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在数学通报1964年7月号呂烈翰同志写的“談談培养学生空間想象力”一文第22頁中,举了两个例題。例題举得不錯,很說明問題。但笔者对两个图形的画法有一点不同意見,提出来供大家参考。問題一,一个三面角的三个面角都是60°,从頂点起在一条棱上截取一条6厘米长的线段,求这线段的另一端点到棱所对的面的距离。文中所談图12左图是比較好的,这一点我是同意的。但作这个图时,須先明确下列三点: 1.由于SF是SA在平面BSC內的投影(图4);SA与SB,SC所成的角相等,所以SF是∠BSC的平分线。但由于图形所在位置的影响,水平面內的∠BSF与∠CSF是不一定相等的。 2.若将SF成水平綫,則∠ASF在正垂平面內,∠AF应垂直于SF,且各綫段的长度不改变。 3.因为SE=SG,∠SEG为等腰三角形;而SF是
On the 22nd page of the article “Talk about Training Students’ Spatial Imagination” written by Comrade Lu Liehan in the July 1964 issue of the Mathematics Bulletin, two examples are given. The examples are well-expressed and illustrate the problem. However, the author has a little different opinion on the drawing of the two figures and proposes it for your reference. For problem 1, the three face angles of a trihedron are all 60°. From the apex, a 6-cm segment is taken on an edge. Find the distance from the other end of the segment to the face of the edge. The figure on the left of figure 12 in the article is better. I agree with this point. However, when making this diagram, the following three points must be clarified: 1. Since SF is the projection of SA in plane BSC (Fig. 4); SA forms the same angle with SB, SC, so SF is the bisector of BSC. However, due to the influence of the position of the graphics, the BSF and ∠CSF in the horizontal plane are not necessarily equal. 2. If SF is horizontal, ∠ASF is in the positive vertical plane, ∠AF is perpendicular to SF, and the length of each line segment does not change. 3. Since SE=SG, ∠SEG is an isosceles triangle; and SF is