On Friday I posted this puzzle….

If you have not tried to solve it, have a go now.  For everyone else, the answer is after the break.

Did you solve it? Any other answers?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


    1. For some reason, I absolutely could not see this answer when I was trying to figure it out. Which, now that I see what it is, amazes me.

  1. It would have been a better puzzle if the text was tilted so that it was parallel to the screen. Now the square is parallel to the screen. That makes it too easy.

  2. That was a fiendish problem but fun to solve. The only problem here is that Richard chaged the goal-posts. So much for “Mickey Mouse” in the URL?

    As posed with the original picture, there is no solution. With the picture shown above, of course, the solution is rather straightforward.

    For the original problem, here are some slides with my ruminations on it. Alright, maybe I went a little overboard 🙂

    Basically, the strict answer to the question as originally asked is “No, you can’t.” Considering that Richard’s original photo is quite distorted, one could attempt to rectify the picture. But the distortion is too complex and not just a simple perspective distortion, and one still can’t find a solution.

    In any case, I looked further, at the generalized problem of finding a square circumscribing any general quadrangle. There is the interesting case that if the diagonals are orthogonal and of equal length, any circumscribed rectangle is in fact a square. The locus of the corner points are circles as follows from Thales’ theorem.

    1. I’m totally in agreement – see my post (with overlays). I started editing my response before I saw it.

      What I think happened was that Richard hadn’t had the original puzzle held flat to the glass when it was scanned.

      This is an overlay of the answer and the problem (scaled for the difference in size)

    2. The images were not scanned but taken with an iPhone4, according to the EXIF data. I would agree that the distortion is not intentional – but then again …. 😉

      Perhaps the images are from a book where the pages laid nicely flat for the solutions chapter but not for the problem chapter which is likely early in the book. Hmm, perhaps the problem appeared on a left, even-numbered, page.

    3. I’d started to type “or photographed” but thought nobody would do that with a geometrical puzzle. There are just some many ways of introducing distortion. It’s pretty well impossible hand-held, you need a rig of some sort.

      I didn’t think of looking at the EXIF. I see that the answer has not been imaged with an iPhone. From the limited information available I’m guessing it’s been through a computer as the JPEG was produced by the GD graphics library. I suspect that’s been scanned.

      It was quite good fun going through the formal geometry. However, how much better would the answer have been if the impossibility had been pointed out and just how quickly people jump to conclusions. That’s the really interesting bit.

    4. I agree on these meta-questions. I myself thought I had a good solution and for all intents and purposes of the original puzzle, that was the solution.

      But details matter too. Having done the math, it’s surprising to see how sensitive the geometry is and why. So yes, a rather interesting puzzle, socially and mathematically, and on several levels.

    5. My main post on this is marked “awaiting moderation”. Heaven knows what triggered that. Anyway, here is a link to my geometric construction for a rectangle through four points.

      The formulas for calculating the side lengths is just too long to include on the diagram, but it’s easy enough to derive.

      Anyway, for those that care, this is not wasting time, it’s recreational activity.

    6. Steve,

      Actually, the formula for the side length can be rather short — term w on my slide 5, that is, with an inspired parametrization. I got mine from a 1914 paper (slide 4). Placing the coordinate system along a diagonal (or side, for that matter), implies 3 zero cartesians, leaving 5 other cartesians (for 4 points). Satisfyingly, 5 matches the number of givens you need for a general quadrilateral. The implied zeros account for position and orientation. One could knock down one more parameter by ignoring size, and renormalize lengths, such as set the other point on the x-axis to be at x = 1.

      Having the axis on the diagonal is particularly useful since then the orientation angle of the rectangle (my alpha) is zero when the solution is degenerate.

      And of course, I agree on the recreational aspects. To each one’s own.

    7. Steve,
      This isn’t really a geometric construction with a compass and straightedge. But it still represents an exact (and impressive) solution, but still requires accurate numerical measurement. BTW, is there a formula for the angle in terms of the points?

    8. I’m sure you are right – the Greeks would have no time for it. The solution is very sensitive to co-ordinates – I simply used Photoshop with the pointer and read off the X&Y coords (although the Y co-ord is inverted for some reason). However, judging where the centre of the dot is isn’t so easy when it’s fairly large and blurred.

      As far as producing a formula for the angle is is concerned, it ought just be a matter of taking the formula for he lengths of two sides at right angles to one another (a bit of Pythagorus on the differences between the co-ordinates of the intersections) and hoping, and praying that a lot of terms drop out or it will cover several lines. I suspect it will involve solving “x” for a “tan(x) – tan(x+90)” element – or worse – so I might have to hunt out some trignometrical identities.

      Of course now that you’ve mentioned it, I’m going to have to have a go at it. I have an idea it won’t be pretty.

      I simply used a spreadsheet and homed in on the “correct” angle for the square when two sides at right angles to one another were equal.

    9. @fluffy – you did what I did too. It was that nagging doubt over the dots not being bisected which got to me. It shows up even more if you use a thin line. However, you clearly know when to stop – I clearly don’t, but well done for trying and noticing something is not quite right – this is just a trivial puzzle, but it’s sometimes spotting the tiny anomalies that lead to great discoveries.

    10. Steve,
      I think figuring out the angle will be quite a challenge. If I were to do it analytically, I would start with the geometric construction and then start converting to equations and find intersections of the various circles and lines.

    11. I’ll give it a go – one thing that might help is that the co-ordinates of the corner where the two orthogonal sides intersect will appear in the equations for both sides. Off the top of my head, I think they’s cancel. That will help.

    12. @deepfield

      I realise it doesn’t say through the centre, but I’ve never come across a geometric problem where that wasn’t an implied condition. Several of us had worked out solutions that grazed the edges – but that didn’t seem right. In any event, it gave some of us the opportunity to run through some long-dormant mathematical skills.

    13. Steve,
      I think the best approach is to write four linear equations through the points, two with slope m, two with -1/m, set the separation distances between the parallel lines equal and solve for m. Then the intersection points for the square can be determined.

    14. Gentlemen, that work is done. 🙂 May I point you to my slides, http://public.iwork.com/document/?a=p123607446&d=dots.key specifically slide 5.

      I derived a formula for the tangent of the angle that makes the circumscribed rectangle a square, as a function of the Cartesian coordinates of the points. The formula is sensitive at or near certain special cases (see, ahem, my last slides). I figure that any Euclidean construction would be similarly sensitive in those cases.

      Actually, since I get the tan of the angle as a quotient of two lengths (length differences), the angle should be easily constructible from a triangle with those two lengths as cathedes. This construction will expectedly degenerate if the length differences get small.

    15. Michael,
      I tried the purely algebraic solution but could see it was going to get messy. Your trig method is simpler. But I don’t see how q=e-b-c=0, (e=b+c) means the diagonals are equal. Is there a mistake or am I missing something?

    16. The “diagonals equals” bit comes into play only when “diagonals orthogonal”, p = d – a = 0, i.e., when B’ and D’ coincide along BD, and BD being parallel to the y-axis.

      Then read the expression as q = (e – b) – c = |BD| – |AC|. Recall that I defined b not as length but Cartesian, so b ≤ 0 by choice of the axes. Thus, e – b = yD – yB, the length of BD.

      (Hoping the WordPress engine accepts my HTML …)

    17. Michael,
      Doh. I should have remembered b0.

      I am still trying reconcile the infinite solutions picture with the construction method. It seems the inside circle intersections (other than at the diagonal intersection) have to be 90 degrees from A/B/C/D on the various circles. All the square diagonals have to pass through them. How does that follow from equal perpendicular diagonals?

    18. Regarding b, I initially took it as positive, which just meant a sign change in the formulas. But I thought it’s more consistent to treat it as cartesian coordinate (negative).

      4 circles having the sides of a quadrilateral as diameters all pass through the intersection of the diagonals if the latter are perpendicular. The reverse is also true. This follows directly from Thales’ Theorem and considerations of adjacent angles adding to 180°. This holds for any lengths of the diagonals.

      By the same theorem, the vertices of any circumscribed rectangle or square (i.e., for any α) must lie on the same circles. That is in fact used in the Princeton construction paper cited elsewhere here in the thread.

      In my discussion on the number of possible squares, orthogonal diagonals (p = 0) is the over-arching criterion. The lengths only come into play because of the equation in slide 12. Since p makes the right-hand side 0, there are solution squares when either factor on the left-hand side is also zero. If p ≠ 0, we can simply divide by it (not being zero), and can find (exactly one finite) α from arc cot.

  3. i got it stright away,but i’m the same as the comments above, it’s too easy, i thought had a little twist in there somewhere…

    1. But then I did not expect a complicated geometric construction to be the answer for the kinds of puzzles posted here.

    2. Nor did I, but what I half expected was that the dots had been cunningly arranged so that it wasn’t possible to draw a square with one dot on each side. However, that didn’t turn out to be the case but on the original (distorted) puzzle it turned out it wasn’t possible. However, it does throw up a point – how many people just assume that something that looks roughly square is, actually a square.

    3. As Steve commented on Friday, perfect square is actually a term from number theory, and “perfect” is redundant when applied to a square in geometry.

      @Steve: Yes, we are easy to fool about geometrical figures. Not only squares vs. general rectangles, but also simply right angles. Without context, it’s sometimes easy to mistake 85° or 80° for a right angle.

    4. @edwardv: Sure, a “rough” and quick answer would have done, but I got my curiosity piqued as to why I couldn’t quite get a square to truly bisect all four dots, i.e., the finite dots taken as abstracted to zero-dimensional points, as Steve also commented earlier.

      I like a challenge and had fun digging out some 100 year-old math paper and building on it to figure out the solution in analytic geometry.

    5. I was pretty sure the quick answer had to be pretty close geometrically constructed answer. I’m not sure how the problem could be worded differently to better suggest the quick answer. Maybe just drop the word “perfect”.

  4. I’d solved it in seconds when I realised my block of sticky-notes was the perfect size to overlay the on-screen image!

    The ‘not as easy as it may first appear’ probably refers to the square not being orientated in line with the writing.

    1. You can do it in line with the writing as well though, it just goes outside the bounds of the picture as photographed. But there’s nothing limiting you from doing that. There’s also no instruction limiting you to going precisely through the centre of the dots. All in all another poor puzzle I feel. Quite disappointed this week 😦

    2. Cally,

      No, you can’t, that is, if you connect the dots with lines parallel to the bounding box of the text (even extending outside the photo’s area) you will not get a square but a rectangle – that’s the actual puzzle, and why it is “not as easy as it may first appear”.

  5. Good grief!! I can’t believe how anal the above correspondence is. Get over it folks – move on – or get at least get a life

    1. Is it necessary to be so insulting? I would suggest you confine your comments to the puzzle. Be thankful that there are people in the world that take care or you’d be sitting in the dark in a cave without any of the conveniences of modern life.

    2. @Amanda,

      That you ignore how easy it is to come up with the EXIF data of a picture (literally just a right-click and a left-click for me) is no excuse to behave like a condescending jackass. In fact, you typing that took more effort for you than getting the EXIF for me, and probably for them.

    1. Nope, I took the image (the original one from friday’s post, without the solution) to an image processing program and drew a square parallel and permendicular to the text. Perfectly possible to make a perfect square (approx 333×333 pixels) that goes through each dot that way. The lines just don’t go through the centers of the dots, but nowhere in the puzzle does it say they should.

  6. This was a bad puzzle. the combination of a geometrical puzzle involving a perfect square and a half-skewed picture of some large dots is not a good combo. even if I had gone through the trouble of printing it out (or putting it into an application) the solution wouldnt be right, and “Thinking” the solution (like I did) isnt really a full solution (“yeah, that would be about square-ish”) because it wouldnt be a _perfect_ square.

    1. That’s basically it – the human eye is not very good at judging if something is actually a square or something that’s just close to being one. Basically don’t trust eyesight alone. Perhaps Richard will say why it’s one of his private puzzles – it might be for that reason.

    1. Unfortunately it’s not quite a square – when I print it, it’s 92mm one side and about 88mm the other (and the two lateral dots aren’t quite bisected so it would be slightly longer than 92mm). I think PowerPoint might do a snap-to-grid so that makes it tricky.

      Of course the human eye doesn’t really see a difference of only about 4% in the lengths – which is why I think this problem was described as more difficult than it looks. Also, for those measuring on screen, be aware that the geometry of those is often less than perfect and they can stretch one direction more than another (most printers seem to be better).

  7. I started out that way, but I thought people might cry cheat if I rotated the points. I know you can just rotate the axis, but it might look confusing.

    In fact once I’d generalised the intersections for the lines that didn’t pass through the origins, I might as well just have stuck to it as the formulas are all of identical form – just the subscripts change.

    I put the whole thing into a spreadsheet and, with naming the boxes, it transcribed super-easy.

    1. Oops – different computer so this went in anonymous. It was meant to be a reply from SteveGJ to Michael Sternberg.

  8. Really fantastic job Michael Sternberg, Steve Jones, etc;. I only wish Richard Wiseman would put half the thought you have into the puzzles he posts. Yet again he has wasted our time and patience with an ill-conceived puzzle, supposedly his “favorite ever”, but how could it be if he does not appreciate the subtleties. What a hack, Im outta here….

    1. It doesn’t matter what Richard posts; people will still argue over its validity, complexity, difficulty, etc. That’s half the fun. Bye.

      Fascinating analysis, Michael — thanks.

    2. “I’m taking my toys and going home, and I’m NEVER COMING BACK!” Don’t let the door hit you on the way out.

    3. I should say that I’m not too worried about the odd bit of ambiguity in the puzzles. It’s not a maths exam (except for those of us who turn it into one). However, I like a challenge and little inconsistencies nag at me so it’s worth exercising a few geometric and algebraic muscles I’ve forgotten how to use.

      As for any time wasted – well, that’s my decision alone.

  9. What the hell? So it’s exactly “as easy as it appears at first glance”. Was reverse psychology the trick or something?

    Here’s another one:

    Draw a stickman. It is not as easy as it appears at first glance! Can you do it?

  10. My issue with the solution is that the rules don’t specify that the line go through the center of each circle, just that each dot be included on a line. In which case a perfect square could be drawn to match the square the text takes up and each of those dots could be on the line and not the vertex (which is also not specified). I think this problem is made more complicated by inferred rules.

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