One 8x8 checkerboard is placed directly on top of another one so that their edges are aligned. The top checkerboard is then rotated 45 degrees about the common center of the two boards. What is the total area of the region where the black squares of the two boards overlap?
Attempt to simplify the problem of total BB overlap:
We depend heavily on symmetry for two assumptions:
Assume there is equality of black overlapping black area and white overlapping white area. Further assume this is equal to the area of white overlapping black and equal to the area of black overlapping white. Thus BB is 1/4 the total overlap.
The overall shape of the outline of the combined boards is an eight-pointed star with the 8 points having no overlap.
(I corrected this answer later).We just get the area (h^2) of the eight isosceles right triangles that hang out (not overlapping anything). Here: h = 4 sqrt (2) - 4. Area of a single triangle is T=h^2
The total area of two boards is 2 x 64. Overlapped area is counted twice by definition. So: 2 * Overlapped + 8 T = 2 x 64
2 x (4 x BB) + 8 T = 2 x 64
BB = 16 - h^2 = 13.255. (approx.)
A simulation was used to check that BB = WW = BW = WB. It also found 10.981 for the total area of the 8 triangles.
che
xstar xbb xww xbw xwb xall
10.981 13.256 13.254 13.255 13.254 64.0
A picture is seen here, (ignore the red and blue triangles) and the simulation to check is here:
program che
implicit none
integer ulx(8,8),urx(8,8),llx(8,8),lrx(8,8),
1 uly(8,8),ury(8,8),lly(8,8),lry(8,8),
2 color(8,8),ix,iy,ibig,iseed,i,
3 cntstar,cntbb,cntww,cntbw,cntwb,
4 cnt(8,8),cnttot,ix1,iy1,color0,color1,ixx,iyy,ixx1,iyy1
real xx,yy,xxa,yya,xxb,yyb,xx1,yy1,rat(8,8),r45,pi,rad0,rad1,
1 xstar,xbb,xww,xwb,xbw,xall,c
pi=3.14159
r45=45.*pi/180.
iseed=time8()
call srand(iseed)
c
c color 0 = black, 1 = white
c
do iy=1,8
do ix=1,8
cnt(ix,iy)=0
llx(ix,iy)=ix-1
lly(ix,iy)=iy-1
ulx(ix,iy)=ix-1
uly(ix,iy)=iy
urx(ix,iy)=ix
ury(ix,iy)=iy
lrx(ix,iy)=ix
lry(ix,iy)=iy-1
color(ix,iy)=mod(ix+iy,2)
c print1,ix,iy,llx(ix,iy),lly(ix,iy),ulx(ix,iy),uly(ix,iy),
c 1 urx(ix,iy),ury(ix,iy),lrx(ix,iy),lry(ix,iy),
c 2 color(ix,iy)
c1 format(i1,x,i1,x,4('(',i1,',',i1,') '),i1)
enddo
enddo
ibig=10**9
c ibig=10
c=64./ibig
cntstar=0
cntbb=0
cntww=0
cntwb=0
cntbw=0
cnttot=0
do i=1,ibig
xx=8*rand()
yy=8*rand()
xxa=xx-4
yya=yy-4
xxb=xxa*cos(r45)-yya*sin(r45)
yyb=xxa*sin(r45)+yya*cos(r45)
xx1=xxb+4
yy1=yyb+4
rad0=sqrt(xxa**2+yya**2)
rad1=sqrt(xxb**2+yyb**2)
c print 10,i,xx,yy,xx1,yy1,rad0,rad1
10 format(i3,1x,6(f5.2,1x))
c
c cntstar
c
if(xx1.lt.0.or.xx1.gt.8.or.yy1.lt.0.or.yy1.gt.8)then
cntstar=cntstar+1
c print*,'cntstar = ',cntstar
else
ixx=xx+1
iyy=yy+1
ixx1=xx1+1
iyy1=yy1+1
color0=mod(ixx +iyy ,2)
color1=mod(ixx1+iyy1,2)
if(color0.eq.0.and.color1.eq.0)cntbb=cntbb+1
if(color0.eq.1.and.color1.eq.1)cntww=cntww+1
if(color0.eq.0.and.color1.eq.1)cntbw=cntbw+1
if(color0.eq.1.and.color1.eq.0)cntwb=cntwb+1
c print 20,cntbb,cntww,cntbw,cntwb
20 format('bb ww bw wb ',4(i3,1x))
endif
enddo
xstar=c*cntstar
xbb=c*cntbb
xww=c*cntww
xbw=c*cntbw
xwb=c*cntwb
xall=c*(cntstar+cntbb+cntww+cntbw+cntwb)
print 30,xstar,xbb,xww,xbw,xwb,xall
30 format('xstar xbb xww xbw xwb xall ',6(f10.3,2x))
end
Edited on August 27, 2019, 7:31 am
simplifying method?
