Curious Property of a Regular HeptagonDate: 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/ |
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