四面体的外心－内心公式

数学星空

今天看到《东方论坛》上陈都先生又贴出了此问题相关历史记录：

数学星空

根据 http://blogs.ams.org/visualinsig ... ielsson-inequality/

In the comments on this blog, Greg Egan has proposed a generalized Grace–Danielsson inequality in n  dimensions. More precisely, he has shown the following. Suppose we have a smaller ball of radius r  inside a larger ball of radius \(R\)  in \(IR^n\)  . Suppose the distance between the centers of these balls is \(d\) . Then we can fit an n -simplex inside the larger ball and outside the smaller one if

\[d^2 \leqslant (R+(n−2)r)(R–nr)\]

where if the inequality is an equation the simplex may touch the surface of these balls. He conjectures that this sufficient condition is also necessary.
- See more at: http://blogs.ams.org/visualinsig ... thash.C6sYsa5Z.dpuf


To flesh out my conjecture a little more, it’s helpful to start with an equation that is satisfied by the inradius \( r\)  of an isosceles triangle with base 2b  and height h :

\[b^2 (h−2r)–hr^2=0\]

This is easy to prove from Pythagoras and similar triangles. The inradius is the radius of a circle that is tangent to all three sides of the triangle.

Now in the context of two nested spheres in R n  , the large one with radius R , the small one with radius r , and a distance of \(d\) between their centres, suppose we place a regular \((n−1)-\)simplex tangent to the small sphere, with its centre at one of the points of intersection with the axis that runs between the centres of both spheres. If we choose a scale for this \((n−1) -\)simplex so that its vertices lie on the large sphere, its circumradius will be given by the radius of the sphere of intersection between the large sphere and the hyperplane tangent to the small sphere, which is:

\[\sqrt{R^2 −(r\pm d)^2}\]

Here the sign depends on which point of intersection with the axis we use as the point of tangency. The inradius of a regular \((n−1) -\)simplex is smaller than the circumradius by a factor of \(n−1\) , so if we call that inradius b , we will have:

\[b^2 =\frac{R^2 −(r\pm d)^2}{(n−1)^2}\]

Now, suppose we turn the regular \((n−1) -\)simplex into an \(n -\)simplex by adding a point at the intersection of the axis and the large sphere that lies on the opposite side of the small sphere to the\((n−1) -\)simplex. The “altitude” of this \(n -\)simplex measured from the \((n−1) -\)simplex as its “base” will be given by:

\[h=R+(r\pm d)\]

We can construct an isosceles triangle by taking the centre of the regular \((n−1) -\)simplex as the centre of the triangle’s base, the centre of any of the \((n−2) -\)simplex faces of the regular \((n−1) -\)simplex as an endpoint of the triangle’s base, and the added point we use to create the\( n \)-simplex as the triangle’s apex. This isosceles triangle with have a half-base of b  and a height of\( h\) . What’s more, if every face of the \(n -\)simplex is tangent to the small sphere, then the radius of the small sphere, \(r\) , will be equal to the inradius of the isosceles triangle. So we have:

\[b^2 (h−2r)–hr^2 =0\]

It’s convenient to divide this through by \(h\)  and multiply by \((n−1)^2\)  , to obtain:

\[\frac{b^2(n−1)^2 (h−2r)}{h}–(n−1)^2 r^2 =0\]

Substituting in for \(b^2\)   and \(h\)  we get:

\[(R–(r\pm d))(R+(r\pm d)–2r)–(n−1)^2 r^2 =0\]

If we solve this for \(d^2\)  , we get:

\[d^2 =(R+(n−2)r)(R–nr)\]

So, this is the condition for an n -simplex with a regular (n−1) -simplex as its “base” to have all its faces tangent to the small sphere and all its vertices lying on the large sphere, given that we choose to make the “base” orthogonal to the axis running between the centres of the spheres.
- See more at: http://blogs.ams.org/visualinsig ... thash.C6sYsa5Z.dpuf


当\(n=2\)时

有Euler inequality holds:：

\[d^2 \leq R(R-2r)\]

当\(n=3\)时

有Grace–Danielsson inequality holds：

\[d^2 \leq  (R+r)(R-3r)\]

数学星空

数学星空

现在的问题如何简化：

\[f=\frac{S_1S_2a_{12}^2+S_1S_3a_{13}^2+S_1S_4a_{14}^2+S_2S_3a_{23}^2+S_2S_4a_{24}^2+S_3S_4a_{34}^2}{(S_1+S_2+S_3+S_4)^2}-r(2R+3r)\]

数学星空

又根据32#结果

设四面体$ABC-D$的四个面$BCD,ACD,ABD,ABC$的面积和外接圆半径分别为$S_i,R_i (i=1,2,3,4),$且四面体各棱长$AB=a_{12},AC=a_{13},BC=a_{23},AD=a_{14},BD=a_{24},CD=a_{34}$， 则有

$S_1S_2a_{12}^2+S_1S_3a_{13}^2+S_1S_4a_{14}^2+S_2S_3a_{23}^2+S_2S_4a_{24}^2+S_3S_4a_{34}^2$

$=(a_{14}a_{24}a_{34})/4(a_{23}a_{13}a_{12})/4(a_{12} a_{34} (1/(R_1R_2)+1/(R_3R_4))+a_{13} a_{24} (1/(R_1R_3)+1/(R_2R_4))+a_{23} a_{14} (1/(R_1R_4)+1/(R_2R_3)))$

且

\(S_1+S_2+S_3+S_4=\frac{V}{3r}\)

陈九章

陈九章

数学青年

《数学教学研究》 1995年06期

Euler公式及其不等式的三维推广
孔令恩 王敬秀

【摘要】：Ｅｕｌｅｒ公式及其不等式的三维推广
孔令恩（山东省枣庄师范学校２７７１００）     王敬秀（山东省枣庄市一中２７７１００）
在△ＡＢＣ中，有著名的Ｅｕｌｅｒ公式其中Ｏ，Ｉ为△ＡＢＣ的外、内心，Ｒ、ｒ分别是其外接圆、内切圆半径．由（１）立得Ｅｕｌｅｒ不等式对既有外...

【作者单位】： 山东省枣庄师范学校   山东省枣庄市一中
【关键词】： Euler公式四面体欧拉不等式     中学数学教学      数学教学研究      三维欧氏空间   几何不等式    内切球半径   山东省枣庄市师专学报
【分类号】：G634.6

【正文快照】：
Ｅｕｌｅｒ公式及其不等式的三维推广

孔令恩（山东省枣庄师范学校２７７１００）    王敬秀（山东省枣庄市一中２７７１００）

在△ＡＢＣ中，有著名的Ｅｕｌｅｒ公式  其中Ｏ，Ｉ为△ＡＢＣ的外、内心，Ｒ、ｒ分别是其外接圆、内切圆半径．由（１）立得  Ｅｕｌｅｒ不等式对既有外接

陈九章

陈九章