計算「西瓦點」的座標。
👉 什麼是「西瓦點」❓
PPP 為三條西瓦線的交點,三條西瓦線分別交 BC↔,CA↔,AB↔\overleftrightarrow{BC},\overleftrightarrow{CA},\overleftrightarrow{AB}BC,CA,AB 於 D,E,FD,E,FD,E,F 三點。
假設:
BD→DC→=α\dfrac{\overrightarrow{B{\color{red}D}}}{\overrightarrow{{\color{red}D}C}}=\alphaDCBD=α , CE→EA→=β\dfrac{\overrightarrow{C{\color{red}E}}}{\overrightarrow{{\color{red}E}A}}=\betaEACE=β , AF→FB→=γ\dfrac{\overrightarrow{A{\color{red}F}}}{\overrightarrow{{\color{red}F}B}}=\gammaFBAF=γ
則:
P=(11+1β+γ)A+(11+1γ+α)B+(11+1α+β)CP=\left(\dfrac{1}{1+\frac{1}{\beta}+\gamma}\right)A + \left(\dfrac{1}{1+\frac{1}{\gamma}+\alpha}\right)B + \left(\dfrac{1}{1+\frac{1}{\alpha}+\beta}\right)CP=(1+β1+γ1)A+(1+γ1+α1)B+(1+α1+β1)C
或寫成:
P=11+(A,P,D)A+11+(B,P,E)B+11+(C,P,F)CP= \dfrac{1}{1+(A,{\color{red}P},D)}A + \dfrac{1}{1+(B,{\color{red}P},E)}B + \dfrac{1}{1+(C,{\color{red}P},F)}CP=1+(A,P,D)1A+1+(B,P,E)1B+1+(C,P,F)1C
💡 下圖 A,B,C,PA,B,C,PA,B,C,P 點可拖曳。
(11+1β+γ)+(11+1γ+α)+(11+1α+β)=1\left(\dfrac{1}{1+\frac{1}{\beta}+\gamma}\right) + \left(\dfrac{1}{1+\frac{1}{\gamma}+\alpha}\right) + \left(\dfrac{1}{1+\frac{1}{\alpha}+\beta}\right) = 1(1+β1+γ1)+(1+γ1+α1)+(1+α1+β1)=1
原式
推論
說明
AP→PD→\dfrac{\overrightarrow{A{\color{red}P}}}{\overrightarrow{{\color{red}P}D}}PDAP
=AE→EC→+AF→FB→= \dfrac{\overrightarrow{A{\color{red}E}}}{\overrightarrow{{\color{red}E}C}}+ \dfrac{\overrightarrow{A{\color{red}F}}}{\overrightarrow{{\color{red}F}B}}=ECAE+FBAF
西瓦線分點比
=1β+γ=\dfrac{1}{\beta}+\gamma=β1+γ
向量比「性質 7」
=αγ+γ=\alpha\gamma+\gamma=αγ+γ
根據「西瓦定理」: αβγ=1\alpha\beta\gamma=1αβγ=1
=γ(α+1)=\gamma(\alpha+1)=γ(α+1)
因此:
1️⃣ P{\color{red}P}P
=A+γ(α+1)D1+γ(α+1)=\dfrac{A+{\color{blue}\gamma(\alpha+1)}D}{1+{\color{blue}\gamma(\alpha+1)}}=1+γ(α+1)A+γ(α+1)D
分點「性質 1」
另外:
BD→DC→\dfrac{\overrightarrow{B{\color{red}D}}}{\overrightarrow{{\color{red}D}C}}DCBD
=α=\alpha=α
原條件
2️⃣ DDD
=B+αC1+α=\dfrac{B+{\color{blue}\alpha}C}{1+{\color{blue}\alpha}}=1+αB+αC
代 2️⃣ 回 1️⃣ 可得:
P{\color{red}P}P
=A+γ(α+1)⋅B+αC1+α1+γ(α+1)=\dfrac{A+{\color{blue}\gamma(\cancel{\alpha+1})}\cdot \dfrac{B+{\color{blue}\alpha}C}{\cancel{1+{\color{blue}\alpha}}}}{1+{\color{blue}\gamma(\alpha+1)}}=1+γ(α+1)A+γ(α+1)⋅1+αB+αC
=A+γ(B+αC)1+γ(α+1)=\dfrac{ A+{\color{blue}\gamma}(B+{\color{blue}\alpha}C) }{ 1+{\color{blue}\gamma(\alpha+1)} }=1+γ(α+1)A+γ(B+αC)
其中:
🔸 AAA 的係數:
=11+γ(α+1)=\dfrac{ 1 }{ 1+{\color{blue}\gamma(\alpha+1)} }=1+γ(α+1)1
=11+γα+γ=\dfrac{ 1 }{ 1+{\color{blue}\gamma\alpha+\gamma} }=1+γα+γ1
=11+1β+γ=\dfrac{ 1 }{ 1+{\color{blue}\frac{1}{\beta}+\gamma} }=1+β1+γ1 ▨
根據「西瓦定理」:
αβγ=1\alpha\beta\gamma=1αβγ=1
🔸 BBB 的係數:
=γ1+γα+γ=\dfrac{ {\color{blue}\gamma} }{ 1+{\color{blue}\gamma\alpha+\gamma} }=1+γα+γγ
=11γ+α+1=\dfrac{ 1 }{ {\color{blue}\frac{1}{\gamma}+\alpha}+1 }=γ1+α+11 ▨
同除 γ\color{blue}\gammaγ
🔸 CCC 的係數:
=γα1+γα+γ=\dfrac{ {\color{blue}\gamma\alpha} }{ 1+{\color{blue}\gamma\alpha+\gamma} }=1+γα+γγα
=11γα+1+1α=\dfrac{1}{{\color{blue}\frac{1}{\gamma\alpha}}+1+{\color{blue}\frac{1}{\alpha}}}=γα1+1+α11
同除 γα\color{blue}\gamma\alphaγα
=1β+1+1α=\dfrac{1}{{\color{blue}\beta}+1+{\color{blue}\frac{1}{\alpha}}}=β+1+α11 ▨
由「西瓦線分點比」可知:
(A,P,D)=(A,E,C)+(A,F,B)(A,{\color{red}P},D)=(A,{\color{red}E},C)+(A,{\color{red}F},B)(A,P,D)=(A,E,C)+(A,F,B)
(A,P,D)=1β+γ1(A,{\color{red}P},D)=\dfrac{1}{\beta}+\dfrac{\gamma}{1}(A,P,D)=β1+1γ
同理:
(B,P,E)=1γ+α1(B,{\color{red}P},E)=\dfrac{1}{\gamma}+\dfrac{\alpha}{1}(B,P,E)=γ1+1α
(C,P,F)=1α+β1(C,{\color{red}P},F)=\dfrac{1}{\alpha}+\dfrac{\beta}{1}(C,P,F)=α1+1β
將上三式代回「第一個公式」即可得證 ▨
Last updated 4 years ago
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