Cut a cross made up of five identical squares into three polygons, equal in area and perimeter.
Problem
Source: Sharygin 2005 VIII-IX CR 2
Tags: geometry, perimeter, area, polygon
455264
03.09.2020 12:39
NikoIsLife wrote: There is probably a better solution than this one.
Madlad this is brilliant
Marinchoo
03.11.2020 14:48
If we divide every of the unit squares into 9 smaller squares and label every each of them as $A,B,C,D,E,F,G,H,I$ for the lines (top to bottom) and 1 to 9 for the columns, then we can find the following example: The first figure consists of squares $D2,D1,E1,F1,F2,F3,F4,F5,G4,H4,H5,H6,I4,I5,I6$. The second one is basically reflect the first one through the center of the cross (squares $A4,A5,A6,B4,B5,B6,C6,D5$ to $D9,E9, F8, F9$). The last polygon is formed by the rest of the squares:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\
\hline
A & & & & $\bigstar$ & $\bigstar$ & $\bigstar$ & & &\\
\hline
B & & & & $\bigstar$ & $\bigstar$ & $\bigstar$ & & &\\
\hline
C & & & & $\checkmark$ & $\checkmark$ & $\bigstar$ & & &\\
\hline
D & $\spadesuit$ & $\spadesuit$ & $\checkmark$ & $\checkmark$ & $\bigstar$ & $\bigstar$ & $\bigstar$ & $\bigstar$ & $\bigstar$\\
\hline
E & $\spadesuit$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\bigstar$\\
\hline
F & $\spadesuit$ & $\spadesuit$ & $\spadesuit$ & $\spadesuit$ & $\spadesuit$ & $\checkmark$ & $\checkmark$ & $\bigstar$ &$\bigstar$\\
\hline
G & & & & $\spadesuit$ & $\checkmark$ & $\checkmark$ & & & \\
\hline
H & & & & $\spadesuit$ & $\spadesuit$ & $\spadesuit$ & & &\\
\hline
I & & & & $\spadesuit$ & $\spadesuit$ & $\spadesuit$ & & &\\
\hline
\end{tabular}
prime_coprime
07.08.2024 12:05
I guess the reason behind we are choosing 3×3 becoz if the sq side is a then every polygon will have area 5a^2/3 , to deal we should take a divisible by 3 then we taken smallest as a = 3