美數藝廊

利用「向量比」或「分點比」,可以產生許多簡潔有力的推論。這個章節,我們收集了許多可以利用這種方法來證明的美麗定理。

圖示

摘要

PDDA+PEEB+PFFC=1\dfrac{\overrightarrow{P{\color{red}D}}}{\overrightarrow{{\color{red}D}A}}+ \dfrac{\overrightarrow{P{\color{red}E}}}{\overrightarrow{{\color{red}E}B}}+ \dfrac{\overrightarrow{P{\color{red}F}}}{\overrightarrow{{\color{red}F}C}}=-1

[APD]+[BPE]+[CPF]=2[A{\color{red}P}D]+[B{\color{red}P}E]+[C{\color{red}P}F] = {\color{blue}2}

APPD=AEEC+AFFB\dfrac{\overrightarrow{A{\color{red}P}}}{\overrightarrow{{\color{red}P}D}}= \dfrac{\overrightarrow{A{\color{red}E}}}{\overrightarrow{{\color{red}E}C}}+ \dfrac{\overrightarrow{A{\color{red}F}}}{\overrightarrow{{\color{red}F}B}}

(A,P,D)=(A,F,B)+(A,E,C)(A,{\color{red}P},D)=(A,{\color{red}F},B)+(A,{\color{red}E},C)

P=11+1β+γA+11+1γ+αB+11+1α+βCP= \frac{1}{1+\frac{1}{\beta}+\gamma}A + \frac{1}{1+\frac{1}{\gamma}+\alpha}B + \frac{1}{1+\frac{1}{\alpha}+\beta}C

11+(A,P,D)A+11+(B,P,E)B+11+(C,P,F)C \frac{1}{1+(A,{\color{red}P},D)}A + \frac{1}{1+(B,{\color{red}P},E)}B + \frac{1}{1+(C,{\color{red}P},F)}C

DMMA=11+α(BPPA+αCQQA)\dfrac{\overrightarrow{D\color{red}{M}}}{\overrightarrow{{\color{red}M}A}}= \dfrac{ 1 }{ 1+{\color{blue}\alpha} } \left( \dfrac{\overrightarrow{B\color{red}{P}} }{\overrightarrow{{\color{red}P}A}} + {\color{blue}\alpha} \cdot\dfrac{\overrightarrow{C\color{red}{Q}}}{\overrightarrow{{\color{red}Q}A}} \right)

1[AMD]=11+α(1[APB]+α1[AQC])\dfrac{1}{[A{\color{red}M}D]}=\dfrac{ 1 }{ 1+{\color{blue}\alpha} } \left( \dfrac{1}{[A{\color{red}P}B]} + {\color{blue}\alpha} \cdot \dfrac{1}{[A{\color{red}Q}C]} \right)

X,Y,ZX,Y,Z 三點共線

Last updated

Was this helpful?