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[原创] 三角形中的最值问题

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 楼主| 发表于 2015-7-10 20:09:46 | 显示全部楼层
当\(n=4\)时

设各边长依次为\(a,b,c,d\),\(P\)对各边的张角依次为

\(\alpha,\beta,\gamma,\delta\),且\(PA=x,PB=y,PC=z,PD=u\)

对面积:

\(2s=xy\sin(\alpha)+yz\sin(\beta)+zu\sin(\gamma)+ux\sin(\delta)\)

求导易得:

\(xy\cos(\alpha)=yz\cos(\beta)=zu\cos(\gamma)=ux\cos(\delta)\)

则有:

\(\alpha=\beta=\gamma=\delta=\frac{\pi}{2}\)

时\(s=\frac{1}{2}(xy+yz+zu+ux)\)取最大值


对周长:

\[L=\sqrt{x^2+y^2-2xy\cos(\alpha)}+\sqrt{y^2+z^2-2yz\cos(\beta)}+\sqrt{z^2+u^2-2zu\cos(\gamma)}+\sqrt{x^2+u^2-2xu\cos(\delta)}\]

求导易得:

\[\frac{xy\sin(\alpha)}{a}=\frac{yz\sin(\beta)}{b}=\frac{zu\sin(\gamma)}{c}=\frac{ux\sin(\delta)}{d}\]

由面积公式得:

\[\frac{s_{APB}}{a}=\frac{s_{BPC}}{b}=\frac{s_{CPD}}{c}=\frac{s_{DPA}}{d}=r\]

即\(P\)为内心,且得到\(a+c=b+d=\frac{L}{2}\)

不妨设

\[a=m+n,b=n+p,c=p+q,d=q+m\]



\[\sqrt{x^2-r^2}+\sqrt{y^2-r^2}=a\]

\[\sqrt{y^2-r^2}+\sqrt{z^2-r^2}=b\]

\[\sqrt{z^2-r^2}+\sqrt{u^2-r^2}=c\]

\[\sqrt{u^2-r^2}+\sqrt{x^2-r^2}=d\]

易知当\(r\to 0\)时,\(L \to  2(x+y+z+u)\) 取极大值
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 2015-7-10 20:20:41 | 显示全部楼层
对于\(n=4\)时,且\(PA,PB,PC,PD\)不共面时,即\(ABCD\)为四面体的各棱长和取最大值的条件见:

/forum.php ... 86&fromuid=1455

当 定长\(PA ,PB, PC, PD\) 构成的四面体各条棱边长最大时即\((AB+AC+BC+DA+DB+DC)_{\max}\),设\(\triangle ABC,\triangle DBC,\triangle DAC,\triangle DAB\)的内心分别为\(P_0,P_1,P_2,P_3\)

\(P\)点即为\(AP_1\cap BP_2 \cap CP_3  \cap DP_0\)  
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 2015-7-14 08:37:25 | 显示全部楼层
根据楼上结论可设:

\(A(x_1, y_1, z_1), B(x_2, y_2, z_3), C(x_3, y_3, z_3), D(x_0, y_0, z_0)\)

\(I_0(x_{11}, y_{11}, z_{11}), I_1(x_{10}, y_{10}, z_{10}), I_2(x_{20}, y_{20}, z_{20}), I_3(x_{30}, y_{30}, z_{30}), P(0, 0, 0)\)

\(AB = c, AC = b, AD = a_1, BC = a, BD = b_1, CD = c_1\)

\(\triangle ABC\) 内心 \(I_0\), \(\triangle ABD\) 内心 \(I_3\), \(\triangle ACD\) 内心 \(I_2\), \(\triangle BCD\) 内心 \(I_1\)

可得到下列方程:

\(k_0x_0 = \frac{ax_1+bx_2+cx_3}{a+b+c}\)

\(k_0y_0 = \frac{ay_1+by_2+cy_3}{a+b+c}\)

\(k_0z_0 = \frac{az_1+bz_2+cz_3}{a+b+c}\)

\(k_1x_1 = \frac{ax_0+b_1x_3+c_1x_2}{a+b_1+c_1}\)

\(k_1y_1 = \frac{ay_0+b_1y_3+c_1y_2}{a+b_1+c_1}\)

\(k_1z_1 = \frac{az_0+b_1z_3+c_1z_2}{a+b_1+c_1}\)

\(k_2x_2 = \frac{a_1x_3+bx_0+c_1x_1}{a_1+b+c_1}\)

\(k_2y_2 = \frac{a_1y_3+by_0+c_1y_1}{a_1+b+c_1}\)

\(k_2z_2 = \frac{a_1z_3+bz_0+c_1z_1}{a_1+b+c_1}\)

\(k_3x_3 = \frac{a_1x_2+b_1x_1+cx_0}{a_1+b_1+c}\)

\(k_3y_3 = \frac{a_1y_2+b_1y_1+cy_0}{a_1+b_1+c}\)

\(k_3z_3 = \frac{a_1z_2+b_1z_1+cz_0}{a_1+b_1+c}\)

\(x_0^2+y_0^2+z_0^2 = u^2\)

\(x_1^2+y_1^2+z_1^2 = x^2\)

\(x_2^2+y_2^2+z_2^2 = y^2\)

\(x_3^2+y_3^2+z_3^2 = z^2\)

\(a^2 = (x_2-x_3)^2+(y_2-y_3)^2+(z_2-z_3)^2\)

\(b^2 = (x_1-x_3)^2+(y_1-y_3)^2+(z_1-z_3)^2\)

\(c^2 = (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\)

\(a_1^2 = (x_0-x_1)^2+(y_0-y_1)^2+(z_0-z_1)^2\)

\(b_1^2 = (x_0-x_2)^2+(y_0-y_2)^2+(z_0-z_2)^2\)

\(c_1^2 = (x_0-x_3)^2+(y_0-y_3)^2+(z_0-z_3)^2\)

若取:

\(x = 3, y = 4, z = 5, u = 6\)

解得上述方程组,得到:

\(a = 7.296240846, a_1 = 7.374787895, b = 6.436574223, b_1 = 8.359762010, c = 5.633079407, c_1 = 9.552187140\)

\( k_0 = -0.2538712725, k_1 = -0.4295419047, k_2 = -0.3606779756, k_3 = -0.3020537869\)

\(x_0 = -5.367888653, x_1 = 1.027073438, x_2 = -.2983002272, x_3 = 3.695524296\)

\(y_0 = 2.524437675, y_1 = 1.230397184, y_2 = -3.927146938, y_3 = .6903579963\)

\(z_0 = -0.9016572689, z_1 = 2.535989535, z_2 = .6989520041, z_3 = -3.296438383\)
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 2015-7-14 16:39:42 | 显示全部楼层
1.如何检验上述计算结果是否正确?画图或者给出一组更好的结果?

2.如何简化上述32#结论?
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 2015-7-15 20:50:41 | 显示全部楼层
123456.png

TO wayne and chyanog 帮忙看看哪里有问题

  1. Factor[GroebnerBasis[Numerator[Together[ax[0]y[1]-ax[1]y[0]+bx[0]y2-bx[2]y[0]+cx[0]y[3]-cx[3]y[0], ax[0]z[1]-ax[1]z[0]+bx[0]z2-bx[2]z[0]+cx[0]z[3]-cx[3]z[0], ax[0]y[1]-ax[1]y[0]-b[1]x[1]y[3]+b[1]x[3]y[1]-c[1]x[1]y2+c[1]x[2]y[1], ax[0]z[1]-ax[1]z[0]-b[1]x[1]z[3]+b[1]x[3]z[1]-c[1]x[1]z2+c[1]x[2]z[1], a[1]x[2]y[3]-a[1]x[3]y2-bx[0]y2+bx[2]y[0]-c[1]x[1]y2+c[1]x[2]y[1], a[1]x[2]z[3]-a[1]x[3]z2-bx[0]z2+bx[2]z[0]-c[1]x[1]z2+c[1]x[2]z[1], a[1]x[2]y[3]-a[1]x[3]y2+b[1]x[1]y[3]-b[1]x[3]y[1]+cx[0]y[3]-cx[3]y[0], a[1]x[2]z[3]-a[1]x[3]z2+b[1]x[1]z[3]-b[1]x[3]z[1]+cx[0]z[3]-cx[3]z[0], a^2 - (x[2]-x[3])^2+(y2-y[3])^2+(z2-z[3])^2, a[1]^2 - (x[0]-x[1])^2+(y[0]-y[1])^2+(z[0]-z[1])^2, b^2 - (x[1]-x[3])^2+(y[1]-y[3])^2+(z[1]-z[3])^2, b[1]^2 - (x[0]-x[2])^2+(y[0]-y2)^2+(z[0]-z2)^2, c^2 - (x[2]-x[1])^2+(y2-y[1])^2+(z2-z[1])^2, c[1]^2 - (x[0]-x[3])^2+(y[0]-y[3])^2+(z[0]-z[3])^2, x[0]^2+y[0]^2+z[0]^2 - u^2, x[1]^2+y[1]^2+z[1]^2 - x^2, x[2]^2+y2^2+z2^2 - y^2, x[3]^2+y[3]^2+z[3]^2 - z^2]],{a,x,y,z,u},{x[0],y[0],z[0],x[1],y[1],z[1],x[2],y[2],z[2],x[3],y[3],z[3],b,c,a[1],b[1],c[1]},{MonomialOrder->EliminationOrder}]
复制代码

点评

不是花括弧的原因啊?  发表于 2015-7-15 21:42
Together里面的表达式要用花括弧  发表于 2015-7-15 21:29
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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