C1
A rectangle 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 R are either all odd or all even.
Re: IMO SL 2017 C1 (Easy or fakesolve)
Posted: Thu Jul 01, 2021 8:52 am
by Asif Hossain
Solution(expected solve with coloring a solution with coloring would be generous still afraid that this may be a fakesolve )
Update: It is a partial solve i.e half solved (Trying to fix it )
Denote the small rectangles as sub-rectangles and their sides be $a_i$ and $b_i$ also let the side of the rectangle R be $m$ and $n$ and both are odd.
It is enough to prove at least one of the $a_i$ and $b_i$ are both odd.
FTSOC,assume every $a_i$ is even and every $b_i$ is odd.
The rectangle R is sub-ractangled into $n$ rectangles
So the area of R$=a_1 \times b_1+...+a_n \times b_n=2k$
$\Rightarrow 2x+1=2k$ Contradiction and we are done.
Note:adding even doesn't change parity so the case $a_i$ and $b_i$ are both even isn't needed.
Re: IMO SL 2017 C1 (Easy or fakesolve)
Posted: Sun Jul 04, 2021 10:18 pm
by Mehrab4226
The solution is quite wordy. My combinatorics solutions are always like that.
Let us divide $R$ in unit squares. We can color the first square(Bottom left) black and the squares adjacent to white. In this manner, we can create a chessboard with $R$. Now for obvious reasons, the first corner of $R$ is black so the other corners must also be black(since the rectangle has odd sides).
Now each of the black squares is one of our desired rectangles. As our first black square(bottom left) is good(name of our desired rectangle). To get to any other black square we need to change the distances from the sides of $R$ by an even number. You can check it out yourself.
Note:
1. For obvious reasons a rectangle with corners of the same color is a rectangle with odd sides.
2. On a rectangle with odd sides the distance from the horizontal sides of $R$ is of the same parity(Either both odd or both even)
3. (2) for the vertical sides as well.
4. In $R$ there the number of black squares is more than white squares.
Now to our proof,
We claim that any rectangle with black unit square as its borders is good.
By $(1)$ the rectangle has odd sides. And since its border is black, let us take the rectangle(small one), it has four sides let us call them up, left, down, right. Now since the upper right square is black, the Distance from the sides of $R$ to up and right are either both odd or both even. And again by $(2)$ and $(3)$ down and left have the same parity as well. So our rectangle is good.
All is left is to prove that there must be at least one small rectangle with black unit square borders. This is very easy. If the borders are not black either all are white or $2$ of them are black and the other $2$ are white. Either case the $\text{white} \geq \text{black}$. If all the small rectangles are like that then, There should be more squares in $R$ white than black. But $(4)$ opposes that. A contradiction. $\square$
still afraid that this may be a fakesolve )
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