三角形内点分成三个三角形的极值问题
有关三角形内点P与\(\triangle ABC\)三个顶点构成的三个\(\triangle PBC,\triangle PAC,\triangle PAB\)的外接圆半径,内切圆半径分别为\(R_1,R_2,R_3,r_1,r_2,r_3\),且记\(PA=x,PB=y,PC=z,AB=c,AC=b,BC=a\),\(\triangle ABC\)外接圆,内切圆半径及面积分别记为\(R,r,s\),
\(\triangle PBC,\triangle PAC,\triangle PAB\)的面积分别记为\(s_1,s_2,s_3,\alpha=\angle PBC,\beta=\angle PAC,\gamma=\angle PAB\),
\(u=\cos(\alpha),v=\cos(\beta),w=\cos(\gamma)\)
我们现在讨论下面两个问题:
1.\(R_0=R_1+R_2+R_3\)的最小值如何?
2.\(r_0=r_1+r_2+r_3\)的最小值如何?
对于第1个问题的讨论见:
https://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=15716&fromuid=1455
我们易知:\(R_0=\frac{a}{2\sin(\alpha)}+\frac{b}{2\sin(\beta)}+\frac{c}{2\sin(\gamma)}\),且有\(\cos(\alpha)^2+\cos(\beta)^2+\cos(\gamma)^2-2\cos(\alpha)\cos(\beta)\cos(\gamma)-1=0\)
根据拉格朗日乘子法可以得到:
\(\frac{a\cos(\alpha)}{1-\cos(\alpha)^2}=k\sin(\alpha)(\cos(\alpha)-\cos(\beta)\cos(\gamma))\)
\(\frac{b\cos(\beta)}{1-\cos(\beta)^2}=k\sin(\beta)(\cos(\beta)-\cos(\alpha)\cos(\gamma))\)
\(\frac{c\cos(\gamma)}{1-\cos(\gamma)^2}=k\sin(\gamma)(\cos(\gamma)-\cos(\alpha)\cos(\beta))\)
我们现在证明:
\(\frac{a\cos(\alpha)}{1-\cos(\alpha)^2}=\frac{b\cos(\beta)}{1-\cos(\beta)^2}=\frac{c\cos(\gamma)}{1-\cos(\gamma)^2}\)
即只需要证明:
\(\sin(\alpha)(\cos(\alpha)-\cos(\beta)\cos(\gamma))-\sin(\beta)(\cos(\beta)-\cos(\alpha)\cos(\gamma))=0\)
展开左边有
\(\frac{\sin(2\alpha)-\sin(2\beta)}{2}-\cos(\gamma)(\sin(\alpha)\cos(\beta)-\sin(\beta)\cos(\alpha))\)
\(=\sin(\alpha-\beta)\cos(\alpha+\beta)-\cos(\alpha+\beta)\sin(\alpha-\beta)=0\)
同理可证:
\(\sin(\alpha)(\cos(\alpha)-\cos(\beta)\cos(\gamma))-\sin(\gamma)(\cos(\gamma)-\cos(\alpha)\cos(\beta))=0\)
即可以得到\(R_0\)取极值条件为:
\(\frac{au}{1-u^2}=\frac{bv}{1-v^2}=\frac{cw}{1-w^2}=\frac{1}{h}\)\(h\)为恒等参数
我们进一步有
\(1-u^2=auh\)
\(1-v^2=bvh\)
\(1-w^2=cwh\)
\(u^2+v^2+w^2-2uvw-1=0\)
消元易得:
\(-(a^2+2ab+2ac+b^2-2bc+c^2)(a^2-2ab+2ac+b^2+2bc+c^2)(a^2-2ab-2ac+b^2-2bc+c^2)(a^2+2ab-2ac+b^2+2bc+c^2)+8abc(a^6-a^4b^2-a^4c^2-a^2b^4-22a^2b^2c^2-a^2c^4+b^6-b^4c^2-b^2c^4+c^6)h+2a^2b^2c^2(9a^4+46a^2b^2+46a^2c^2+9b^4+46b^2c^2+9c^4)h^2+4abc(a^6b^2+a^6c^2-2a^4b^4+3a^4b^2c^2-2a^4c^4+a^2b^6+3a^2b^4c^2+3a^2b^2c^4+a^2c^6+b^6c^2-2b^4c^4+b^2c^6)h^3+a^2b^2c^2(3a^6+5a^4b^2+5a^4c^2+5a^2b^4-43a^2b^2c^2+5a^2c^4+3b^6+5b^4c^2+5b^2c^4+3c^6)h^4+4a^3b^3c^3(a^4+7a^2b^2+7a^2c^2+b^4+7b^2c^2+c^4)h^5+a^2b^2c^2(a^6b^2+a^6c^2-2a^4b^4-8a^4b^2c^2-2a^4c^4+a^2b^6-8a^2b^4c^2-8a^2b^2c^4+a^2c^6+b^6c^2-2b^4c^4+b^2c^6)h^6+2a^3b^3c^3(b^2+c^2)(a^2+c^2)(a^2+b^2)h^7=0\)
\((-2a^4bc-2a^2b^3c-2a^2bc^3-2b^3c^3)u^7+(a^6-2a^4b^2-2a^4c^2+a^2b^4-5a^2b^2c^2+a^2c^4+b^4c^2+b^2c^4)u^6+(-10a^4bc+2a^2b^3c+2a^2bc^3+6b^3c^3)u^5+(14a^2b^2c^2-3b^4c^2-3b^2c^4)u^4+(-4a^2b^3c-4a^2bc^3-6b^3c^3)u^3+(-3a^2b^2c^2+3b^4c^2+3b^2c^4)u^2+2b^3c^3u-c^2b^4-c^4b^2=0\)
\((-2a^3b^2c-2a^3c^3-2ab^4c-2ab^2c^3)v^7+(a^4b^2+a^4c^2-2a^2b^4-5a^2b^2c^2+a^2c^4+b^6-2b^4c^2+b^2c^4)v^6+(2a^3b^2c+6a^3c^3-10ab^4c+2ab^2c^3)v^5+(-3a^4c^2+14a^2b^2c^2-3a^2c^4)v^4+(-4a^3b^2c-6a^3c^3-4ab^2c^3)v^3+(3a^4c^2-3a^2b^2c^2+3a^2c^4)v^2+2a^3c^3v-c^2a^4-c^4a^2=0\)
\((-2a^3b^3-2a^3bc^2-2ab^3c^2-2abc^4)w^7+(a^4b^2+a^4c^2+a^2b^4-5a^2b^2c^2-2a^2c^4+b^4c^2-2b^2c^4+c^6)w^6+(6a^3b^3+2a^3bc^2+2ab^3c^2-10abc^4)w^5+(-3a^4b^2-3a^2b^4+14a^2b^2c^2)w^4+(-6a^3b^3-4a^3bc^2-4ab^3c^2)w^3+(3a^4b^2+3a^2b^4-3a^2b^2c^2)w^2+2a^3b^3w-b^2a^4-b^4a^2=0\)
为了便于消元得到\(x,y,z\)的单变量代数方程,我们可以利用下面方程
\(a^4b^3x-a^4bx^3-a^4bxz^2-a^3b^4y+2a^3b^2x^2y+2a^3b^2yz^2-a^3x^4y+2a^3x^2yz^2-a^3yz^4-2a^2b^3xy^2-2a^2b^3xz^2+2a^2bx^3y^2+2a^2bx^3z^2+2a^2bxy^2z^2+2a^2bxz^4+ab^4y^3+ab^4yz^2-2ab^2x^2y^3-2ab^2x^2yz^2-2ab^2y^3z^2-2ab^2yz^4+ax^4y^3+ax^4yz^2-2ax^2y^3z^2-2ax^2yz^4+ay^3z^4+ayz^6+b^3xy^4-2b^3xy^2z^2+b^3xz^4-bx^3y^4+2bx^3y^2z^2-bx^3z^4-bxy^4z^2+2bxy^2z^4-bxz^6=0\)
\(a^4c^3x-a^4cx^3-a^4cxy^2-a^3c^4z+2a^3c^2x^2z+2a^3c^2y^2z-a^3x^4z+2a^3x^2y^2z-a^3y^4z-2a^2c^3xy^2-2a^2c^3xz^2+2a^2cx^3y^2+2a^2cx^3z^2+2a^2cxy^4+2a^2cxy^2z^2+ac^4y^2z+ac^4z^3-2ac^2x^2y^2z-2ac^2x^2z^3-2ac^2y^4z-2ac^2y^2z^3+ax^4y^2z+ax^4z^3-2ax^2y^4z-2ax^2y^2z^3+ay^6z+ay^4z^3+c^3xy^4-2c^3xy^2z^2+c^3xz^4-cx^3y^4+2cx^3y^2z^2-cx^3z^4-cxy^6+2cxy^4z^2-cxy^2z^4=0\)
\((-a^2+y^2+z^2)^2x^2+(-b^2+x^2+z^2)^2y^2+(-c^2+x^2+y^2)^2z^2-(-a^2+y^2+z^2)(-b^2+x^2+z^2)(-c^2+x^2+y^2)-4x^2y^2z^2=0\)
最终的消元结果待续~~~
若有好的消元方案可以一起讨论,从数值消元可知消元结果为28次并不算太高,应该可以得到一般表达式
例如:
取{a=3,b=4,c=5},我们可以算得
\({R_0=6.898795286,x=3.485335210,y=2.395859245,z=0.8909132886,\cos(\alpha)=-0.5776829955,\cos(\beta)=-0.4925253609,\cos(\gamma)=-0.4258670424}\)
\(38804609882869022500x^{28}-5769485756827601847500x^{26}+393138060222049195200625x^{24}-16320682813152076823035000x^{22}+463653143417568250333340000x^{20}-9609746953227137796801640000x^{18}+151293078416735563653003660000x^{16}-1856668935099645524429163033600x^{14}+18003098828684466037695596755200x^{12}-138027089888944019201338176563200x^{10}+825209182319366667404685126867200x^8-3726676721096716852024679509657600x^6+11986384188587402057670916015058944x^4-24507560983281671057859293612441600x^2+23976576307688052249324314296385536=0\)
\(38804609882869022500y^{28}-3189787823184865402500y^{26}+116548721482561286450625y^{24}-2505946398458082355777500y^{22}+35765851378489362027933750y^{20}-362014105412889878024610000y^{18}+2711099362083700258637769375y^{16}-15461814518509623274829568600y^{14}+68376026565615242810947789800y^{12}-236093780268418695628141685700y^{10}+632439032082478712664590192175y^8-1281841319595705601897417265100y^6+1858661914759093591827305648094y^4-1711325757223338608021202426600y^2+738848778065271518071350228561=0\)
\(3700102357484040912746042377022500z^{28}-222141019615667934081165982765687500z^{26}+7526379952815463594033533436613000625z^{24}-178155426640721124773822346161719875000z^{22}+3221561270664112445228132548975229580000z^{20}-46470697664778962444371357451826333000000z^{18}+546123613768454868224799034207941442380000z^{16}-5279369429419513180733959475719314282393600z^{14}+41953846158198634281896856808872590551200000z^{12}-271580844698385579322691664563327363976345600z^{10}+1403517178219554049698950839674955257693600000z^8-5583942726231558785011323350990690254705920000z^6+16079914923753276429043726090109273293654585344z^4-29925656328696304226382721508764224573193420800z^2+27664690363821889473057186325714671075590209536=0\)
关于第2问题:
我们易知:\(r_0=\frac{2s_1}{y+z+a}+\frac{2s_2}{x+z+b}+\frac{2s_3}{x+y+c}\),且有\(s_1+s_2+s_3-s=0\)
根据拉格朗日乘子法可以得到:
\(\frac{x(b^2+z^2-x^2)}{2s_2(x+z+b)}+\frac{x(c^2+y^2-x^2)}{2s_3(x+y+c)}-\frac{4s_2}{(x+z+b)^2}-\frac{4s_3}{(x+y+c)^2}=kx(\frac{b^2+z^2-x^2}{4s_2}+\frac{c^2+y^2-x^2}{4s_3})\)
\(\frac{y(a^2+z^2-y^2)}{2s_1(y+z+a)}+\frac{y(c^2+x^2-y^2)}{2s_3(x+y+c)}-\frac{4s_1}{(y+z+a)^2}-\frac{4s_3}{(x+y+c)^2}=ky(\frac{a^2+z^2-y^2}{4s_1}+\frac{c^2+x^2-y^2}{4s_3})\)
\(\frac{z(a^2+y^2-z^2)}{2s_1(y+z+a)}+\frac{z(b^2+x^2-z^2)}{2s_2(x+z+b)}-\frac{4s_1}{(y+z+a)^2}-\frac{4s_2}{(x+z+b)^2}=kz(\frac{a^2+y^2-z^2}{4s_1}+\frac{b^2+x^2-z^2}{4s_2})\)
上面三式相加可以得到:
\(k\)值
其中
\(16s_1^2=-a^4+2a^2y^2+2a^2z^2-y^4+2y^2z^2-z^4\)
\(16s_2^2=-b^4+2b^2x^2+2b^2z^2-x^4+2x^2z^2-z^4\)
\(16s_3^2=-c^4+2c^2x^2+2c^2y^2-x^4+2x^2y^2-y^4\)
谁有兴趣简化上面的极值条件并检验是否正确?? 经过近一个星期的计算，终于获得问题1的单变量代数方程，次数为28次，结果保存在TXT中有3.19MB（仅关于x的代数方程）,
有兴趣可见下面的附件
对于问题1的逆问题：
平面共点的三条线段\(PA=x,PB=y,PC=z\),可自由的绕\(P\)点旋转，求三个三角形\(\triangle PBC,\triangle PAC,\triangle PAB\)的三个外接圆半径和\(R_0=R_1+R_2+R_3\)何时取得极值？
根据拉格朗日乘子法可以得到：
\(R_0=R_1+R_2+R_3=\frac{a}{2\sin(\alpha)}+\frac{b}{2\sin(\beta)}+\frac{c}{2\sin(\gamma)}=\frac{\sqrt{y^2+z^2-2yz\cos(\alpha)}}{2\sin(\alpha)}+\frac{\sqrt{x^2+z^2-2xz\cos(\beta)}}{2\sin(\beta)}+\frac{\sqrt{x^2+y^2-2xy\cos(\gamma)}}{2\sin(\gamma)}\)
取极值的条件为：
\(\frac{yz}{a}-\frac{a\cos(\alpha)}{1-\cos(\alpha)^2}=\frac{xz}{b}-\frac{b\cos(\beta)}{1-\cos(\beta)^2}=\frac{xy}{c}-\frac{c\cos(\gamma)}{1-\cos(\gamma)^2}\)
我们可以得到如下方程组：
\(-ahu^2+u^2yz+a^2u+ah-yz=0\)
\(-bhv^2+v^2xz+b^2v+bh-xz=0\)
\(-chw^2+w^2xy+c^2w+ch-xy=0\)
\(-2uvw+u^2+v^2+w^2-1=0\)
\(a^2+(2uyz-y^2-z^2)=0\)
\(b^2+(2vxz-x^2-z^2)=0\)
\(c^2+(2wxy-x^2-y^2)=0\)
例如我们取\(x=3,y=4,z=5\) 可以解得：
\(a = 7.665328635, b = 7.004928257, c = 6.184167480\)
\(h = 6.847258521, u = -0.4439315770, v = -0.5023006628, w = -0.5518303093\)
\(\alpha = 116.3550016^\circ, \beta = 120.1523277^\circ, \gamma = 123.4926706^\circ\)
\(R_1=4.277237438, R_2=4.050528903, R_3=3.707728206, R_0=12.03549455\)
关于第二个问题的逆问题：
平面共点的三条线段\(PA=x,PB=y,PC=z\),可自由绕P点旋转，求三个三角形\(\triangle PBC,\triangle PAC,\triangle PAB\)的三个内切圆半径和\(r_0=r_1+r_2+r_3\)何时取得极值?
根据拉格朗日乘子法可以得到：
\(r_0=r_1+r_2+r_3=\frac{yz\sin(\alpha)}{y+z+a}+\frac{xz\sin(\beta)}{x+z+b}+\frac{xy\sin(\gamma)}{x+y+c}=\frac{yz\sin(\alpha)}{y+z+\sqrt{y^2+z^2-2yz\cos(\alpha)}}+\frac{xz\sin(\beta)}{x+z+\sqrt{x^2+z^2-2xz\cos(\beta)}}+\frac{xy\sin(\gamma)}{x+y+\sqrt{x^2+y^2-2xy\cos(\gamma)}}\)
取极值条件为：
\(\frac{yz\cos(\alpha)}{a+y+z}-\frac{y^2z^2\sin(\alpha)^2}{(a+y+z)^2a}=\frac{xz\cos(\beta)}{b+x+z}-\frac{x^2z^2\sin(\beta)^2}{(b+x+z)^2b}=\frac{xy\cos(\gamma)}{c+x+y}-\frac{x^2y^2\sin(\gamma)^2}{(c+x+y)^2c}=k\)
即可以得到如下方程组：
\(a^4+4a^3k+2a^3y+2a^3z+8a^2ky+8a^2kz+4aky^2+8akyz+4akz^2-2ay^3-2ay^2z-2ayz^2-2az^3-y^4+2y^2z^2-z^4=0\)
\(b^4+4b^3k+2b^3x+2b^3z+8b^2kx+8b^2kz+4bkx^2+8bkxz+4bkz^2-2bx^3-2bx^2z-2bxz^2-2bz^3-x^4+2x^2z^2-z^4=0\)
\(c^4+4c^3k+2c^3x+2c^3y+8c^2kx+8c^2ky+4ckx^2+8ckxy+4cky^2-2cx^3-2cx^2y-2cxy^2-2cy^3-x^4+2x^2y^2-y^4=0\)
\((-a^2+y^2+z^2)^2x^2+(-b^2+x^2+z^2)^2y^2+(-c^2+x^2+y^2)^2z^2-(-a^2+y^2+z^2)(-b^2+x^2+z^2)(-c^2+x^2+y^2)-4x^2y^2z^2=0\)
例如：
取\(x = 3, y = 4, z = 5\) 可以解得：
\(a = 7.480144776, \alpha = 111.9510388^\circ, b = 7.047189126, \beta = 121.4729402^\circ, c = 6.269077444, \gamma= 126.5760210^\circ\)
\(k = -0.6230328092, r_0= 2.702073658, r_1= 1.125601313, r_2=0 .8502121180, r_3=0 .7262602267\) 进一步，我们可以得到四边形，五边形，N边形的外接圆半径和及内切圆半径和的极值条件？
但是，如何进一步向三维推广，值得我们继续探讨！！
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