正多边形

fzy · 发表于 2005-3-25 22:21:17

多边形Ａ1Ａ2...Ａn中有一个正多边形Ｂ1Ｂ2...Ｂn。Ｂ1在Ａ1Ａ2上，Ｂ2在Ａ2Ａ3上，...，Ｂn在ＡnＡ1上。已知Ａ1Ｂ1＝Ａ2Ｂ2＝...＝ＡnＢn。问Ａ1Ａ2...Ａn是否也是正多边形？ www.ddhw.com

Background: Some time ago the 4 sided version of this problem appeared at 情趣生活( https://www.heidisun.com/qqsh/viewtopic.php?t=208 ). And the answer is yes. Later the three sided version appeared at WXC, known as IBM puzzle: 初等几何 ( https://web.wenxuecity.com/BBSView.php?SubID=netiq&MsgID=29622 ). The answer is also Yes.

I posted both problems here but nobody is interested. Probably you all have seen them. Now I made up this problem and they should not have appeared anywhere.

Give it a try?

怀疑 · 发表于 2005-3-26 04:41:15

For larger n, the answer is not always positive. One guy generalized the IBM puzzle in 2000. He claimed to have counterexamples for n=5, 6, 8, 10.

I remember somebody here tried your problem.

Good problems anyway.

fzy · 发表于 2005-3-26 04:58:29

OK thanks. But I believe I have the complete solution and it is quite surprising.

怀疑 · 发表于 2005-3-26 05:03:26

At least you can submit your article to some college math magazine.

fzy · 发表于 2005-3-26 05:09:06

It might. But I have not done it in many years. They charge a lot of money for each article and I no longer have grant to pay for it. (Maybe Wild Cauliflower or another professor here would agree to cosign it with me and pay from their grant? :) )

新用户 · 发表于 2005-3-26 07:18:09

hmmm, good idea

fzy · 发表于 2005-4-6 23:39:48

I changed my mind. This is NOT good enough to publish. So I post the answer here. Whoever wants to take it and publish it somewhere, please feel free to do so.

1. N odd, >= 7. The answer is yes.

Assume A1... is not regular. Then one of the angles, say Ai, is < 180*(N-2)/N. We also have Ai > 180*(N-4)/N, because angles AiBiB(i-1) and AiB(i-1)Bi are > 360/N. Now let C be a point on AiB(i-1) so that B(i-1)CBi = 180*(N-2)/N. Then we have CBi > AiBi, because Ai > AiCBi. Let D be a point on CBi so that DBi = AiBi. Then B(i-1)DBi = A(i-1) > 180*(N-2)/N.

Similarly, if Ai > 180*(N-2)/N, then A(i-1) < 180*(N-2)/N. This contradicts to that N is odd.

Notice that these two are true for any N>= 6, not just odd. All the folowing counter examples are based on them.

2. N even, >= 6. The answer is no.

First we construct a regular N-gon B1B2...BN. Draw an arc of degree 720/N from B1 to B2. Choose points C, D on the arc, with C between A1 and D. The line segment A1D and A2C intersect at E. The extension of A1C and A2D intersect at F. Now we adjust C and D so that A2E = A2F. This is possible because when C and D are symmetric, A2F > A2E, and when D is near A1, A2F < A2E.

Now Add the triangles B1B2E and B1B2F alternatively to the N sides of the polygon. We have a polygon which is not regular but satisfies the condition.

3. N = 5. The answer is no.

I used my computer to find a counter example. The dimensions:

A1 = 70 (It needs to be < 72 from the above argument)
A1B1B5 = 47.25343 (This is what the computer found)
A1B1 = 0.946043 (Assuming the inner regular polygon has a side length 1)

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正多边形

FYI: I did see this problem. [;)]

回复：FYI: I did see this problem. [;)]

Then it is worth publishing. [@};-]

回复：Then it is worth publishing. [@};-]

hmmm, good idea[;)]

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