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法同著名数学家费马提出关于三角形的一个有趣问题:在三角形所在平面上求一点,使该点到三角形三个顶点距离之和最小．人们称这个点为费马点．1．如果三角形每一内角都小于120°,在三角形内必存在一点,它对三条边所张的角都是120°,它到三顶点的距离之和最小．证明如图1所示,任作-△ABC使其内角均小于120°,然后分别以AB与AC为一边向外作正△ABE与正△ACD,连结CE与BD,它们的交点为P．
Law and the famous mathematician Fermat raised an interesting question about triangles: Find a little on the plane where the triangle is, and minimize the distance between the vertex and the three vertices of the triangle. People call this point Fermat. 1. If each interior angle of a triangle is less than 120 °, there must be a point in the triangle that has an angle of 120 ° to each of the three sides and minimizes the distance to the three vertices. Proof as shown in Figure 1, Ren - △ ABC to make their internal angles are less than 120 °, then AB and AC aside for the positive side of the positive △ ABE and △ ACD, CE and BD link, the intersection of their P.