Triangle ABC has vertices A=(0, 1), B=(1, 0) and C=(0, 0) in the coordinate plane.
The insides of the sides are lined with mirrors, and then a laser beam is fired from the origin with a slope of 314/379
Which corner of the triangle will the laser beam hit first?
While I have not yet quite grokked all the math involved, I wrote a ray-tracing simulation for this triangle, starting from the origin with any slope. I experimented with rational slopes.
The program results confirm that the analysis in the postings heretofore are all correct, except for my own original post, where I was dead wrong! :-).
Also, while confirming them, I can add a bit to Steve H.’s rules:
If numerator and denominator are both odd, the answer is (0,0) after (2d + n - 2) reflections;
If numerator is even, the answer is (1,0), after (2d + n - 3) reflections;
If denominator is even, the answer is (0,1), after (2d +n - 3) reflections,
where n/d is a reduced fraction.
A plot of 1/5 is here, 13/19 is here, and the first 30 rays of 314/379 are here.
(This also confirms Charlie’s result of 1069 reflections for the original puzzle.)
The ray-trace program follows. As Charlie pointed out (I think) there are only
4 possible slopes. Since CA and CB in reflections negate the slope and
AB inverts it - all rays have slope +/-(m) or +/-(1/m) where m is the original n/d.
Examples and program:
input num, denom
314 379
1 0 after 1069 reflections
input num, denom
2 67
1 0 after 133 reflections
input num, denom
3 91
0 0 after 183 reflections
input num, denom
13 22
0 1 after 54 reflections
rabbit-3:~ lord$ more mir.f
program mir
implicit none
integer ori,dest,i,cnt,max
real*8 x3,y3,x4,y4,xxx,yyy,dxxx,dyyy,delta,oldm,newm,n,d
call delold('mirror.txt')
open(20,file='mirror.txt',status='new')
write(20,4)0.,0.,1,0.,1.,2,1.,0.,3,0.,0.,4
4 format(2(f7.3,1x),i4)
delta=0.1
3 print*,' input num, denom'
read*,n,d
if(n.eq.0)stop
newm=n/d
cnt=-1
max=100000
c max=30
x3=0
y3=0
x4=delta
y4=x4*newm
ori=1
1 cnt=cnt+1
c print*,' going in - cnt = ',cnt
if(cnt.eq.max)then
print*,'reached max ',cnt,' reflections'
stop
endif
oldm=newm
call intersect(ori,dest,x3,y3,x4,y4,xxx,yyy)
write(20,4)xxx,yyy,cnt+4
if(abs(xxx).lt.1e-5.and.abs(yyy).lt.1.e-5)then
print*,' 0 0 after ',cnt,' reflections'
go to 3
endif
if(abs(xxx-1.).lt.1e-5.and.abs(yyy).lt.1.e-5)then
print*,' 1 0 after ',cnt,' reflections'
go to 3
endif
if(abs(xxx).lt.1e-5.and.abs(yyy-1.).lt.1.e-5)then
print*,' 0 1 after ',cnt,' reflections'
go to 3
endif
if(dest.eq.3)newm=1./oldm
if(dest.eq.1.or.dest.eq.2)newm=-oldm
ori=dest
x3=xxx
y3=yyy
c print*,cnt,xxx,yyy
x4=x3+delta
y4=y3+delta*newm
go to 1
end
subroutine intersect(ori,dest,x3,y3,x4,y4,xxx,yyy)
implicit none
integer i,ori,dest,select(2,3),i1,i2,step,havedest
real *8
1 x1,y1,x2,y2,x3,y3,x4,y4,xx,yy,p(2,2,3),term1,term2,termd,
1 xxx,yyy
data p/0,0,0,1,0,0,1,0,0,1,1,0/
havedest=0
i1=mod(ori,3)+1
i2=mod(ori+1,3)+1
step=i2-i1
do i=i1,i2,step
x1=p(1,1,i)
y1=p(2,1,i)
x2=p(1,2,i)
y2=p(2,2,i)
c print 2,ori, i,x1,y1,x2,y2,x3,y3,x4,y4
2 format(' ori des x1 y1 x2 y2',
1 ' x3 y3 x4 y4',
1 /, i2,1x,i2,1x,8(f5.2,2x))
term1=x1*y2-y1*x2
term2=x3*y4-y3*x4
termd=(x1-x2)*(y3-y4)-(y1-y2)*(x3-x4)
c print*,'term1 term2 termd ',term1,term2,termd
xx=(term1*(x3-x4)-(x1-x2)*term2)/termd
yy=(term1*(y3-y4)-(y1-y2)*term2)/termd
c print*,'ori dest xx, yy ',ori,i,xx,yy
if( (i.eq.1.and. 0.le.yy.and.yy.le.1) .or.
1 (i.ne.1.and.( (x1.le.xx.and.xx.le.x2).or.
1 (x2.le.xx.and.xx.le.x1) ))) then
if(havedest.eq.0)then
dest=i
havedest=1
xxx=xx
yyy=yy
c print*,dest,xxx,yyy
else
dest=i
xxx=xx
yyy=yy
c print*,'two destinations!'
endif
endif
enddo
if(havedest.eq.0)then
print*,'found no intersection!'
stop
endif
c print*,ori,i1,i2,step
return
end
subroutine delold(name)
character*10 name
logical ex
inquire(file=name,exist=ex)
if(ex.eqv..true.)then
open(10,file=name,status='old')
close(10,status='delete')
endif
return
end
Edited on February 22, 2019, 7:57 pm
Confirming and counting reflections....
