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!!

]]>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.

Bob wrote:

But I'm no nearer a proof.

I hope you can find it!

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

]]>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°.

]]>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)!!

]]>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).

]]>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

]]>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...

]]>

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 ]]>