be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
.
s.t. 
.
,![$K=\mathbb{Q}[\sin^2(\dfrac{2}{3}\arctan(3\sqrt{3}))]$](http://latex.artofproblemsolving.com/5/b/8/5b87dd74c6659350c000071d5a9228f2a80b1466.png)
.
.
and not its conjugate transpose.
be a finite abelian group. There is a magic box 
.
contains only the group identity, 
.
be a positive real number. Divide the positive real axis into intervals 
,
,
,
,
,
satisfying the following differential equations:![\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]](http://latex.artofproblemsolving.com/2/6/8/268b7bc6150b13eed666d566b540df7e643d8344.png)
![\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]](http://latex.artofproblemsolving.com/b/4/6/b46d0d00b34cd3fb72ec310625384547b14871c5.png)
with the initial conditions:![\[
u(0) = 1, \quad u'(0) = 1,
\]](http://latex.artofproblemsolving.com/4/1/6/416754efbff23b7efbf02b32149b1e1fd5cb9905.png)
and the continuity conditions:![\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]](http://latex.artofproblemsolving.com/b/3/0/b309f7b8c70cb27683a96af8452ebf9e36f67a4c.png)
Show that![\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]](http://latex.artofproblemsolving.com/f/0/b/f0b9105523f0561e0a4d151a2695cbf6917e5cc7.png)
![$x\in [0,4]$](http://latex.artofproblemsolving.com/a/8/9/a89e045d4c66cdb788e64e613bd6c3c9b5931cc2.png)
at which the function 
takes its maximum value
be an algebraic extension of 
,
be an algebraic closure of 
is an embedding such that 
for all 
,
extends to an automorphism of 
is an embedding. Then, there exists an embedding 
that extends 
,
is an isomorphism.
and 
that extends 
is also an algebraically closed field.
,
be the field of algebraic numbers, 
.
defined by
Then, according to the exercise above, 
How should we interpret 
?
,
.
such that![\[\mathcal{E}(B \cup C) = B \cup C.\]](http://latex.artofproblemsolving.com/e/c/3/ec34115c9e6e169c19c739220065f647934a1a5f.png)
Y by megarnie, deduck
Let 
and 
be positive integers. Let 
be the set of integer points 
with 
and 
. A configuration of 
rectangles is called happy if each point in is a vertex of exactly one rectangle, and all rectangles have sides parallel to the coordinate axes. Prove that the number of happy configurations is odd.
Proposed by Serena An and Claire Zhang
and 
be positive integers. Let 
be the set of integer points 
with 
and 
. A configuration of 
rectangles is called happy if each point in is a vertex of exactly one rectangle, and all rectangles have sides parallel to the coordinate axes. Prove that the number of happy configurations is odd.Proposed by Serena An and Claire Zhang
This post has been edited 1 time. Last edited by KevinYang2.71, Mar 20, 2024, 4:42 PM
and 
operations which flip the grid on the vertical and horizontal axis. To compute the number of happy arrangements 
,
,
.
.
.
be the group (acting on configurations) generated by the actions ``swap the 
t
t
,
,
,
,
,
)
denote the number of happy configurations on a 
grid. If 
,
,
grid would be paired, where each rectangle is represented by four of the same number.![\[\begin{matrix}
\textcolor{red}{1} & \textcolor{red}{2} & \textcolor{red}{2} & \textcolor{red}{3} & \textcolor{red}{1} & \textcolor{red}{3}\\
\textcolor{blue}{4} & \textcolor{blue}{2} & \textcolor{blue}{2} & \textcolor{blue}{5} & \textcolor{blue}{4} & \textcolor{blue}{5}\\
1 & 6 & 6 & 5 & 1 & 5\\
4 & 6 & 6 & 3 & 4 & 3
\end{matrix}\iff
\begin{matrix}
\textcolor{blue}{4} & \textcolor{blue}{2} & \textcolor{blue}{2} & \textcolor{blue}{5} & \textcolor{blue}{4} & \textcolor{blue}{5}\\
\textcolor{red}{1} & \textcolor{red}{2} & \textcolor{red}{2} & \textcolor{red}{3} & \textcolor{red}{1} & \textcolor{red}{3}\\
1 & 6 & 6 & 5 & 1 & 5\\
4 & 6 & 6 & 3 & 4 & 3
\end{matrix}\]](http://latex.artofproblemsolving.com/f/6/6/f66ec292446172bafb17fc18fc9f4175aaeb4f8e.png)
The happy configurations that are paired with themselves, i.e. unchanged under this operation, are the ones in which each rectangle has all four vertices contained entirely within either the first 2 rows or the last 
rows. This gives 
configurations, so![\[f(2m, 2n)\equiv f(2, 2n)\cdot f(2m - 2, 2n)\pmod 2.\]](http://latex.artofproblemsolving.com/2/e/7/2e74f70e75e6c492a9554c0e500972b82b8b2355.png)
Then by induction, we have![\[f(2m, 2n)\equiv f(2, 2n)^m\pmod 2.\]](http://latex.artofproblemsolving.com/d/f/2/df295b6b05eb07fdcc23b150adcd16035aa9ae65.png)
grid when 
gives 
,![\[f(2, 2n)\equiv f(2, 2)\equiv 1\pmod 2.\]](http://latex.artofproblemsolving.com/6/8/a/68a504007a40e68563bdd598b8ba7eb5ccf24299.png)
,
)![\[f(2m, 2n) = \frac{1!!\cdot 3!!\cdots (2m + 2n - 3)!!}{(1!! \cdot 3!!\cdots (2m - 3)!!)(1!! \cdot 3!!\cdots (2n - 3)!!)}.\]](http://latex.artofproblemsolving.com/f/c/0/fc0f060a01bb02c1beac23814d9ba17a8b54d659.png)
Let us know if you prove this!!
.
case. Thus the amount is odd.
case is trivial. Now, we induct proving that 
is odd if 
t![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), black);
dot((3,0), purple);
dot((4,0), black);
dot((5,0), black);
dot((0,1), red);
dot((1,1), black);
dot((2,1), black);
dot((3,1), red);
dot((4,1), black);
dot((5,1), black);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((5,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), black);
dot((3,3), red);
dot((4,3), black);
dot((5,3), black);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((5,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), black);
dot((3,5), purple);
dot((4,5), black);
dot((5,5),black);
[/asy]](http://latex.artofproblemsolving.com/b/1/4/b147c3592d18385013298e97720574f5e33fe125.png)
ways to pair up the 2n points. From here, there are 
to partition the remaining points into rectangles. Both of these quanitities are odd so total ways in this case are odd.![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), purple);
dot((3,0), black);
dot((4,0), black);
dot((5,0), black);
dot((0,1), red);
dot((1,1), black);
dot((2,1), black);
dot((3,1), black);
dot((4,1), red);
dot((5,1), black);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((5,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), black);
dot((3,3), black);
dot((4,3), red);
dot((5,3), black);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((5,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), purple);
dot((3,5), black);
dot((4,5), black);
dot((5,5), black);
[/asy]](http://latex.artofproblemsolving.com/5/9/6/5965fa66d20c09cf6a622aebf1bebe9392421668.png)
![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), black);
dot((3,0), black);
dot((4,0), purple);
dot((5,0), black);
dot((0,1), red);
dot((1,1), black);
dot((2,1), red);
dot((3,1), black);
dot((4,1), black);
dot((5,1), black);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((5,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), red);
dot((3,3), black);
dot((4,3), black);
dot((5,3), black);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((5,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), black);
dot((3,5), black);
dot((4,5), purple);
dot((5,5), black);
[/asy]](http://latex.artofproblemsolving.com/0/1/9/019d37253bfb07489ae46035b82a752909428d06.png)
by 
sub-rectangle.![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), black);
dot((3,0), purple);
dot((4,0), black);
dot((0,1), red);
dot((1,1), black);
dot((2,1), black);
dot((3,1), red);
dot((4,1), black);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), black);
dot((3,3), red);
dot((4,3), black);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), black);
dot((3,5), purple);
dot((4,5), black);
[/asy]](http://latex.artofproblemsolving.com/0/a/0/0a0f1315b470a3d8e869a3ab9f42213c9eb073f9.png)
![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), purple);
dot((3,0), black);
dot((4,0), black);
dot((0,1), red);
dot((1,1), black);
dot((2,1), black);
dot((3,1), black);
dot((4,1), red);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), black);
dot((3,3), black);
dot((4,3), red);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), purple);
dot((3,5), black);
dot((4,5), black);
[/asy]](http://latex.artofproblemsolving.com/5/2/6/5269a298d2e35c2160ebf98c5dcb2bd9ca62db1c.png)
![[asy]
unitsize(0.5 cm);
dot((0,0), purple);
dot((1,0), black);
dot((2,0), black);
dot((3,0), black);
dot((4,0), purple);
dot((0,1), red);
dot((1,1), black);
dot((2,1), red);
dot((3,1), black);
dot((4,1), black);
dot((0,2), orange);
dot((1,2), black);
dot((2,2), black);
dot((3,2), orange);
dot((4,2), black);
dot((0,3), red);
dot((1,3), black);
dot((2,3), red);
dot((3,3), black);
dot((4,3), black);
dot((0,4), orange);
dot((1,4), black);
dot((2,4), black);
dot((3,4), orange);
dot((4,4), black);
dot((0,5), purple);
dot((1,5), black);
dot((2,5), black);
dot((3,5), black);
dot((4,5), purple);
[/asy]](http://latex.artofproblemsolving.com/4/e/a/4ea9408e232150eb75eabcc3784880d7e93ada05.png)
t
grid.
It suffices to show that 
,
we have shown that there is an odd number of happy tilings of a 
grid. It suffices to show that the number of tilings for a 
grid is odd. To do this, label the 
and 
row by the set 
,
case.
![[asy]
real eps = 0.6; for (int x=1; x<=10; ++x) { for (int y=1; y<=6; ++y) { pair P = (x,y); dot(P); } } dotfactor *= 1; real eps = 0.3; void rect(real a, real b, real u, real v, pen p) { draw( (a,b)--((a+u)/2, b-eps)--(u,b)--(u+eps,(b+v)/2)--(u,v) --((a+u)/2, v+eps)--(a,v)--(a-eps, (b+v)/2)--cycle, p); fill(circle((a,b), 0.2), p); fill(circle((u,b), 0.2), p); fill(circle((a,v), 0.2), p); fill(circle((u,v), 0.2), p); }
rect(1,1,8,3, blue + 1.5); rect(1,2,5,6, deepgreen + 1.5); rect(1,4,8,5, red + 1.5);
filldraw(box((4.5, 0.6), (5.5, 6.4)), invisible, black+dashed); filldraw(box((7.5, 0.6), (8.5, 6.4)), invisible, black+dashed);
[/asy]](http://latex.artofproblemsolving.com/b/2/8/b2879df01a19cb98abb4aef9e94954802c6ec221.png)
on configurations by saying that 
if they differ by any permutation of the highlighted columns. This is an equivalence relation. And in general, if there are 
highlighted columns, its equivalence class under 
elements.
,
.
ways to pick a single highlighted column, and then 
ways to use the left column and highlighted column. So the count in the claim is exactly given by ![\[ (2m-1) \cdot (2n-1){!}{!} f(2m-2, 2n). \]](http://latex.artofproblemsolving.com/4/9/2/4922d884f070187d8c0e0a32d233d66b8ad54e7e.png)
This implies 
and proceeding by induction solves the problem.
to 
mohs, it's not extremely difficult but from a different perspective it's easy to fakesolve; I don't really know if that contributes to the actual difficulty of solving the problem though. All my opinion though.
mohs, it's not extremely difficult but from a different perspective it's easy to fakesolve; I don't really know if that contributes to the actual difficulty of solving the problem though. All my opinion though.
denote the corresponding number of happy configurations. We first resolve when 
.
,
and 
must be vertices of the same rectangle, so it's equivalent to find the number of ways to form unordered pairs from 
.
that 
.
is odd, by an inductive argument 
is odd for any positive integer 
.
(denote this by 
)
be some point in 
be the unique other point in 
-
,
be the other vertices of the rectangle s.t. 
have the same 
-
is symmetric across 
is asymmetric across 
be their corresponding reflections over 
is also a rectangle, and that 
are on the same side of 
.
points in 
)
in a couple, reflect them over 
and draw rectangle 
in a quadruple, reflect them over 
points in total. Then, using these 
happy configs, so it now suffices to show that the number of happy configs with only couples is odd. But by considering the points in 
for 
,
ways to do this, which is odd. So, we are done.
be the number of happy configurations on a 
be the set of points 
.
and vice versa. Note that swiping is an involution, so we want to show the number of configurations that remain the same after being swiped (say, unswipeable configurations) is odd.
is also present in the configuration. It is natural to consider the operation
on any rectangles with vertices 
and its reflection. We will call it crossing. Since it is also an involution, define all configurations that remain the same under the operation uncrossable. It suffices to show the number of such configurations is odd.
subgrid. There are 
ways to choose the rectangles, and 
ways to arrange rectangles in the subgrid.
.
and we are done.
.![\[\frac{\binom{2n}{2} \cdot \binom{2n-2}{2} \cdot \dots \cdot \binom{2}{2}}{n!}.\]](http://latex.artofproblemsolving.com/8/f/0/8f0f18f55086d781e95e03746097f097c986e681.png)
![\[\prod_{i=1}^n \frac{\frac{2i(2i-1)}{2}}{i} = \prod_{i=1}^n (2i-1),\]](http://latex.artofproblemsolving.com/e/8/6/e8647a4e349fb5bf544e014fb14595c06d5eee05.png)
are paired up such that 
and 
are together for 
.
grid with 
and 
:![[asy]
size(10cm);
import geometry;
for(int i = 0; i<12; ++i){
for(int j = 0; j<6; ++j){
dot((i,j));
}
}
draw((0.5,-0.5)--(0.5,5.5),dotted);
draw((3,1)--(0,1)--(0,0)--(3,0),red+dotted);
draw((3,1)--(3,0),red);
draw((10,3)--(0,3)--(0,2)--(10,2),blue+dotted);
draw((10,3)--(10,2),blue);
draw((2,5)--(0,5)--(0,4)--(2,4),green+dotted);
draw((2,5)--(2,4),green);
[/asy]](http://latex.artofproblemsolving.com/a/8/8/a8816ceee46d43245847cbf6efdf6ec2f848067b.png)
produces any happy configurations, then 
produces an equal amount of happy configurations due to symmetry. The only time where the symmetry fails is when 
.
,
in the original problem. Hence, we can use induction on 
grid; on a 
?
points have the same configuration) from the top-left most to the bottom-right most set of the 4 points, we have only the configurations where all rectangles are Type 1 left.
,
iff corresponding configuration 
is in following relationship;
can be made from 
by xooking some non
,
if 
)
)
symmetric is odd. And this is trivial, since there is only one.
ways to do that, and we multiply by the number of ways to do it in a 2(m-1)x2n grid. Note that a 2x2 grid is trivially odd so by induction we are done.
given in the problem statement.
.
ways to pick this. Then, simply shift the lattices down. This gives the recursion 
which finishes.
![\[f(m + 1,n) = f(m, n)\cdot f(1, n) + 2\cdot(\text{happy configurations with some rectangles having vertices in both the last two columns and first }2m\text{ columns}) \equiv f(m,n)\pmod2.\]](http://latex.artofproblemsolving.com/6/4/1/6414050bd958b9c74f73c195c490bcebf4339bdf.png)
