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        riddles >> easy >> Nonagon diagonals
        
(Message started by: NickH on Sep 4^th, 2004, 9:16am)

Title: Nonagon diagonals
Post by NickH on Sep 4^th, 2004, 9:16am

In regular nonagon ABCDEFGHI, show that AB + AC = AE.

Title: Re: Nonagon diagonals
Post by Barukh on Sep 5^th, 2004, 4:00am

The attached figure includes everything needed for the proof. Note: [hide]the blue triangle was built equilateral[/hide].

Title: Re: Nonagon diagonals
Post by Sir Col on Sep 5^th, 2004, 12:16pm

You, sir, are a genius; such a lovely solution!

However, you have inspired me to (perhaps) an even simpler solution...

As each interior angle is 140^o, ABC=140^o, BCA=20^o (isosceles), ACD=140-20=120^o, and so JCD=180-120=60^o. We know that triangle JAF is at least isosceles (AJ=AF), and as triangle JCD is equilateral, so too is triangle JAF; hence AJ=AF=AE. As AJ=AC+CJ=AC+AB, we prove that AE=AC+AB.

Title: Re: Nonagon diagonals
Post by NickH on Sep 5^th, 2004, 4:56pm

Excellent proof, Sir Col!

Here's a follow-up. In regular heptagon ABCDEFG, show that 1/AB = 1/AC + 1/AD.

Title: Re: Nonagon diagonals
Post by TenaliRaman on Sep 6^th, 2004, 9:02am

::[hide]Apply Ptolemy's Theorem on the quad ACDE or on any other suitable quad[/hide]::

Title: Re: Nonagon diagonals
Post by Sir Col on Sep 7^th, 2004, 4:56pm

The best I could prove was that AD = (AC²–AB²)/AB.

TenaliRaman, that is an inspired approach! I would have never have thought of using AE. In fact, it took me a while to figure that AD=AE! :-[

Using the theorem we get AC.DE+CD.AE=AD.CE, but as AB=CD=DE, AD=AE, and AC=CE, it gives AC.AB+AB.AD=AD.AC. Therefore AB(AC+AD)=AD.AC, so 1/AB=1/AC+1/AD.

I must say that I had fun trying to prove Ptolemy's theorem too!

Title: Re: Nonagon diagonals
Post by Barukh on Sep 8^th, 2004, 11:26pm

Here’s another solution of heptagon diagonals that does not use any additional theorems.

Construct CX || AB. Clearly, CX is another big diagonal, therefore ABCX is a rhombus. Also, construct CY = CX = AB.

We have: ACD = 4[pi]/7; triangle YCX is isosceles, YCX = [pi]/7, so AYX = 4[pi]/7. Therefore, YX || CD, and triangles AYX, ACD are similar.

But then AY/AX = AC/AD => (AC-AB)/AB = AC/AD, and the result follows.

Title: Re: Nonagon diagonals
Post by NickH on Sep 12^th, 2004, 12:19pm

Very cool, Barukh!

Note that the nonagon puzzle can be solved using Ptolemy by considering cyclic quadrilateral ABDG, but I much prefer Sir Col's argument by symmetry.