三角形两内点间距

数学星空

若已知\(\triangle ABC\)内部有两点\(P,Q\),
且\(AB=c,AC=b,BC=a,PA=u_1,PB=v_1,PC=w_1,QA=u_2,QB=v_2,QC=w_2\),
求\(PQ\)的表达式

creasson

\( PQ^2=\Bigl(a^4u_1^4-a^4u_1^2u_2^2-2a^2b^2u_1^2v_1^2+a^2b^2u_2^2v_1^2+a^2u_1^2u_2^2v_1^2+b^4v_1^4-b^2u_2^2v_1^4+a^2b^2u_1^2v_2^2-a^2u_1^4v_2^2-b^4v_1^2v_2^2+b^2u_1^2v_1^2v_2^2-2a^2c^2u_1^2w_1^2+a^2c^2u_2^2w_1^2+a^2u_1^2u_2^2w_1^2-2b^2c^2v_1^2w_1^2-2a^2u_2^2v_1^2w_1^2+b^2u_2^2v_1^2w_1^2+c^2u_2^2v_1^2w_1^2+b^2c^2v_2^2w_1^2+a^2u_1^2v_2^2w_1^2-2b^2u_1^2v_2^2w_1^2+c^2u_1^2v_2^2w_1^2+b^2v_1^2v_2^2w_1^2+c^4w_1^4-c^2u_2^2w_1^4-c^2v_2^2w_1^4+a^2c^2u_1^2w_2^2-a^2u_1^4w_2^2+b^2c^2v_1^2w_2^2+a^2u_1^2v_1^2w_2^2+b^2u_1^2v_1^2w_2^2-2c^2u_1^2v_1^2w_2^2-b^2v_1^4w_2^2-c^4w_1^2w_2^2+c^2u_1^2w_1^2w_2^2+c^2v_1^2w_1^2w_2^2 \Bigr) \biggm/ \Bigl(2a^2b^2c^2+a^4u_1^2-a^2b^2u_1^2-a^2c^2u_1^2-a^2b^2v_1^2+b^4v_1^2-b^2c^2v_1^2-a^2c^2w_1^2-b^2c^2w_1^2+c^4w_1^2 \Bigr) \)

数学星空

能否给出计算过程？

creasson

\(PQ^2=\alpha AQ^2+\beta BQ^2+\gamma CQ^2-(\alpha\beta AB^2+\beta\gamma BC^2+\gamma\alpha CA^2)\)
\(PA^2=\beta AB^2+\gamma AC^2-(\alpha\beta AB^2+\beta\gamma BC^2+\gamma\alpha CA^2)\)
\(PB^2=\gamma BC^2+\alpha AB^2-(\alpha\beta AB^2+\beta\gamma BC^2+\gamma\alpha CA^2)\)
\(PC^2=\alpha AC^2+\beta BC^2-(\alpha\beta AB^2+\beta\gamma BC^2+\gamma\alpha CA^2)\)
\(\alpha+\beta+\gamma=1\)
五式联立即可得。
注，由后四式可得 \(\alpha PA^2+\beta PB^2+\gamma PC^2=\alpha\beta AB^2+\beta\gamma BC^2+\gamma\alpha CA^2\)

creasson

将PABC,QABC看做四面体，体积须等于0，由欧拉公式知9个变量中有两个隐含的条件式

northwolves

对于\(\triangle ABC\)，设\(B(0,0)，C(0,a)\), 则：
\(\displaystyle x_A=c\cos(B)=\frac{a^2+c^2-b^2}{2a},y_A=\sqrt{c^2-\left(\frac{a^2+c^2-b^2}{2a}\right)^2}\)
\(x_P,y_P,x_Q,y_Q\)的计算同理，
\(\displaystyle x_P=\frac{a^2+v_1^2-w_1^2}{2a}, y_P=\sqrt{v_1^2-\left(\frac{a^2+v_1^2-w_1^2}{2a}\right)^2}\\
\displaystyle x_Q=\frac{a^2+v_2^2-w_2^2}{2a},y_Q=\sqrt{v_2^2-\left(\frac{a^2+v_2^2-w_2^2}{2a}\right)^2}\)
所以
\(PQ^2=(x_P-x_Q)^2+(y_P-y_Q)^2=...\)

northwolves

似乎 b,c,u1,u2四个条件可由其他5个推出来

northwolves

连接任意一个顶点的四条线段的长度都可由其他5个线段的长度推导出来

数学星空

本题中共有3*3＝9个已知量，其实可以除掉任何两个，均可求出\(PQ\)，为了让结果更对称，故保留了多余项
对于三角形内点P，设\(PA=x,PB=y,PC=z\),可以得到

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

此结果即5#所述的，此结果也多次被我们计算过，具体可见

完全四点形的边长约束
http://bbs.emath.ac.cn/forum.php ... 87&fromuid=1455

当然本题可以通过互换A,B,C,P,Q位置得到三个四边形，也能求出答案，似乎计算也很麻烦

数学星空

对于2#的方程，我们可以得到
\(\alpha = -(a^4u_1^2-a^2b^2v_1^2-a^2c^2w_1^2-a^2u_1^4+a^2v_1^2w_1^2+b^2v_1^4-b^2v_1^2w_1^2-c^2u_1^2v_1^2+c^2u_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

\(\beta = (a^2b^2u_1^2-a^2u_1^4+a^2u_1^2w_1^2-b^4v_1^2+b^2c^2w_1^2-b^2u_1^4+2b^2u_1^2v_1^2-b^2u_1^2w_1^2+c^2u_1^4-c^2u_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

\(\gamma = (a^2c^2u_1^2-a^2u_1^4+a^2u_1^2v_1^2+b^2c^2v_1^2+b^2u_1^2v_1^2-b^2v_1^4-c^4w_1^2-2c^2u_1^2v_1^2+c^2u_1^2w_1^2+c^2v_1^2w_1^2)/(a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2)\)

可是\(\alpha +\beta+\gamma\)并不能化简为1

并且最终答案：
\(PQ^2=\frac{M}{a^4u_1^2+2a^2b^2c^2-a^2b^2u_1^2-a^2b^2v_1^2-2a^2c^2u_1^2+b^4v_1^2-b^2c^2u_1^2-b^2c^2v_1^2+c^4u_1^2}\)

\(M=a^4u_1^4-a^4u_1^2u_2^2-2a^2b^2u_1^2v_1^2+a^2b^2u_1^2v_2^2+a^2b^2u_2^2v_1^2-a^2c^2u_1^4-a^2c^2u_1^2w_1^2+a^2c^2u_1^2w_2^2+a^2c^2u_2^2w_1^2+a^2u_1^4u_2^2-a^2u_1^4v_2^2-a^2u_1^4w_2^2+a^2u_1^2v_1^2w_2^2+a^2u_1^2v_2^2w_1^2-a^2u_2^2v_1^2w_1^2+b^4v_1^4-b^4v_1^2v_2^2-b^2c^2u_1^2v_1^2-b^2c^2v_1^2w_1^2+b^2c^2v_1^2w_2^2+b^2c^2v_2^2w_1^2-b^2u_1^4v_2^2+2b^2u_1^2v_1^2v_2^2+b^2u_1^2v_1^2w_2^2-b^2u_1^2v_2^2w_1^2-b^2u_2^2v_1^4+b^2u_2^2v_1^2w_1^2-b^2v_1^4w_2^2+c^4u_1^2w_1^2-c^4w_1^2w_2^2+c^2u_1^4v_2^2+c^2u_1^2u_2^2v_1^2-c^2u_1^2u_2^2w_1^2-2c^2u_1^2v_1^2w_2^2-c^2u_1^2v_2^2w_1^2+c^2u_1^2w_1^2w_2^2+c^2v_1^2w_1^2w_2^2\)

取\(a=b=c=1,u_1=\frac{\sqrt{3}}{2},v_1=w_1=\frac{1}{2},u_2=v_2=w_2=\frac{\sqrt{3}}{3}\),可知\(PQ=r=\frac{\sqrt{3}}{6}\)

但是将上面代入得到\(PQ^2=-\frac{11}{12}\)也不正确