1、數(shù)形結(jié)合論文翻譯文章勾股定理不是，或者一個古老的數(shù)學(xué)定理的證明勾股定理是一個古老的數(shù)學(xué)定理，它一直占據(jù)著重要位置，即使現(xiàn)在它仍然是靈感的源泉。在本文中我們嘗試給出一個有前途的新的證明方法，我們的結(jié)果會更加直觀。關(guān)鍵詞：勾股定理 幾何變換 歐幾里得幾何學(xué)最古老的數(shù)學(xué)定理之一勾股定理指出：直角三角形斜邊（直角所對的邊）所畫正方形的面積等于另外兩條邊（直角邊）所畫正方形的面積的和。就是。如果我們分別用a,b，c表示三角形三遍的長度。盧米斯收集了數(shù)百種勾股定理的證明，并將其分類。馬奧爾追溯了勾股定理的演變以及它對數(shù)學(xué)和我們的文化的影響。網(wǎng)絡(luò)上包含了約93種勾股定理的證明。我們的目標(biāo)是通過一種新方法給出

2、證明，準(zhǔn)確的說是定理的另一類型。定理1：如果在簡單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。當(dāng)然我們必須給出證明說明我們的假設(shè)是正確的。這就涉及到幾何變換的屬性，我們將用到：在給定的方向上收縮或拉伸歐幾里得平面（見圖1）。圖1.轉(zhuǎn)換讓我們假設(shè)T是這樣一個收縮/拉伸系數(shù)為a的轉(zhuǎn)變。我們可以認為這種轉(zhuǎn)化為在三維空間內(nèi)平面之間的平行投影的效果，如圖2所示。圖2.拉伸和收縮轉(zhuǎn)換T明顯具有的性質(zhì)：·轉(zhuǎn)換T通過系數(shù)a收縮/拉伸一個平面圖形的面積，如果P是一個平面圖形，那么·它與不相交的圖形的面積的可加性是兼容的，如果P,Q是兩個不相交的圖形，那么·它保留了以下

3、森斯的填充率：讓P是面積為A的正多邊形，并修復(fù)他的邊緣L。讓我們假設(shè)T把它轉(zhuǎn)換為面積為aA的多邊形Q = T（P），并假設(shè)圖形L轉(zhuǎn)換為K。然后將填充率定義為與多邊形邊的長度和收縮/拉伸方向無關(guān)（見圖3）。圖3.不變性現(xiàn)在讓我們考慮一個直角三角形的兩條直角邊的長度為a和b。由勾股定理斷言該性質(zhì)是均勻的對于三角形邊的長度，因此我們可以選擇我們喜歡的計量單位。現(xiàn)在讓我們來考慮圖4的設(shè)置，也就是規(guī)則的正方形的平面排列。我們將選擇測量單位為藍色的正方形（1區(qū)）的長度。是有原因的，將是明確的很快。圖4.特殊情況很明顯的兩個較小的正方形的面積之和恰好是第三個正方形的面積，因此勾股定理在這種特殊情況下成立。現(xiàn)

4、在讓我們應(yīng)用下面的轉(zhuǎn)換。·按系數(shù)a水平收縮，見圖5。圖5.水平收縮·在水平切高x，將高度為1-x的區(qū)域的等價形式添加到另一邊，根據(jù)圖6，我們得到這里y不是一個參數(shù)，是x的函數(shù)。圖6.割補·按系數(shù)b垂直拉伸，見圖7。圖7.垂直拉伸·面積的合適不變的，所以對于任意x都成立。·通過兩次變換，大正方形的面積變?yōu)?#183;我們可以假設(shè)選擇a，b到第三條邊長度為s的三角形中，所以·因此，s一邊的正方形的填充率為·現(xiàn)在同樣的填充率對兩個小的正方形是有效的，分別在a，b邊，因此我們可以計算x。首先我們設(shè)置x，使得所得到的矩形是相似的。我們

5、有x的條件：·這個等式給出現(xiàn)在以a和b為邊的矩形是相似的。其邊長的公比是，這是共同的填充率（小于1）的兩個較小的正方形的面積總和改變，第三個正方形面積也改變。·經(jīng)由收縮系數(shù)a和拉伸系數(shù)b，所有的面積都被系數(shù)a，b改變，正如我們看到的，所以有因此這一方程簡單的意味著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

6、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,

7、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 o

8、ther 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 mathem

9、atics 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

10、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.

11、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 obvious

12、ly 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 (

13、 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

14、 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

15、 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 th

16、e 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 p

17、iece 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 al

18、l 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: Thi