一个平面折线问题

hbghlyj

平面上的点$A_1,A_2,\dots,A_n$，作折线$PP_1P_2\cdots P_n$，其中$P_1$在$PA_1$上，$P_2$在$P_1A_2$上，且$\frac{PA_1}{PP_1},\frac{P_1A_2}{P_1P_2},\cdots$成等差数列。
证明：不论$A_1,A_2,\dots,A_n$的次序如何变更，折线的终点$P$都是同一个点。

Roger A. Johnson Advanced Euclidean Geometry page 79 §106 Theorem. Given in the plane $k+1$ points, $A_1, A_2, \ldots A_k$, $P$, and a fraction $m / n$. If we construct a broken line of it: then each successive segment toward one of the $A$'s, with multipliers in harmonic progression, viz:\begin{array}c \overline{PP_1}=\frac{m}{n}\overline{PA_1},& \overline{P_1P_2}=\frac{m}{m+n}\overline{P_1A_2},& \overline{P_2P_3}=\frac{m}{2 m+n}\overline{P_2A_3},&\text{etc.} \end{array}then the same point $P$ is the end of the broken line, in whatever order the points $A_1, A_2, \ldots A_k$ are taken.

如果只有两个点，结论成立。如果点数超过两个，可以通过连续地交换一对相邻的点来实现，而每一次交换不影响最后的结果。

If there are only two points the preceding theorem establishes the result. If there are more than two, any change of order can be effected by a succession of interchanges of pairs of $A$'s, no one of which affects the result.

hbghlyj

hbghlyj

\begin{align*}
P-P_1&=\frac{m}{n} (P-A_1)&⟹P_1&=\frac{n-m}nP+\frac mnA_1\\
P_1- P_2&=\frac{m}{m+n}(P_1-A_2)&⟹P_2&=\frac n{m+n}P_1+\frac m{m+n}A_2\\
P_2-P_3&=\frac{m}{2 m+n} (P_2-A_3)&⟹P_3&=\frac{m+n}{2m+n}P_2+\frac{m}{2m+n}A_3
\end{align*}