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Curious Property of a Regular Heptagon


Date: 04/06/2001 at 13:12:34
From: Todd McCready
Subject: Properties of a Regular Heptagon

I am stuck on a problem I was given for a math contest. It involves a 
regular heptagon. We are allowed to use any resources or people that 
we choose, so if you have any ideas, they would really help. 

Here's the problem: In a regular heptagon A1A2A3A4A5A6A7, prove that:

     (1/A1A2) = (1/A1A3) + (1/A1A4)

I began by breaking up the heptagon into triangles and looking at 
them. I think it has something to do with an isosceles trapezoid with 
one base A1A4 and the rest of the sides being sides on the heptagon, 
and I've thought about using trig functions too, but I'm not coming up 
with much.


Date: 04/11/2001 at 18:50:02
From: Doctor Greenie
Subject: Re: Properties of a Regular Heptagon

Hi, Todd -

I was intrigued by this problem (also submitted by another student) 
when I first saw it; I have been hoping that another doctor with more 
skill in this area would jump in and provide a solution so I could see 
how he arrived at it. But since a week has passed without a response, 
I will post my thoughts and leave your question where other doctors 
can jump in and help out. Or maybe my thoughts will help you to find 
the solution on your own (if so, by all means let me know how you 
finished the problem!).

I too looked a long time at trapezoid A1A2A3A4...

But I think the key to the solution may lie in triangle A1A3A4. In 
this triangle, A3A4 is the same as A1A2, so the lengths of the sides 
of this triangle are A1A2, A1A3, and A1A4 - the three lengths involved 
in the equation we are supposed to verify.

And in this triangle, the angle at A1 is pi/7; the angle at A3 is 
2pi/7; and the angle at A4 is 4pi/7.

It seems that perhaps using the law of sines we should be able to show 
that in a triangle whose angles are in the ratio A:B:C = 1:2:4, the 
sides a, b and c satisfy the relation 1/a = (1/b)+(1/c)...

I hope this helps.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/17/2001 at 18:33:34
From: Doctor Greenie
Subject: Re: Properties of a Regular Heptagon

Hello again, Todd --

Still no input from other doctors on this problem, and I'm still 
puzzling over it. I keep thinking I'm getting close, but I can never 
finish off the proof! I now think maybe neither the equilateral 
trapezoid A1A2A3A4 nor the triangle A1A3A4 is the key.

If we let the length of A1A2 be 1 (which we can do, without affecting 
the proposition we are trying to prove, then:

   (1) By the law of cosines on triangle A1A2A3,

          (A1A3)^2 = 2-2cos(5pi/7)

              A1A3 = sqrt(2-2cos(5pi/7))
                   = sqrt(2+2cos(2pi/7))

   (2) By examination of isosceles trapezoid A1A2A3A4,

          A1A4 = 1+2cos(2pi/7)

I have shown using a calculator that with A1A2 = 1, and with A1A3 and 
A1A4 given by the expressions above, that it is indeed true that 
1/A1A2 = 1/A1A3 + 1/A1A4. And it seems that it should be possible to 
show that is so, but so far, the details of the algebra have escaped 
me.

I hope you are still thinking about this problem... I'd like to see 
somebody complete the details of the proof!

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/25/2001 at 11:28:58
From: Doctor Greenie
Subject: Re: Properties of a Regular Heptagon

Hello once more, Todd...!

I am still working on this problem -- well, at least I am thinking 
about it.

One of the other math doctors sent me an e-mail indicating he had 
managed an analytical proof, but the algebra/trig involved was so 
messy he felt the solution was not worth passing on.

So I'm still looking for an "elegant" analytical solution.

Of course, the way the question was presented to you, perhaps the 
numerical method using a calculator, which I described in a previous 
response, would have been adequate.

Here is my latest approach that I thought looked very promising (and 
it may be the key to an elegant solution - I just haven't yet been 
able to find how):

If we call the regular heptagon ABCDEFG, and if we use a, b, and c to 
designate the lengths, respectively, of the side of the heptagon, the 
shorter diagonal, and the longer diagonal, then:

  (1) Looking at isosceles triangle ABC and the altitude from vertex B 
      to side AC, we have:

         cos(pi/7) = (b/2)/a = b/2a

  (2) Looking at isosceles triangle ACE and the altitude from vertex C
      to side AE, we have:

         cos(2pi/7) = (c/2)/b = c/2b

I thought perhaps using these two expressions with the double angle 
formula for cosine would lead somewhere, but again the algebra 
required to complete the proof has escaped me, or perhaps again I am 
looking down the wrong path.

I hope that you are still playing with this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/01/2001 at 11:50:19
From: Doctor Greenie
Subject: Re: Properties of a Regular Heptagon

Hi, Todd -

AHA! I finally got it!

As I suspected, there is a very simple solution to this problem, but 
I had a devil of a time finding it. Many thanks for sending in this 
problem! There is nothing quite like the feeling of finding a neat, 
simple, and elegant solution to a problem that has been giving you 
headaches for weeks.

As I think you did, and as I know Dr. Schwa here at the Math Forum 
did, I played for a long time with the law of sines and the law of 
cosines on the several different triangles you can form by joining 
various combinations of vertices of the heptagon. Dr. Schwa said he 
was able to slog through the horrific algebra to eventually complete 
the required proof; I was never able to finish my proof.

Finally, after playing with the problem for four weeks, I drew a 
couple of new lines in my figure and immediately came to a proof that 
requires only some basic ideas of geometry. After all those struggles 
with sines and cosines and double angle formulas, I discovered that 
all that trigonometry was not required!

So here is the proof of the proposition:

I'm going to let a, b and c represent the lengths, respectively, of 
the side of the regular heptagon, the short diagonal of the regular 
heptagon, and the long diagonal of the regular heptagon.  Then the 
proposition we are to prove is that:

      1     1     1
     --- = --- + ---
      a     b     c
or
      1     b+c
     --- = -----
      a      bc
or
             bc
       a = -----
            b+c

Let the regular heptagon be ABCDEFG. (In my figure, vertex A is at "12 
o'clock" and the vertices are labeled alphabetically in a clockwise 
direction, but the orientation of the figure is not critical.)

Draw diagonals FA (short diagonal, length b) and FC (long diagonal, 
length c). Then draw perpendiculars from vertices E and A to diagonal 
FC, intersecting that diagonal in points P and Q, respectively.

Now look at triangles EFP and AFQ. Both are right triangles; and the 
angles at vertex F in both triangles have measure (2pi/7). So 
triangles EFP and AFQ are similar. We then have, by corresponding 
parts of similar triangles:

      FP     FE
     ---- = ----
      FQ     FA
or
      (c-a)/2     a
     --------- = ---
        c/2       b

From this, some very simple algebra leads us to

         ac = bc-ab
      ab+ac = bc
     a(b+c) = bc

and, FINALLY...!

           bc
     a = -----
          b+c

QED.

Once again, many thanks for this problem. You have provided this old 
brain with some delightful exercise.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   


Date: 05/05/2001 at 21:55:14
From: Doctor Greenie
Subject: Re: Properties of a Regular Heptagon

Hello one more time, Todd -

Another math doctor here, with whom I have been discussing this 
problem, has come up with an even more elegant solution than mine. His 
proof uses Ptolemy's theorem.

I wasn't familiar with that theorem. What the theorem says is that, in 
a cyclic quadrilateral, the sum of the products of the lengths of the 
opposite sides is equal to the product of the lengths of the 
diagonals.

If you look again at the regular heptagon as being ABCDEFG, then 
consider the quadrilateral ABCE. If a, b, and c again represent the 
lengths of the side, the short diagonal, and the long diagonal, then 
the lengths of the sides of ABCE are, in order, a, a, b and c; and the 
diagonals of ABCE are b and c.

Ptolemy's Theorem then immediately gives you:

   sum of products of opposite sides = product of diagonals

                             ab + ac = bc

which with only a very little bit of algebra gives you:

     a(b+c) = bc

          a = bc/(b+c)

         1     b+c     1     1
        --- = ----- = --- + ---
         a      bc     b     c

Pretty slick.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Triangles and Other Polygons
College Trigonometry
High School Triangles and Other Polygons
High School Trigonometry

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