三条特殊线共点的三角形

数学星空

若\(\triangle  ABC\)内存在一点\(P\),且\(\angle A\)的角平分线,\(AC\)边上的中线,\(AB\)边上的垂线，三线的交点为\(P\)，求满足此条件的三角形\(\triangle ABC\)三边\(a,b,c\)应满足的条件？
注：对于\(正\triangle ABC\),\(P\)即为中心  

上面问题已解决．

若记\(\triangle  ABC\)中\(\angle A\)的角平分线,\(AC\)边上的中线,\(AB\)边上的垂线，三线两两相交的三个交点构成\(\triangle DEF\),且面积记为\(s\),\(\triangle ABC\)面积记为\(S\),试找到\(\frac{s}{S}\)的最大值？

再进一步：

若记\(\triangle  ABC\)中\(\angle A\)的角平分线,\(AC\)边上的中线,\(AB\)边上的垂线，三线两两相交的三个交点构成\(\triangle DEF\),且周长记为\(l\),\(\triangle ABC\)面积记为\(L\),试找到\(\frac{l}{L}\)的最大值？

creasson

\[{b^2} + {c^2} - {a^2} - bc = 0\]

数学星空

设\(A(-\frac{c}{2},0),B(\frac{c}{2},0),C(x_0,y_0)\)，则有：

\(\frac{y}{x+\frac{a}{2}}=\tan(\frac{A}{2})=\sqrt{\frac{1-k}{1+k}}\)

\((y-{y_0}/2)/(x-(x_0-a/2)/2)\)\(=\frac{y}{x-\frac{a}{2}}\)

\(k=\cos(A)=\frac{b^2+c^2-a^2}{2bc}\)

\((x_0+\frac{c}{2})^2+y_0^2=b^2\)

\((x_0-\frac{c}{2})^2+y_0^2=a^2\)

最终得到：

\((a^2b+a^2c-b^3-b^2c+bc^2-c^3)(a^4-2a^2b^2-a^2bc+a^2c^2+b^4+b^3c-b^2c^2-bc^3-2c^4)＝0\)

容易验证：

\(a^2b+a^2c-b^3-b^2c+bc^2-c^3＝0\)

例：

\(\{a = 4, b = 3, c = 5.150638865, x_0 = -0.6795273540, y_0 = 2.325074708\}\)

creasson

令\[P = \alpha A + \beta B + \gamma C\left( {\alpha  + \beta  + \gamma  = 1} \right)\]
由PA平分角A知  \[{S_{\Delta PAB}}:{S_{\Delta PAC}} = \gamma :\beta  = c:b        \left( 1 \right)\]
由P在AC中线上知   \[\alpha  = \gamma                                       \left( 2 \right)\]
令D是C在AB的垂足  \[D = \lambda A + \left( {1 - \lambda } \right)B\]
由\[C{D^2} = \lambda {b^2} + \left( {1 - \lambda } \right){a^2} - \lambda \left( {1 - \lambda } \right){c^2}\]取最小值可求得
\[\lambda  = \frac{{{c^2} + {a^2} - {b^2}}}{{2{c^2}}}\]
\[D = \frac{{{c^2} + {a^2} - {b^2}}}{{2{c^2}}}A + \frac{{{c^2} + {b^2} - {a^2}}}{{2{c^2}}}B\]
P在CD上知，\[\alpha :\beta  = \frac{{{c^2} + {a^2} - {b^2}}}{{2{c^2}}}:\frac{{{c^2} + {b^2} - {a^2}}}{{2{c^2}}} = \left( {{c^2} + {a^2} - {b^2}} \right):\left( {{c^2} + {b^2} - {a^2}} \right) \left( 3 \right)\]
三式立即得到  \[\frac{c}{b} = \frac{{{c^2} + {a^2} - {b^2}}}{{{c^2} + {b^2} - {a^2}}}\]

再看第二问，三条线分别满足
\[\begin{array}{l}
{l_1}:  = {\alpha _1}A + {\beta _1}B + {\gamma _1}C \\
\left( 1 \right)\left\{ \begin{array}{l}
{\alpha _1} + {\beta _1} + {\gamma _1} = 1 \\
{\gamma _1}:{\beta _1} = c:b \\
\end{array} \right. \\
\end{array}\]

\[\begin{array}{l}
{l_2}: = {\alpha _2}A + {\beta _2}B + {\gamma _2}C \\
\left( 2 \right)\left\{ \begin{array}{l}
{\alpha _2} + {\beta _2} + {\gamma _2} = 1 \\
{\alpha _2} = {\gamma _2} \\
\end{array} \right. \\
\end{array}\]

\[\begin{array}{l}
{l_3}: = {\alpha _3}A + {\beta _3}B + {\gamma _3}C \\
\left( 3 \right)\left\{ \begin{array}{l}
{\alpha _3} + {\beta _3} + {\gamma _3} = 1 \\
{\alpha _3}:{\beta _3} = \left( {{c^2} + {a^2} - {b^2}} \right):\left( {{c^2} + {b^2} - {a^2}} \right) \\
\end{array} \right. \\
\end{array}\]

于是，可求得三线的两两交点分别为:
\[D = \frac{c}{{b + 2c}}A + \frac{b}{{b + 2c}}B + \frac{c}{{b + 2c}}C\]

\[E = \frac{{b\left( {{c^2} + {a^2} - {b^2}} \right)}}{{c\left( {{c^2} + {b^2} - {a^2} + 2bc + {c^2}} \right)}}A + \frac{{b\left( {{c^2} + {b^2} - {a^2}} \right)}}{{c\left( {{c^2} + {b^2} - {a^2} + 2bc + {c^2}} \right)}}B + \frac{{{c^2} + {b^2} - {a^2}}}{{{c^2} + {b^2} - {a^2} + 2bc}}C\]

\[F = \frac{{{c^2} + {a^2} - {b^2}}}{{3{c^2} + {a^2} - {b^2}}}A + \frac{{{c^2} + {b^2} - {a^2}}}{{3{c^2} + {a^2} - {b^2}}}B + \frac{{{c^2} + {a^2} - {b^2}}}{{3{c^2} + {a^2} - {b^2}}}C\]

于是
\[\frac{{{S_{DEF}}}}{{{S_{ABC}}}} = |\det \left( {\begin{array}{*{20}{c}}
   {\frac{c}{{b + 2c}}} & {\frac{b}{{b + 2c}}} & {\frac{c}{{b + 2c}}}  \\
   {\frac{{b\left( {{c^2} + {a^2} - {b^2}} \right)}}{{c\left( {{c^2} + {b^2} - {a^2} + 2bc + {c^2}} \right)}}} & {\frac{{b\left( {{c^2} + {b^2} - {a^2}} \right)}}{{c\left( {{c^2} + {b^2} - {a^2} + 2bc + {c^2}} \right)}}} & {\frac{{{c^2} + {b^2} - {a^2}}}{{{c^2} + {b^2} - {a^2} + 2bc}}}  \\
   {\frac{{{c^2} + {a^2} - {b^2}}}{{3{c^2} + {a^2} - {b^2}}}} & {\frac{{{c^2} + {b^2} - {a^2}}}{{3{c^2} + {a^2} - {b^2}}}} & {\frac{{{c^2} + {a^2} - {b^2}}}{{3{c^2} + {a^2} - {b^2}}}}  \\
\end{array}} \right)|\]

\[ = \frac{{{{\left( {{b^3} + {b^2}c - b{c^2} + {c^3} - {a^2}b - {a^2}c} \right)}^2}}}{{c\left( {b + 2c} \right)\left( {{a^2} - {b^2} + 3{c^2}} \right)\left( {{b^2} + {c^2} + 2bc - {a^2}} \right)}}\]

数学星空

\(a : b : c={m+n}:{1+n}:{1+m}\)

\(=2.109567946:2.750303128:1.359264818\)时，面积比值取最大值\(0.4656902104\)

其中\(m,n\)分别是下列方程的正实根：

\(m^6+12m^5+53m^4+102m^3+69m^2-10m-11=0\)

\(n^6-12n^5+53n^4-102n^3+69n^2+10n-11=0\)

数学星空

同3＃建立坐标系有：

\[2s=\begin{vmatrix}
x_0 & x_0 t+\frac{tc}{2} & 1\\
x_0 & \frac{y_0(-2x_0+c)}{-2x_0+3c} &1\\
-\frac{c(3tc-2y_0-2tx_0)}{2(3tc+2y_0-2tx_0)}& \frac{2tcy_0}{3tc+2y_0-2x_0 t} & 1
\end{vmatrix}\]

\[2S=cy_0\]

得到：

\(p=\frac{s}{S}=\frac{(4x_0 tc+4y_0 x_0-4x_0^2 t+3tc^2-2cy_0)^2}{4cy_0(3c-2x_0)(3tc+2y_0-2x_0 t)}\)

消元后得到：

\(p=\frac{(a^4-2a^2b^2-a^2bc+a^2c^2+b^4+b^3c-b^2c^2-bc^3-2c^4)^2}{c^2(b+c+a)(-b-c+a)(a^2-b^2-bc+c^2)(a^2-b^2+3c^2)}\)

\(p=\frac{(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^2}{c(2c+b)(b+c+a)(-b-c+a)(a^2-b^2+3c^2)}\)

易检验：

\(p=\frac{(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^2}{c(2c+b)(b+c+a)(-b-c+a)(a^2-b^2+3c^2)}\)

然后令\(a=y+z,b=x+z,c=x+y\),代入上式后求导得到：

\(6px^6+34px^5y+2px^5z+74px^4y^2+28px^4yz-6px^4z^2+78px^3y^3+66px^3y^2z+2px^3yz^2-2px^3z^3+40px^2y^4+56px^2y^3z+18px^2y^2z^2+8pxy^5+16pxy^4z+10pxy^3z^2+2pxy^2z^3-x^6-2x^5y-4x^5z-x^4y^2-6x^4z^2+8x^3y^2z+8x^3yz^2-4x^3z^3+4x^2y^3z+6x^2y^2z^2+8x^2yz^3-x^2z^4-8xy^3z^2+2xyz^4-4y^4z^2-4y^3z^3-y^2z^4=0\)

\(36px^5+170px^4y+10px^4z+296px^3y^2+112px^3yz-24px^3z^2+234px^2y^3+198px^2y^2z+6px^2yz^2-6px^2z^3+80pxy^4+112pxy^3z+36pxy^2z^2+8py^5+16py^4z+10py^3z^2+2py^2z^3-6x^5-10x^4y-20x^4z-4x^3y^2-24x^3z^2+24x^2y^2z+24x^2yz^2-12x^2z^3+8xy^3z+12xy^2z^2+16xyz^3-2xz^4-8y^3z^2+2yz^4=0\)

\(34px^5+148px^4y+28px^4z+234px^3y^2+132px^3yz+2px^3z^2+160px^2y^3+168px^2y^2z+36px^2yz^2+40pxy^4+64pxy^3z+30pxy^2z^2+4pxyz^3-2x^5-2x^4y+16x^3yz+8x^3z^2+12x^2y^2z+12x^2yz^2+8x^2z^3-24xy^2z^2+2xz^4-16y^3z^2-12y^2z^3-2yz^4=0\)

\(2px^5+28px^4y-12px^4z+66px^3y^2+4px^3yz-6px^3z^2+56px^2y^3+36px^2y^2z+16pxy^4+20pxy^3z+6pxy^2z^2-4x^5-12x^4z+8x^3y^2+16x^3yz-12x^3z^2+4x^2y^3+12x^2y^2z+24x^2yz^2-4x^2z^3-16xy^3z+8xyz^3-8y^4z-12y^3z^2-4y^2z^3=0\)

进一步消元,并设\(y=mx,z=nx\)得到：

\(8m^7n+20m^6n^2+18m^5n^3+7m^4n^4+m^3n^5+72m^6n+140m^5n^2+90m^4n^3+21m^3n^4+m^2n^5+12m^6+282m^5n+378m^4n^2+138m^3n^3+3m^2n^4-3mn^5+60m^5+550m^4n+442m^3n^2+42m^2n^3-15mn^4+n^5+119m^4+541m^3n+180m^2n^2-34mn^3+4n^4+117m^3+237m^2n-24mn^2-6n^3+57m^2+17mn-20n^2+11m-11n=0\)

\(8m^6n+12m^5n^2+6m^4n^3+m^3n^4+24m^5n-4m^4n^2-30m^3n^3-15m^2n^4-2mn^5-12m^5-54m^4n-186m^3n^2-120m^2n^3-15mn^4+2n^5-60m^4-290m^3n-384m^2n^2-96mn^3+9n^4-119m^3-420m^2n-276mn^2-8n^3-117m^2-258mn-62n^2-57m-58n-11=0\)

\(16m^6+32m^5n+24m^4n^2+8m^3n^3+m^2n^4+80m^5+104m^4n+36m^3n^2-2m^2n^3-2mn^4+148m^4+88m^3n-18m^2n^2-10mn^3+n^4+112m^3-30m^2n-36mn^2+4n^3+9m^2-66mn-6n^2-30m-20n-11=0\)

最终消元得到：

\(m^6+12m^5+53m^4+102m^3+69m^2-10m-11=0\)

\(n^6-12n^5+53n^4-102n^3+69n^2+10n-11=0\)

解得：

\(\{m =0 .3592648181, n = 1.750303128，a = 2.109567946, b = 2.750303128, c = 1.35926481，p=0.4656902104\}\)

数学星空

数学星空

\(l=(a^4-2a^2b^2-a^2bc+a^2c^2+b^4+b^3c-b^2c^2-bc^3-2c^4)(\frac{\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}}{c(a^2-b^2-2bc-c^2)(a^2-b^2+3c^2)}+\frac{\sqrt{b}}{\sqrt{c(-a^2+b^2+2bc+c^2)}(-a^2+b^2+bc-c^2)}+\frac{\sqrt{2a^2-b^2+2c^2}}{(a^2-b^2+3c^2)(-a^2+b^2+bc-c^2)})\)

\(L=a+b+c\)

化成代数方程：

\((b-c+a)^2(-b+c+a)^2(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^8+4c(b-c+a)(-b+c+a)(a^4b-2a^4c-2a^2b^3+4a^2b^2c-18a^2bc^2+4a^2c^3+b^5-2b^4c+6b^3c^2+4b^2c^3-15bc^4-2c^5)(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^6l^2+2c^2(3a^8b^2-4a^8bc+12a^8c^2-12a^6b^4+16a^6b^3c+12a^6b^2c^2-80a^6bc^3-16a^6c^4+18a^4b^6-24a^4b^5c-96a^4b^4c^2+224a^4b^3c^3+426a^4b^2c^4+72a^4bc^5+40a^4c^6-12a^2b^8+16a^2b^7c+108a^2b^6c^2-208a^2b^5c^3-604a^2b^4c^4+272a^2b^3c^5+892a^2b^2c^6+112a^2bc^7-80a^2c^8+3b^{10}-4b^9c-36b^8c^2+64b^7c^3+226b^6c^4-216b^5c^5-568b^4c^6+256b^3c^7+459b^2c^8-100bc^9+44c^{10})(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^4l^4+4c^3(b+2c)^2(b+c+a)(-b-c+a)(a^4b-2a^4c-2a^2b^3+4a^2b^2c+14a^2bc^2+4a^2c^3+b^5-2b^4c-10b^3c^2+4b^2c^3+17bc^4-2c^5)(a^2-b^2+3c^2)^2(a^2b+a^2c-b^3-b^2c+bc^2-c^3)^2l^6+c^4(b+2c)^4(b+c+a)^2(-b-c+a)^2(a^2-b^2+3c^2)^4l^8=0\)

若令\(a=1+x,b=1+y,c=x+y,l=2k(x+y+1)\)代入后分别对\(x,y\)求导，可以得到\(x,y,k\)值，进而也能得到\(a,b,c\)值，谁有兴趣利用数值计算得到最终答案？

1. (b-c+a)^2*(-b+c+a)^2*(a^2*b+a^2*c-b^3-b^2*c+b*c^2-c^3)^8+4*c*(b-c+a)*(-b+c+a)*(a^4*b-2*a^4*c-2*a^2*b^3+4*a^2*b^2*c-18*a^2*b*c^2+4*a^2*c^3+b^5-2*b^4*c+6*b^3*c^2+4*b^2*c^3-15*b*c^4-2*c^5)*(a^2*b+a^2*c-b^3-b^2*c+b*c^2-c^3)^6*k^2*(b+c+a)^2+2*c^2*(3*a^8*b^2-4*a^8*b*c+12*a^8*c^2-12*a^6*b^4+16*a^6*b^3*c+12*a^6*b^2*c^2-80*a^6*b*c^3-16*a^6*c^4+18*a^4*b^6-24*a^4*b^5*c-96*a^4*b^4*c^2+224*a^4*b^3*c^3+426*a^4*b^2*c^4+72*a^4*b*c^5+40*a^4*c^6-12*a^2*b^8+16*a^2*b^7*c+108*a^2*b^6*c^2-208*a^2*b^5*c^3-604*a^2*b^4*c^4+272*a^2*b^3*c^5+892*a^2*b^2*c^6+112*a^2*b*c^7-80*a^2*c^8+3*b^10-4*b^9*c-36*b^8*c^2+64*b^7*c^3+226*b^6*c^4-216*b^5*c^5-568*b^4*c^6+256*b^3*c^7+459*b^2*c^8-100*b*c^9+44*c^10)*(a^2*b+a^2*c-b^3-b^2*c+b*c^2-c^3)^4*k^4*(b+c+a)^4+4*c^3*(b+2*c)^2*(b+c+a)^7*(-b-c+a)*(a^4*b-2*a^4*c-2*a^2*b^3+4*a^2*b^2*c+14*a^2*b*c^2+4*a^2*c^3+b^5-2*b^4*c-10*b^3*c^2+4*b^2*c^3+17*b*c^4-2*c^5)*(a^2-b^2+3*c^2)^2*(a^2*b+a^2*c-b^3-b^2*c+b*c^2-c^3)^2*k^6+c^4*(b+2*c)^4*(b+c+a)^10*(-b-c+a)^2*(a^2-b^2+3*c^2)^4*k^8=0

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