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I have this problem :

Let $H$ be a right angled hexagon with three non consecutive sides of the same length $a > 0$.
Show, without computing their lengths, that the lengths of the other three sides are also all the same.

I do not know how to solve problems like this.
Could someone clarify this to me please?

What if I want to actually compute the lengths? What should I do?

Prem
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Intuition
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    Even when you're completely stumped, you can provide useful context. If this is a class exercise, tell us the topic(s) covered so far that seem relevant. If it's an online challenge/contest, indicate what the target audience is expected to know. Sometimes, just "thinking out loud" about things you know can get you going, and it can help others avoid explaining things you already understand. – Blue Aug 10 '24 at 04:20
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    ((1)) It is almost impossible that the hexagon is "euclidean-geometry" & "hyperbolic-geometry" at the same time. ((2)) What does "right angled hexagon" mean ? – Prem Aug 10 '24 at 04:26
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    By a "right-angled hexagon", do you mean a rectilinear hexagon (i.e., a concave hexagon with five $90^{\circ}$ angles and one $270^{\circ}$ angle)? – Geoffrey Trang Aug 10 '24 at 05:03
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    It means all angles are $90^\circ$ , @GeoffreyTrang , which might occur in hyperbolic geometry – Prem Aug 10 '24 at 05:40
  • @Blue I am studying sth called right angled hexagons in hyperbolic geometry, in preparation to learn Teichmuller theory. – Intuition Aug 10 '24 at 13:31
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    @Prem I am in hyperbolic geometry in this exercise – Intuition Aug 10 '24 at 13:32
  • @GeoffreyTrang yes, exactly as Prem said. – Intuition Aug 10 '24 at 13:33

1 Answers1

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I got some Data to share with OP.

This makes sense only in hyperbolic geometry , where hexagon can/will have less than $6 \times 180 - 360 = 720^\circ$ , while Euclidean geometry will not allow $90 \times 6 = 540^\circ$ around the hexagon.

Using this question which was answered by user aes , I modified the Diagram there to high-light what the question here is talking about :

hexagon

Here , all angles are $90^\circ$ , giving the right-angled hexagon.

We can then see that while we might show the other sides are equal , it is not necessary calculate the $b$ value.

Since OP wanted clarification , which I have given here , I think I should not proceed to the Solution itself.

Still , I will give the very high-level "Proof Out-line" here.

According to standard text books & other sources listed here , every triplet $(a,b,c)$ has a "Single" hexagon with all right-angles with sides $(a,x,b,y,c,z)$ in that order.

Hence , when we have triplet $(a,a,a)$ , that "Single" hexagon must be $(a,x,a,y,a,z)$ , which we can rotate to get $(a,y,a,z,a,x)$ & $(a,z,a,x,a,y)$
We should naturally get back the Original hexagon.
In other words , $x=y=z$

Prem
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    how do you rotate it? – Intuition Aug 11 '24 at 01:11
  • Just like rotating a Circle around the Center , @Intuition , the hexagons will not coincide when we turn them by small angles , though at some suitable angle , the rotation will make the hexagons coincide. – Prem Aug 11 '24 at 06:02
  • Is it the rotation by a right angle that will make them coincide? – Intuition Aug 11 '24 at 23:13
  • We have to rotate the Diagram by one-third of a Circle around the Center , in either Direction. In other words , we should take the Sphere , then rotate it by one-third of a Circle along the Axis. – Prem Aug 12 '24 at 04:25
  • so my job in this proof is to specify exactly what is the angle that we should rotate by or what? I am still confused about the symmetry argument and how to prove it rigorously? In this context are we always in the Poincaré model or what? – Intuition Aug 12 '24 at 18:06
  • Which text book you are following ? What "course content" has been covered till this Exercise ? I could try responding given that Context. – Prem Aug 12 '24 at 18:37
  • there is no course content, I am just reading Buser book by myself with the help of Bram Petri notes, I am a self learner – Intuition Aug 12 '24 at 18:51
  • I also tried Anderson's book and university of Manchester lectures – Intuition Aug 12 '24 at 18:52