数形结合论文翻译文章.doc

数形结合论文翻译文章勾股定理不是，或者一个古老的数学定理的证明勾股定理是一个古老的数学定理，它一直占据着重要位置，即使现在它仍然是灵感的源泉。在本文中我们尝试给出一个有前途的新的证明方法，我们的结果会更加直观。关键词：勾股定理 几何变换 欧几里得几何学最古老的数学定理之一勾股定理指出：直角三角形斜边（直角所对的边）所画正方形的面积等于另外两条边（直角边）所画正方形的面积的和。就是。如果我们分别用a,b，c表示三角形三遍的长度。卢米斯收集了数百种勾股定理的证明，并将其分类。马奥尔追溯了勾股定理的演变以及它对数学和我们的文化的影响。网络上包含了约93种勾股定理的证明。我们的目标是通过一种新方法给出证明，准确的说是定理的另一类型。定理1：如果在简单的直角三角形中勾股定理成立，那么在任意直角三角形中都成立。当然我们必须给出证明说明我们的假设是正确的。这就涉及到几何变换的属性，我们将用到：在给定的方向上收缩或拉伸欧几里得平面（见图1）。图1.转换让我们假设T是这样一个收缩/拉伸系数为a的转变。我们可以认为这种转化为在三维空间内平面之间的平行投影的效果，如图2所示。图2.拉伸和收缩转换T明显具有的性质：转换T通过系数a收缩/拉伸一个平面图形的面积，如果P是一个平面图形，那么它与不相交的图形的面积的可加性是兼容的，如果P,Q是两个不相交的图形，那么它保留了以下森斯的填充率：让P是面积为A的正多边形，并修复他的边缘L。让我们假设T把它转换为面积为aA的多边形Q = T（P），并假设图形L转换为K。然后将填充率定义为与多边形边的长度和收缩/拉伸方向无关（见图3）。图3.不变性现在让我们考虑一个直角三角形的两条直角边的长度为a和b。由勾股定理断言该性质是均匀的对于三角形边的长度，因此我们可以选择我们喜欢的计量单位。现在让我们来考虑图4的设置，也就是规则的正方形的平面排列。我们将选择测量单位为蓝色的正方形（1区）的长度。是有原因的，将是明确的很快。图4.特殊情况很明显的两个较小的正方形的面积之和恰好是第三个正方形的面积，因此勾股定理在这种特殊情况下成立。现在让我们应用下面的转换。按系数a水平收缩，见图5。图5.水平收缩在水平切高x，将高度为1-x的区域的等价形式添加到另一边，根据图6，我们得到这里y不是一个参数，是x的函数。图6.割补按系数b垂直拉伸，见图7。图7.垂直拉伸面积的合适不变的，所以对于任意x都成立。通过两次变换，大正方形的面积变为我们可以假设选择a，b到第三条边长度为s的三角形中，所以因此，s一边的正方形的填充率为现在同样的填充率对两个小的正方形是有效的，分别在a，b边，因此我们可以计算x。首先我们设置x，使得所得到的矩形是相似的。我们有x的条件：这个等式给出现在以a和b为边的矩形是相似的。其边长的公比是，这是共同的填充率（小于1）的两个较小的正方形的面积总和改变，第三个正方形面积也改变。经由收缩系数a和拉伸系数b，所有的面积都被系数a，b改变，正如我们看到的，所以有因此这一方程简单的意味着勾股定理在任意直角三角形中都有效。If 1+1=2 then the Pythagorean theorem holds, or one more proof of the oldest theorem of mathematicsABSTRACTThe Pythagorean theorem is one of the oldest theorems of mathematics. It gained during the time a central position and even today it continues to be a source of inspiration. In this note we try to give a proof which is based on a hopefully new approach. Our treatment will be as intuitive as it can be. Keywords: Pythagorean theorem, geometric transformation, Euclidean geometry One of the oldest theorems of mathematics the Pythagorean theorem states that the area of the square drawn on the hypotenuse(the side opposite the right angle)of a right angled triangle, is equal to the sum of the areas of the squares constructed on the other two sides,(the catheti).That is, ，if we denote the length of the sides of the triangle by a, b, c respectively. Loomis collected several hundreds of different proofs of this theorem, and classified them in categories. Maor traces the evolution of the Pythagorean theorem and its impact on mathematics and on our culture in general. The internet page contains a collection of 93 approaches to proving the theorem.Our aim is to give a proof by a new approach, more precisely a theorem of the next type. Theorem1. If the Pythagorean theorem holds in one single right angled triangle, then it holds in any right angled triangle! Of course we must start the proof by giving some facts we assume to be true. These will be related to the properties of a geometric transformation which we will use: the shrinking or stretching the Euclidean plane by a factor in a given direction (see gure1). Figure 1. The transformationLet us assume T is such a transformation having a as the shrinking/stretching factor. We can think of such a transformation as an effect of parallel projection between planes in the three-dimensional space, like in gure2.Figure 2. Streching and shrinkingThe transformation T obviously has the properties: T shrinks/stretches the area of a plane figure by the factor a ,that is if P is a plane figure, thenArea ( T ( P ) ) = a Area( P ) . It is compatible with the additivity of areas of disjoint figures, that is if P, Q are two disjoint figures thenArea ( T ( P Q ) = Area ( T ( P ) + Area ( T ( Q ) . It preserves the filling ratio in the following sens: let P be a regular polygon with area A , and fix an edge L of it. Let us suppose T transforms it in a polygon Q = T ( P ) with area a A ,and suppose the image of L is K .Then the filling ratio defined as is independent to the length of polygons side and the shrinking/stretching direction(see gure3).Figure 3. The invariance Let us consider now a right angled triangle having the length of the two catheti a and b .The property asserted by the Pythagorean theorem is homogeneous in the length of the triangle edges, so we can choose the measurement unit as we like.Now let us consider the setup of the gure4, i.e. a regular square tiling of the plane. We will choose the measurement unit as the length of a blue square (of area 1), for a reason which will be clear soon. Figure 4. The special caseIt is obvious that the sum of the areas of the two smaller squares is exactly the area of the third square, so the Pythagorean theorem holds in this special case.Let us now apply the following transformations. Shrink horizontally by the factor a, see gure5. Figure 5. Shrink horizontally Cut at the level x，and join the piece of height 1 x in an area equivalent form to the other side, according to gure6,so we have.By this y is not a parameter, it is a function of x.Figure 6. Cut and join Stretch vertically by the factor b, see gure7. Figure 7. Stretch vertically The sum of areas is preserved, so which is true for all x . By the two transformations the area of the big square changes to We can assume a, b are chosen so to have the third side of the triangle the same length s, so Consequently, the lling ratio counted for the square of side s is Now the same lling ratio is valid for the two small squares, of sides a and b respectively, so now we can compute x. We rst set the value of the x such that the resulting rectangles be similar. We have the condition for x: This equalit