内接的投影四面体体积
杨学枝关于四面体的猜想http://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=5771&fromuid=1455
(出处: 数学研发论坛)
在上面的问题中，提到了四面体\(D-ABC\)内任一点\(P\),由\(P\)点向四面体的各个面\(BCD,ACD,ABD,ABC\)作垂线并交各个面于\(A_1,B_1,C_1,D_1\),现在的问题是如何计算\(D_1-A_1B_1C_1\)的体积\(V_0\)？
已知：\(AB=c,AC=b,BC=a,AD=a_1,BD=b_1,CD=c_1,AP=x,BP=y,CP=z\). 我先尝试用数值计算的方法：
先记\(D-ABC\)的体积为\(V\),外接圆半径为\(R\),分别过四个顶点\(A,B,C,D\)向各个面作垂线，其垂线长度分别为\(h_1,h_2,h_3,h_4\),
\(h_{01}=P A_1,h_{02}=P B_1,h_{03}=P C_1,h_{04}=P D_1\)
\(\angle A_1 P B_1=\gamma,\angle A_1 P C_1=\beta,\angle B_1 P C_1=\alpha,\angle A_1 P D_1=\alpha_1,\angle B_1 P D_1=\beta_1,\angle C_1 P D_1=\gamma_1\)
\(\angle (A－BC－D)=\eta,\angle (B－AC－D) =\theta,\angle (C－AB－D)=\delta,\angle (B－AD－C)=\eta_1,\angle (A－BD－C)=\theta_1,\angle A-CD-B=\delta_1\)
可设各个点的坐标：\(A,B,C,D,P\)
容易算得：
\(x=3, y= 2\sqrt{3}, w=\sqrt{11} , z=3, a = 5, a_1 = \sqrt{38}, b = 4, b_1 = \sqrt{35}, c = 3, c_1 = \sqrt{30},V=10,V_1=\frac{10}{3},V_2=\frac{2}{3},V_3=2,V_4=4,R=\frac{\sqrt{41}}{2}\)
\( h_{01}=\frac{2\sqrt{26}}{26} , h_{02}=\frac{\sqrt{29}}{29} , h_{03}=\frac{2\sqrt{34}}{17} , h_{04}=2 , h_1=\frac{6\sqrt{26}}{13} , h_2=\frac{15\sqrt{29}}{29} , h_3=\frac{10\sqrt{34}}{17} , h_4=5 \)
\(D-ABC\)的六个二面角分别为\(\sin(\eta)=\frac{5\sqrt{34}}{34} , \sin(\theta)=\frac{5\sqrt{29}}{29} , \sin(\delta)=\frac{5\sqrt{19\*17\*29}}{493} \)
\( \sin(\eta_1)=\frac{3\sqrt{15\*29\*13}}{377} , \sin(\theta_1)=\frac{2\sqrt{13\*17\*35}}{221} , \sin(\delta_1)=\frac{5\sqrt{26}}{26} \)
\(\alpha＝\pi-\eta,\beta=\pi-\theta,\gamma=\pi-\delta,\alpha_1=\pi-\eta_1,\beta_1=\pi-\theta_1,\gamma_1=\pi-\delta_1\)
\(A_1 B_1=x_{00} =\frac{\sqrt{26693485+76908\sqrt{13\*29\*17}}}{6409}=0.8942941444,C_1 D_1=x_{01} =\frac{2\sqrt{45907667+281996\sqrt{13\*29\*17}}}{6409}=2.582449030\)
\( A_1 C_1=y_{00} = \frac{4\sqrt{2787915+12818\sqrt{13\*29\*17}}}{6409}=1.218889617,\ B_1 D_1=y_{01} =\frac{3\sqrt{18413057+25636\sqrt{13\*29\*17}}}{6409}=2.117585107\)
\(B_1 C_1=z_{00} =\frac{15\sqrt{493}}{493}=0.6755660237, A_1 D_1=z_{01} =2\)
\(V_0=\frac{10\sqrt{352\sqrt{13\*17\*29}+66250}}{19227}=0.1598243668\)
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