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在數學雜誌一卷一期中,有路見可先生一文:多邊形的三角分裂。其中包括任意n角形的內角和等於(n-2)π的定理。可以一入初等幾何學參考書中(在初等幾何敎本中,亦可以提到)。但其中所用的話語及尤拉公式,則出於初等幾何範圍以外。而非中學生所能懂。本文目的、想用初等方法來證明。又本文定理一的證明經編者建議修改後,較原來的證明更簡單,特此感謝。在本文中,首先要提到下面的基本知識:(甲) 多邊形不相鄰的兩邊沒有交點(不許延長)。(乙) 多邊形內任一點與其外一點的連結線段,至少與其周有一交點。以下用P_n表多邊形。n乃共邊數,J表其周界,n爲大於3的整數。定理一。在多邊形的諸頂點中,至少有二頂點,其連結線段上的點都是內點。(端
In the first issue of the Mathematical magazine, there was a message from Mr. Lu to see the triangle of polygons. Including any n-angle internal angle and theorem equal to (n-2)π. Can be entered in the elementary geometry reference book (in the elementary geometry script, it can also be mentioned). However, the utterances and Euler’s formula used therein are outside the primary geometric range. Instead of middle school students can understand. The purpose of this article is to prove it with elementary methods. Also, the proof of the theorem of this paper is recommended by the editors. It is simpler than the original proof. Thanks. In this article, we must first mention the following basic knowledge: (A) polygons do not have two intersections on either side (no extension). (b) At least one point of intersection with a point outside the polygon and at least one point in the polygon. The following uses the P_n table polygon. n is the co-edge number, J is its perimeter, and n is an integer greater than 3. Theorem one. Among the vertices of a polygon, there are at least two vertices whose points on the joining line segment are all internal points. (end