Teaser 2503

[Cactus]

Sunday Times Teaser 2503 by Danny Roth
---------------------------------------

Two vertical mirrors that are touching along a vertical edge are
placed at an angle that gives multiple images of a small object
between them. The angle in degrees and the number of images formed are
each a whole number and their product is a perfect cube. What is the
angle between the mirrors?

Apologies for the earlier erroneous descriptions.

Cactus

[Teasemaster]

Does anyone else get 108 degrees and 2 reflections?
TM

[Claude]

The nearest solution I could get was 18 degrees, but this is not an
exact answer.

I used the physics formula:
Number of images = 360/angle - 1

For an angle of 18 deg, this gives 19 images.
So, 18 * 19 = 342, which is nearly 7^3 (343)

Could there be a mistake in the question?
Claude

[Cactus]

Hi Claude and Bruce,

I'm guessing that the lack of early responses to this teaser is a
symptom of the difficulty of working out the number of images formed
when the angle in degrees between the two mirrors is not a divisor of
180. This is quite complex. For example 90 degrees gives 3 images but
89 and 91 degrees each give four images. Equally an angle above 120
degrees gives a maximum of two images but below 120 degrees these two
images each split into two to give a maximum of four images in
total.

But it gets worse because the number of these images that can be seen
depends on the observer's viewpoint and the size of the mirrors. For
those who have a java enabled browser, this physics site posting:

http://preview.tinyurl.com/2uoujt8

has a simulation that can be used to experiment with different mirror
angles.

At TM's 108 degrees there are viewpoints and mirror sizes where only
two images are visible so his answer, 2 * 108 = 216 = 6^3, is valid at
some locations. But at 49 degrees, the maximum number of images is 8
so, if we pick a location where only 7 of these are visible, we can
get a value of 49 * 7 = 7^3, which is the maximum cube value.

And there are mirror angles and viewpoints that give other integer
cube values as well.

Cactus

[beermagnet]

I guess there is a solution around 7 degrees where 49 images can be
seen.
(Mind you I can't prove it)

[Peter]

Hello Claude and Brian,
I think angle 90° gives 2 images, not 3. The directly viewed object is
not an "image". The formula for the number of images becomes n = 360/
angle - 2. 72° gives then n = 3 and angle x n will be 216 = 6^3.
Peter

> > TM- Zitierten Text ausblenden -
>
> - Zitierten Text anzeigen -

[Cactus]

Hello Peter,

If we put the object between the two mirrors on the angle bisector and
put the mirrors at plus and minus 45 degrees from this line, we get an
image from a single reflection off each mirror at plus and minus 90
degrees. But we also get an image from a reflection off both mirrors
at 180 degrees. So we get three images.

It took me some time to figure out how many images were formed by
mirrors at d degrees - and I may still have this wrong - but I think
the answer is:

If degrees divides 180:

no of images = 360 / degrees - 1

otherwise:

no of images = 2 * floor((360 / degrees + 1) / 2)

Brian

[nick f]

Actually, the number of images when the angle is a divisor of 360 is
arguable! In the example given in this Teaser question - ie. at 90
degrees - there are actually four images: two from two single
reflections, one to either side; and two from the two different double
reflections - but from any single viewpoint, these last two happen to
coincide so that they appear to be one single image. Arguably, the
question gets around this by specifically saying "counted how many
images there are of it in the mirrors". So the example is perhaps
sufficient to teach us that multiple coincident images must be counted
as a single image.

It is easy to construct a table of the product of angles and images
for all the different integer divisors of 360
(2,3,4,5,6,8.9,10,12,15,18,20,24,30,36,40,45,60,72,90,120,180) and it
quickly becomes clear that none of these will satisfy the
requirement. So we must move to integers that are not divisors of
360. The question does not specify that the degree integers are
divisors of 360, so there is no reason to assume it. It's just a bit
harder to figure out the number of images.

The question does not specify whether the number of images seen and
counted are just those images that can be seen from a single viewpoint
or whether the observer (George) might be moving around in order to
see all possible images. Unfortunately, the question doesn't help
because in the quoted special example of the angle being a divisor of
360, three images will always be seen regardless of the viewpoint or
the location of the object.

But, whichever way you look at the problem, I believe the number of
images will always be only one of two numbers. For example, for
angles between 72 degrees and 90 degrees, there will be three images
visible from a single viewpoint and four if varying viewpoints are
used. From 60 degree to 72 degrees, there will be four images, or
five. And so on.

It is easy to construct a table, for all integer degree angles between
0 degrees and 180 degrees, using both numbers of images. The number
of images is either (a) 360/angle, rounded up; or (b) [360/angle]-1,
rounded up. The cube root of the product of images and angles varies
between a little from under 6 and little over 8. But only one angle,
49 degrees, and 7 images provides a perfect cube, 343.

[nick f]

[Cactus]

Hello Nick,

Much of our analysis is similar but I don't agree with your figures
for the number of images.

For example, at 65 degrees, I find the maximum number of images to be
6, not 5. Moreover I don't see only a difference of 1 between
different viewpoints.

Brian

[Peter]

Hello Brian,
I discovered my mistake soon after I sent my first message. Please
ignore it. An Thanks for your comment.
Peter

>      Brian- Zitierten Text ausblenden -

[Ben Alligin]

A really interesting one, this, and I'm not sure yet that I have got
it right.

Assume that George is due south of the mirror junction, looking
northwards, and that the mirrors are symmetric about north, one lying
roughly towards the southeast and the other towards the southwest. I
also assume that George is a considerable distance away from the
mirrors (to avoid the complications to do with the observer's
viewpoint).

I found it easiest to think in terms of a light beam leaving George's
eye and travelling due north, although in reality the reverse happens.
Anyway, after this light beam hits the left mirror, it is deflected so
that it is now travelling at an angle A east of north, where A is the
angle between the mirrors. The beam then hits the right mirror, after
which it is travelling at an angle 2A west of north. After the next
reflection, it is travelling 3A east of north, and so on.

As long as the angle away from north is less than 90, there will be
scope for another reflection to give a new possible path towards the
cube, but once the angle exceeds 90 you will be duplicating another
path. After the first reflection, there will be another if A is less
than 90, and another is 2A is less than 90 etc. So, if A is 89, there
will be two possible routes for the beam reflecting off the left
mirror to reach the cube, and George will see two images of the object
in the left mirror. He will also see two images in the right mirror,
giving 4 in total. The same number of images is seen for values of A
down to 46, so if A is 54, there are 4 images and the product, 216, is
a perfect cube, so I make the answer 54 degrees.

John

[nick f]

Brian, you may be right. My reasoning allowed for the number of
images at 65 degrees to be either 5 or 6. And at 49 degrees, it could
be 7 or 8 - depending on the observer's view point, etc. etc. But if
you consider both possibilities for all integer angles, there's still
only one solution that fits.

[Cactus]

Hi Nick,

I have checked my number of images formula with a couple of physics
sites and I am fairly confident that I have this right. I see the
possible solutions as (this table needs a fixed width font):

angle image cube_root angle image cube_root
1 1(359) 1.00 1 8(359) 2.00
1 27(359) 3.00 1 64(359) 4.00
1 125(359) 5.00 1 216(359) 6.00
1 343(359) 7.00 2 4(179) 2.00
2 32(179) 4.00 2 108(179) 6.00
3 9(119) 3.00 3 72(119) 6.00
4 2(89) 2.00 4 16(89) 4.00
4 54(89) 6.00 5 25(71) 5.00
6 36(59) 6.00 7 49(52) 7.00
8 1(46) 2.00 8 8(46) 4.00
8 27(46) 6.00 9 3(39) 3.00
9 24(39) 6.00 12 18(29) 6.00
16 4(22) 4.00 18 12(19) 6.00
24 9(16) 6.00 25 5(14) 5.00
27 1(14) 3.00 27 8(14) 6.00
32 2(12) 4.00 36 6(9) 6.00
49 7(8) 7.00 54 4(6) 6.00
64 1(6) 4.00 72 3(6) 6.00
108 2(4) 6.00 125 1(2) 5.00

where an image term a(b) means that 'a' out of a possible 'b' images
are seen. This list includes all suggestions so far made: TM: 108
degrees and 2 images; nick_f: 49 degrees and 7 images; beefmagnet: 7
degrees and 49 images; John: 54 degrees and 4 images.

As specified I don't believe that there is a 'right' answer.

Brian

[nick f]

Try it another way. Compile a table with every integer angle between
0 and 180 down one side, and every perfect cube, from 1 to 1000 across
the top (only ten of them). Then compute cube/angle for each
combination and identify all integer solutions. There aren't that
many, 56. Most can be intuitively ruled out, eg. 4 images at 128
degrees; 10 images at 100 degrees. To my mind, the most viable
solution in the whole table of 1,790 numbers is 49 degrees and 7
images.

[Cactus]

Hi John,

At 54 degrees there are three images on each mirror even at a long
distance - I have uploaded fiftyfour.png to show the paths involved.

Brian



>
> John

[Ben Alligin]

Brian,

Thanks for the diagram. In my analysis, I should have said that so
long as the angle away from north is less than 180, then the light
beam can intercept the object. So, in your diagram, the angle of the
blue line away from north after the third reflection is 3*54, i.e.
162, but after a fourth reflection, the angle would be 216 so this
could not intercept the object. Therefore, 3 is the number of images
seen in the left mirror (and obviously 3 in the right mirror). In this
analysis, I have assumed that the object and observer are on the
bisector with the observer a considerable distance away (I agree that
the number of images can vary according to the position of the
observer), and that the mirrors are large enough to see all possible
images.

Anyway, I find that I agree with 360/A - 1 for the number of images if
A divides 180, but otherwise I get 2 x floor (180/A), which is
slightly different from you. For instance, if A is 51, I get 6 images
while you get 8. I haven't yet worked out what is wrong with my
formula.

Using either formula fails to yield an answer to the teaser, so either
he's made a mistake or he is allowing different viewpoints or mirror
sizes. I'd be surprised at the latter since this is a Sunday Times
teaser, not a rigorous analysis of a physical problem, so I would
assume all possible simplifications to be made.

John

[Cactus]

Hi John

I am going to find it hard to explain the extra images and I have to
admit that I only worked out what happens after writing a ray tracing
program.

When the mirror angle is 360 / n with n odd, it turns out that a ray
passes exactly along the surface of one of the two mirrors. If the
angle increases nothing much happens but if the angle decreases this
ray then hits the mirror and hence forms another image (each
reflection adds a new image). In effect, at such angles, a single
image splits into two. This happens at 360 / 7 (~51.4 degrees) and at
any angle that is 360 divided by an odd integer.

This can (just possibly) be explained as follows. If we put the
observer at 0 degrees and the mirrors at +d/2 and -d/2 degrees, we get
images in, say, the left mirror at d, 2.d, ..., n.d while n < 180 - d/
2 (while we are left of both mirrors when extended northwards). So we
get n = floor(180 / d - 1/2).

But what happens between 180 - d/2 and 180 + d/2? When d divides 180
exactly we get one image in this zone but otherwise we get two images.
one in EACH mirror. So the total is 2 * [floor(180 / d - 1/2) + 1] = 2
* floor(180 / d + 1/2).

I agree with you that it seems most unlikely that Danny Roth expected
Sunday Times readers to work out the full subtleties of this physical
problem. Until I had worked out the physics, I had put the images in
the wrong positions and got a wrong image count.

Brian

[Ben Alligin]

Brian,

Perhaps I'm being stupid, but I don't completely follow your argument.
In particular, I don't agree with the statement that each reflection
adds a new image. Take your 54 degree image and reduce the angle to 50
degrees. Considering the blue line, the directions of travel are 50
degrees east of north after the first reflection, 100 west of north
after the second and 150 east of north after the third. The right-hand
mirror lies at an angle of 155 degrees east of north, so indeed there
will be a further reflection. However, the angle of travel after the
4th reflection will be 200 degrees west of north, i.e. greater than
180, so this ray can never hit the object which is placed centrally on
the bisector. I therefore believe that there will still only be 6
images with a 50 degree angle.

It is a different matter, of course, if you allow the object or
observer to be non-central.

John

[nick f]

This may go against the grain, but I suggest that instead of messing
around with the mathematics and principals of reflection, you just get
two mirrors! Positioning them at 90 degrees, 72 degrees, 60 degrees
and 45 degrees can be done without a protractor - just use your eye to
tell you when the images are right. Then put an object between them,
anywhere, and move it around a bit and move yourself around a bit, and
count! It's not rocket science...

[Cactus]

Hi John

I'm sorry that I didn't make it clear that I was trying to describe
the circumstances in which a single image splits into two, which
happens when the sight line lies along one of the two mirrors. As the
mirror angle decreases this new image will eventually become visible
on the centre line.

Another way of counting images when d does not divide 180 is to
recognise that whatever happens on one side of either of the mirror
planes must be mirrored exactly on the other side of this plain.

The angles of the images relative to a mirror plane are at d/2, 3d/
2, ..., (2.n - 1)d/2 while (2.n - 1)d/2 < 180, which gives n =
floor(180 / d + 1/2). So doubling this for the other half plane (or
mirror if you prefer) we get 2 * floor(180/d + 1/2).

It will be interesting to ask Victor Bryant for Danny Roth's solution
when the answer is published.

Brian

[Cactus]

Hi Nick,

I did this as one of several steps I took to verify the numbers from
my ray tracing because, at the time, I didn't believe the numbers I
was getting.

Brian

[Ben Alligin]

Brian,

At last I think I've got the solution that the author intended. I
start with the following assumptions. George is far away from the
mirror junction, on the angle bisector, and due south of it, and he
doesn't move around looking for extra images. The mirrors are large
enough to show all images and the cube is very small.

Using my idea of a light ray travelling northwards from George, after
the first reflection it is travelling A degrees east of north, and so
on. Once the angle away from north exceeds 180-A/2, there will be no
further reflections. The question is whether the final reflection will
allow the beam to intercept the cube. If the angle after final
reflection is less than 180, the beam can certainly intercept the
cube. If it is greater than 180, it cannot intercept a centrally
placed cube. However, if the cube is displaced just a tiny amount to
one side, then the beam can intercept the cube if the beam comes from
the mirror on that side. In other words, George will see an image in
one of the mirrors but not in the other.

Consider our example of A=50. After, 1, 2 and 3 reflections, beams
will be travelling at 50, 100 and 150 degrees away from north, and all
of these can intercept the cube. However, after a fourth reflection,
the beam is travelling at 200 degrees from north. This cannot
intercept a cube on the bisector, so only 6 images will be seen (3 in
each mirror), but a slight displacement of the cube to one side will
allow George to see four images in one mirror and three in the other,
so a total of 7 images. The same argument will apply for a centrally
placed cube but a slight displacement of George off the bisector.
Either way, he can see 7 images. I think I'm right in saying that 7
images will be seen even if the displacements from the bisector are
large.

The number of images works out to be floor(360/A), so I'm agreeing
with nick f. The only value of A for which A*floor(360/A) is a cube is
49. I'm sure that this is what the author intended.

To sum up, if George is not allowed to move around to search for
images, and if he and the cube are not precisely lined up on the
bisector, then he will see floor(360/A) images. The intended answer is
then 49.

John

[Cactus]

On Sep 16, 10:32 am, Ben Alligin <jao...@jaowen.plus.com> wrote:

Hello John,

Yes, at large distances for a mirror angle of 49 degrees and a
centred cube, 6 out of 8 images are simultaneously visible in a
central region and 7 out of 8 are visible in two regions either side
of this.

The 49 degrees solution has been my leading candidate from the outset
(see my second posting above) but I don't think it is the only
candidate for the teaser as written.

And, of course, the need for the cube or the observer to be off centre
by some margin conflicts with your reasonable 'all possible
simplifications' thesis, which would put the cube and the observer on
the centre line.

In terms of challenge this was by far the most interesting in some
years.

Brian

Note: Edited to remove an error

[Ben Alligin]

Brian,

Actually, the observer only needs to be just off the centre line to
see the extra image, rather than "by some margin". The tiniest move to
the side will reveal the extra image which can be seen very close to
the mirror junction, provided the cube is small enough (and the teaser
does specify a small cube). You could argue that there is always a 7th
image to be seen with an angle of 49 degrees, even when observer and
cube are central, but in this case the cube gets in the way of seeing
it. Anyway, I will be most surprised if the answer is anything other
than 49.

John

[Cactus]

Hi John

At long distance, he has to be 16 degrees (or more) off the centre
line to see 7 images of a cube on the centre line.

With the mirrors at 24.5 degrees off the centre line, this is not by
any means small - in nearly 2/3 of the angular space between the
mirrors he only sees 6 images.

Brian

[Cactus]

For those who are interested, I have uploaded an analysis of this
teaser to our files area as t2503.pdf.

Brian

[Peter]

Hallo John,

one can also look into the mirrors from the angle bisector plane at
some angle upwards or downwards to ovod any obstruction by the object.
The analysis of angles remains unaltered when dealing with vertical
projections.

Peter

[Cactus]

Hi All,

This has been an unusual teaser in that we have not reached a strong
consensus on the 'intended' answer.

Unless I have made a mistake, here are the votes cast so far:

degrees images supporters
108 2 TM
7 49 beermagnet
54 4 Ben Alligin (initial)
49 7 Nick_f, Ben Alligin, Cactus

What do others think about these (or other) solutions? Should any be
ruled out? Is there a clear winner among them? Has the long debate
changed anyone's mind about the intended solution?

Brian

[Peter]

Hello Brian,

I calculated manually the reflection angles using not-to-scale
sketches, viewing from far distance on the bisector plane. Result:
Degrees images
108 2
7 50
54 6
49 6

I could not construe an odd number of images except for 90 degrees.

Peter

[Garry]

Hello all

Having been on holiday last week, I have just come to this teaser and
so took no part in the detailed discussion. My opinion, based on
diagrams and actual mirror-holding, is that for an object on the
bisector the only perfect-cube product comes from the 7 images which
exist (not necessarily visible from all observation points) when the
angle between the mirrors is 49 degrees.

Garry

[Cactus]

On Sep 19, 7:00 am, Peter <pe...@noll.org.uk> wrote:

Hello Peter,

You should see an odd number of images when an integer divides 180,
i.e. 90, 60, 45 ...

Brian

[Cactus]

Hi Garry,

Although quite a few people here think that this is the intended
solution, why do you think it is the only solution?

Brian

[Garry]

Hi Brian

For the case of the object being halfway between the mirrors, I
calculated that the number of images (maybe not all visible to a
particularly-placed observer) is related to the number of integral
degrees thus:

Degrees Images
------- ------
180 1
120-179 2
91-119 4
90 3
72-89 4
60-71 5
52-59 6
45-51 7
40-44 8
36-39 9
33-35 10
30-32 11

etc

From this, the only product of angle and number of images giving a
perfect cube which I could find was 49X7 = 343, from which I assumed
that the intended solution was 49 degrees. Of course, the compiler
didn't specify the position of the object or the observer, or define
what was meant by number of images, so under different rules 2 visible
images at angular separation of 108 degrees would multiply to a cube,
and there may well be other similar ones. So I wasn't saying the only
solution, just the only one with certain conditions. A more precise
specification of the problem would still have left an interesting
exercise without all of the uncertainty, as discussed in the posts
above.

Note that nothing was said restricting the angle to being less than
180 degrees, so valid "solutions" could arise from the single image
existing, or visible to an observer in a suitable region, for mirrors
inclined at 216 or 343 degrees. The essential daftness of this makes
me think it likely that Danny meant the position as I interpreted it.
But as ever, we don't know, in the absence of an eventual worked
analysis by the compiler. An occasional book of these problems, with
worked solutions at the back, could well have a viable sale -
newspapers do this with old crosswords.

Garry

[Cactus]

Thanks Garry,

I queried this because I wasn't sure about your criterion for the
observer's viewpoint.

From what you said, I felt that it was probably "the total of all
distinct images that can be seen from any viewpoint".

But this didn't work for me since I calculate that 8 images can be
seen for an angle of 49 degree between the mirrors (6 always visible
and 1 more on each side above a 16 degree observer angle).

In fact, unless the mirrors are at an angle that allows two images to
coincide (i.e. the angle is a divisor of 180), I don't think that the
total number of images can be odd since any image that can be seen to
the left of centre will be matched by one at the same angle to the
right of centre.

Brian

[Ben Alligin]

Hi Garry,

The last time the Sunday Times published an anthology of teasers was
in 1994 (published by Times Books - ISBN 0-7230-0679-2 if you're
interested), and there was one prior to that in 1981, so another one
is overdue. Although it doesn't say so any longer on the Teaser page,
you can get the solution (after the closing date for entries) by
sending a stamped addressed envelope to the address for sending
answers to. The teaser editor, Barbara Hall, still prefers to work
with snail mail, so I don't think you'd be able to get an electronic
copy of the solution.

John

[john newton]

Hi, showing my age here, there have been five anthologies published. First in 1974, edited by Ronald Postill, then two almost together, 1980 and 1981, then a fourth in 1994 and the most recent, in 2002.
The ISBN nos are(1)0-1400-3689-X
                          (2) 0-0479-3045-4
                          (3) 0-0479-3046-2
                           (4) 0-7230-0679-2
                           (5) 0-0071-2749-9
They can usually be obtained via Abe Books but beware of the American versions or books with 'brain teasers' in the titles (Books 2 and 3 I think were published in one volume in the USA)
Regards, John Newton.
 
> Date: Mon, 20 Sep 2010 13:27:07 -0700
> Subject: [ST Teaser] Re: Teaser 2503
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> To: sunday-times-t...@googlegroups.com

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[Ben Alligin]

Thanks for that, John. I was aware that there had been an earlier one
than 1981 but didn't know any details. I'm surprised I didn't know
about the 2002 one; it can't have been well publicised. Of course, the
most interesting teasers don't necessarily get selected for the book,
as they may have long solutions.

John

[Ben Alligin]

I've just had a reply from Dr Bryant about the teaser. He said:

The solution intended was the 'pure' one in which all images are
visible, with superimposed ones only counting as one. I am away at the
moment, but I seem to recall that the assumption was that the cube was
tiny and centrally placed.

I think it may be time to request the official solution from the ST.

John

[Peter]

Hello Brian,

Sorry for replying so late, I was abroad.

While away I checked again more thoroughly the rays for different
angles. I assumed that both, the small object and the observer are on
the bisector plane and that the observer is far away so that the view
is essentially parallel to the bisector.

I agree that there is an odd number of images if the angle divides
360, except (I think) for an angle of 40 degrees where I found 8
images. But for this teaser it is irrelevant.

However, as I said previously, I found that 108 degrees gives exactly
2 images (not 4 as implied by Garry's list of 19th September). This
means that, in my opinion, the correct answer was presented by
Teasemaster (2*108 = 6^3). Easy to check experimentally.

Also I'd like to mention again that I could only find 6 images in full
view for angle 49 degrees, not 7. I wonder if anybody can demonstrate
there are 7 images, with object/observer positions as I assumed above.

Peter

[Cactus]

On Sep 26, 12:31 am, Peter <pe...@noll.org.uk> wrote:
> Hello Brian,
>

> Sorry for replying so late, I was abroad.
>
> While away I checked again more thoroughly the rays for different
> angles. I assumed that both, the small object and the observer are on
> the bisector plane and that the observer is far away so that the view
> is essentially parallel to the bisector.
>
> I agree that there is an odd number of images if the angle divides
> 360, except (I think) for an angle of 40 degrees where I found 8
> images. But for this teaser it is irrelevant.

Hi Peter,

The criterion for an odd number of images is that the mirror angle
divides 180, which 40 doesn't, so it gives an even number of images.

> However, as I said previously, I found that 108 degrees gives exactly
> 2 images (not 4 as implied by Garry's list of 19th September). This
> means that, in my opinion, the correct answer was presented by
> Teasemaster (2*108 = 6^3). Easy to check experimentally.

Garry was giving the total number of images that can be seen from all
angles, not just those on the bisector.

> Also I'd like to mention again that I could only find 6 images in full
> view for angle 49 degrees, not 7. I wonder if anybody can demonstrate
> there are 7 images, with object/observer positions as I assumed above.

The short answer is 'no', since, for the configuration you give, there
ARE only 6 images.

To see 7 images with a cube on the bisector, the observer has to be 16
or more degrees away from the centre line.

The official answer was given today as 49, which was our 'consensus'
answer. John has already discussed this with Victor Bryant but, as he
has said, it will be interesting to see how the puzzle setter
justifies this answer since it is certainly not unique.

Brian

[Peter]

Hi Brian,

many thanks for your thorough comment to my recent reply. It is indeed
a self-imposed and unjustified restriction to place the observer and
the object on the bisector.

I also missed superficially that you said that an odd number of images
could be seen if the mirror angle divides 180 degrees (two coinciding
images), not 360 degrees as I noted.

Peter

[Ben Alligin]

Hi John (Newton),

I managed to obtain a copy of the 2002 Teaser anthology. I was most
surprised to see one of my teasers in it! At least I would have
expected the ST to let me know that they were publishing another
collection, although I don't suppose they would have needed my
permission as they would already have the copyright.

John

[Ben Alligin]

I've obtained a copy of the author's "solution". With no explanation,
justification or proof, he states that:
An angle of 90 degrees gives 3 images.
An angle of 89, 88, 87, ... 73 or 72 degrees gives 4 images, etc.
In general the number of images with angle A is the highest integer
less than 360/A.

What astonishes me about this is that Dr Bryant allowed him to state
the formula for the number of images without any proof at all. Also,
there is no explanation of the assumptions made in order to obtain his
result. I'll be asking further questions of Dr Bryant.

John

[Cactus]

Hi John

Thank you for the further information. I have also asked for a copy
of the solution and I was awaiting it before approaching Dr Bryant.

I suspect that this teaser is a great deal more subtle than either
Danny Roth or Victor Bryant realised.

Brian

[Ben Alligin]

Hi Brian,

I think you're right. As far as I can recall, Danny Roth's previous
teasers have been generally straightforward, so I also suspect that he
thought this one was a relatively simple one.

John