四面体的外心－内心公式

数学星空

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$s_i,R_i (i=1,2,3,4),$四面体的内切球半径为$r$,外接球半径为$R$， 则有

$r=(sqrt(R^2-R_1^2)*s_1+sqrt(R^2-R_2^2)*s_2+sqrt(R^2-R_3^2)*s_3+sqrt(R^2-R_4^2)*s_4)/(s_1+s_2+s_3+s_4)$

数学星空

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$AB=c,AC=b,BC=a,AD=a_1,BD=b_1,CD=c_1$， 则有

$sqrt(s_1s_2)c+sqrt(s_1s_3)b+sqrt(s_1s_4)a_1+sqrt(s_2s_3)a+sqrt(s_2s_4)b_1+sqrt(s_3s_4)c_1$

$=((1/sqrt(R_1 R_2)+1/sqrt(R_3 R_4))sqrt(a a_1)+(1/sqrt(R_1R_3)+1/sqrt(R_2 R_4))sqrt(b b_1)+(1/sqrt(R_1 R_4)+1/sqrt(R_2 R_3))sqrt(c c_1))sqrt(abc a_1 b_1 c_1)/4$


$s_1s_2c^2+s_1s_3b^2+s_1s_4a_1^2+s_2s_3a^2+s_2s_4b_1^2+s_3s_4c_1^2$

$=(abc)/4(a_1b_1c_1)/4(c c_1 (1/(R_1R_2)+1/(R_3R_4))+b b_1 (1/(R_1R_3)+1/(R_2R_4))+a a_1 (1/(R_1R_4)+1/(R_2R_3)))$

mathe

第一个命题光R和r应该是无法确定OI的。
我们可以考虑特殊的四面体，四条边相等为a,另外两条对边(不相邻的边)长度相等为b。那么OI应该总是等于0,而R和r之间这时是无法相互确定的。也就是我们至少还要引入第三变量

mathe

不过R>=3r的证明还是比较容易的，可以直接将平面上三角形问题的方法直接推广过来

数学星空

mathe 发表于 2013-8-17 15:07
第一个命题光R和r应该是无法确定OI的。
我们可以考虑特殊的四面体，四条边相等为a,另外两条对边(不相邻的 ...

的确，当$BC=a,AC=a,AB=b,AD=a,BD=a,CD=b$时

$V=(sqrt(2)/12)n^2sqrt(2m^2-n^2),r= (sqrt(2)/4)nsqrt(2m^2-n^2)/sqrt(4m^2-n^2),d=0,R=(sqrt(2)/4)sqrt(2m^2+n^2)$

即所mathe所述此时d与r,R无关

当$BC=m,AC=n,AB=p,AD=m,BD=n,CD=p$时

$V=1/12sqrt(-2m^6+2m^4n^2+2m^4p^2+2m^2n^4-4m^2n^2p^2+2m^2p^4-2n^6+2n^4p^2+2n^2p^4-2p^6) $

$R=1/4sqrt(2m^2+2n^2+2p^2)$

$r= 1/4sqrt(-2m^6+2m^4n^2+2m^4p^2+2m^2n^4-4m^2n^2p^2+2m^2p^4-2n^6+2n^4p^2+2n^2p^4-2p^6)/sqrt(-m^4+2m^2n^2+2m^2p^2-n^4+2n^2p^2-p^4)$

$d=1/2sqrt(m^2+n^2+p^2)$

即此时$d=sqrt(2)R$

数学星空

若d最少能用三个特征参数（V,R,r)能表示，则

可以设$BC=a,AC=b,AB=c,AD=BD=CD=t$求出

$V=1/12sqrt(-a^4t^2-a^2b^2c^2+2a^2b^2t^2+2a^2c^2t^2-b^4t^2+2b^2c^2t^2-c^4t^2)$

$R=1/2sqrt((-a^4t^4+2a^2b^2t^4+2a^2c^2t^4-b^4t^4+2b^2c^2t^4-c^4t^4)/(-a^4t^2-a^2b^2c^2+2a^2b^2t^2+2a^2c^2t^2-b^4t^2+2b^2c^2t^2-c^4t^2))$

$r=sqrt(-a^4t^2-a^2b^2c^2+2a^2b^2t^2+2a^2c^2t^2-b^4t^2+2b^2c^2t^2-c^4t^2)/(sqrt(-a^4+4a^2t^2)+sqrt(-b^4+4b^2t^2)+sqrt(-c^4+4c^2t^2)+sqrt(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4))$

$d=1/2sqrt((-a^4t^4+2a^2b^2t^4+2a^2c^2t^4-b^4t^4+2b^2c^2t^4-c^4t^4)/(-a^4t^2-a^2b^2c^2+2a^2b^2t^2+2a^2c^2t^2-b^4t^2+2b^2c^2t^2-c^4t^2)-$

$(4(sqrt((-a^4+4a^2t^2)(-b^4+4b^2t^2))c^2+sqrt((-a^4+4a^2t^2)(-c^4+4c^2t^2))b^2+sqrt((-a^4+4a^2t^2)(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4))t^2+$

$sqrt((-b^4+4b^2t^2)(-c^4+4c^2t^2))a^2+sqrt((-b^4+4b^2t^2)(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4))t^2+sqrt((-c^4+4c^2t^2)(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4))t^2))$

$/(sqrt(-a^4+4a^2t^2)+sqrt(-b^4+4b^2t^2)+sqrt(-c^4+4c^2t^2)+sqrt(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4))^2)$

即$d^2=R^2-M/N^2 $,

$M=abc(sqrt((-a^2+4t^2)(-b^2+4t^2))c+sqrt((-a^2+4t^2)(-c^2+4t^2))b+sqrt((-c^2+4t^2)(-b^2+4t^2))a)+4t^2sqrt(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4)(asqrt(-a^2+4t^2)+bsqrt(-b^2+4t^2)+csqrt(-c^2+4t^2))$

$N=asqrt(-a^2+4t^2)+bsqrt(-b^2+4t^2)+csqrt(-c^2+4t^2)+sqrt(-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4)$

数学星空

下面我们需要做的是有最少的特征参数表示d,请问有几个？ （我猜想是4个）
若特征参数必须包含R,r,那么另外几个是什么？
当然我们要求的基本条件是表达式简洁对称

creasson 第一个用6个参数(即各棱长）表达了d

zuijianqiugen

数学星空 发表于 2013-7-28 10:51
能否给出具体的表达式？？

参见《关于四面体的内外心距问题》
http://zuijianqiugen.blog.163.co ... 062201422801918454/

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：

\[d^2 \leq  (R+r)(R-3r)\]