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 
.
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
is selected inside acute 
so that 
and 
.
is chosen on ray 
so that 
.
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
and 
be positive integers. The cells of an 
grid are colored amber and bronze such that there are at least 
amber cells and at least 
bronze cells. Prove that it is possible to choose 
chosen cells lie in the same row or column.
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
-
of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let 
?
of identical dominoes is a maximal grid-aligned configuration on the board if 
dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: 
be the number of distinct maximal grid-aligned configurations obtainable from 
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)
