An entry in Fortnightly Topic Challenge #37: Rare and Endangered 1


  • Pencil;
  • Eraser;
  • Blank A4 sheet of paper; and
  • A ruler (also to use as a straightedge).


  1. Draw a square with a side-length of $a=16$cm.


  1. Then, draw eight circles inside. The circles must have $2$cm in diameter; they must all be the same size; they must not touch each other; their distances must all be different from each other (i.e. their positions must be randomised); and they must not touch the perimeter of the square. $$\bigcirc\quad\bigcirc\quad\bigcirc\quad\bigcirc\quad\bigcirc\quad\bigcirc\quad\bigcirc\quad\bigcirc$$

  2. Now, draw two points on each side of the square. Begin with the top side, and from left to right, make points $A_1$ and $B_1$. The distance from these two points is arbitrary, but none of them can be on the corners of the square. Now, rotate the square $90^\circ$ anti-clockwise. You will now have a new top side, where you can put points $A_2$ and $B_2$ from left to right. Their distance is also arbitrary and cannot be on the corners of the square, but it also cannot be the same distance as any other points placed (i.e. the previously placed points).

    Continue with this method to make points $A_3$ and $B_3$, and then $A_4$ and $B_4$, making sure that the distance between each two points on a side is unique. Now, connect the points with a line in the following fashion. $$A_1\to A_2\to A_3\to A_4\to B_1\to B_2\to B_3\to B_4\to A_1$$

But, ensure that the lines do not touch nor intersect any of the circles!

Note: You might have to place your points carefully.


$$\verb|Ensure the lines do not touch nor intersect any of the circles!|$$


Removed Part 2 of the puzzle as it is much too difficult and incorrect (or perhaps impossible) in very specific cases of drawing circles. Thus, I am adding a bonus:

Bonus: What is the minimum value of $a$ for your specific case you have chosen (i.e. how you have randomly positioned the circles)? Note that you cannot move the circles in different positions after plotting them.

I used the tag connections-puzzle because you have to connect points with lines.

  • 4
    $\begingroup$ What's to stop us from drawing the lines first, putting eight tiny circles in the open spaces, and then declaring the scale to be such that the circles are 2 cm? Or is it actually an optimization puzzle where $a$ is to be as small as possible? $\endgroup$ Jul 10, 2018 at 5:52
  • 3
    $\begingroup$ It's not a matter of honesty. By introducing (fakeable, as Jaap mentioned) randomization, you have essentially created instead of 1, INFINITE riddles. For each random distribution, the min(α) will be different. Except, of course, if you are asking for an answer that COVERS ALL PROBABLE DISTRIBUTIONS. This, except for sounding - in my humble opinion - very difficult, is NOT stated in your post's body. $\endgroup$ Jul 10, 2018 at 7:53
  • 3
    $\begingroup$ Is there a reason the title is latex-styled? $\endgroup$ Jul 10, 2018 at 13:30
  • 3
    $\begingroup$ @user477343: Its a tricky thing. This puzzle suffers from the multiple variables in it (square size and circle placement). It also suffers from having a trivial solution (the one highlighted by the answer by "just a student" with your points being arbitrarily close to the corners causing your line to be a box just inside the perimeter. My thinking is that it would be better if it had more restrictions on it - eg once you have drawn the line then no two circles can be in the same region. This would remove the trivial solution of just a box round the whole lot.... (to be continued) $\endgroup$
    – Chris
    Jul 11, 2018 at 11:02
  • 2
    $\begingroup$ Also I think that the variable a is unneeded. If the puzzle is sovlable with a=15 then what advantage is there to allowing different values of a? One thing worthy of note though is of course that with random circle placement if you insisted on a non-trivial line then I don't know if you would necessarily always be able to solve the problem with randomly placed circles. $\endgroup$
    – Chris
    Jul 11, 2018 at 11:04

1 Answer 1


Maybe I am misunderstanding the puzzle, but it seemed straightforward to

take the requirement of not touching the circles into account when placing the points in order.

I literally plopped down circles as random as I could (but aligned to a grid, for ease of working) in a convenient drawing tool, in a 16x16 square. Then I

added the points while tracing the line in order $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4, A_1$, ensuring the lines do not intersect circles along the way. I moved the first point when it turned out to be positioned in an inconvient spot. Finally I ensured distances to be unique.

The result is:

Eight circles with diameter 2 in a 16 by 16 square. The square has 2 points on each side, that are connected by lines as the puzzle prescribes. The lines do not touch or intersect the circles.

As $a = 16$, I wonder if we are allowed to break that requirement for the bonus? Of course, circle placement will not be random anymore when the square barely provides enough room to hold them.

I allowed myself to go all out and move the circles and the points while trying to make everything fit. It then was not hard to construct the below example, which matches all the requirements in an 8x8 square. Of course, this is not in the spirit of the puzzle, but it shows that obtaining $a < 16$ for the bonus question is possible if one places the circles lucky.

Same setup as previous picture, but in an 8x8 square.

Side note: when looking at this puzzle from a mathematical perspective, one can place the points arbitrarily close to the corners (but with sliiiightly different distances to fullfill the requirements). The lines are then arbitrarily close to the sides of the square, transforming this in a question of packing 8 circles with diameter 2 in a square that is as small as possible. According to Wikipedia, that can be done in a square with sides $2 + \sqrt{2} + \sqrt{6} \approx 5.863\ldots$, so let's say 6.

Packing of 8 unit circles in a square with side length $2 + \sqrt{2} + \sqrt{6} \approx 5.863\ldots$, extended to a square with side length 6, with the lines as in previous configurations going in the space between the two squares.
You can imagine that the blue lines can move arbitrarily close to the side of the square; in the limit, the square will have side lengths $2 + \sqrt{2} + \sqrt{6}$ as mentioned above.

  • $\begingroup$ Your "intersection" interpretation was correct. I will include that in the post. Also, I am interested in how you created the image! How did you do it? And finally, well done for finding a solution! :) $\endgroup$
    – Mr Pie
    Jul 10, 2018 at 8:54
  • $\begingroup$ Thank you for the puzzle, @user477343! I did it as I describe in the third spoiler block. I am uncertain what other details you'd like me to share? $\endgroup$ Jul 10, 2018 at 9:06
  • 1
    $\begingroup$ No, not really, and no problem either. If you want to make more solutions, you can. It is a bit of a broad puzzle. I was originally going to put up this puzzle of mine after finding out the aim was achievable, but I realised it was very similar to this other puzzle unfortunately. If you want, you can attempt at that puzzle I created (but just don't look at the answers!). :) $\endgroup$
    – Mr Pie
    Jul 10, 2018 at 9:11
  • $\begingroup$ I cannot upvote as I have reached my daily voting limit. I will also accept answers once the fortnight is over, for the sake of the challenge. $\endgroup$
    – Mr Pie
    Jul 10, 2018 at 9:12
  • 2
    $\begingroup$ I just went over some of your other puzzles, and I must say I love your riddles. This puzzle is much more technical; if you are going to do another one, try to let it sit for some time, read it again and see if anything is unclear before posting. It can also help to try to write everything down as succinctly as possible. Final bit of advice would be to just do it a lot, that's the best way to learn! $\endgroup$ Jul 10, 2018 at 9:32

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