三线共点的另一类极值问题

数学星空

我们已讨论过如下问题:

1.若三角形\(\triangle ABC\)内有一点\(P\),求\(AP^n+BP^n+CP^n\)取最小值的条件？

对于\(n =1,2\)我们已知道了完美的结论，那对于\(n=3,4,5…\)取最小值的条件是什么？

具体讨论及结果见:http://bbs.emath.ac.cn/forum.php ... 09&fromuid=1455


2.已知平面上有三个动点$A,B,C$及一个定点$P$,且满足$AP=x,BP=y,CP=z$.
(1)求三角形ABC的最大面积值$S$?
(2)求三角形各边长之和($L=a+b+c$)的最大值?

具体讨论及结果见:http://bbs.emath.ac.cn/forum.php ... 27&fromuid=1455



现在我们来讨论问题2的推广问题

3.已知平面上有三个动点$A,B,C$及一个定点$P$,且满足$AP=x,BP=y,CP=z$.
(1)求三角形\(\triangle ABC\)各边长倒数和\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)的最小值?
(2)求三角形\(\triangle ABC\)各边长之幂和$L(n)=a^n+b^n+c^n$取最小值的条件?(其中\(n \in Z\) )

creasson

数学星空

楼上给出的恒等式似乎有误,正确的恒等式可参见:

http://bbs.emath.ac.cn/forum.php ... 05&fromuid=1455

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

备注:

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

\(＝(2yz\cos(\alpha))^2x^2+(2xz\cos(\beta))^2y^2+(2xy\cos(\gamma))^2z^2-2yz\cos(\alpha)2xz\cos(\beta)2xy\cos(\gamma)-4x^2y^2z^2\)

\(=4x^2y^2z^2((\cos(\alpha))^2+(\cos(\beta))^2+(\cos(\gamma))^2-2\cos(\alpha)\cos(\beta)\cos(\gamma)-1)\)

即本质上就是\( \alpha+\beta+\gamma=2\pi\) 条件下

\((\cos(\alpha))^2+(\cos(\beta))^2+(\cos(\gamma))^2-2\cos(\alpha)\cos(\beta)\cos(\gamma)-1\)

\(＝(\cos(\alpha))^2+(\cos(\beta))^2+(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))^2-2\cos(\alpha)\cos(\beta)(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))-1\)

\(=(\cos(\alpha))^2+(\cos(\beta))^2+(\cos(\alpha))^2(\cos(\beta))^2+(1-(\cos(\alpha))^2)(1-(\cos(\beta))^2)-2(\cos(\alpha))^2(\cos(\beta))^2-1\)

\(=0\)

数学星空

对于第1问题,利用拉格朗日乘子法可以得到:

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

\(4a^5\lambda x^2+2a^3b^2c^2\lambda -2a^3b^2\lambda x^2-2a^3b^2\lambda y^2-2a^3c^2\lambda x^2-2a^3c^2\lambda z^2+2a^3\lambda x^4-2a^3\lambda x^2y^2-2a^3\lambda x^2z^2+2a^3\lambda y^2z^2-1=0\)

\(2a^2b^3c^2\lambda -2a^2b^3\lambda x^2-2a^2b^3\lambda y^2+4b^5\lambda y^2-2b^3c^2\lambda y^2-2b^3c^2\lambda z^2-2b^3\lambda x^2y^2+2b^3\lambda x^2z^2+2b^3\lambda y^4-2b^3\lambda y^2z^2-1=0\)

\(2a^2b^2c^3\lambda -2a^2c^3\lambda x^2-2a^2c^3\lambda z^2-2b^2c^3\lambda y^2-2b^2c^3\lambda z^2+4c^5\lambda z^2+2c^3\lambda x^2y^2-2c^3\lambda x^2z^2-2c^3\lambda y^2z^2+2c^3\lambda z^4-1=0\)

\(sabc-ab-ac-bc=0\)

消去变元\(a,b,c,\lambda\)即可得到我们需要的结果

数学星空

仿照问题1的求解方法:

已知平面上有三个动点$A,B,C$及一个定点$P$,且满足$AP=x,BP=y,CP=z$.

(1)求三角形\(\triangle ABC\)各边长倒数和\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)的最小值?

(2)求三角形\(\triangle ABC\)各边长之幂和$L(n)=a^n+b^n+c^n$取最小值的条件?(其中\(n \in Z\) )

对于第(2)有一般性的取最小值条件:

\[\frac{\sin(\alpha)}{x}a^{n-1}=\frac{\sin(\beta)}{y}b^{n-1}=\frac{\sin(\gamma)}{z}c^{n-1}\]

\[(\frac{\sin(\alpha)}{x})^2(y^2+z^2-2yz\cos(\alpha))^{n-1}=(\frac{\sin(\beta)}{y})^2(x^2+z^2-2xz\cos(\beta))^{n-1}=(\frac{\sin(\gamma)}{z})^2(x^2+y^2-2xy\cos(\gamma))^{n-1}\]

对于第(1),只需取\(n=-1\)即可得到

\[\frac{\frac{\sin(\alpha)}{x}}{y^2+z^2-2yz\cos(\alpha)}=\frac{\frac{\sin(\beta)}{y}}{x^2+z^2-2xz\cos(\beta)}=\frac{\frac{\sin(\gamma)}{z}}{x^2+y^2-2xy\cos(\gamma)}\]

\[\alpha+\beta+\gamma=2\pi\]

\[2s=xy\sin(\gamma)+xz\sin(\beta)+yz\sin(\alpha)\]

数学星空

对于楼上的结论有:

\(\frac{-m^2+1}{x^2}-k(-2myz+y^2+z^2)^2=0\)

\(\frac{-n^2+1}{y^2}-k(-2nxz+x^2+z^2)^2=0\)

\(\frac{-p^2+1}{z^2}-k(-2pxy+x^2+y^2)^2=0\)

\(m-\sqrt{(-n^2+1)(-p^2+1)}+np=0\)


消元结果:

\((x^2y-x^2z+xy^2-6xyz+xz^2-y^2z+yz^2)(x^2y+x^2z-xy^2-6xyz-xz^2+y^2z+yz^2)(x^2y-x^2z-xy^2+6xyz-xz^2-y^2z+yz^2)(x^2y+x^2z+xy^2+6xyz+xz^2+y^2z+yz^2)-4z^2y^2x^2(x^8y^4-18x^8y^2z^2+x^8z^4-18x^6y^6+82x^6y^4z^2+82x^6y^2z^4-18x^6z^6+x^4y^8+82x^4y^6z^2+2170x^4y^4z^4+82x^4y^2z^6+x^4z^8-18x^2y^8z^2+82x^2y^6z^4+82x^2y^4z^6-18x^2y^2z^8+y^8z^4-18y^6z^6+y^4z^8)k-16x^4y^4z^4(x^8y^4+14x^8y^2z^2+x^8z^4+14x^6y^6-270x^6y^4z^2-270x^6y^2z^4+14x^6z^6+x^4y^8-270x^4y^6z^2-1542x^4y^4z^4-270x^4y^2z^6+x^4z^8+14x^2y^8z^2-270x^2y^6z^4-270x^2y^4z^6+14x^2y^2z^8+y^8z^4+14y^6z^6+y^4z^8)k^2+64x^6y^6z^6(5x^8y^4-26x^8y^2z^2+5x^8z^4-26x^6y^6-166x^6y^4z^2-166x^6y^2z^4-26x^6z^6+5x^4y^8-166x^4y^6z^2-414x^4y^4z^4-166x^4y^2z^6+5x^4z^8-26x^2y^8z^2-166x^2y^6z^4-166x^2y^4z^6-26x^2y^2z^8+5y^8z^4-26y^6z^6+5y^4z^8)k^3+1024y^8z^8x^8(y^2+z^2)^2(x^2+z^2)^2(x^2+y^2)^2k^4=0\)

\(2(y^2+z^2)(-z+y)^2(y+z)^2(x^2+z^2)(x^2+y^2)+(-7x^4y^4-2x^4y^2z^2-7x^4z^4-13x^2y^6-3x^2y^4z^2-3x^2y^2z^4-13x^2z^6-7y^6z^2-2y^4z^4-7y^2z^6)a^2+(8x^4y^2+8x^4z^2+32x^2y^4+32x^2y^2z^2+32x^2z^4+8y^4z^2+8y^2z^4)a^4+(-4x^4-36x^2y^2-36x^2z^2-4y^2z^2)a^6+16x^2a^8=0\)

\(2(y^2+z^2)(x^2+z^2)(x-z)^2(x+z)^2(x^2+y^2)+(-13x^6y^2-7x^6z^2-7x^4y^4-3x^4y^2z^2-2x^4z^4-2x^2y^4z^2-3x^2y^2z^4-7x^2z^6-7y^4z^4-13y^2z^6)b^2+(32x^4y^2+8x^4z^2+8x^2y^4+32x^2y^2z^2+8x^2z^4+8y^4z^2+32y^2z^4)b^4+(-36x^2y^2-4x^2z^2-4y^4-36y^2z^2)b^6+16y^2b^8=0\)

\(2(y^2+z^2)(x^2+z^2)(x^2+y^2)(-y+x)^2(x+y)^2+(-7x^6y^2-13x^6z^2-2x^4y^4-3x^4y^2z^2-7x^4z^4-7x^2y^6-3x^2y^4z^2-2x^2y^2z^4-13y^6z^2-7y^4z^4)c^2+(8x^4y^2+32x^4z^2+8x^2y^4+32x^2y^2z^2+8x^2z^4+32y^4z^2+8y^2z^4)c^4+(-4x^2y^2-36x^2z^2-36y^2z^2-4z^4)c^6+16z^2c^8=0\)