嵌套三角形的心

数学星空

设 由锐角 \(\triangle ABC\) 的垂心\(P\) 分别连接三个顶点交边于\(A_1,B_1,C_1\),然后依次连接\(B_1C_1,A_1C_1,A_1B_1\)分别交\(AP,BP,CP\)于\(A_2,B_2,C_2\),同样作法得到\(A_k,B_k,C_k\),\(k=1..5\),最终得到一个嵌套\(\triangle A_kB_kC_k\) .





1. 若已知\(\triangle ABC\) 各边长依次为\(a,b,c\),求\(\triangle A_kB_kC_k\) 各边长依次为\(a_k,b_k,c_k\) ?


2.请给出点\(P\)在\(\triangle A_kB_kC_k\)中所具备的性质?

数学星空

1.由\(P\)为\(\triangle ABC\)的垂心.我们可以得到:

   \(a_1=R\sin(2A), b_1=R\sin(2B), c_1=R\sin(2C)\)


2.我们容易得到\(P\)为\(\triangle A_1B_1C_1\)的内心.





3.我们容易得到\(P\)为\(\triangle A_2B_2C_2\)的？心.


4.我们得到\(P\)为\(\triangle A_3B_3C_3\)的 ? 心.


5.我们得到\(P\)为\(\triangle A_4B_4C_4\)的 ? 心.


6.我们得到\(P\)为\(\triangle A_5B_5C_5\)的 ? 心.

葡萄糖

数学星空 发表于 2015-8-11 19:09
1.由\(P\)为\(\triangle ABC\)的垂心.我们可以得到:

  \(a_1=R\sin(2A)\), \(b_1=R\sin(2B)\), \(c_1=R\sin(2C)\) ...

数学星空 发表于 2015-8-11 19:09
3.我们容易得到\(P\)为\(△A_2B_2C_2\)的\(\color{red}{外心}\). ...

\(P\)\(\color{red}{不是}\)\(△A_2B_2C_2\)的\(\color{red}{外心}\). ...


\(有点问题！\)

迭代

bbs.cnool.net/cthread-102993804.html

数学星空

当\(k=1\)时

\(-a^2(a^2-b^2-c^2)^2+4a_1^2b^2c^2=0\)

\(-b^2(a^2-b^2+c^2)^2+4b_1^2a^2c^2=0\)

\(-c^2(a^2+b^2-c^2)^2+4c_1^2b^2a^2=0\)


当\(k=2\)时

\(-(a^6-3a^4b^2-3a^4c^2+3a^2b^4+3a^2b^2c^2+3a^2c^4-b^6+b^4c^2+b^2c^4-c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(a^4-a^2b^2-2a^2c^2-b^2c^2+c^4)^2(a^4-2a^2b^2-a^2c^2+b^4-b^2c^2)^2a_2^2=0\)

\((a^6-3a^4b^2-a^4c^2+3a^2b^4-3a^2b^2c^2-a^2c^4-b^6+3b^4c^2-3b^2c^4+c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(a^4-2a^2b^2-a^2c^2+b^4-b^2c^2)^2(a^2b^2+a^2c^2-b^4+2b^2c^2-c^4)^2b_2^2=0\)

\((a^6-a^4b^2-3a^4c^2-a^2b^4-3a^2b^2c^2+3a^2c^4+b^6-3b^4c^2+3b^2c^4-c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(a^4-a^2b^2-2a^2c^2-b^2c^2+c^4)^2(a^2b^2+a^2c^2-b^4+2b^2c^2-c^4)^2c_2^2=0\)

当\(k=3\)时

\(-(a^6-5a^4b^2-5a^4c^2+7a^2b^4+11a^2b^2c^2+7a^2c^4-3b^6+3b^4c^2+3b^2c^4-3c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(a^4-3a^2b^2-2a^2c^2+2b^4-3b^2c^2+c^4)^2(a^4-2a^2b^2-3a^2c^2+b^4-3b^2c^2+2c^4)^2a_3^2=0\)

\((3a^6-7a^4b^2-3a^4c^2+5a^2b^4-11a^2b^2c^2-3a^2c^4-b^6+5b^4c^2-7b^2c^4+3c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(2a^4-3a^2b^2-3a^2c^2+b^4-2b^2c^2+c^4)^2(a^4-2a^2b^2-3a^2c^2+b^4-3b^2c^2+2c^4)^2b_3^2=0\)

\((3a^6-3a^4b^2-7a^4c^2-3a^2b^4-11a^2b^2c^2+5a^2c^4+3b^6-7b^4c^2+5b^2c^4-c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(a^4-3a^2b^2-2a^2c^2+2b^4-3b^2c^2+c^4)^2(2a^4-3a^2b^2-3a^2c^2+b^4-2b^2c^2+c^4)^2c_3^2=0\)

当\(k=4\)时

\(-(9a^6-33a^4b^2-33a^4c^2+39a^2b^4+43a^2b^2c^2+39a^2c^4-15b^6+15b^4c^2+15b^2c^4-15c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(3a^4-6a^2b^2-5a^2c^2+3b^4-5b^2c^2+2c^4)^2(3a^4-5a^2b^2-6a^2c^2+2b^4-5b^2c^2+3c^4)^2a_4^2=0\)

\((15a^6-39a^4b^2-15a^4c^2+33a^2b^4-43a^2b^2c^2-15a^2c^4-9b^6+33b^4c^2-39b^2c^4+15c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(3a^4-6a^2b^2-5a^2c^2+3b^4-5b^2c^2+2c^4)^2(2a^4-5a^2b^2-5a^2c^2+3b^4-6b^2c^2+3c^4)^2b_4^2=0\)

\((15a^6-15a^4b^2-39a^4c^2-15a^2b^4-43a^2b^2c^2+33a^2c^4+15b^6-39b^4c^2+33b^2c^4-9c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(2a^4-5a^2b^2-5a^2c^2+3b^4-6b^2c^2+3c^4)^2(3a^4-5a^2b^2-6a^2c^2+2b^4-5b^2c^2+3c^4)^2c_4^2=0\)

当\(k=5\)时

\(-(25a^6-105a^4b^2-105a^4c^2+135a^2b^4+171a^2b^2c^2+135a^2c^4-55b^6+55b^4c^2+55b^2c^4-55c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(5a^4-11a^2b^2-10a^2c^2+6b^4-11b^2c^2+5c^4)^2(5a^4-10a^2b^2-11a^2c^2+5b^4-11b^2c^2+6c^4)^2a_5^2=0\)

\((55a^6-135a^4b^2-55a^4c^2+105a^2b^4-171a^2b^2c^2-55a^2c^4-25b^6+105b^4c^2-135b^2c^4+55c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(5a^4-10a^2b^2-11a^2c^2+5b^4-11b^2c^2+6c^4)^2(6a^4-11a^2b^2-11a^2c^2+5b^4-10b^2c^2+5c^4)^2b_5^2=0\)

\((55a^6-55a^4b^2-135a^4c^2-55a^2b^4-171a^2b^2c^2+105a^2c^4+55b^6-135b^4c^2+105b^2c^4-25c^6)(a^2+b^2-c^2)^2(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2+4(5a^4-11a^2b^2-10a^2c^2+6b^4-11b^2c^2+5c^4)^2(6a^4-11a^2b^2-11a^2c^2+5b^4-10b^2c^2+5c^4)^2c_5^2=0\)

当\(k=6\)时

\(-(121a^6-473a^4b^2-473a^4c^2+583a^2b^4+683a^2b^2c^2+583a^2c^4-231b^6+231b^4c^2+231b^2c^4-231c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(11a^4-22a^2b^2-21a^2c^2+11b^4-21b^2c^2+10c^4)^2(11a^4-21a^2b^2-22a^2c^2+10b^4-21b^2c^2+11c^4)^2a_6^2=0\)

\((231a^6-583a^4b^2-231a^4c^2+473a^2b^4-683a^2b^2c^2-231a^2c^4-121b^6+473b^4c^2-583b^2c^4+231c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(11a^4-22a^2b^2-21a^2c^2+11b^4-21b^2c^2+10c^4)^2(10a^4-21a^2b^2-21a^2c^2+11b^4-22b^2c^2+11c^4)^2b_6^2=0\)

\((231a^6-231a^4b^2-583a^4c^2-231a^2b^4-683a^2b^2c^2+473a^2c^4+231b^6-583b^4c^2+473b^2c^4-121c^6)(a^2-b^2+c^2)^2(a^2-b^2-c^2)^2(a^2+b^2-c^2)^2+4(11a^4-21a^2b^2-22a^2c^2+10b^4-21b^2c^2+11c^4)^2(10a^4-21a^2b^2-21a^2c^2+11b^4-22b^2c^2+11c^4)^2c_6^2=0\)

zeus

记
\[\left\{ \begin{array}{l}
{a_k} = {B_k}{C_k} \\
{b_k} = {C_k}{A_k} \\
{c_k} = {A_k}{B_k} \\
\end{array} \right.\]
\[\left\{ \begin{array}{l}
{A_k}  =  {x_{k - 1}}{B_{k - 1}} + \left( {1 - {x_{k - 1}}} \right){C_{k - 1}} \\
{B_k}  =  {y_{k - 1}}{C_{k - 1}} + \left( {1 - {y_{k - 1}}} \right){A_{k - 1}} \\
{C_k} = {z_{k - 1}}{A_{k - 1}} + \left( {1 - {z_{k - 1}}} \right){B_{k - 1}} \\
\end{array} \right.\]
那么，有递推式
\[\left\{ \begin{array}{l}
{x_k}  = \frac{{a_{k - 1}^2 + b_{k - 1}^2 - c_{k - 1}^2}}{{2a_{k - 1}^2}} \\
{y_k}  = \frac{{b_{k - 1}^2 + c_{k - 1}^2 - a_{k - 1}^2}}{{2b_{k - 1}^2}} \\
{z_k}  = \frac{{c_{k - 1}^2 + a_{k - 1}^2 - b_{k - 1}^2}}{{2c_{k - 1}^2}} \\
\end{array} \right.\]
\[\left\{ \begin{array}{l}
{b_k}^2 = \left( {1 - {x_{k - 1}} - {y_{k - 1}}} \right)\left\{ {\left( {1 - {y_{k - 1}}} \right)b_{k - 1}^2 - {x_{k - 1}}a_{k - 1}^2} \right\} + {x_{k - 1}}\left( {1 - {y_{k - 1}}} \right)c_{k - 1}^2 \\
{c_k}^2 = \left( {1 - {y_{k - 1}} - {z_{k - 1}}} \right)\left\{ {\left( {1 - {z_{k - 1}}} \right)c_{k - 1}^2 - {y_{k - 1}}b_{k - 1}^2} \right\} + {y_{k - 1}}\left( {1 - {z_{k - 1}}} \right)a_{k - 1}^2 \\
{a_k}^2 = \left( {1 - {z_{k - 1}} - {x_{k - 1}}} \right)\left\{ {\left( {1 - {x_{k - 1}}} \right)a_{k - 1}^2 - {z_{k - 1}}c_{k - 1}^2} \right\} + {z_{k - 1}}\left( {1 - {x_{k - 1}}} \right)b_{k - 1}^2 \\
\end{array} \right.\]
及
\[P{\rm{ = }}\frac{{\left( {{b^2} + {c^2} - {a^2}} \right)\left( {{c^2} + {a^2} - {b^2}} \right)\left( {{a^2} + {b^2} - {c^2}} \right)}}{{16{S^2}}}\left\{ {\frac{A}{{\left( {{b^2} + {c^2} - {a^2}} \right)}} + \frac{B}{{\left( {{c^2} + {a^2} - {b^2}} \right)}} + \frac{C}{{\left( {{a^2} + {b^2} - {c^2}} \right)}}} \right\}\]
各心的重心坐标公式有专门的收集，慢慢地检查，也许能发现点巧合