周末趣题，如果知道方法，此题应该很容易

新用户 · 发表于 2005-3-26 05:41:20

见下图，每相邻两个点的距离（水平，上下）都相等（=1），

1。求多变形abcdef的面积。

2。给出一个简易的方法，使得我们可以算出任意点相连组成的多边形的面积

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a f

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e

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d

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b c

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怀疑 · 发表于 2005-3-26 05:52:00

Pick 定理要去数格点。也许用有向三角形面积之和来求也简单？

新用户 · 发表于 2005-3-26 06:42:53

哈哈，我指的就是你说的这个定理，我想可能有的人不知道这种办法。

fzy · 发表于 2005-3-28 22:24:29

This is something I saw in an American high school math book, called Shoe Lace Theorem or something similar. It goes like this:

Suppose a polygon has vertices (a0, b0), (a1, b1), ..., (an, bn). Write the vertices as follows:

a0, b0
a1, b1
...
an, bn
a0, b0

Then add up products of all primary diagonals (a0b1 + a1b2 + ... + anb0), and subtract products of all secondary diagonals (- a1b0 - a2b1 - ... - a0bn). The area of the polygon is one half of the absolute value. It works for non integer points too, and the sign means the direction of the rotation.

There was no proof in the book. The students were asked to use it to calculate the area of some polygons. It can be proved using outer products (that is probably what you mean by 有向三角形面积之和), but it requires some argument to deal with non-convex polygons. If not restricted to elementary math, Green's theorem gives an elegant proof.

独木桥 · 发表于 2005-3-29 02:06:07

用梯形面积计算公式就可以证明这个定理. 显然这个公式关于平移不变,不妨假设多边型在第一限象. 从每个顶点向X轴(或Y轴)引一条垂线,计算这些梯形面积,求和,就可以得到这个公式.

fzy · 发表于 2005-3-29 17:57:51

This is the same as the outer product approach. The fact that you have to discuss the direction and sign made it less appealing. In contrast, the Green's theorem proof is only one line.

独木桥 · 发表于 2005-3-30 04:44:16

I just meant that to prove this theorem it doesn't need "outer product" or Green's theorem, but middle school mathematics.

fzy · 发表于 2005-3-30 18:23:18

Actually that is the way (one way at least) to prove the outer product result.

怀疑 · 发表于 2005-3-31 03:27:43

从计算复杂度的角度讲，你这方法更简单。

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周末趣题，如果知道方法，此题应该很容易

希望看到一个比Pick 定理更简单的办法。

回复：希望看到一个比Pick 定理更简单的办法。

回复：希望看到一个比Pick 定理更简单的办法。

回复：回复：希望看到一个比Pick 定理更简单的办法。

回复：回复：回复：希望看到一个比Pick 定理更简单的办法。

回复：回复：回复：回复：希望看到一个比Pick 定理更简单的办法。

OK I got it.

从计算复杂度的角度讲，你这方法更简单。

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