Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 225

convex hexagon

The opposite sides of a convex hexagon

are parallel, the distances separating the parallel pairs of sides are equal,
and the angles at vertices A and D are right angles. Prove that the diagonals BE
and CF enclose an angle of

Last edited by tony123 (2023-03-17 06:28:49)

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi tony123,

Another one of your puzzles that I can't solve!

I drew it up in Geogebra (see image). That gave me some angle pairs I hoped might help, but they haven't triggered anything...

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 9,723

Re: convex hexagon

I haven't got it yet, either.

AB extended will meet DC extended in a right angle. Similarly AF and DE.  That makes a square.

You could move GH so that G=A and JK so J=A to get this square. (AJDH)

You can make another (square? of the same size?) with  lines from B and D, parallel to LM,  to meet BC extended and FE extended. 

With your points that would be move LM until L=B and for a fourth side draw the line parallel to LM through D.
It's certainly a rectangle as it has right angled corners but, so far, I cannot prove all four sides are equal.  Why would you want to do this?

Well one diagonal from each square intersect at the point you've labelled N, and such diagonals make triangles 45-45-90.  Does that get anywhere?  Well I'm still working on it.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi Bob,

I suspect that making a square with those extensions will help!

I've redrawn the diagram:

It also 'happens' that PO runs through N...and, of course, PO divides right angles APD and DOA into 45s.

...and there are umpteen similar triangles (but I haven't seen anything yet).

Last edited by phrontister (2023-03-30 11:14:20)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi Bob,

I've returned to this puzzle again, hoping to get somewhere with it this time...

I think I've found a proof (with the help of Geogebra), but only in the case of symmetrical shapes as per this image:

The symmetry only occurs when N is precisely in the centre of square APDO.

I've typed my workings onto the drawing, and I hope it's all clear enough.

Btw, 12 for AP, PD & BF (all equal, per post #1) is just an arbitrary figure I chose to get some numbers into my drawing so I could use Geogebra.

The hide box below has two more configurations, each obeying tony123's rules, with Geogebra showing that the enclosed angle is 45° in both cases:

For the last 2 drawings I just slid Point B up/down the AP axis (the stops were random locations). The interconnectivity between components (eg, 'Perpendicular Line', 'Point on Object') enabled the whole BCFE figure to move and change shape automatically while still obeying all OP rules and the 45° angle requirement.

I have no idea how to prove a non-symmetrical solution, though squillions (maybe limitless?) of such solutions exist (that is, as I understand it)!!

Last edited by phrontister (2023-05-13 01:20:35)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi Bob,

Here's a little video of an animation of the range of non-symmetrical scenarios obtained by running Point B along line AP:
Non-symmetrical range (video)

Edit: Sorry, I lied.   The range includes the symmetrical scenario I mentioned in my previous post. The symmetry appears 3 times: at the start of the video, at the 14-second mark, and at the end of the video. Each time, N is smack in the middle of the square and the 4 vertices are 67.5°.

Last edited by phrontister (2023-05-15 00:15:24)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 9,723

Re: convex hexagon

hi phrontister

That's a real work of art; I love it.  But I'm no nearer a proof.  I've had a lot of other things on my plate so haven't had the time.  But I'm not giving up.  One of my early posts was about a problem that took me 30 years!

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi Bob,

I added an instrumental soundtrack to my video, and I'm sure you'll recognise the tune. This piece is played by a brilliant guitarist, in an amazing finger-picking style that is his own arrangement. I think you'll like it!

The original @ 62 seconds is much slower than my soundtrack, which I sped up to squeeze the whole of the first verse into the 46-second Geogebra recording.

But I'm no nearer a proof.

I hope you can find it!

Last edited by phrontister (2023-05-15 11:51:25)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 9,723

Re: convex hexagon

I do like it.  Who is the guitarist?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,758

Re: convex hexagon

Hi Bob...the guitarist is Gerhard Gschossmann.

And here's the link to the recording I used: Gerhard Gschossmann - 'Summertime' (George Gershwin) guitar solo fingerstyle

That version is played at the song's correct tempo, with good feeling and expression...which makes my Geogebra soundtrack sound rather chipmunkeyish.

Here's a vocal clip of the song: GEORGE & IRA GERSHWIN 'Summertime' HAROLYN BLACKWELL ('Porgy & Bess'). Very moving!!

Last edited by phrontister (2023-05-16 13:58:59)

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson