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We have 5 dots connected by lines like in the image. We have 5 points, so we have a pentagon. For a regular pentagon, we can connect all the vertices and there are 10 lines, with 5 groups of parallel lines. None of these lines are perpendicular to other. In the irregular pentagon (polygon) of the figure, there are 10 lines and each one of these is either parallel or perpendicular to another one. Is this figure unique? I.e. are there other polygons where we have 10 lines having parallel and/or perpendicular? At least one right relation. In a square with a central point, there also would be 10 parallel/perpendicular relationships, but that is not a convex polygon (a concave polygon change shape if we connect vertices). Any easy way of demonstrating using analytical geometry. I can demonstrate for a rectangular grid by using Pythagorans several times, but not for a general case. Can you find another 5 side polygon with this property?

enter image description here

Proof with another pentagon, slightly different and only 9 relationships, using a rectangle and demonstrating it must be a 1:sqrt(2) rectangle by iterative use of Pythagoras theorem.

enter image description here

Cesar
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    Looks like being unique. A tiny detail : why have you chosen to take this moon's surface background :) ? – Jean Marie Sep 13 '23 at 16:38
  • Because it is a real feature on the surface of another planet. Do you know how to prove uniqueness? 4 or 5 Pytagoras theorem but from a given array. Surely the general case (real uniqueness) must not be too difficult but I cannot complete the proof. – Cesar Sep 13 '23 at 17:39
  • What is your motivation for having this pattern and moreover proving it's unique ? – Jean Marie Sep 13 '23 at 20:21
  • Proving my first intuition on the polygon and its properties. It is a very basic property but cannot find anything on geometry books. I proved for a rectangle with sides 1:sqrt(2), that there is only one geometry object with that property using the rectangle, but I am not able to generalize the proof. I tried a numerical simulation using Montecarlo and did not find any other object but the simulation is a little cumbersome and would prefer analytical proof. – Cesar Sep 13 '23 at 20:41
  • Please see the proof for a slightly different pentagon with only 9 relationships. Using a rectangle, I demonstrate that the rectangle must be 1:sqrt(2). The 10 relationships of the original post pentagon also use the rectangle as a starting point. I would like to find a general proof of the intuition that the pentagon with those maximum relationships (at least one right angle and some parallels, with all 10 lines fullfilling criterion) is unique. – Cesar Sep 13 '23 at 21:19
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    An angle of attack could be to consider the star inscribed in the pentagon and see whether there are really different cases occuring for the parallelism or orthogonality of its sides vs. the pentagon's sides... – Jean Marie Sep 13 '23 at 21:41
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    (Four vertices of a square plus its center can form a polygon. Not a convex polygon.) – aschepler Sep 13 '23 at 21:53
  • I will give it a thought. My initial reasoning was using ABD being a right triangle, using orthogonality from A to G to have an extra orthogonal direction, and from that using parallelism, but cannot reach the result. I will try your suggestion. Thanks. I cannot understand why such an apparently hyper simple problem can resist so much to proof. – Cesar Sep 13 '23 at 21:54
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    In the image, there are 4 pairs of parallel lines and 5 pairs of perpendicular lines. This is only 9 relationships. – Daniel Mathias Sep 13 '23 at 22:26
  • Thanks for all the comments. It seems that the definition of the problem is quite poor from my part. I need a convex polygon, if it is a square with center point, if I join the verteces by lines I got a different figure. I don't what that. Also, there are 10 lines having the property of parallel - perpendicular, not 10 relationships. With that already defined (I hope it is well defined already), is the figure unique? Are there other figures with the same max lines having the properties? To clarify, in computer simulations I only got the initial figure and the square with dot center, excluded. – Cesar Sep 13 '23 at 23:27
  • A "house" made from a square topped by an isosceles right triangle has 4 parallelisms and 8 perpendicularities. However, two segments (joining the apex of the "roof" to the endpoints of the "floor") aren't parallel or perpendicular to anything, so I guess this doesn't count. – Blue Sep 14 '23 at 01:01
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    Another satisfying pentagon, with more symmetry: image – Daniel Mathias Sep 14 '23 at 03:19
  • Bravo, you have found another one and more symmetrical. Are there many more? As a counterexample of uniqueness, the question is settled. Thanks. – Cesar Sep 14 '23 at 05:35
  • @Daniel Mathias You should give your image as a solution that I will be happy to upvote. – Jean Marie Sep 14 '23 at 08:11
  • I guess that @DanielMathias has noticed the golden ratio in some of the proportions of sides!! – Cesar Sep 14 '23 at 11:08

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Another satisfying pentagon, with more symmetry:

enter image description here

I have also found that pentagons such as your original solution can be constructed on an arbitrary right triangle. See this GeoGebra link for details. Drag point $C$ around the circle, then position point $D$ so that the gray points coincide.

Daniel Mathias
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  • Is this the same pentagon as in this source? https://www.cut-the-knot.org/do_you_know/NoRegularPentagon.shtml – Cesar Sep 14 '23 at 11:16
  • @Cesar That one has no right angles. – Daniel Mathias Sep 14 '23 at 11:19
  • It would be very interesting to know how you arrived to this figure (involving golden ratio?). I cannot see a clear drawing technique for getting the result except for the initial square. I am sorry, I am too dumb. – Cesar Sep 14 '23 at 11:58
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    Answer to the question of Cesar : Just apply a vertical stretch $h$ with matrix $\pmatrix{1&0\0&h}$ to the vertices of a regular pentagon, in such a way that the angle between (transformed) green segments is a right angle. – Jean Marie Sep 14 '23 at 22:26
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    @Cesar See edit for more pentagons. – Daniel Mathias Sep 15 '23 at 03:25
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    @DanielMathias: In your GeoGebra sketch, the gray dots will align (that is, four parallelisms will hold) when $D=B+\frac23 C$ (so that $E=A+\frac23 C$). To get, say, $DA\perp EA$ (and also $AD\perp BD$, $CE\perp AE$, $BD\perp CE$) requires $\cos\angle AOC = 1/3$. This makes OP's pentagon unique (up to symmetry) for its type. – Blue Sep 15 '23 at 05:11
  • @Blue The additional perpendicular relationships, which are not required, are also attained when $AB\perp BD$ – Daniel Mathias Sep 15 '23 at 08:46
  • @DanielMathias: I should've been more clear. By "unique for its type", I was thinking in terms of a specific arrangement of parallelisms and perpendicularities. I was just lazy not to have also mentioned the $AB\perp BD$ case. Anyway, my intention was simply to show how to bring the total number of parallelisms and perpendicularities in that figure up to OP's $9$. ... Cheers! – Blue Sep 15 '23 at 09:19
  • The GeoGebra construct of @DanielMathias is indeed very interesting. And as Blue says, my initial figure is unique if we maximize the number of perpendicular relationships. In fact, if we add the restriction of all right angles being a vertex, the figure would be unique. So far, it seems we have two different constructions: one from a regular pentagon and one from a rectangle and both seem unique. – Cesar Sep 15 '23 at 15:27
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    @Cesar This pentagon has 8 perpendicular pairs along with 5 parallel pairs. – Daniel Mathias Sep 16 '23 at 04:09
  • @DanielMathias, we have a clear winner here. I guess that starting with the regular pentagon, which had already all the 10 lines parallel each other, and stretching it to make the right angles appear, was the strategy from the beginning. Thank you very much for sharing your finds. – Cesar Sep 16 '23 at 08:45