As the figure is shown, place a $2\times 5$ grid table in horizontal or vertical direction, and then remove arbitrary one $1\times 1$ square on its four corners. The eight different shapes consisting of the remaining nine small squares are called banners. [asy][asy] defaultpen(linewidth(0.4)+fontsize(10));size(50); pair A=(-1,1),B=(-1,3),C=(-1,5),D=(-3,5),E=(-5,5),F=(-7,5),G=(-9,5),H=(-11,5),I=(-11,3),J=(-11,1),K=(-9,1),L=(-7,1),M=(-5,1),N=(-3,1),O=(-5,3),P=(-7,3),Aa=(-1,7),Ba=(-1,9),Ca=(-1,11),Da=(-3,11),Ea=(-5,11),Fa=(-7,11),Ga=(-9,11),Ha=(-11,11),Ia=(-11,9),Ja=(-11,7),Ka=(-9,7),La=(-7,7),Ma=(-5,7),Na=(-3,7),Oa=(-5,9),Pa=(-7,9); draw(B--C--H--J--N^^B--I^^D--N^^E--M^^F--L^^G--K); draw(Aa--Ca--Ha--Ja--Aa^^Ba--Ia^^Da--Na^^Ea--Ma^^Fa--La^^Ga--Ka); [/asy][/asy] [asy][asy] defaultpen(linewidth(0.4)+fontsize(10));size(50); pair A=(-1,1),B=(-1,3),C=(-1,5),D=(-3,5),E=(-5,5),F=(-7,5),G=(-9,5),H=(-11,5),I=(-11,3),J=(-11,1),K=(-9,1),L=(-7,1),M=(-5,1),N=(-3,1),O=(-5,3),P=(-7,3),Aa=(-1,7),Ba=(-1,9),Ca=(-1,11),Da=(-3,11),Ea=(-5,11),Fa=(-7,11),Ga=(-9,11),Ha=(-11,11),Ia=(-11,9),Ja=(-11,7),Ka=(-9,7),La=(-7,7),Ma=(-5,7),Na=(-3,7),Oa=(-5,9),Pa=(-7,9); draw(B--Ca--Ea--M--N^^B--O^^C--E^^Aa--Ma^^Ba--Oa^^Da--N); draw(L--Fa--Ha--J--L^^Ga--K^^P--I^^F--H^^Ja--La^^Pa--Ia); [/asy][/asy] Here is a fixed $9\times 18$ grid table. Find the number of ways to cover the grid table completely with 18 banners.
Problem
Source: CSMO 2019 Grade 10 Problem 4
Tags: combinatorial geometry, combinatorics
31.07.2019 16:08
31.07.2019 16:10
MarkBcc168 wrote:
Isn't it 512?
31.07.2019 17:12
I don't know how to draw in asymptote, but if you take two horizontal banners and connect the cut out corners, you can form a 9*2. By reflecting it vertically you can get 2 different orientations Since there will be 9 pairs of these banner, there are 2^9 different ways, but then you can also rotate the whole thing to have vertical banners, so there arae 2^10=1024 possible tilings. I don't see any other way to tile it, but there might be more.
31.07.2019 17:25
natmath wrote: I don't know how to draw in asymptote, but if you take two horizontal banners and connect the cut out corners, you can form a 9*2. By reflecting it vertically you can get 2 different orientations Since there will be 9 pairs of these banner, there are 2^9 different ways, but then you can also rotate the whole thing to have vertical banners, so there arae 2^10=1024 possible tilings. I don't see any other way to tile it, but there might be more. The $9\times 18$ grid table is fixed, so you can't rotate the whole thing. (Maybe my translation is not that clear. If so, sorry for that.)