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hbghlyj posted 2024-4-25 04:23 |Read mode

MathWorld写道,可以从\begin{align}
d_1^2+h^2 & =h_1^2 \\
d_2^2+h^2 & =h_2^2 \\
r_1^2+\left(l_1-h_2\right)^2 & =h_1^2 \\
r_2^2+\left(l_2-h_1\right)^2 & =h_2^2 \\
d_1+d_2 & =d \\
r_1^2+l_1^2 & =d^2 \\
r_2^2+l_2^2 & =d^2 \\
s_1 h_1 & =h r_1 \\
s_2 h_2 & =h r_2
\end{align}这九个方程中,消去$ d_1, d_2, h, h_1, h_2, l_1, l_2$,这个怎样办到呢?会得到$s_1,s_2$满足相同的8次方程?

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kuing posted 2024-4-26 22:46
啥意思,直接由
\[\frac{s_1}{r_1}=\frac{r_2}d,~\frac{s_2}{r_2}=\frac{r_1}d\]
不就得到
\[s_1=s_2=\frac{r_1r_2}d\]
了吗?

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不知道Mathworld说的8次多项式是什么呢🤔  posted 2024-4-26 23:30

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original poster hbghlyj posted 2024-4-26 23:25

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