杨学枝关于四面体的猜想

数学星空

杨学枝2009年3月提出

猜想:   设\(P\)为四面体\(ABCD\)内部或边界上任意一点，\(P\)点在面\(BCD,ACD,ABD,ABC\)上的正投影分别为\(A_1,B_1,C_1,D_1,|PA|=R_1,|PB|=R_2,|PC|=R_3,|PD|=R_4\),

          四面体\(PBCD,APCD,ABPD,ABCP,A_1B_1C_1D_1\)的有向体积（右手定则）分别为\(V_1,V_2,V_3,V_4,V_0\),四面体\(ABCD\)外接球的半径为\(R\),则：

\[V_1 R_1^2+V_2 R_2^2+V_3 R_3^2+V_4 R_4^2=27 V_0 R^2\]

上述猜想由全国初等数学研究会理事长杨学枝特级教师提出，尚未被证实或证否。

杨学枝邮箱：yangxuezhi1121@163.com

转自：http://www.cdmath.org/Article/ShowArticle.asp?ArticleID=909

zeus

数学星空

我们可以先退一步讨论平面三角形内一点的情形，类似地我们可以猜测：

        设\(P\)为三角形\(ABCD\)内部任意一点，\(P\)点在三边上\(BC,AC,AB\)上的正投影分别为\(A_1,B_1,C_1,D_1,|PA|=R_1,|PB|=R_2,|PC|=R_3\),

         三角形\(PBC,APC,ABP,ABC,A_1B_1C_1D_1\)的面积分别为\(S_1,S_2,S_3,S_0\),三角形\(ABC\)外接圆的半径为\(R\),则：

\[S_1 R_1^2+S_2 R_2^2+S_3 R_3^2=4 S_0 R^2\]

数学星空

为了便于计算，我们从特殊点开始考虑验证：

我们记三角形内切圆半径为\(r\),半周长为\(p\),\(S_{ABC}=S,S_{C_1 B_1 A}=s_1,S_{A_1 C_1 B}=s_2,S_{B_1 C_1 A}=s_3\),则\(s_1+s_2+s_3+S_0=S\)

1.若\(P\)点为外心\(O\),则有

    \(R_1=R_2=R_3=R\)

    又\(s_1=\frac{1}{2}\frac{b}{2}\frac{c}{2}\sin(A)=\frac{S}{4}\),同理可以得到\(s_2=s_3=\frac{S}{4}\)

    则\(S_0=S-3\frac{S}{4}=\frac{S}{4}\)

    最终可以得到：

   \(R_1^2 S_1+R_2^2 S_2+R_3^2 S_3=4 R^2 S_0\)


2.若\(P\)为内心\(I\)，则有\(R_1=\frac{r}{\sin(\frac{A}{2})},R_2=\frac{r}{\sin(\frac{B}{2})},R_3=\frac{r}{\sin(\frac{C}{2})}\)

   \(S_1=\frac{1}{2}ar, S_2=\frac{1}{2}br, S_3=\frac{1}{2}cr,S_0=\frac{1}{2}r^2(\sin(A)+\sin(B)+\sin(C))\)

   \(R_1^2 S_1+R_2^2 S_2+R_3^2 S_3-4 R^2 S_0=4Rr^2(r(\cot(\frac{A}{2})+\cot(\frac{B}{2})+\cot(\frac{C}{2}))-R(\sin(A)+\sin(B)+\sin(C)))\)

   \(=4Rr^2(p-a+p-b+p-c-p)=0\)

数学星空

记三角形的三条中线长分别为\(m_a,m_b,m_c\)

3. 若\(P\)点为重心\(G\),则 \(R_1=\frac{2m_a}{3}, R_2=\frac{2m_b}{3}, R_3=\frac{2m_c}{3}\)

  \( S_1=S_2=S_3=\frac{S}{3}\)

  \(S_0=\frac{1}{2}(\frac{2S}{3})^2(\frac{\sin(A)}{bc}+\frac{\sin(B)}{ac}+\frac{\sin(C)}{ab})\)

  \(\frac{S^2}{9R}\frac{1}{abc}(a^2+b^2+c^2)=\frac{S}{36R^2}(a^2+b^2+c^2)\)

    我们可以得到：

   \(R_1^2 S_1+R_2^2 S_2+R_3^2 S_3-4R^2 S_0=\frac{4S}{27}(m_a^2+m_b^2+m_c^2)-4R^2 \frac{S}{36R^2}(a^2+b^2+c^2)\)

    \( =\frac{4S}{27} \frac{1}{4} (2a^2+2b^2-c^2+2a^2+2c^2-b^2+2b^2+2c^2-a^2) - \frac{S}{9} (a^2+b^2+c^2)=0 \)




4.若\(P\)点为三角形的垂心\(H\)，则 \(R_1=2R\cos(A),R_2=2R\cos(B),R_3=2R\cos(C)\)

  \(S_1=1-\frac{R^2\sin(2A)}{S}, S_2=1-\frac{R^2\sin(2B)}{S},S_3=1-\frac{R^2\sin(2C)}{S}\)

   \(S_0=S(1-\cos(A)^2-\cos(B)^2-\cos(C)^2)\)

  我们可以得到：

  \( R_1^2 S_1+R_2^2 S_2+R_3^2 S_3-4R^2 S_0=4R^2(\cos(A)^2(S-R^2\sin(2A))+\cos(B)^2(S-R^2\sin(2B))+\cos(C)^2(S-R^2\sin(2C)))-4R^2S(1-\cos(A)^2-\cos(B)^2-\cos(C)^2)\)

  \(=4R^2S(2S(\cos(A)^2+\cos(B)^2+\cos(C)^2)-S-R^2(\cos(A)^2\sin(2A)+\cos(B)^2\sin(2B)+\cos(C)^2\sin(2C)))\)

  \(=4R^2S(1-4\cos(A)\cos(B)\cos(C))-R^4(2(\sin(2A)+\sin(2B)+\sin(2C))+(\sin(4A)+\sin(4B)+\sin(4C)))\)

\(=R^4(8\sin(A)\sin(B)\sin(C)(1-4\cos(A)\cos(B)\cos(C))-(2(\sin(2A)+\sin(2B)+\sin(2C))+(\sin(4A)+\sin(4B)+\sin(4C)))\)..............................(1)

对于\(A+B+C=n\pi,n \in\mathbb{N}\) ,我们可以得到：

\(\sin(2nA)+\sin(2nB)+\sin(2nC)=2\sin(nA+nB)\cos(nA-nB)-\sin(2nA+2nB)\)

\(=2\sin(nA+nB)(\cos(nA-nB)-\cos(nA+nB))=4\sin(nA)\sin(nB)\sin(nC)\)..........................................(2)

根据（2）我们可以得到：

\(\sin(2A)+\sin(2B)+\sin(2C)＝4\sin(A)\sin(B)\sin(C)\)

\(\sin(4A)+\sin(4B)+\sin(4C)＝4\sin(2A)\sin(2B)\sin(2C)=32\sin(A)\sin(B)\sin(C)\cos(A)\cos(B)\cos(C)\)

进一步，对于（1）式可以得到：

\(8\sin(A)\sin(B)\sin(C)(1-4\cos(A)\cos(B)\cos(C))-(2(\sin(2A)+\sin(2B)+\sin(2C))+(\sin(4A)+\sin(4B)+\sin(4C)))＝0\)


即猜测可能是正确的，我们需要证明一般情形！

数学星空

对于三角形内任一点 \(P\)， 我们设\(\angle APB=\gamma, \angle APC=\beta, \angle BPC=\alpha.PA_1=h_1,PB_1=h_2,PC_1=h_3\)

   我们可以得到：

   \(S_0= \frac{1}{2}(h_1 h_2\sin(C)+h_1 h_3\sin(B)+h_2 h_3\sin(A))\)

  \(=\frac{xyz\sin(\alpha)\sin(\beta)\sin(\gamma)}{2abc}(\frac{cz\sin(C)}{\sin(\gamma)}+\frac{by\sin(B)}{\sin(\beta)}+\frac{ax\sin(A)}{\sin(\alpha)})\)

  \(=\frac{xyz\sin(\alpha)\sin(\beta)\sin(\gamma)R}{abc}(\frac{x\sin(A)^2}{\sin(\alpha)}+\frac{y\sin(B)^2}{\sin(\beta)}+\frac{z\sin(C)^2}{\sin(\gamma)})\)

  \(=\frac{xyz\sin(\alpha)\sin(\beta)\sin(\gamma)}{8R^2\sin(A)\sin(B)\sin(C)}(\frac{x\sin(A)^2}{\sin(\alpha)}+\frac{y\sin(B)^2}{\sin(\beta)}+\frac{z\sin(C)^2}{\sin(\gamma)})\)


   \( R_1^2 S_1+ R_2^2 S_2+ R_3^2 S_3=\frac{xyz}{2}(x\sin(\alpha)+y\sin(\beta)+z\sin(\gamma))\)

  进一步我们可以得到：

  \(R_1^2 S_1+ R_2^2 S_2+ R_3^2 S_3－4R^2 S_0\)

\(=\frac{xyz}{16R^3\sin(A)\sin(B)\sin(C)}(abc(x\sin(\alpha)+y\sin(\beta)+z\sin(\gamma))-2R\sin(\alpha)\sin(\beta)\sin(\gamma)(\frac{xa^2}{\sin(\alpha)}+\frac{yb^2}{\sin(\beta)}+\frac{zc^2}{\sin(\gamma)}))\)

  即猜测成立的条件为：

\[\frac{2S}{\frac{xa^2}{\sin(\alpha)}+\frac{yb^2}{\sin(\beta)}+\frac{zc^2}{\sin(\gamma)}}=\frac{\sin(\alpha)\sin(\beta)\sin(\gamma)}{x\sin(\alpha)+y\sin(\beta)+z\sin(\gamma)}\]

数学星空

现在我们来证明楼上的条件是成立的：

\[\frac{2S}{\frac{xa^2}{\sin(\alpha)}+\frac{yb^2}{\sin(\beta)}+\frac{zc^2}{\sin(\gamma)}}=\frac{\sin(\alpha)\sin(\beta)\sin(\gamma)}{x\sin(\alpha)+y\sin(\beta)+z\sin(\gamma)}\]

又因为

\(2S=xy\sin(\gamma)+xz\sin(\beta)+yz\sin(\alpha)\)

\(a^2=y^2+z^2-2yz\cos(\alpha)\)

\(b^2=x^2+z^2-2xz\cos(\beta)\)

\(c^2=x^2+y^2-2xy\cos(\gamma)\)

将上式代入并移项得到：

\((xy\sin(\gamma)+xz\sin(\beta)+yz\sin(\alpha))(x\sin(\alpha)+y\sin(\beta)+z\sin(\gamma))-\)

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

\(=xyz(2(\sin(\alpha)\sin(\beta)\cos(\gamma)+\sin(\beta)\sin(\gamma)\cos(\alpha)+\sin(\alpha)\sin(\gamma)\cos(\beta))+(\sin(\alpha)^2+\sin(\beta)^2+\sin(\gamma)^2))\)...........................(2)

又因为：

\(2(\sin(\alpha)\sin(\beta)\cos(\gamma)+\sin(\beta)\sin(\gamma)\cos(\alpha)+\sin(\alpha)\sin(\gamma)\cos(\beta))\)

\(=2(\sin(\alpha)\sin(\beta+\gamma)+\cos(\alpha)\sin(\beta)\sin(\gamma))\)

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

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

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



\(\sin(\alpha)^2+\sin(\beta)^2+\sin(\gamma)^2\)

\(=1-\frac{1}{2}(\cos(2\alpha)+\cos(2\beta))+1-\cos(\gamma)^2\)

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

\(=2-2\cos(\alpha)\cos(\beta)\cos(\gamma)\)

即有：

\(2(\sin(\alpha)\sin(\beta)\cos(\gamma)+\sin(\beta)\sin(\gamma)\cos(\alpha)+\sin(\alpha)\sin(\gamma)\cos(\beta))+(\sin(\alpha)^2+\sin(\beta)^2+\sin(\gamma)^2))＝0\)

猜测对一般情形也是成立的！

数学星空

我们可以用数值计算的方法验证猜测是不成立的！

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：\(A[0,0,0],B[0,3,0],C[4,0,0],D[3,2,5],P[2,1,2]\)

容易算得：

\(R_1=3, R_2= 2\sqrt{3}, R_4=\sqrt{11} , R_3=3, a = 5, a_1 = \sqrt{38}, b = 4, b_1 = \sqrt{35}, c = 3, c_1 = \sqrt{30},V=10,V_1=\frac{10}{3},V_2=\frac{2}{3},V_3=2,V_4=4,R=\frac{\sqrt{41}}{2}\)

\( h_{01}=\frac{2\sqrt{26}}{26} , h_{02}=\frac{\sqrt{29}}{29} , h_{03}=\frac{2\sqrt{34}}{17} , h_{04}=2 , h_1=\frac{6\sqrt{26}}{13} , h_2=\frac{15\sqrt{29}}{29} , h_3=\frac{10\sqrt{34}}{17} , h_4=5 \)

\(D-ABC\)的六个二面角分别为\(\sin(\eta)=\frac{5\sqrt{34}}{34} , \sin(\theta)=\frac{5\sqrt{29}}{29} , \sin(\delta)=\frac{5\sqrt{19\*17\*29}}{493} \)

\( \sin(\eta_1)=\frac{3\sqrt{15\*29\*13}}{377} , \sin(\theta_1)=\frac{2\sqrt{13\*17\*35}}{221} , \sin(\delta_1)=\frac{5\sqrt{26}}{26} \)

\(\alpha＝\pi-\eta,\beta=\pi-\theta,\gamma=\pi-\delta,\alpha_1=\pi-\eta_1,\beta_1=\pi-\theta_1,\gamma_1=\pi-\delta_1\)

\(A_1 B_1=x_{00} =\frac{\sqrt{26693485+76908\sqrt{13\*29\*17}}}{6409}=0.8942941444,C_1 D_1=x_{01} =\frac{2\sqrt{45907667+281996\sqrt{13\*29\*17}}}{6409}=2.582449030\)

\( A_1 C_1=y_{00} = \frac{4\sqrt{2787915+12818\sqrt{13\*29\*17}}}{6409}=1.218889617,\ B_1 D_1=y_{01} =\frac{3\sqrt{18413057+25636\sqrt{13\*29\*17}}}{6409}=2.117585107\)

\(B_1 C_1=z_{00} =\frac{15\sqrt{493}}{493}=0.6755660237, A_1 D_1=z_{01} =2\)

\(V_0=\frac{10\sqrt{352\sqrt{13\*17\*29}+66250}}{19227}=0.1598243668\)

最终可以验证杨学枝猜测并不正确：

\(R_1^2 V_1+R_2^2 V_2+R_3^2 V_3+R_4^2 V_4-27 R^2 V_0=\frac{-1845\sqrt{352\sqrt{13\*29\*17}+66250}}{12818}+100=55.76860648\)