Tiling rectangle with smaller rectangles.

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1595 posts

#1 h Jul 10, 2018, 11:25 AM • 13 Y

Y by Amir Hossein, Davi-8191, rustam, Kayak, Tenee, Purple_Planet, centslordm, jhu08, megarnie, Adventure10, Mango247, lian_the_noob12, Sedro

A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Proposed by Jeck Lim, Singapore

This post has been edited 2 times. Last edited by MarkBcc168, Jul 15, 2018, 12:57 PM

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MarkBcc168

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MarkBcc168

#2 h Jul 10, 2018, 11:25 AM • 31 Y

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Solution

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RMO-prep

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RMO-prep

#3 h Jul 10, 2018, 6:54 PM • 5 Y

Y by centslordm, jhu08, sabkx, Adventure10, Mango247

It was also India TST

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juckter

323 posts

juckter

#4 h Jul 10, 2018, 7:45 PM • 4 Y

Y by centslordm, jhu08, Adventure10, Mango247

Let $\mathcal{R}$ 's dimensions be $m$ and $n$ , with $m$ and $n$ odd. Label the square $(x,y)$ with $A$ if $x \equiv y \equiv 1 \bmod{2}$ , $B$ if $x \equiv y \equiv 0 \bmod{2}$ , and $C$ if $x \not \equiv y \bmod{2}$ . Then a rectangle satisfies the problem's condition iff all of its corners are labeled with $A$ or all of them are labeled with $B$ . Notice that because $m$ and $n$ are odd, there exists one less square labeled $C$ than the total number of squares labeled $A$ of $B$ , and thus there must exist at least one small rectangle in which more squares are labeled $A$ or $B$ . But this is easily seen to imply that all four corners of the rectangle are labeled $A$ or $B$ .

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orthocentre

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orthocentre

#6 h Jul 14, 2018, 5:47 PM • 4 Y

Y by centslordm, jhu08, Adventure10, Mango247

Edit: made a really dumb mistake.

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joyce_tan

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joyce_tan

#7 h Jul 14, 2018, 6:01 PM • 4 Y

Y by centslordm, jhu08, sabkx, Adventure10

jdevine wrote:

There's a much easier way to do this than with colouring arguments.

Click to reveal hidden text

There's a reason why everyone else is using coloring arguments - your solution proves a weaker statement than the problem. You've proven that a rectangle with both odd dimensions exists, and an odd dimension ensures that the opposite distances have equal parity, but you haven't shown that the parity of the distances in the two different axes are equal, i.e. you could have a rectangle that is an odd distance from the top and the bottom, but an even distance from the left and the right.

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mofumofu

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mofumofu

#8 h Jul 14, 2018, 6:02 PM • 15 Y

Y by Ankoganit, 62861, DVDthe1st, Kayak, oneplusone, rkm0959, AnArtist, AlastorMoody, skt, rashah76, centslordm, jhu08, megarnie, sabkx, Adventure10

This problem was proposed by oneplusone.

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test20

988 posts

test20

#9 h Jul 14, 2018, 6:03 PM • 4 Y

Y by joyce_tan, centslordm, jhu08, Adventure10

jdevine wrote:

There's a much easier way to do this than with colouring arguments.

Consider the length of $\mathcal{R}$ . Since it's odd, we let it be $2n+1$ . Consider a smaller rectangle $\mathcal{S}$ . Let the distance from the left side of $\mathcal{S}$ to the left side of $\mathcal{R}$ be $a$ , and let the distance from the right side of $\mathcal{S}$ to the right side of $\mathcal{R}$ be $b$ . Iff $a$ and $b$ are the same parity, then $a+b$ is even and so $2n+1-a-b$ is odd. Repeat in the other direction (up and down instead of left and right).

This means that we want to prove that there exists a rectangle whose sides are of odd length. Suppose, for sake of contradiction, that there isn't one. Then all the small rectangles have even area, but the large rectangle has an odd area. Contradiction.

I'm not sure why they put that on the IMO shortlist... it would have been much more suitable in something like this and would have been a really good problem to put in, probably near the end of the paper.

Naah, that doesn't quite work.

You have only established the existence of a rectangle $S$ whose sides are of odd length.
But $S$ might have odd distances to the northern and southern side of $\cal R$ ,
and even distances to the eastern and western side of $\cal R$ .

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anantmudgal09

1980 posts

anantmudgal09

#10 h Jul 14, 2018, 8:12 PM • 4 Y

Y by Saikat002, centslordm, jhu08, Adventure10

By induction on chessboard polygons, it is easy to show that each rectangle in such a tiling can be indexed with vertices as lattice points.

Now for each rectangle of area $A$ apply the following operation. For $A \equiv 0 \pmod{2}$ tile it with dominoes and for $A \equiv 1\pmod{2}$ mark the leftmost corner and tile the remaining rectangle with dominoes. Observe that each tiling $\mathcal{T}$ maps to a tiling $\mathcal{T}’$ consisting only of dominoes and boxes such that $\mathcal{T}$ has a desired rectangle iff $\mathcal{T}’$ does.

Apply chessboard colouring with upper left corner coloured red. For $\mathcal{T}’$ notice that any box placed at red square is desired. If not then all such red squares are covered in dominoes. Since no domino contains two monochrome squares and we have more red squares, we get the desired contradiction.

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orthocentre

72 posts

orthocentre

#11 h Jul 14, 2018, 9:56 PM • 4 Y

Y by centslordm, jhu08, Adventure10, Mango247

test20 wrote:

jdevine wrote:

There's a much easier way to do this than with colouring arguments.

Consider the length of $\mathcal{R}$ . Since it's odd, we let it be $2n+1$ . Consider a smaller rectangle $\mathcal{S}$ . Let the distance from the left side of $\mathcal{S}$ to the left side of $\mathcal{R}$ be $a$ , and let the distance from the right side of $\mathcal{S}$ to the right side of $\mathcal{R}$ be $b$ . Iff $a$ and $b$ are the same parity, then $a+b$ is even and so $2n+1-a-b$ is odd. Repeat in the other direction (up and down instead of left and right).

This means that we want to prove that there exists a rectangle whose sides are of odd length. Suppose, for sake of contradiction, that there isn't one. Then all the small rectangles have even area, but the large rectangle has an odd area. Contradiction.

I'm not sure why they put that on the IMO shortlist... it would have been much more suitable in something like this and would have been a really good problem to put in, probably near the end of the paper.

Naah, that doesn't quite work.

You have only established the existence of a rectangle $S$ whose sides are of odd length.
But $S$ might have odd distances to the northern and southern side of $\cal R$ ,
and even distances to the eastern and western side of $\cal R$ .

Oh... I thought it was way too easy for IMO shortlist...

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Math.Is.Beautiful

850 posts

Math.Is.Beautiful

#12 h Jul 15, 2018, 8:14 AM • 4 Y

Y by centslordm, jhu08, Adventure10, Mango247

MarkBcc168 wrote:

A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

Don't know, but I feel like I have seen this problem long ago.

EDIT: No, it wasn't this one, it was the following theorem:
Theorem: If a rectangle can tiled with rectangles, each of which have at least one side of integeral length, then the rectangle which was tiled also has at least one side of integeral length.

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oneplusone

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oneplusone

#14 h Jul 15, 2018, 12:54 PM • 27 Y

Y by 62861, CZRorz, DVDthe1st, Kayak, mofumofu, anantmudgal09, Wizard_32, Salman.elourdi, ring_r, AnArtist, Idea_lover, djmathman, rashah76, OlympusHero, Eliot, centslordm, jhu08, PRMOisTheHardestExam, sabkx, ihatemath123, Adventure10, Mango247, RTB, EpicBird08, Sedro, OronSH, Rijul saini

I discovered this problem when I was trying to solve the problem "there exists only finitely many ways to divide a rectangle into smaller rectangles of equal area". A subproblem is that there is no smooth deformation of the smaller rectangles that preserves all their area (turns out there is a completely different elegant proof for this). A slightly stronger statement is that if we perturb each edge slightly up/down or left/right, there is a rectangle that strictly contains its original form, or is strictly contained in its original form. This is equivalent to this C1. I was hoping to find some kind of "Brouwer fixed-pt" solution, but couldn't find one.

My original solution is to colour the rectangle black/white in a checkerboard manner, with the 4 corners black, then it suffices to find a smaller rectangle with 4 black corners. We count the number of black and white corners among all the rectangles. For each corner vertex that is not the corner of $\mathcal{R}$ , there are an equal number of black and white rectangle corners at that vertex. So there are more black corners than white, hence there is a smaller rectangle with at least 3 black corners. But a rectangle with 3 black corners necessarily has the 4th corner black.

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yayups

1614 posts

yayups

#16 h Aug 21, 2018, 7:43 AM • 4 Y

Y by centslordm, jhu08, Adventure10, Mango247

ugh this took way too long, but I suck at combo anyways....

Call $(i,j)$ white if $2\mid i-j$ , and call it black otherwise. We wish to show the existence of a rectangle with all white corners. However, a rectangle has all white corners if and only if it has more white squares than black squares. But since $\mathcal{R}$ has more whites than blacks, by PHP we are done.

Remark: In my "defense", for most of the time I was trying to show that the number of all white cornered squares was odd (this is the usual strategy), but experimentation very quickly reveals this to be FALSE. Therefore, everything I was trying beforehand was bound to fail. But I was too lazy to actually experiment till the very end.
Lesson: experiment

One more remark:

oneplusone wrote:

My original solution is to colour the rectangle black/white in a checkerboard manner, with the 4 corners black, then it suffices to find a smaller rectangle with 4 black corners. We count the number of black and white corners among all the rectangles. For each corner vertex that is not the corner of $\mathcal{R}$ , there are an equal number of black and white rectangle corners at that vertex. So there are more black corners than white, hence there is a smaller rectangle with at least 3 black corners. But a rectangle with 3 black corners necessarily has the 4th corner black.

I think this solution needs slight rewording since $1\times x$ squares have at most 2 corners.

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DynamoBlaze

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DynamoBlaze

#17 h Dec 15, 2018, 2:39 PM • 3 Y

Y by centslordm, jhu08, Adventure10

I need some clarification. Are the rectangles in a grid? I mean, does the large rectangle have rows and columns which intersect to make the smaller rectangles, or are the small rectangles independent? In other words, if there is a rectangle $PQRS$ , will $QR$ be the side of one other small rectangle as well?

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Durjoy1729

221 posts

Durjoy1729

#18 h Dec 17, 2018, 2:48 PM • 8 Y

Y by Digonta1729, potentialenergy, DonaldJ.Trump, ike.chen, centslordm, jhu08, lian_the_noob12, Adventure10

Official solution

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ihatemath123

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ihatemath123

#55 h Jun 27, 2023, 7:46 PM • 1 Y

Y by bjump

WTF this is such an elegant problem

Color the rectangle black and white such that the corners are all black. Clearly, a small rectangle that satisfies the parity condition must have all black corners - equivalently, it has one more black square than white square. Other rectangles either contain an equal number of black squares and white squares, or more white squares than black. Since the large rectangle contains more black squares than white squares, we must have a small rectangle that satisfies the parity condition.

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noppi_kun

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noppi_kun

#56 h Jul 30, 2023, 10:21 PM

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Nice problem.

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Reason: Sssssss

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ihatemath123

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ihatemath123

#57 h Jul 30, 2023, 10:44 PM

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noppi_kun wrote:

Each small rectangle must have an even area, and therefore $\mathcal{R}$ must also have an even area, which is a contradiction.

I believe this idea is correct, but there is no one else doing the same. Can someone please tell me if this is right or wrong

If a rectangle has odd area, we know for sure that the distance from the vertical sides of this rectangle to $\mathcal R$ are both even or odd; we also know for sure that the distance from the horizontal sides of this rectangle to $\mathcal R$ are both even or odd; however, we do not know that all four parities are the same.

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noppi_kun

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noppi_kun

#58 h Jul 30, 2023, 10:49 PM

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ihatemath123 wrote:

noppi_kun wrote:

Each small rectangle must have an even area, and therefore $\mathcal{R}$ must also have an even area, which is a contradiction.

I believe this idea is correct, but there is no one else doing the same. Can someone please tell me if this is right or wrong

If a rectangle has odd area, we know for sure that the distance from the vertical sides of this rectangle to $\mathcal R$ are both even or odd; we also know for sure that the distance from the horizontal sides of this rectangle to $\mathcal R$ are both even or odd; however, we do not know that all four parities are the same.

I was wrong.. Thanks.

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LLL2019

834 posts

LLL2019

#59 h Mar 28, 2024, 12:08 PM

Y by

Same as above. Here is my motivation for the coloring.

If the rectangle is $[0,a]\times [0,b]$ and we are considering $[x,x+c]\times [y,y+d]$ , we wish to prove $c$ , $d$ are odd and $x\equiv y\pmod 2$ is possible. Let us just focus on the first half. We can note that by weighting each cell by $1$ , we see $\sum cd$ must be odd, so there must be some rectangle with $cd$ odd.

We can then think of doing a similar thing with $x\equiv y \pmod 2$ . It is quite natural to think of labeling each $(x,y)$ with $x+y$ . However, we see this does not work. Then, our next idea is to label each $(x,y)$ with $\delta(x+y\equiv 0\pmod 2)$ , which works.

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bjump

1013 posts

bjump

#60 h Aug 15, 2024, 1:25 AM

Y by

Fun combo

Tile the grid like a checkerboard with corners shaded. We can observe that a small rectangle works iff its corners are all shaded, or in other words it has one more shaded squares than non shaded squares. Now since the sum of the areas of the rectangles inside $\mathcal{R}$ is odd. There must exist an odd by odd rectangle. If all odd rectangles had blank corners then, there would be more blank squares total than shaded squares, however the rectangle has more shaded squares a contradiction. So there must exist at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.

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RedFireTruck

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RedFireTruck

#61 h Sep 22, 2024, 1:03 AM

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Color the cells of the grid black and white in checkerboard pattern so that it has $4$ black corners. Iff the rectangle's distances from the four sides are all odd or all even, then it has $4$ black corners. Assume FTSOC that we do not use any rectangle with $4$ black corners in our dissection. Then we can only use rectangles with $2$ white and $2$ black corners, or $4$ white corners. In both cases, there are at least as many white squares as black squares, but in $\mathcal G$ , we have more black squares than white squares. This is a contradiction, as desired.

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ezpotd

1262 posts

ezpotd

#62 h Sep 22, 2024, 2:18 AM

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Checkerboard tiling, with white at the corners. Clearly some rectangle must have white squares than black corners, if the rectangle has even area it is forced to be evenly split between white and black, so it is odd and all four corners are white. Taking distances from the white corners (every two adjacent directions have the same parity, so they all have the same parity) finishes.

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AshAuktober

1001 posts

AshAuktober

#63 h Oct 2, 2024, 9:24 AM

Y by

Solution with the following flavortext:

Quote:

A rectangular grid $\mathcal G$ has an odd number of square unit cells. The grid is dissected along its gridlines into several rectangles (with sides parallel to the grid). Prove that one of the rectangles $R$ in the dissection has the following property: the distance from $R$ to the four sides of $\mathcal G$ are either all odd or all even.

Color the unit squares of $\mathcal G$ in a chessboard fashion, with the bottom-left square being black.For a rectangle $\mathcal R \in \mathcal G$ , define $B(\mathcal R)$ to be the number of black squares contained in it, and $W(\mathcal G)$ to be the number of white squares in it. Then the condition on our required rectangle $R$ is equivalent to $B(R) > W(R)$ . So assume $B(R) \le W(R)$ for every rectangle. Then $B(G) = \sum_{R \in P} B(R) \le \sum_{R \in P} W(R) = W(G),$ where $G$ refers to the rectangle covering the entire of $\mathcal G$ , and $P$ refers to the partition. But this is impossible due to $G$ itself satisfying that $B(G) > W(G)$ , contradiction. $\square$

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shendrew7

794 posts

shendrew7

#64 h Dec 31, 2024, 11:34 PM

Y by

Checkerboard color the grid using black and white, where the corners are colored black. Notice a desired rectangle $R$ is equivalent to having odd dimensions and black corners, which must exist as there are more black squares than white squares in the grid. $\blacksquare$

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Maximilian113

574 posts

Maximilian113

#65 h Jan 9, 2025, 9:17 PM

Y by

Color the rectangle like a chessboard with alternating black and white colors, with the corners being black. If the desired result is not true, clearly each rectangle either has an equal amount of black and white squares, or one more white compared to black. Summing this up, it follows that the number of white squares is at least the number of black squares in the entire board, which is clearly a contradiction. QED

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Turtwig113

128 posts

Turtwig113

#66 h Mar 8, 2025, 1:00 PM

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Split the large rectangle into unit squares and color them black and white in a checkerboard, with each corner being black. Then, any rectangle with more black squares than white squares would satisfy the condition because:

$\bullet$ Such a small rectangle would have odd area and thus odd side lengths, meaning the distances from each pair of parallel sides have the same parity (i.e. the distance from the top and bottom would both be odd or both be even)

$\bullet$ Such a small rectangle would also have the sum of the distances to the top and left sides of the large rectangle be even, since the top-left corner of the small rectangle would be black.

This means that the distances to the top and left sides of the large rectangle would either be both odd or both even, implying all the distances to the sides are all odd or all even. This clearly holds for some small rectangle, as the large rectangle has one more black square than white square, so there must be some small rectangle with more black squares than white squares, hence satisfying the condition.

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Bonime

36 posts

Bonime

#67 h Apr 20, 2025, 11:42 PM

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$\textbf{C1 IMOSL 2017}$
Nice problem. I think I have an overcomplicated local solution:

The idea here is to apply an algorithm that transforms any general partition of the board into one consisting of only $2\times 2$ , $2\times 1$ , $1\times 2$ and $1\times 1$ rectangles. Suppose such a rectangle doesn't exists. Then, for any rectangle $a\times b$ which is contained in some division of $\mathcal{R}$ , it follows the lemma

$\textbf{Lemma:}$ If a smaller rectangle $a\times b$ contained in $\mathcal{R}$ does not have the desired property, dividing it in a rectangle $a\times (b-2)$ and $a\times 2$ or in a $(a-2) \times b$ and $2 \times b$ will never generate rectangles with that.
$\textit{Proof:}$ By a $90^\circ$ rotation, it suffices to consider the first case. Note that the piece $a\times 2$ will never have, because $\mathcal{R}$ has odd sides, the distances of this piece from the right edge and the left edge of $\mathcal{R}$ have different parities. Now, note that since the original rectangle $a\times b$ don't have and the parity of the distance from the four $\mathcal{R}$ sides is preserved, the new piece $a\times (b-2)$ won't have it too $\blacksquare$ .

Now, divide again all the rectangles of the board as much as possible using this process. Note that we'll get at that desired state. From here, it's easier to finish. Color $\mathcal{R}$ in a checkerboard pattern, such that all four corners are black. Note that the $2\times 2$ , and the dominoes pieces cover an equal number of squares of each color. So, we need to have a $1\times 1$ covering a black square, but this implies all its distances to the sides of $\mathcal{R}$ have the same parity, then we're done. $\blacksquare$

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cursed_tangent1434

612 posts

cursed_tangent1434

#68 h Apr 23, 2025, 1:15 AM

Y by

Solution which is slightly sub-optimal than the official solution but the key idea is pretty much the same. It is pretty easy to see that all the rectangles must have sides along the lines dissecting $\mathcal{R}$ into a grid of $1 \times 1$ squares. Color the cells chessboard-style with the corners all white and number the white cells $1$ and $2$ with the top row $1$ and alternating numbers between rows. It is not hard to check that there are 6 types of rectangles (determined uniquely by their opposite corners) with the following parities of their areas.

$\bullet$ Type 1 - $1-B-B-1$ with even area
$\bullet$ Type 2 - $2-B-B-2$ with even area
$\bullet$ Type 3 - $1-B-2-B$ with even area
$\bullet$ Type 4 - $1-1-1-1$ with odd area
$\bullet$ Type 5 - $2-2-2-2$ with odd area
$\bullet$ Type 6 - $B-B-B-B$ with odd area

Now, assuming the contrary the dissection of $\mathcal{R}$ contains no rectangles of types 4 and 5 so it has an odd number of Type 6 rectangles. However, we now make the observation that for any $B$ square which is a corner of a sub-rectangle of $\mathcal{R}$ , one of the cells neighboring it is a corner square of a sub-rectangle as white and this square is white. Thus, (counting with multiplicity in case of a row/column rectangle) the number of black corner cells in total is at most the number of white corner cells as black corner cells are never corners of $\mathcal{R}$ . However, each rectangle of types 1,2 and 3 have at least as many black corners as white while Type 6 rectangles have more. This means our assumption implies strictly more black corner squares than white which is impossible so it is impossible to have solely Type 6 rectangles implying we must have atleast one rectangle of types 4 and 5 as desired.

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YaoAOPS

1537 posts

YaoAOPS

#69 h Apr 23, 2025, 3:32 PM • 1 Y

Y by Bonime

Take a coordinate grid. Then note that for a rectangle $\mathcal{D}$
$f(\mathcal{D}) = \int\int_\mathcal{D} e^{\pi i(x+y)}$ is nonzero iff it has all odd side lengths. Since $f(\mathcal{R})$ is nonzero, the $f$ of one of the rectangles in the partition must also be nonzero.

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