The Number of Halving Circles
The Number of Halving Circles
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减半圈数
DOI：
10.1080/00029890.2004.11920117
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发表时间：
2004
期刊：
影响因子：
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通讯作者：
Federico Ardila
中科院分区：
文献类型：
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作者：
Federico Ardila
对于第1节和第2节的其余部分，n是一个固定的正整数，S表示平面中一般位置上的任意一组2n + 1个点。问题1有几个解决方案。一种可能的办法如下。设A和B是S的凸船体的两个连续顶点。我们声称经过A和B的某个圆正在减半。所有通过A和B的圆的圆心都在线段A B的垂直平分线e上。在f上选取一点O，该点与S位于A B的同一侧，并且距离A B足够远，使得以O为圆心并通过A和B的圆F完全包含S。这显然可以做到。现在慢慢地“推”O沿沿着e，移动它向AB。圆F随着O不断变化。当我们这样做时，F不再包含S的一些点。事实上，它一次失去一个S的点：如果它同时失去P和Q，那么点P、Q、A和B将是共圈的。我们可以将O移到足够远的地方，使其超过AB，最终圆不包含S的任何点。最初，F包含S的所有点。现在，由于在这个过程中它每次丢失S的一个点，我们可以决定我们希望它包含多少个点。特别地，如果我们在圆即将失去S的第n个点P时停止移动O，则
For the rest of sections 1 and 2, n is a fixed positive integer and S signifies an arbitrary set of 2n + 1 points in general position in the plane. There are several solutions to Problem 1. One possible approach is the following. Let A and B be two consecutive vertices of the convex hull of S. We claim that some circle going through A and B is halving. All circles through A and B have their centers on the perpendicular bisector e of the segment A B. Pick a point O on f that lies on the same side of A B as S and is sufficiently far away from A B that the circle F with center O and passing through A and B completely contains S. This can clearly be done. Now slowly "push" O along e, moving it towards AB. The circle F changes continuously with O. As we do this, F stops containing some points of S. In fact, it loses the points of S one at a time: if it lost P and Q simultaneously, then points P, Q, A, and B would be concyclic. We can move O sufficiently far away past AB that, in the end, the circle does not contain any points of S. Originally, F contained all the points of S. Now, as it loses one point of S at a time in this process, we can decide how many points we want it to contain. In particular, if we stop moving O when the circle is about to lose the nth point P of S, then the