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楼主: 数学星空

[讨论] 关于三角形的几个问题

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 楼主| 发表于 2014-2-7 20:15:23 | 显示全部楼层
当 \( s \) 点为 \( \triangle ABC \) 的内心时
1.即28# 第一列的第2幅图有( \( s,S,r \) 的含义同27#)
\[\begin{split} r_1:r_2:r_3&=\frac{a}{\frac{1}{\sin\frac B2}+\frac{1}{\sin\frac C2}+\frac{1}{\cot\frac B2+\cot\frac C2}}:\frac{b}{\frac{1}{\sin\frac A2}+\frac{1}{\sin\frac C2}+\frac{1}{\cot\frac A2+\cot\frac C2}}:\frac{c}{\frac{1}{\sin\frac A2}+\frac{1}{\sin\frac B2}+\frac{1}{\cot\frac A2+\cot\frac B2}}\\
&=\frac{a}{\frac{1}{\sin\frac B2}+\frac{1}{\sin\frac C2}+\frac{\sin\frac B2\sin\frac C2}{\cos\frac A2}}:\frac{b}{\frac{1}{\sin\frac A2}+\frac{1}{\sin\frac C2}+\frac{\sin\frac A2\sin\frac C2}{\cos\frac B2}}:\frac{c}{\frac{1}{\sin\frac A2}+\frac{1}{\sin\frac B2}+\frac{\sin\frac A2\sin\frac B2}{\cos\frac C2}}\\
&=\frac{a}{\sqrt{\frac{a}{s-a}}\left( \sqrt{\frac{b}{s-b}}+\sqrt{\frac{c}{s-c}} \right)+\frac{r}{a}}:\frac{b}{\sqrt{\frac{b}{s-b}}\left( \sqrt{\frac{a}{s-a}}+\sqrt{\frac{c}{s-c}} \right)+\frac{r}{b}}:\frac{c}{\sqrt{\frac{c}{s-c}}\left( \sqrt{\frac{a}{s-a}}+\sqrt{\frac{b}{s-b}} \right)+\frac{r}{c}}\end{split} \]
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-8 21:50:41 | 显示全部楼层
关于27#的问题,即马尔法蒂问题(Malfatti's Problem):在一个已知三角形内画三个圆,每个圆与其他两个圆以及三角形的两边相切。
容易得到
\( r_2 \cot(B/2)+2\sqrt{r_2r_3}+r_3 \cot(C/2)=a \)
\( r_1 \cot(A/2)+2\sqrt{r_1r_3}+r_3 \cot(C/2)=b\)
\( r_1 \cot(A/2)+2\sqrt{r_1r_2}+r_2 \cot(B/2)=c \)

\( r_2s(s-b)+2S\sqrt{r_2r_3}+r_3s(s-c)=aS \)
\( r_1s(s-a)+2S\sqrt{r_1r_3}+r_3s(s-c)=bS \)
\( r_1s(s-a)+2S\sqrt{r_1r_2}+r_2s(s-b)=cS\)
消元得到:
\( (-b-c+a)(b-c+a)^6(-b+c+a)^6+16(b+c+a)(-b-c+a)(b-c+a)^5(-b+c+a)^5r_1+128(a^5-a^3b^2-3a^3bc-a^3c^2-a^2b^3+6a^2b^2c+6a^2bc^2-a^2c^3-3ab^3c+6ab^2c^2-3abc^3+b^5-b^3c^2-b^2c^3+c^5)(b-c+a)^3(-b+c+a)^3r_1^2+128(-b-c+a)(5a^3+3a^2b+3a^2c+3ab^2-14abc+3ac^2+5b^3+3b^2c+3bc^2+5c^3)(b-c+a)^3(-b+c+a)^3r_1^3+128(-b-c+a)(17a^4-2a^2b^2-2a^2c^2+17b^4-2b^2c^2+17c^4)(b-c+a)^2(-b+c+a)^2r_1^4+1024(5a^2+2ab+2ac+5b^2+2bc+5c^2)(-b-c+a)^2(b-c+a)^2(-b+c+a)^2r_1^5+8192(b-c+a)(-b+c+a)(a^3+abc+b^3+c^3)(-b-c+a)^2r_1^6+8192(b-c+a)(b+c+a)(-b+c+a)(-b-c+a)^3r_1^7+4096(b+c+a)^2(-b-c+a)^3r_1^8=0 \)

\((-b+c+a)(-b-c+a)^6(b-c+a)^6-16(b+c+a)(-b+c+a)(-b-c+a)^5(b-c+a)^5r_2+128(a^5-a^3b^2-3a^3bc-a^3c^2-a^2b^3+6a^2b^2c+6a^2bc^2-a^2c^3-3ab^3c+6ab^2c^2-3abc^3+b^5-b^3c^2-b^2c^3+c^5)(-b-c+a)^3(b-c+a)^2r_2^2-128(-b+c+a)(5a^3+3a^2b+3a^2c+3ab^2-14abc+3ac^2+5b^3+3b^2c+3bc^2+5c^3)(-b-c+a)^3(b-c+a)^3r_2^3+128(-b+c+a)(17a^4-2a^2b^2-2a^2c^2+17b^4-2b^2c^2+17c^4)(-b-c+a)^2(b-c+a)^2r_2^4-1024(5a^2+2ab+2ac+5b^2+2bc+5c^2)(-b-c+a)^2(b-c+a)^2(-b+c+a)^2r_2^5+8192(b-c+a)(-b-c+a)(a^3+abc+b^3+c^3)(-b+c+a)^2r_2^6-8192(b-c+a)(b+c+a)(-b-c+a)(-b+c+a)^3r_2^7+4096(b+c+a)^2(-b+c+a)^3r_2^8=0 \)

\( (b-c+a)(-b-c+a)^6(-b+c+a)^6-16(b+c+a)(b-c+a)(-b-c+a)^5(-b+c+a)^5r_3+128(a^5-a^3b^2-3a^3bc-a^3c^2-a^2b^3+6a^2b^2c+6a^2bc^2-a^2c^3-3ab^3c+6ab^2c^2-3abc^3+b^5-b^3c^2-b^2c^3+c^5)(-b-c+a)^3(-b+c+a)^3r_3^2-128(b-c+a)(5a^3+3a^2b+3a^2c+3ab^2-14abc+3ac^2+5b^3+3b^2c+3bc^2+5c^3)(-b-c+a)^3(-b+c+a)^3r_3^3+128(b-c+a)(17a^4-2a^2b^2-2a^2c^2+17b^4-2b^2c^2+17c^4)(-b-c+a)^2(-b+c+a)^2r_3^4-1024(5a^2+2ab+2ac+5b^2+2bc+5c^2)(-b-c+a)^2(b-c+a)^2(-b+c+a)^2r_3^5+8192(-b-c+a)(-b+c+a)(a^3+abc+b^3+c^3)(b-c+a)^2r_3^6-8192(b+c+a)(-b+c+a)(-b-c+a)(b-c+a)^3r_3^7+4096(b+c+a)^2(b-c+a)^3r_3^8=0 \)

注: \( r_1 \) 取第三个正实根(根从小到大排列), \( r_2 \) 取第二个正实根(根从小到大排列), \( r_3 \) 取第一个正实根(根从小到大排列)
例:
\( a = 3, b = 4, c = 5, r_1 = 0.7519993534, r_2 = 0.6648941885, r_3 =0 .5079339624 \)
\( a = 3, b = 4, c = 6, r_1 =0.6660007375, r_2 = 0.6161503060, r_3 = 0.3399085088 \)
\( a = 4, b = 5, c = 6, r_1 =0 .9619865980, r_2 = 0.8635452050, r_3 = 0.7086379575 \)
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-8 23:22:03 | 显示全部楼层
若 \( \triangle ABC \) 内接四个等圆,且有三个圆分别相切于三角形角 \(A,B,C \) 的两条边,且第四个圆相切于三个圆,求此等圆的半径 \( r_0 \) ?(三角形三边为 \( a,b,c \) )

此问题来自于http://bbs.cnool.net/cthread-104971475.html

设分别相切于角A,B,C的两边的圆心分别为 \( O_1,O_2 ,O_3, \) 第四个圆心为 \( O_4 \) (\( \odot O_4\) 相切于 \( \odot O_1, \odot O_2,\odot O_3 \) ),
且 \( O_1O_2=z,O_1O_3=y,O_2O_3=x \)

\[ \begin{equation}\left\{ \eqalign{ x=a-r_0\left( \cot\frac B2+\cot\frac C2 \right)=a\left( 1-\frac{s\*r_0}{S} \right) \\
y=b-r_0\left( \cot\frac C2+\cot\frac A2 \right)=b\left( 1-\frac{s\*r_0}{S} \right) \\
z=c-r_0\left( \cot\frac A2+\cot\frac B2 \right)=c\left( 1-\frac{s\*r_0}{S} \right) } \right. \end{equation} \]

又由于 \( O_4 \) 为 \( \triangle O_1O_2O_3 \) 的外心,且 \( R_0=2r_0 \)

即 \[ \begin{equation} 2r_0=\frac{xyz}{\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}} \end{equation} \]

由(1~2)得到

\[ r_0=\frac{abc\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}{2(2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4+abc(a+b+c))}=\frac{4Sabc}{2(16S^2+2sabc)}=\frac{RS}{2S+sR}=\frac{Rr}{2r+R}=\frac{R}{2+\frac{R}{r}} \]

下面的图片取自于链接网页3#

20140206185238078.png
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-9 13:53:20 | 显示全部楼层
若 点\( P \) 为 \( \triangle ABC \)(三边长分别为 \( a,b,c \) ) 内一点,且 \( \triangle BPC,\triangle APC,\triangle APB \) 的内切圆圆心和半径分别为 \( \odot O_1,\odot O_2,\odot O_3 \) 和 \( r_1,r_2,r_3 \) ,
求 \( S_n=\max(r_1^n+r_2^n+r_3^n) \)及 \( L_n=\min(r_1^n+r_2^n+r_3^n) \)?
注,我们可以先讨论 \( n=1,2 \) ?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-9 14:43:27 | 显示全部楼层
可以发现:当 \( P \) 点位于 \( \triangle ABC \) 的任一个顶点时 \( S_n=\max(r_1^n+r_2^n+r_3^n)=r^n=(\frac{S}{s})^n
\) 看来只需要讨论 \( L_n=\min(r_1^n+r_2^n+r_3^n)\) ? 猜测仅当 \( r_1=r_2=r_3 \) 时成立??
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-9 16:12:21 | 显示全部楼层
当 \( P \) 点在 \( \triangle ABC  \) 的三边时,相当于求两个圆 \( \odot O_1,\odot O_2 \) 的情形

A1.当两圆均相切于 \( BC \) 边时

\( (-8a^4+16a^3b+16a^3c-32a^2bc-16ab^3+16ab^2c+16abc^2-16ac^3+8b^4-16b^2c^2+8c^4)r_2r_1+6a^4bc-4a^3b^2c-4a^3bc^2+c^6+b^6-2bc^5+4b^3c^3-b^2c^4-2b^5c-b^4c^2-2ac^5+(4a^4-8a^2b^2-8a^2c^2+64a^2r_2^2+4b^4-8b^2c^2+4c^4)r_1^2+(4a^4-8a^2b^2-8a^2c^2+4b^4-8b^2c^2+4c^4)r_2^2+a^6-2ab^5-a^2c^4+4a^3c^3-a^2b^4+4a^3b^3-2a^5b-2a^5c-a^4b^2-a^4c^2+6abc^4-4ab^2c^3-4ab^3c^2+6ab^4c-4a^2bc^3-4a^2b^3c+10a^2b^2c^2=0 \)

A2.当两圆均相切于 \( AC \) 边时

  \( 6a^4bc-4a^3b^2c-4a^3bc^2+c^6+b^6+(8a^4-16a^3b+16a^2bc-16a^2c^2+16ab^3-32ab^2c+16abc^2-8b^4+16b^3c-16bc^3+8c^4)r_2r_1-2bc^5+4b^3c^3-b^2c^4-2b^5c-b^4c^2-2ac^5+(4a^4-8a^2b^2-8a^2c^2+4b^4-8b^2c^2+4c^4)r_2^2+a^6-2ab^5-a^2c^4+4a^3c^3-a^2b^4+4a^3b^3-2a^5b-2a^5c-a^4b^2-a^4c^2+6abc^4-4ab^2c^3-4ab^3c^2+6ab^4c-4a^2bc^3-4a^2b^3c+10a^2b^2c^2+(4a^4-8a^2b^2-8a^2c^2+4b^4-8b^2c^2+64b^2r_2^2+4c^4)r_1^2=0 \)

A3.当两圆均相切于 \( AB \) 边时

\( 6a^4bc-4a^3b^2c-4a^3bc^2+c^6+(4a^4-8a^2b^2-8a^2c^2+4b^4-8b^2c^2+4c^4+64c^2r_2^2)r_1^2+b^6-2bc^5+4b^3c^3-b^2c^4-2b^5c-b^4c^2-2ac^5+(4a^4-8a^2b^2-8a^2c^2+4b^4-8b^2c^2+4c^4)r_2^2+a^6-2ab^5-a^2c^4+4a^3c^3-a^2b^4+4a^3b^3-2a^5b-2a^5c-a^4b^2-a^4c^2+(8a^4-16a^3c-16a^2b^2+16a^2bc+16ab^2c-32abc^2+16ac^3+8b^4-16b^3c+16bc^3-8c^4)r_2r_1+6abc^4-4ab^2c^3-4ab^3c^2+6ab^4c-4a^2bc^3-4a^2b^3c+10a^2b^2c^2=0 \)

当 \( r_1=r_2=r_0 \) 时
      
B1.当两等圆均相切于 \( BC \) 边时
\(64a^2r_0^4+16(b+c)(b-c+a)(-c+a-b)(a-b+c)r_0^2+(b-c+a)^2(-c+a-b)^2(a-b+c)^2=0 \)

B2.当两等圆均相切于 \( AC \) 边时
\( 64b^2r_0^4+16(a+c)(b-c+a)(-c+a-b)(a-b+c)r_0^2+(b-c+a)^2(-c+a-b)^2(a-b+c)^2=0 \)

B3.当两等圆均相切于 \( AB \) 边时
\( 64c^2r_0^4+16(a+b)(b-c+a)(-c+a-b)(a-b+c)r_0^2+(b-c+a)^2(-c+a-b)^2(a-b+c)^2=0 \)


360截图20140209155400659.png

毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-9 18:37:19 | 显示全部楼层
关于本贴的第二问:
通过相当长时间的消元计算得到,具体结果见附件,有谁有兴趣试着分解一下,我
觉得答案可以简洁很多。
001 - 副本.rar (558.95 KB, 下载次数: 13)
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-14 22:13:46 | 显示全部楼层
楼上的结果由下列方程消元得到:

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

经过分解楼上的结果得到了如下简洁的答案:

\((a+b+c)^4(-c+a+b)^6(-c+a-b)^6(c+a-b)^6+8(9a^2+10ab+10ac+9b^2+10bc+9c^2)(a+b+c)^3(-c+a+b)^5(-c+a-b)^5(c+a-b)^5r^2+16(125a^4+268a^3b+268a^3c+382a^2b^2+580a^2bc+382a^2c^2+268ab^3+580ab^2c+580abc^2+268ac^3+125b^4+268b^3c+382b^2c^2+
268bc^3+125c^4)(a+b+c)^2(-c+a+b)^4(-c+a-b)^4(c+a-b)^4r^4+1024(a+b+c)(24a^6+73a^5b+73a^5c+152a^4b^2+248a^4bc+152a^4c^2+
190a^3b^3+447a^3b^2c+447a^3bc^2+190a^3c^3+152a^2b^4+447a^2b^3c+630a^2b^2c^2+447a^2bc^3+152a^2c^4+73ab^5+
248ab^4c+447ab^3c^2+447ab^2c^3+248abc^4+73ac^5+24b^6+73b^5c+152b^4c^2+190b^3c^3+152b^2c^4+73bc^5+24c^6)(-c+a+b)^3
(-c+a-b)^3(c+a-b)^3r^6+1024(101a^8+374a^7b+374a^7c+1244a^6b^2+2018a^6bc+1244a^6c^2+2426a^5b^3+5814a^5b^2c+
5814a^5bc^2+2426a^5c^3+2974a^4b^4+9202a^4b^3c+13188a^4b^2c^2+9202a^4bc^3+2974a^4c^4+2426a^3b^5+9202a^3b^4c+
17012a^3b^3c^2+17012a^3b^2c^3+9202a^3bc^4+2426a^3c^5+1244a^2b^6+5814a^2b^5c+13188a^2b^4c^2+17012a^2b^3c^3+
13188a^2b^2c^4+5814a^2bc^5+1244a^2c^6+374ab^7+2018ab^6c+5814ab^5c^2+9202ab^4c^3+9202ab^3c^4+5814ab^2c^5+
2018abc^6+374ac^7+101b^8+374b^7c+1244b^6c^2+2426b^5c^3+2974b^4c^4+2426b^3c^5+1244b^2c^6+374bc^7+101c^8)(-c+a+b)^2
(-c+a-b)^2(c+a-b)^2r^8-16384(-c+a+b)(c+a-b)(-c+a-b)(6a^9+33a^8b+33a^8c-129a^7b^2+24a^7bc-129a^7c^2-383a^6b^3-474a^6b^2c-
474a^6bc^2-383a^6c^3-551a^5b^4-1808a^5b^3c-1944a^5b^2c^2-1808a^5bc^3-551a^5c^4-551a^4b^5-2718a^4b^4c-4109a^4b^3c^2-
4109a^4b^2c^3-2718a^4bc^4-551a^4c^5-383a^3b^6-1808a^3b^5c-4109a^3b^4c^2-4920a^3b^3c^3-4109a^3b^2c^4-1808a^3bc^5-
383a^3c^6-129a^2b^7-474a^2b^6c-1944a^2b^5c^2-4109a^2b^4c^3-4109a^2b^3c^4-1944a^2b^2c^5-474a^2bc^6-129a^2c^7+33ab^8+
24ab^7c-474ab^6c^2-1808ab^5c^3-2718ab^4c^4-1808ab^3c^5-474ab^2c^6+24abc^7+33ac^8+6b^9+33b^8c-129b^7c^2-383b^6c^3-
551b^5c^4-551b^4c^5-383b^3c^6-129b^2c^7+33bc^8+6c^9)r^{10}+(262144a^{10}+262144a^9b+262144a^9c-3866624a^8b^2-262144a^8bc-
3866624a^8c^2-20054016a^7b^3-9306112a^7b^2c-9306112a^7bc^2-20054016a^7c^3+3604480a^6b^4-2490368a^6b^3c+65732608a^6b^2c^2-
2490368a^6bc^3+3604480a^6c^4+39583744a^5b^5+11796480a^5b^4c+91881472a^5b^3c^2+91881472a^5b^2c^3+11796480a^5bc^4+
39583744a^5c^5+3604480a^4b^6+11796480a^4b^5c+80216064a^4b^4c^2+362676224a^4b^3c^3+80216064a^4b^2c^4+11796480a^4bc^5+
3604480a^4c^6-20054016a^3b^7-2490368a^3b^6c+91881472a^3b^5c^2+362676224a^3b^4c^3+362676224a^3b^3c^4+91881472a^3b^2c^5-
2490368a^3bc^6-20054016a^3c^7-3866624a^2b^8-9306112a^2b^7c+65732608a^2b^6c^2+91881472a^2b^5c^3+80216064a^2b^4c^4+
91881472a^2b^3c^5+65732608a^2b^2c^6-9306112a^2bc^7-3866624a^2c^8+262144ab^9-262144ab^8c-9306112ab^7c^2-2490368ab^6c^3+
11796480ab^5c^4+11796480ab^4c^5-2490368ab^3c^6-9306112ab^2c^7-262144abc^8+262144ac^9+262144b^{10}+262144b^9c-3866624b^8c^2-20054016b^7c^3+3604480b^6c^4+39583744b^5c^5+3604480b^4c^6-20054016b^3c^7-3866624b^2c^8+262144bc^9+
262144c^{10})r^{12}+(9437184a^6b^2+9437184a^6c^2+9437184a^5b^2c+9437184a^5bc^2-18874368a^4b^4-9437184a^4b^3c-
113246208a^4b^2c^2-9437184a^4bc^3-18874368a^4c^4-9437184a^3b^4c-452984832a^3b^3c^2-452984832a^3b^2c^3-9437184a^3bc^4+
9437184a^2b^6+9437184a^2b^5c-113246208a^2b^4c^2-452984832a^2b^3c^3-113246208a^2b^2c^4+9437184a^2bc^5+9437184a^2c^6+
9437184ab^5c^2-9437184ab^4c^3-9437184ab^3c^4+9437184ab^2c^5+9437184b^6c^2-18874368b^4c^4+9437184b^2c^6)r^{14}+339738624a^2b^2c^2r^{16}=0 \)



毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-2-14 22:35:36 | 显示全部楼层
现举例计算来验证楼上的结论的是否正确(与23#计算结果对比即可)
1. 取   \( a = 5, b = 12, c = 8 \)
  \(78275778969600r^{16}-25088032663142400r^{14}+2026054624326451200r^{12}-26077679761320345600r^{10}+100412030199497241600r^8-135768794249894400000r^6+81948404811881250000r^4-22739654569921875000r^2+2364627008056640625=0 \)
取最小正实根 \( r =0.557928817455220\)

2.取 \( a = 5, b = 12, c = 9\)
\(99067782758400r^{16}-35002884487643136r^{14}+3139645196732465152r^{12}-71948685392562094080r^{10}+493460529030713311232r^8-1212323906176294060032r^6+1335383159057948868608r^4-677148573597475799040r^2+128627308957328736256=0\)
取最小正实根得\(r=0.745703010559392\)

3.取 \( a = 5, b = 12, c = 10\)
\(122305904640000r^{16}-47148034424832000r^{14}+4648204904354021376r^{12}-143102337111313219584r^{10}+1298972942870607922176r^8-4275658653825212688384r^6+6325482740192497077840r^4-4310784974826787654008r^2+1100181539196023252409=0\)
取最小正实根得\(r=0.859897963713285\)

计算结果正确
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-4-19 11:19:35 | 显示全部楼层
关于在东论上‘天下无毒史’给出的一个图:

20140415215849750.gif

并不具备本题一般特性的说明:

A. 若三角形一内点\(P\)与各顶点构成的\(\triangle APC,\triangle APB,\triangle BPC\)的内切圆均两两相切于外公切线\(AP,BP,CP\),且切于同一点? (并不正确)

我们现在做的就是若A成立,能求出三个内切圆的半径\(r_1,r_2,r_3\) 吗?

其实我们可设\(PA=t+x,PB=t+y,PC=t+z\),则有结论:

\(x=\frac{b+c-a}{2},y=\frac{a+c-b}{2},z=\frac{b+c-a}{2}\)

\(r_1=\sqrt{\frac{yzt}{y+z+t}},r_2=\sqrt{\frac{xzt}{x+z+t}},r_3=\sqrt{\frac{xyt}{x+y+t}}\)

很容易得到如下结果:

\((b+a-c)^2(a+c-b)^2(-b-c+a)^2-(4(b+a-c))(a+c-b)(-b-c+a)(a^2-2ab-2ac+b^2-2bc+c^2)t+(20a^4-16a^3b-16a^3c-8a^2b^2+16a^2bc-8a^2c^2-16ab^3+16ab^2c+16abc^2-16ac^3+20b^4-16b^3c-8b^2c^2-16bc^3+20c^4)t^2=0\)

\((-b-c+a)^2(b+a-c)^4(a+c-b)^4+16(5a^3-3a^2b-3a^2c-ab^2+2abc-ac^2-b^3+b^2c+bc^2-c^3)^2r_1^4-(8(-b-c+a))(3a^3-5a^2b-5a^2c+ab^2-2abc+ac^2+b^3-b^2c-bc^2+c^3)(b+a-c)^2(a+c-b)^2r_1^2=0\)

\((a+c-b)^2(-b-c+a)^4(b+a-c)^4+16(a^3+a^2b-a^2c+3ab^2-2abc-ac^2-5b^3+3b^2c+bc^2+c^3)^2r_2^4+(8(a+c-b))(a^3+a^2b-a^2c-5ab^2-2abc-ac^2+3b^3-5b^2c+bc^2+c^3)(-b-c+a)^2(b+a-c)^2r_2^2=0\)

\((b+a-c)^2(-b-c+a)^4(a+c-b)^4+16(a^3-a^2b+a^2c-ab^2-2abc+3ac^2+b^3+b^2c+3bc^2-5c^3)^2r_3^4+(8(b+a-c))(a^3-a^2b+a^2c-ab^2-2abc-5ac^2+b^3+b^2c-5bc^2+3c^3)(-b-c+a)^2(a+c-b)^2r_3^2=0\)


360截图20140419111910347.png
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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