Mirrors around an equilateral triangle is not two mirrors???

I don't know an quick way to get a formula. Sorry.

Bob

]]>To start with, that's exactly what I did. See diagram below.

I set the mirrors at 72 degrees and chose an F shape because it has no lines or rotational symmetry. That was shape 1 on the diagram,

Reflecting in BC gives 2.

Reflecting 1 and 2 in AB gives 3 and 4.

Reflecting 4 in BC gives 5 and so on. Successive reflections give images up to 10.

Thereafter, no further images are generated by reflecting any shape in either mirror.

So I thought ten images. But then I realised that 1 and 7 together can be treated as a single 'motif' in which case the number of images is reduced to five.

So I moved 1 until it joined to 7 to make a single shape.

I think this happens whatever shape you start with.

Bob

]]>Try to do that pic with only half of the figure you posted there. I think you are assuming the object to be symmetrical and halfway between the mirrors.

]]>i still don't get it... since the image can be placed anywhere between them i am unable to make any relation with the number of images formed can you please explain me how to analyse the diagram that you have posted i am still having trouble dealing with the resulting proof]]>

n even formula looks right to me but:

When the angle is 72, I'm getting 5 images. To prove the property I think you'll have to analyse the angle between the red lines (see diagram) for different angles.

I can prove it for any particular angles like 90 and 60. I cannot think of a way to prove the general result in a single proof.

Bob

]]>i just found that the number of images(n) is given by n=[360/θ] for n being even and [360/θ -1] for n being odd and [.] is the greatest integer function but i don't know how this was derived]]>