三角形两内点间距

数学星空

对于2#的方程，我们可以得到
\(\alpha = -(a^4u_1^2-a^2b^2v_1^2-a^2c^2w_1^2-a^2u_1^4+a^2v_1^2w_1^2+b^2v_1^4-b^2v_1^2w_1^2-c^2u_1^2v_1^2+c^2u_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

\(\beta = (a^2b^2u_1^2-a^2u_1^4+a^2u_1^2w_1^2-b^4v_1^2+b^2c^2w_1^2-b^2u_1^4+2b^2u_1^2v_1^2-b^2u_1^2w_1^2+c^2u_1^4-c^2u_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

\(\gamma = (a^2c^2u_1^2-a^2u_1^4+a^2u_1^2v_1^2+b^2c^2v_1^2+b^2u_1^2v_1^2-b^2v_1^4-c^4w_1^2-2c^2u_1^2v_1^2+c^2u_1^2w_1^2+c^2v_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

可是\(\alpha +\beta+\gamma\)并不能化简为1

并且最终答案：
\(PQ^2=\frac{M}{a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2}\)

\(M=a^4u_1^4-a^4u_1^2u_2^2-2a^2b^2u_1^2v_1^2+a^2b^2u_1^2v_2^2+a^2b^2u_2^2v_1^2-a^2c^2u_1^4-a^2c^2u_1^2w_1^2+a^2c^2u_1^2w_2^2+a^2c^2u_2^2w_1^2+a^2u_1^4u_2^2-a^2u_1^4v_2^2-a^2u_1^4w_2^2+a^2u_1^2v_1^2w_2^2+a^2u_1^2v_2^2w_1^2-a^2u_2^2v_1^2w_1^2+b^4v_1^4-b^4v_1^2v_2^2-b^2c^2u_1^2v_1^2-b^2c^2v_1^2w_1^2+b^2c^2v_1^2w_2^2+b^2c^2v_2^2w_1^2-b^2u_1^4v_2^2+2b^2u_1^2v_1^2v_2^2+b^2u_1^2v_1^2w_2^2-b^2u_1^2v_2^2w_1^2-b^2u_2^2v_1^4+b^2u_2^2v_1^2w_1^2-b^2v_1^4w_2^2+c^4u_1^2w_1^2-c^4w_1^2w_2^2+c^2u_1^4v_2^2+c^2u_1^2u_2^2v_1^2-c^2u_1^2u_2^2w_1^2-2c^2u_1^2v_1^2w_2^2-c^2u_1^2v_2^2w_1^2+c^2u_1^2w_1^2w_2^2+c^2v_1^2w_1^2w_2^2\)

取\(a=b=c=1,u_1=\frac{\sqrt{3}}{2},v_1=w_1=\frac{1}{2},u_2=v_2=w_2=\frac{\sqrt{3}}{3}\),可知\(PQ=r=\frac{\sqrt{3}}{6}\)

但是将上面代入得到\(PQ^2=-\frac{11}{12}\)也不正确

数学星空

关于9#提到的内点\(PA=x,PB=y,PC=z\),还有一个很简洁的表达式(值得大家共赏）

\[(-a^2+y^2+z^2)^2x^2+(-b^2+x^2+z^2)^2y^2+(-c^2+x^2+y^2)^2z^2-(-a^2+y^2+z^2)(-b^2+x^2+z^2)(-c^2+x^2+y^2)-4x^2y^2z^2＝0\]

hujunhua

2#和4#被我编辑过，去掉了2#中的图片，将4#的图片变成了Latex公式。
郭老板说过，尽量不要用图片搞公式，本坛公式表达已经很强大了。

建议星空研究一下将这个主题下的公式化成对称而且简明的行列式。

gxqcn

受 hujunhua 启发，可将 \( PABC \) （或 \( QABC \) ）看做是空间四面体，只是其体积 \( V=0 \) ：
\[ \begin{vmatrix}
0 & c^2 & b^2 & u_i^2 & 1 \\
c^2 & 0 & a^2 & v_i^2 & 1 \\
b^2 & a^2 & 0 & w_i^2 & 1 \\
u_i^2 & v_i^2 & w_i^2 & 0 & 1 \\
1 & 1 & 1 & 1 & 0
\end{vmatrix} = 288*V_i^2 = 0 \quad(i=1,2) \]

gxqcn

其实，northwolves 在 6# 里的方法就可以解出来了（里面的公式排版有点小问题），
楼主可能追求的是一种更“美”的表达式。。。

creasson

数学星空 发表于 2014-2-16 20:30
对于2#的方程，我们可以得到
\(\alpha = -(a^4u_1^2-a^2b^2v_1^2-a^2c^2w_1^2-a^2u_1^4+a^2v_1^2w_1^2+b^2 ...

这一定是你计算上的失误所致，即使我在2L所给答案是错的，在4L所给的式子也决不会错.

数学星空

在9#我已提到，其实不需要太多计算

分别在四边形\(APQB,APQC,BPQC\)中利用11#公式可得

\((-v_1^2+v_2^2+x^2)^2u_2^2+(-u_1^2+u_2^2+x^2)^2v_2^2+(-c^2+u_2^2+v_2^2)^2x^2-(-v_1^2+v_2^2+x^2)(-u_1^2+u_2^2+x^2)(-c^2+u_2^2+v_2^2)-4u_2^2v_2^2x^2＝0\)

\((-w_1^2+w_2^2+x^2)^2u_2^2+(-b^2+u_2^2+w_2^2)^2x^2+(-u_1^2+u_2^2+x^2)^2w_2^2-(-w_1^2+w_2^2+x^2)(-b^2+u_2^2+w_2^2)(-u_1^2+u_2^2+x^2)-4u_2^2x^2w_2^2＝0\)

\((-a^2+v_2^2+w_2^2)^2x^2+(-w_1^2+w_2^2+x^2)^2v_2^2+(-v_1^2+v_2^2+x^2)^2w_2^2-(-a^2+v_2^2+w_2^2)(-w_1^2+w_2^2+x^2)(-v_1^2+v_2^2+x^2)-4x^2v_2^2w_2^2＝0\)

将以上三式相加就可以得到我们需要的结果（为了得到比较对称的一种做法）

\(-a^2v_2^2w_1^2+a^2v_1^2w_1^2+a^2v_2^2w_2^2-a^2v_1^2w_2^2-v_1^2v_2^2w_1^2-v_1^2v_2^2w_2^2-v_1^2w_1^2w_2^2-v_2^2w_1^2w_2^2+v_1^2u_2^4+v_2^2u_1^4+v_2^4u_1^2+w_2^4u_1^2+w_1^2u_2^4+u_2^2v_1^4+
u_2^2w_1^4+w_2^2u_1^4+(a^2+b^2+c^2)x^4+(a^4-a^2v_1^2-a^2v_2^2-a^2w_1^2-a^2w_2^2+b^4-b^2u_1^2-b^2u_2^2-b^2w_1^2-b^2w_2^2+c^4-c^2u_1^2-c^2u_2^2-c^2v_1^2-c^2v_2^2+2u_1^2u_2^2-u_1^2v_2^2-
u_1^2w_2^2-u_2^2v_1^2-u_2^2w_1^2+2v_1^2v_2^2-v_1^2w_2^2-v_2^2w_1^2+2w_1^2w_2^2)x^2+v_1^4w_2^2+v_1^2w_2^4+v_2^4w_1^2+v_2^2w_1^4+w_2^2b^2u_2^2+w_1^2b^2u_1^2-w_2^2w_1^2u_1^2-w_1^2b^2u_2^2-
w_1^2u_2^2u_1^2-u_2^2v_2^2v_1^2-w_2^2b^2u_1^2-v_2^2u_2^2u_1^2-w_2^2u_2^2u_1^2-v_2^2u_1^2c^2+u_2^2v_2^2c^2+v_1^2u_1^2c^2-v_1^2u_1^2u_2^2-v_1^2u_1^2v_2^2-v_1^2u_2^2c^2-u_2^2w_2^2w_1^2＝0\)

我们可以利用下面数据验证以上公式的正确性

\(a = 1, b = 1, c = 1, u_1 = \frac{\sqrt{3}}{2}, v_1 = \frac{1}{2}, w_1 = \frac{1}{2}, u_2 = \frac{\sqrt{3}}{3}, v_2 =\frac{\sqrt{3}}{3}, w_2 =\frac{\sqrt{3}}{3}, x =\frac{\sqrt{3}}{6}\)

\(a = 3, b = 4, c = 5, u_1 =\frac{16}{5}, v_1 = \frac{9}{5}, w_1 =\frac{12}{5}, u_2 =\frac{5}{2}, v_2 =\frac{5}{2}, w_2 =\frac{5}{2}, x =\frac{7}{10}\)

上面的公式我们可以称为‘心距万能公式’，下面是具体的应用。

数学星空

我们先计算大家熟知的外心\(O\),内心\(I\),垂心\(H\),重心\(G\)两两之间的心距公式

\(AI =\frac{\sqrt{bc}}{\sqrt{\frac{p-a}{p}}}, BI =\frac{\sqrt{ac}}{\sqrt{\frac{p-b}{p}}}, CI = \frac{\sqrt{ab}}{\sqrt{\frac{p-c}{p}}}\)

\(AH =\frac{a(-a^2+b^2+c^2)}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}, BH =\frac{b(a^2-b^2+c^2)}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}, CH =\frac{c(a^2+b^2-c^2)}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}\)

\(AG = \frac{\sqrt{2(b^2+c^2)-a^2}}{3}, BG =\frac{\sqrt{2(a^2+c^2)-b^2}}{3}, CG =\frac{\sqrt{2(a^2+b^2)-c^2}}{3}\)

\(AO = \frac{abc}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}, BO =\frac{abc}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}, CO =\frac{abc}{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}\)

分别代入‘心距万能公式’得到

1.  \(9(a+b+c)(-b+a-c)(b+a-c)(a-b+c)HG^2+4b^6-4a^4b^2-4b^4c^2-4a^4c^2-4b^2c^4+4c^6+4a^6-4a^2b^4+12a^2b^2c^2-4a^2c^4＝0\)

2.  \((a+b+c)(-b+a-c)(b+a-c)(a-b+c)HO^2+b^6-a^2b^4-a^4b^2+a^6-b^4c^2+3a^2b^2c^2-a^4c^2-b^2c^4-a^2c^4+c^6＝0\)

3.  \((a+b+c)(-b+a-c)(b+a-c)(a-b+c)HI^2-c^5a-a^5c-ba^5-bc^5-b^5c-b^5a-a^4b^2-b^4c^2-a^4c^2-b^2c^4-a^2b^4-a^2c^4+2b^3c^3+2a^3c^3+2b^3a^3+b^6+c^6+a^6+3bca^4-2bc^3a^2+3b^4ac-2a^3b^2c-2ba^3c^2+6a^2b^2c^2-2b^2c^3a-2b^3ac^2-2b^3ca^2+3bac^4＝0\)

4.  \(9(a+b+c)(-b+a-c)(b+a-c)(a-b+c)OG^2+b^6-a^2b^4-a^4b^2+a^6-b^4c^2+3a^2b^2c^2-a^4c^2-b^2c^4-a^2c^4+c^6＝0\)

5.  \((9b+9a+9c)GI^2+b^3-2b^2a-2ba^2+a^3-2b^2c+9bca-2a^2c-2bc^2-2ac^2+c^3＝0\)

6. \((a+b+c)(-b+a-c)(b+a-c)(a-b+c)OI^2+bca(a^3-a^2b-a^2c-ab^2+3abc-ac^2+b^3-b^2c-bc^2+c^3)＝0\)

数学星空

对于以上结果：

我们代入\(a=y+z,b=x+z,c=x+y\),则有

1.  \(-144xzy(y+z+x)HG^2-56z^4yx-40x^3zy^2-40x^3z^2y-40z^3y^2x-40z^3yx^2-40zy^3x^2-56x^4zy-56y^4xz-24z^2y^2x^2-40z^2y^3x+36x^4z^2+72x^3z^3+36x^2z^4+36z^4y^2+72z^3y^3+36y^4x^2+36y^4z^2+36x^4y^2+72x^3y^3=0\)

2.  \(-16xzy(y+z+x)HO^2+9x^4z^2-14z^4yx-10x^3zy^2-10x^3z^2y-10z^3y^2x-10z^3yx^2-10zy^3x^2-14x^4zy-14y^4xz-6z^2y^2x^2-10z^2y^3x+9x^2z^4+9z^4y^2+18z^3y^3+18x^3z^3+9y^4z^2+9x^4y^2+18x^3y^3+9y^4x^2=0\)

3.  \(-16xzy(y+z+x)HI^2-8z^4yx-8x^3zy^2-8x^3z^2y-8z^3y^2x-8z^3yx^2-8zy^3x^2-8x^4zy-8y^4xz+24z^2y^2x^2-8z^2y^3x+4x^4z^2+8x^3z^3+4x^2z^4+4z^4y^2+8z^3y^3+4y^4x^2+4y^4z^2+4x^4y^2+8x^3y^3=0\)

4.  \(-144xzy(y+z+x)OG^2-14z^4yx-10x^3zy^2-10x^3z^2y-10z^3y^2x-10z^3yx^2-10zy^3x^2-14x^4zy-14y^4xz-6z^2y^2x^2-10z^2y^3x+9x^4z^2+18x^3z^3+9x^2z^4+9z^4y^2+18z^3y^3+9y^4x^2+9y^4z^2+9x^4y^2+18x^3y^3=0\)

5.  \((18x+18z+18y)GI^2-6xzy-2x^3-2z^3-2y^3+2x^2z+2xz^2+2z^2y+2zy^2+2x^2y+2xy^2=0\)

6.  \(-16xzy(y+z+x)OI^2+(x+z)(x+y)(z+y)(x^2y+x^2z+xy^2-6xyz+xz^2+y^2z+yz^2)=0\)

若以\(R,r,S\)表示心距公式

\(9HG^2r^2-36R^2r^2-32Rr^3-8r^4+8S^2＝0\)

\(HO^2r^2-9R^2r^2-8Rr^3-2r^4+2S^2=0\)

\(HI^2r^2-4R^2r^2-4Rr^3-3r^4+S^2=0\)

\(9OG^2r^2-9R^2r^2-8Rr^3-2r^4+2S^2=0\)

\(9GI^2r^2+16Rr^3-5r^4-S^2=0\)

\( OI^2-R^2+2Rr=0\)

若以\(R,r,p\)表示心距公式

\(9HG^2-36R^2-32Rr+8p^2-8r^2=0\)

\(HO^2-9R^2-8Rr+2p^2-2r^2=0\)

\(HI^2-4R^2-4Rr+p^2-3r^2=0\)

\(9OG^2-9R^2-8Rr+2p^2-2r^2=0\)

\(9GI^2+16Rr-p^2-5r^2\)

\(OI^2-R^2+2Rr=0\)

注

\(p=\frac{a+b+c}{2}\)为半周长，

\(S=\frac{\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}{4}\)为\(\triangle ABC\)面积

\(R=\frac{abc}{\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}\)为\(\triangle ABC\)外接圆半径

\(r=\frac{\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}{2(a+b+c)}\)为\(\triangle ABC\)内切圆半径

上面的计算利用了下面公式

\(s_1=a+b+c=2p\)

\(s_2=ab+ac+bc=\frac{4Rr^3+r^4+S^2}{r^2}=4Rr+p^2+r^2\)

\(s_3=abc=4RS=4Rpr\)

数学星空

我们可以利用‘万能心距公式’计算‘陈都提出的完全心距公式（他得到的公式都很美）’

1.  三个旁心\(O_1,O_2,O_3\)

2.  费马点\(F\)

3.  九点圆圆心\(L\)  (为防止代号重复取Euler 中的L)

4.  正负等角中心\(E,F\),具体可见三角形正负等角中心间距
http://bbs.emath.ac.cn/thread-5094-1-1.html   1#陈都定义
-------------------------------------------------------------------------------------------
为了防止与费马点\(F\)重复定义，后面将正负等角点重新定义为\(E_1,E_2\),其实\(E_1\)点就是广义的费马点\(F\)

5.  Steiner椭圆焦点\(F_1,F_2\),具体可见
椭圆内接N边形的最大面积
http://bbs.emath.ac.cn/forum.php ... 67&fromuid=1455

6.正负等力点\(S,T\),可见
http://bbs.emath.ac.cn/thread-5094-3-1.html     28#陈都定义

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