3 Perpendicular Mirrors

Suppose you walk into a room where the wall on the left, the wall in front, and the floor are all mirrors. (The walls and the floor are mutually perpendicular.)

If you hold up a ball, then how many images of the ball can you see in the mirrors?


Note : The images shown in the figure are just the primary images.

3 6 7 8 Infinitely many

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2 solutions

Jonathan Quarrie
Jul 20, 2017

Each plane of reflection will not only reflect the ball, but the reflected images in the other mirrors as well.

  • Reflection = primary image
  • Reflection of a primary image = secondary image
  • Reflection of a secondary image = tertiary image

Are we identifying partial images or only 'whole' balls?

Given that there is only one correct answer, we are being asked how many 'whole' balls (perceived whole balls) we will see in the mirrors, rather than counting individual partial secondary/tertiary images, because the number of partial secondary/tertiary images may vary, depending on our observation point.

Partial images may occur in the corners (where mirrors meet), but a single perceived ball will be comprised of the partial images in such cases - As seen below, the left and right balls are primary images, and the middle ball is actually 2 partial secondary images of the left and right primary images, but they are perceived as a single ball - The more you see of one secondary image, the less you see of the other.

Image credit: http://idol.union.edu/malekis/CVision2003/ Image credit: http://idol.union.edu/malekis/CVision2003/

While observing from within the reflection range of 2 perpendicular mirrors (I.e. not behind the mirrors), it is also possible to have an observation point where a secondary image cannot be seen in one of the mirrors (not even partially). But in this case, a whole secondary image will be seen in the other mirror. So the number of perceived whole balls created from secondary images between any 2 given perpendicular mirrors is alway equal to 1 - this is true regardless of whether it is made up of partial secondary images, or a single whole secondary image.


Why are there not an infinite number of balls?

Image credit: https://m.meritnation.com Image credit: https://m.meritnation.com

Once an incident ray has been reflected, it wont return to the mirror again unless there is another mirror on the same axis facing the opposite direction to reflect it back.

Because each mirror is mutually perpendicular (each exclusively existing on a different axis), it will only be possible for an incident ray to be reflected off each mirror once at most.

In addition to the 3 mirrors described in this problem, the rear wall, the ceiling, and the right wall would also need to be mirrors for 'infinitely many' to be a correct answer.


Why are there 7?

If we consider that each mirror doubles the number of perceived balls (including the original ball):

  • 1 mirror = 2 balls
  • 2 perpendicular mirrors = 4 balls
  • 3 mutually perpendicular mirrors = 8 balls

We then need to account for the fact that the original ball counts as one of the 8 observed balls, but isn't counted as a reflection.

Thus, 7 \boxed{7} whole balls are perceived in the mirrors.

I don't understand why the answer is 7. Surely each primary image will be reflected in each perpendicular mirror, which would give us 9 images.

Chris Hartney - 3 years, 10 months ago

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Where are the primary images located? Will a primary image be reflected in all three mirrors?

Pranshu Gaba - 3 years, 10 months ago

Consider an observation point between two perpendicular mirrors: We will only see 2 secondary images of a ball when they are observed near where the mirrors meet, but they will not be whole images. They will become a composite of a single observed ball - The more we see of one secondary image, the less we see of the other secondary image. There are brief and subtle moments in this video that demonstrate this.

It is also possible for one of the secondary images not to be seen from a given observation point. In which case, a full secondary image can be seen in the other mirror.

The total number of whole balls created by secondary images in this scenario is 1, rather than 2.

If a 3rd mutually perpendicular mirror is introduced, we will have 2 such scenarios, which means there will be 7 whole balls observed, instead of 9.

Jonathan Quarrie - 3 years, 10 months ago

This precluded the possibility of standing at a wall in which case one has a line image projection, or in corners which has a double line image. This analysis also has a fault in that the corner has only partial images that are rotationally symmetric. this yields the illusion of a solitary image. If one considers the partials and every path then one gets 12 images (3 primary, 9 secondary {6 from the edges, 3 from the corners}). Also, being not so serious... We have two eyes observing, thus we observe two of everything.

Daniel Langstaff - 3 years, 10 months ago

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Interesting point about the partial images. But this analysis takes its cues from the way the problem and its answers are presented.

It sounds like you would want to raise a report to get a clearer definition on what we are supposed to be identifying - whole images or partial images.

Jonathan Quarrie - 3 years, 10 months ago

Could you elaborate on what you mean by "line image projection" and "double line image"?

Pranshu Gaba - 3 years, 10 months ago

First of all you have 2 eyes which can not look in 3 directions in the same time. So, YOU CAN LOOK ONLY IN TWO MIRRORS SIMULTANEOUSLY. Meaning you see the reflection of the image in two mirrors and if you discount the secondary images reflected from one mirror to the other you see only 2 images of your ball.

Dinu Fotescu - 3 years, 10 months ago

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You are allowed to move your eyes, and the whole point of this problem is to include secondary and tertiary images if they can be seen.

Jonathan Quarrie - 3 years, 10 months ago

6 reflections! Not 7! Where would the 7th be??

Ethan Crowley - 3 years, 10 months ago

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Pictured it in my minds eye! It's 7 because the one in the vertex of the mirrors! Clever

Ethan Crowley - 3 years, 10 months ago

I believe the number of reflections also depend upon where the observer is standing.

Saurabh Shrivastava - 3 years, 10 months ago

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Regardless of the number of partial secondary & tertiary images the can be counted, we will perceive the same number of whole balls.

I will try to add more detail on this in a few days.

Jonathan Quarrie - 3 years, 10 months ago

@Pranshu Gaba I'd really like to see a more comprehensive solution to this problem. Why are there only 7 images? Why aren't there more considering that there are images, and images of images and so on..?

Agnishom Chattopadhyay - 3 years, 10 months ago

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Could you elaborate on what you think would make my solution more comprehensive? I have tried to explain how secondary images (images of images) count towards the number of perceived balls. Would you like more detail on this?

(I'm currently on holiday, and only have access to my mobile phone, so I can't currently provide detailed diagrams to help explain my solution. But I will do so once I return.)

Jonathan Quarrie - 3 years, 10 months ago

Ah, I didn't originally read your comment correctly. You're looking for an explanation of why 'infinitely many' is not an acceptable answer.

I'll try to add that, but it's probably best explained with a diagram (which will take me a few days to be in a position to create one myself.).

Jonathan Quarrie - 3 years, 10 months ago

This can help visualize the positions of the images formed: the seven images and the actual ball together form the eight vertices of a cuboid. The three mirrors form the three planes of symmetry of the cuboid.
The three mirrors form eight octants, the ball is placed in one of the octants, and an image is formed in each of the seven other octants.

Pranshu Gaba - 3 years, 10 months ago

I will see myself reflected in every (3) mirror, every (3) edge and in the (1) corner. That is 7 times.

But why are those the only images?

Agnishom Chattopadhyay - 3 years, 10 months ago

If you regard the mirrors as x = 0, y = 0 and z = 0, then (a,b,c) is reflected to (-a,b,c), (a,-b,c), (a,b,-c), (a,-b,-c), (-a,b,-c), (-a,-b,c) and (-a,-b,-c). Further reflections repeat these. Simpler to say that we have 2x2x2 choices of whether or not to attach a minus sign. (a,b,c) is the original, so that leaves seven.

A Former Brilliant Member - 3 years, 9 months ago

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