Z}[T] / \left( T^4+1, T-1 \right) \cong {\mathbb Z} / 2 = {\mathbb F}_2 \]](http://latex.artofproblemsolving.com/8/9/1/891f618efc341f8fd237341a0972a89a0df9eabf.png)
as desired. 
and the other is at 
, and WLOG rotate about the left endpoint first. Suppose a sequence of rotations works. Right before we switch endpoints to rotate around, draw the needle's current position on the plane. Also draw the line joining and . Then the drawn segments clearly form a cycle/polygon. Furthermore, because the endpoints must be interchanged, there must be an odd number of vertices on this graph. On the other hand, every drawn segment has length and is either horizontal, vertical or has slope 
. Because and 
are linearly independent over 
, it then follows that to end at the same position we started, we need an even number of horizontal and an even number of vertical edges. By rotating the argument 
the same is true for edges with slope and edges with slope 
, so there are an even number of edges and thus vertices: contradiction. 
.. A move consists of taking a complex number and adding 
where is some integer from to 
. (Effectively what this does is that we rotate the needle by some amount, and then jump to the other endpoint.) We now show that the number of moves must be even, if we return to . More formally, if![\[ \sum_{k = 0}^{7} a_k\omega^k = 0 \]](http://latex.artofproblemsolving.com/a/b/c/abcfbd631cd0597cdacf245fbbdf59dc6e9ead09.png)
for some integers 
through 
, we must have that![\[ \sum_{k = 0}^{7} a_k \equiv 0 \pmod{2}.\]](http://latex.artofproblemsolving.com/4/1/c/41c7d30468c67c6fe0d6a84cc77bf2b8e2ac730f.png)
through are zero, by subtracting pairs of 
while preserving the parity.
. As such 
. Likewise, 
. This then forces 
, so the statement is true.
be a primitive eighth root of unity. Suppose otherwise, and consider the locus of all positions of the segment. Modulo the first and last positions, all such segments 
form an equilateral polygon with an odd number of sides.'s that sum to ; equivalently, there exists a polynomial ![$f \in \mathbb Z[X]$](http://latex.artofproblemsolving.com/3/e/3/3e3d314244e5e514fdf63e59038fc713b6a48ed1.png)
such that 
is odd and 
.
, and hence 
, contradiction!
and the other at . We will let 
and 
and after moves, we let 
denote the new endpoint of the needle after a move. Note that if then we can set up the following recurrence relation for 
![\[ z_{n} = \omega^{e_{n}}(z_{n-2}-z_{n-1})-z_{n-2} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2}\]](http://latex.artofproblemsolving.com/4/1/c/41c8e53e652c5e96cd471fdbacb2f92e05d4a376.png)
for an arbitrary sequence of integers 
. This recurrence works because we translate to a new point 
on the unit circle, rotate it by 
and undo the translation, which is exactly the operation we desire. is odd and 
are integers, then ![\[ \omega^{a_1}+\omega^{a_2}+ \cdots + \omega^{a_k} \not = 0 \]](http://latex.artofproblemsolving.com/c/8/3/c8365b7b5921c9a82fe7334cff111fa28799e0f8.png)
then 
is even. Indeed, note that ![\[ \Re(t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = t_1\cdot \Re(\omega^1)+t_2 \cdot \Re(\omega^2) \cdots +t_8 \cdot \Re(\omega^8)=0 \]](http://latex.artofproblemsolving.com/4/9/4/494a80516578c37d743fd357ba5482b01bbee943.png)
![\[ \implies (t_1+t_3-t_5-t_7) \frac{\sqrt{2}}{2} + t_2+t_4 = 0 \implies t_1+t_3-t_5-t_7=0 \]](http://latex.artofproblemsolving.com/f/9/f/f9fbeef15255e342809c0b6d58893034221c02e0.png)
So, 
is even. We also note that ![\[\omega \cdot (t_1\omega^1+t_2\omega^2 \cdots +t_8\omega^8) = 0 \implies t_8\omega^1+t_1\omega^2 \cdots +t_7\omega^8\]](http://latex.artofproblemsolving.com/a/f/0/af0ed3863ec1e2d86c7f2d7597e97424acae65a2.png)
So, 
is even by the same logic. Thus, 
is even as desired. 
denote the number of terms in the expansion of , we can set up the following recursion ![\[ z_{n} = (1- \omega^{e_{n}})z_{n-1}+\omega^{e_{n}}z_{n-2} = z_{n-1} + \omega^{e_n+4} \cdot z_{n-1} + \omega^{e_n}z_{n-2}\]](http://latex.artofproblemsolving.com/c/6/3/c63c4e287786626a944e450a5e7935f11337c8c1.png)
![\[ \implies A_n = 2A_{n-1} + A_{n-2} \]](http://latex.artofproblemsolving.com/2/8/7/28717d4f0ecbab658c885103bee6adbe595c82fc.png)
Where 
and 
. Note that this recursion implies that for all odd , we have 
. Thus, 
has an odd number of terms in it's expansion. Thus, 
for all odd . So, the two ends of the needle cannot swap.
![$\mathbb{Z}\left[\frac{\sqrt{2}}{2}\right]^2$](http://latex.artofproblemsolving.com/9/c/0/9c00968aacadfe8470d61c5f8f7eecaf71153545.png)
. WLOG the needle has length . We will only focus on one endpoint, and prove that it cannot lie on the starting point of the other endpoint.
around another results in both 
and 
being invariant modulo 
. However, since the two endpoints of the needle have distance , the two endpoints have different parities, so they cannot swap.
are vectors with the same magnitude that sum to and the argument of each vector is a multiple of 
degrees, then is even.. The idea here is that since is irrational, both the integer and parts of both 
and 
coordinates must be zero. Thus 
appears the same number of times as 
, and 
appears the same number of times as 
. Furthermore, 
in total occur the same number of times as 
in total due to the coordinate. Thus, the total number of vectors is even. degrees. However, in order for it to get back to its original position with the orientation swapped, the number of such vectors must be odd, contradiction.
. Induct gives 
parity unchanged.
Assume this exists. If we trace out the path as we move along the sequence, it forms a polygon(not necessarily degenerate) with odd number of vertices.![[asy]
size(4cm);
pair Exp(real r){
return (cos(r), sin(r));
}
draw((1, 0) -- (0, 0) -- Exp(pi/4) -- 2 * Exp(pi/4) -- (2 * Exp(pi/4) + Exp(0)) -- (2 * Exp(pi/4) + 2 * Exp(0)));
draw(2 * Exp(pi/4) + (2,0) -- (1, 0), Dotted);
dot((0, 0), black);
dot((1,0), red);
dot(Exp(pi/4), red);
dot(2 * Exp(pi/4), black);
dot(2 * Exp(pi/4) + (1,0), red);
dot(2 * Exp(pi/4) + (2,0), black);
[/asy]](http://latex.artofproblemsolving.com/6/1/2/6124521a2bc1408652ed0d9e0271b22939d9bec1.png)
such that ![\[\sum_{k = 0}^{2\ell} \omega^{a_k} = 0.\]](http://latex.artofproblemsolving.com/4/5/4/45407360bba2600886aab65352ba7608d4c85ee0.png)
Substitute 
and we get that the polynomial ![\[P(x) = \sum_{k = 0}^{2\ell} x^{a_k}\]](http://latex.artofproblemsolving.com/c/2/5/c25c7865d9701a3252501b29d78dfb322c9d06c5.png)
must have a root at 
. Furthermore, ![$P(x)\in\mathbb Z[x].$](http://latex.artofproblemsolving.com/9/6/a/96ab7f9852da14f353985515279469b1abe435e9.png)
Since 
it follows the minimal polynomial of (equivalently 
) must divide 
Thus, 
However, this implies 
However, this is false since is the sum of an odd number of 
