Tuesday Puzzle

Re: Tuesday Puzzle

Nice one.

The idea of working with a nonagon was initially offputting!

180*(n-2) to the rescue.

Re: Tuesday Puzzle

Use equilateral triangles to the maximum and it follows easily.

If you want real entertainment, try to show AB*AC*AD*AE*AF*AG*AH*AI = 9.

Re: Tuesday Puzzle

Interesting. My solution involved two equilateral triangles and two irregular pentagons. It will be interesting to compare these two solutions (and others) when Dan releases it.

RE your followup question, before I work on it, I assume you mean "=9*AB", since no measurements were given? Interesting problem.

Clarification - Correction

For the proof of AB*AC*AD*AE*AF*AG*AH*AI = 9, assume that the polygon is inscribed in the unit circle.

Thanks

Glad I asked rather than assuming!

Interesting...

I see this pretty easily for n=3 and n=4. Is it true of any regular n-gon inscribed in a unit circle?

I still have to work on the n-9 case, then the general. That should be some good mental exercise!

Re: Interesting...

It is true for any regular polygon in the unit circle.

Position vertex A at the point 1+0i on the complex plane and continue from there.

Hmmm

I haven't had time to spend on this yet.

Am I correct to infer from your placement of A in the complex plane that I have a tough row to hoe if I am approaching this with geometry as my main tool?

Re: Hmmm

You can use geometry/trigonometry combination, but the required skill for the trigonometric calculations borders to the insane.

Just stick with what Jacques Hadamard said: every truth about real numbers passes through the complex plane.

couple of ways: #1

Continue DE and HG to meet at X.

By symmetry, AC = EG and AF = DH.

Also by symmetry, EG and DH are parallel to AB.

Line segment BC makes an angle of 360� / 9 = 40� with line segment AB (or its continuation.)

Hence CD makes an angle of 80� with AB, and DE makes an angle of 120� with AB.

Hence HDX = 60�. Similarly XHD = 60�, and so DXH = 60�.

It follows that XHD and XGE are both equilateral.

Hence DH = DX and EG = EX.

So, DX = DE + EX DH = DE + EG.

Therefore, in regular nonagon ABCDEFGHI, AF = AB + AC.

#2

Draw diagonals ID and IG. Let AF and ID intersect at O.

By symmetry, CD, AF, and IG are parallel. Similarly, AC and ID are parallel.

Also by symmetry,

AO = CD = AB

OF = IG = AC

Adding, we obtain

AF = AO + OF = AB + AC

must be lots of solutions

I'll draw mine up and post it later tonight. And I'll be interested in seeing Nikos with the equilateral triangles.

Re: must be lots of solutions

My solution with the equilateral triangles is the solution #1.

As for the product AB*AC*AD*AE*AF*AG*AH*AI = 9 when the nonagon is inscribed in the unit circle:

Assume that A is the complex number 1+0i = 1.

Let B = z1, C=z2, D=z3, E=z4, F=z5, G=z6, H=z7, I=z8, i.e. view the vertices as points in the complex plane.

These 9 vertices of the nonagon are the roots of the equation z^9 - 1 =0.

Therefore z^9 - 1 = (z -1)*(z - z1)*(z - z2)*(z - z3)*(z - z4)*(z - z5)*(z - z6)*(z - z7)*(z - z8).

From this we get (z^9 - 1)/(z -1) = (z - z1)*(z - z2)*(z - z3)*(z - z4)*(z - z5)*(z - z6)*(z - z7)*(z - z8).

Use sum of geometric sequence on the left side to get:

z^8 + z^7 + ... + 1= (z - z1)*(z - z2)*(z - z3)*(z - z4)*(z - z5)*(z - z6)*(z - z7)*(z - z8).

Set z = 1 on both sides and then take absolute values to get:

9 = |1 - z1|*|1 - z2|*|1 - z3|*|1 - z4|*|1 - z5|*|1 - z6|*|1 - z7|*|1 - z8|.

Recalling now the fact that A corresponds to 1+0i and that the z1, z2,...

correspond to B, C, ..., we can see that the numbers |1-z1|, |1-z2|,...

are the distances AB, AC, etc.

The solution can be easily generalized for any regular polygon.

It is a problem which over the years has appeared in several math contests and in problem books for students who prepare for mathematical olympiads. The earliest (and possibly original) source that I am aware of, is the national hungarian math olympiad sometime around 1910. (I was not around in 1910 :), but the American Mathematical Association has published a book with the problems from several years of that contest, since Hungary was notorious for the high quality mathematicians it was producing in the first half of the 20th century.)

By the way, can somebody tell me how I can properly write subscripts and superscripts in this forum? ( I feel funny writing things like z^9 )

thanks

I'll file that away to look at after I have taken a stab at it.

#3 solution

I made lots of use of the fact that the sum of angles of a convex polygon with n sides is 180*(n-2).

So each interior angle of the nonagon is 140 degrees

Consider the pentagram AFGHI. Its angles add to 540. Three are from the nonagon, leaving 120 to be split evenly, so 60 for that part of the angle at A

Consider the triangle ABC. It is isosceles with one 140 degree angle, so the acute angles are 20.

Combining the three facts above, we determine that FAC is 60 degrees.

Place X on AF such that ACX is equilateral.

XCD is also 60 degrees.

Place Y on CX such that CDY is equilateral.

Then consider the pentagon XYDEF, and its symmetries.

DE=EF

Angle D = Angle F

Angle X = Angle Y

So XF = YD, Which is equal to AB

Re: must be lots of solutions

I am attaching a link to a method for doing superscripts and subscripts. Unfortunately, it is a bit messy as you have to do it in html, but you could do the original writing in a notepad or something with some character for the sub/super tags then do a search and replace when done.

One thing to watch: If you use any of these tags, be sure that you close them or it messes up later posts.

An example X² gives you X²

html in messages

You forgot the link.

FWIW, those who use Firefox can use a plugin called BBCode that lets you apply bbcode (not for WC) or html tags to text such as bold, italic, underline, color, and text size, and you can set up your own custom tags, such as I just did with superscript following your example, which if right, will show up as x². That app makes it less likely you will forget to close a tag, since you highlight the desired text then apply the tag.

It is always a good idea to use preview when using these codes.

My vote is for #2

As the most elegant and easiest to see (once you see it!) solution.

Thanks for the puzzle.

The link  *LINK*


html super and subscripts