多面體歐拉定理 的發(fā)現(xiàn)

1、多面體歐拉定理 的發(fā)現(xiàn)研究性課題：一、認識歐拉1 1、數(shù)學奇才歐拉、數(shù)學奇才歐拉歐拉歐拉-瑞士人（瑞士人（Euler,L. 1707-1783）；）；歐拉歐拉-16 歲獲得碩士學位；歲獲得碩士學位；歐拉歐拉-數(shù)學史上數(shù)學史上“高產(chǎn)高產(chǎn)”的數(shù)學家。在世發(fā)表論文的數(shù)學家。在世發(fā)表論文700多篇，多篇， 去世后還留下去世后還留下100多篇待發(fā)表；多篇待發(fā)表；歐拉歐拉-首先使用首先使用f(x)表示函數(shù)，用表示函數(shù)，用e表示自然對數(shù)的底，用表示自然對數(shù)的底，用a、 b、c 表示表示ABC，用，用表示求和，用表示求和，用i表示虛數(shù)單位等；表示虛數(shù)單位等；歐拉歐拉-目前數(shù)學中有歐拉公式、歐拉常數(shù)、歐拉猜想、

2、歐拉目前數(shù)學中有歐拉公式、歐拉常數(shù)、歐拉猜想、歐拉 方法、歐拉方程、歐拉定理。方法、歐拉方程、歐拉定理。2 2 、歐拉眼中的簡單多面體、歐拉眼中的簡單多面體棱棱 柱柱棱棱 錐錐正多面體正多面體凸多面體凸多面體簡單多面體充氣后 在立體幾何多面體的研究中歐拉首在立體幾何多面體的研究中歐拉首先發(fā)現(xiàn)并證明了歐拉定理。今天我們沿先發(fā)現(xiàn)并證明了歐拉定理。今天我們沿著他的著他的正多面體正多面體 頂點數(shù)頂點數(shù)V 面數(shù)面數(shù)F 棱數(shù)棱數(shù)E正四面體正四面體正六面體正六面體正八面體正八面體正十二面體正十二面體正二十面體正二十面體 4 4 6 8 6 12 6 8 12 20 12 30 12 20 30二、觀察歸納

3、1 1、填表、填表2、結(jié)論： V+FE23 3、驗證提高、驗證提高簡單多面體簡單多面體 頂點數(shù)頂點數(shù)V 面數(shù)面數(shù)F 棱數(shù)棱數(shù)EN N棱錐棱錐N N棱柱棱柱V、F、E間的關系間的關系 N+1 N+1 2N 2N N+2 3N V+FE2三、歐拉猜想 簡單多面體的頂點數(shù)簡單多面體的頂點數(shù)V V、棱數(shù)、棱數(shù)E E、及、及面數(shù)面數(shù)F F間有關系：間有關系：V+FV+FE=2E=2 四、嘗試證明1、下面我們以四面體、下面我們以四面體ABCD為例為例 體會歐拉的證法體會歐拉的證法：ABCDV+FE的值暫不知道V+F1E變化情況壓扁壓扁不變V+F1E變化情況不變不變不變V+F1E變化情況不變不變上圖中上圖中

4、V+F1E2+011那么 V+FE= V+（F1+1）E =1+1=2 簡單多面體的頂點數(shù)簡單多面體的頂點數(shù)V V、棱數(shù)、棱數(shù)E E、及面數(shù)及面數(shù)F F間有關系：間有關系：V+FV+FE=2E=22、多面體歐拉定理 ：五、總結(jié)回顧 歐拉公式的發(fā)現(xiàn)和證明是得益于“多面體的表面都是用橡皮薄膜制作的”這一觀念上的創(chuàng)新，是得益于“向它們內(nèi)部沖氣”和“底面剪掉，然后其余各面拉開鋪平”這一方法的創(chuàng)新。 希望同學們能夠象歐拉那樣善于思考、樂希望同學們能夠象歐拉那樣善于思考、樂于創(chuàng)新！于創(chuàng)新！六、作業(yè)六、作業(yè)： 以正六面體為例驗證多面體以正六面體為例驗證多面體歐拉定理歐拉定理4C0z)w&s!pYmU

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