Coloring sells

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Coloring sells

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2017 Belarus Team Selection Test 1.2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

184 posts

#1 h Mar 31, 2019, 5:37 PM • 3 Y

Y by Adventure10, lian_the_noob12, wizixez

An $n\times n$ square table is divided into $n^2$ unit cells. Some unit segments of the obtained grid (i.e. the side of any unit cell) is colored black so that any unit cell of the given square has exactly one black side. Find
a) the smallest
b) the greatest possible number of black unit segments.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1430 posts

#2 h May 11, 2019, 3:52 PM • 1 Y

Y by Adventure10

A(most probably wrong)
Click to reveal hidden text

This post has been edited 1 time. Last edited by leibnitz, May 12, 2019, 5:29 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1430 posts

#3 h May 11, 2019, 4:00 PM • 2 Y

Y by Adventure10, Mango247

Can anyone verify my answers?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

25 posts

#4 h Jun 3, 2019, 9:49 AM • 1 Y

Y by Adventure10

Let $b$ and $i$ be the numbers of segments colored on the boundary and inside the grid respectively. Clearly $b+2i=n^2$ .
a) (i) $n$ is odd. Then $b$ is odd as well, and therefore $b \ge 1$ , so $b+i \ge 1+\frac{n^2-1}{2}=\frac{n^2+1}{2}$ (ii) $n$ is even. $b+i \ge \frac{n^2}{2}$ b) (i) $n$ is odd. Then $b$ is odd as well, and therefore $b \le 4(n-1)-1$ , so $b+i \le 4(n-1)-1+\frac{n^2-(4(n-1)-1)}{2}=\frac{n^2+4n-5}{2}$ (ii) $n$ is even. Then $b \le 4(n-1)$ , so $b+i \le 4(n-1)+\frac{n^2-4(n-1)}{2}=\frac{n^2+4n-4}{2}$ All four values are clearly attainable.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

437 posts

prof.

#5 h Jun 4, 2019, 5:13 AM • 1 Y

Y by Adventure10

Where we can find more Problems from Belarus Math Olyimpiads?
Tnx

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1577 posts

#6 h Jun 1, 2021, 3:26 PM • 1 Y

Y by Mango247

Should the two answers in part (b) of @2above's solution be switched around?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

83 posts

#7 h Nov 20, 2024, 8:02 PM

Y by

a) $n=even$ the answer is $n^2/2$
$n=odd$ the answer is $(n^2+1)/2$

Attachments:

This post has been edited 2 times. Last edited by zaidova, Nov 20, 2024, 8:36 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a