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？
Background: Some time ago the 4 sided version of this problem appeared at 情趣生活( http://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 ( http://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?
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. |
OK thanks. But I believe I have the complete solution and it is quite surprising. |
At least you can submit your article to some college math magazine. |
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? :) ) |
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. 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) |
| 欢迎光临 珍珠湾ART (http://zzwav.com/) | Powered by Discuz! X3 |