5
$\begingroup$

My children have some Duplo fences, these you have to put down on two points, and at both ends they extend a little where you can connect several to surround some area. So a fence is described by a triple, $x, y, z$, where $y$ is an integer, the distance between the two points, while for $x$ and $z$, the lengths of the extensions, $x+z$ has to be a positive integer. Moreover, it is natural to assume that $x$ and $z$ are also integers to allow an axis-parallel rectangle to be constructed. Apparently, Lego knows how to use the Pythagoras theorem, because the two points where you have to put them down are at distance $y=5$ from each other, so it is also possible to put down a fence diagonally, like at (0,0) and (3,4). Also, they have $x=z=2$. From here an easy calculation and checking a few cases should show that it's not possible to bound any interesting area, but only axis-parallel rectangles.

So my question is that supposing you have some identical fences as mentioned above, is it possible to surround a non-rectangular area? Instead of distance 5, the base distance can be something else, and you can also pick how much the fence extends, so I'm interested in any related results, primarily in small, realizable examples, that Lego could make. For example, if $x=y=z=5$, then it's easy to make a rotated square, or a rhombi, of side length 15, as pointed out by Gerhard, but this I consider a trivial solution.

Note on update: In the first version of the problem I've missed the simple construction by jwim, which works if $x=z=0$, as I've posed the problem badly first.

Improve this question
$\endgroup$
10
  • $\begingroup$ I am not accustomed to posting graphics, otherwise I would share with you some nonorthogonal constructions. Here is one using 10 or more 2x2 bricks: place half of them with lower corners at locations (2i,2i) for i from 0 to 4, and place the other half on top with lower corners at (2i+1,2i+1). You get a hinged fence which you can curve around to make a figure with five-fold symmetry, using a large multiple of 5 of 2x2 blocks. There are variations on this, of course. Gerhard "Loves Being Creative With Legos" Paseman, 2016.08.28. $\endgroup$
    – Gerhard Paseman
    Commented Aug 28, 2016 at 20:07
  • 1
    $\begingroup$ I still cannot determine the exact description of the Duplo pieces and their constraints from this description. $\endgroup$
    – Greg Martin
    Commented Aug 29, 2016 at 20:44
  • $\begingroup$ @Greg: I'm not sure what I could write, I think that my linked image should be sufficient alone. Could you please ask what's not clear? $\endgroup$
    – domotorp
    Commented Aug 30, 2016 at 4:51
  • $\begingroup$ I don't understand what you mean by an "interesting" area, nor why you say only rectangles are possible. $\endgroup$
    – Ketil Tveiten
    Commented Aug 30, 2016 at 10:56
  • $\begingroup$ @Ketil: OK, so you understand the description, but not the question. Interesting is relative, but I don't see how non-rectangles are possible. In general, I could have asked what polygons are possible. $\endgroup$
    – domotorp
    Commented Aug 30, 2016 at 12:04

2 Answers 2

Reset to default
1
$\begingroup$

If I understand the geometry of the fences correctly, it should be possible to surrounded a triangle of dimension 15 by 20 by 25, each of which is a multiple of 5. In general, if 5 is replaced with $n$, $n$ must be hypotenuse of a Pythagorean triplet $(a,b,n)$, to get interesting possibilities. Then, a triangle $(na,nb,n^2)$ can be formed. The set of possible shapes that can be surrounded then includes anything built up out of $(n,n)$ squares and this $(na,nb,n^2)$ triangle. These are, of course, the base triangle and square polygons. Its possible that there are base pentagons, hexagons, etc. that can be made as well.

Edit: There are 6 vectors that can be formed with the single fence, ignoring multiples:

  1. (0,5)
  2. (5,0)
  3. (3,4)
  4. (4,3)
  5. (3,-4)
  6. (4,-3)

A fenced in area is a linear combination of these, with coefficients in Z, that add to the zero vector. That probably is the best way to systematically study the problem.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Indeed, I've missed this simple example! Btw, you didn't get the geometry right, I'll update my question to be more precise. $\endgroup$
    – domotorp
    Commented Aug 28, 2016 at 19:53
  • $\begingroup$ In fact, I've realized that this is absolutely not realizable with real Duplo fences, the extension at the ends need to be non-zero. $\endgroup$
    – domotorp
    Commented Aug 29, 2016 at 5:09
0
$\begingroup$

I guess I don't understand the geometry either, as the problem seems simple to me. Take a length which is the distance between two grid points. If this distance $y$ needs to be integral, choose it thusly. Now one can arrange $x$ and $z$ so that $x+z=y$, and also so that the endpoints lie on a horizontal line and on a vertical line, but not necessarily both. Now (assuming nontriviality of the choice), one can form rhombi, polyiamonds, and other nonrectangular areas using just translates and reflections of the original (non-horizontal and non-vertical) fence segment. In order to have nontrivial rotations allowed, one must impose restrictions on $x$ and $z$ which may not be compatible with $x+z$ being integral.

If the question is what manner of shapes can be so constructed, then one needs $x$ and $z$ to be distances from some point to two grid points, and arranged so that there can be more than one possible angle measure between the implied rays. I suspect two such angles will be rationally independent, but do not have even a heuristic at this time to convince others of my suspicion.

Gerhard "Is This Post Rationally Dependent?" Paseman, 2016.08.29.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This is exactly my question and you have described the constraints perfectly what a solution should satisfy. $\endgroup$
    – domotorp
    Commented Aug 30, 2016 at 4:52
  • $\begingroup$ Good. You are welcome to include the posted text (with attribution) in your question. As your question is presently written (probably because of the use of the word rectangular), it seems rhombi are allowed as nonrectangular shapes. If you want more than one interesting angle measure (e.g. not just complementary and supplementary angles) from the resulting equilateral polygon, you should make that clear in the post. Gerhard "Glad I Could Add Clarity" Paseman, 2016.08.29. $\endgroup$
    – Gerhard Paseman
    Commented Aug 30, 2016 at 5:04
  • $\begingroup$ I am thinking again. I think rhombi whose vertices are not grid points are possible within the constraints given, but I admit some carelessness in assuming any distance y is allowed. I will see if I can find integers y and x+z for you. Gerhard "Returning To The Graph Paper" Paseman, 2016.08.29. $\endgroup$
    – Gerhard Paseman
    Commented Aug 30, 2016 at 5:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .