45407 歐拉不等式的另證

### 一、前言

\begin{align} d^2=R(R-2r),\label{1} \end{align}

\begin{align} R\ge 2r.\label{2} \end{align}

### 二、歐拉不等式的證明

\begin{align} \cos A=\frac{b^2+c^2-a^2}{2bc}.\label{3} \end{align}

\begin{align} \sin A=\frac{a}{2R}.\label{4} \end{align}

\begin{align*} \frac{a^2}{4R^2}=\,&1-\frac{(b^2+c^2-a^2)^2}{4b^2 c^2}=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{4b^2 c^2}\\ =\,&\frac{(2bc+b^2+c^2-a^2 )(2bc-b^2-c^2+a^2 )}{4b^2 c^2}\\ =\,&\frac{[(b+c)^2-a^2][a^2-(b-c)^2 ]}{4b^2 c^2}\\ =\,&\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2 c^2}. \end{align*}

\begin{align} R^2=\frac{a^2 b^2 c^2}{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}.\label{5} \end{align}

\begin{align} \triangle =&\sqrt{s(s-a)(s-b)(s-c)}\nonumber\\ =&\frac 14 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)},\label{6} \end{align}

\begin{align} s=\frac 12 (a+b+c).\label{7} \end{align}

\begin{align} R=\frac{abc}{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}}=\frac{abc}{4\triangle}.\label{8} \end{align}

\begin{align} r=\frac{\triangle}s.\label{9} \end{align}

\begin{align} \frac Rr=\,&\frac{sabc}{4\triangle ^2}=\frac{sabc}{4 s(s-a)(s-b)(s-c)}\nonumber\\ =\,&\frac{2abc}{(b+c-a)(c+a-b)(a+b-c)}.\label{10} \end{align}

\begin{align} abc\ge (b+c-a)(c+a-b)(a+b-c).\label{11} \end{align}

\begin{align} \frac{S+T}2\cdot \frac{T+U}2\cdot \frac{U+S}2\ge STU.\label{12} \end{align}

### 三、歐拉定理的向量證明

\begin{align} \vec{XI}=\frac{a}{a+b+c}\!\vec{XA}\!+\,\frac b{a+b+c}\!\vec{XB}\!+\,\frac c{a+b+c}\!\vec{XC}\!.\label{13} \end{align}

\begin{align} d^2=\,&|\!\vec{OI}\!|^2=\,\vec{OI}\!\cdot\!\vec{OI}\nonumber\\ =\,&\frac 1{(a+b+c)^2}(a^2 |\!\vec{OA}\!|^2+b^2 |\!\vec{OB}\!|^2+c^2 |\!\vec{OC}\!|^2+2ab\!\vec{OA}\!\cdot\!\vec{OB}\nonumber\\ &+2bc\vec{OB}\!\cdot\!\vec{OC}\!+\,2ca\!\vec{OC}\!\cdot\!\vec{OA}).\label{14} \end{align}

\begin{align} |\!\vec{OA}\!|=|\!\vec{OB}\!|=|\!\vec{OC}\!|=R,\label{15} \end{align}

\begin{align} \vec{OA}\!\cdot\! \vec{OB}=\,&|\!\vec{OA}\!||\!\vec{OB}\!|\cos \angle AOB=R^2\cos (2\angle ACB)=R^2\cos 2C\nonumber\\ =\,&R^2 (2 \cos^2C-1).\label{16} \end{align}

\begin{align} \vec{OB}\!\cdot\!\vec{OC}=\,&R^2(2 \cos^2 A-1),\label{17}\\ \vec{OC}\!\cdot\!\vec{OA}=\,&R^2(2 \cos^2 B-1).\label{18} \end{align}

\begin{align*} d^2=\,&\frac 1{(a\!+\!b\!+\!c)^2}[(a^2\!+\!b^2\!+\!c^2)R^2\!+\!2abR^2 (2 \cos^2C\!-\!1)\!+\!2bcR^2 (2 \cos^2 A\!-\!1)\\ &+\!2caR^2 (2 \cos^2B\!-\!1)]\\ =\,&\frac{R^2}{(a\!+\!b\!+\!c)^2}[(a^2\!+\!b^2\!+\!c^2\!+\!2ab\!+\!2bc\!+\!2ca)\!+\!2ab (2 \cos^2C\!-\!2)\\ &+\!2bc (2 \cos^2A\!-\!2)\!+\!2ca (2 \cos^2B\!-\!2)]\\ =\,&\frac{R^2}{(a\!+\!b\!+\!c)^2} [(a\!+\!b\!+\!c)^2\!+\!4ab (\cos^2C\!-\!1)\!+\!4bc (\cos^2A\!-\!1)\!+\!4ca (\cos^2B\!-\!1)]\\ =\,&R^2-\frac{4R^2}{(a\!+\!b\!+\!c)^2} (ab \sin^2 C\!+\!bc \sin^2 A\!+\!ca \sin^2B ). \end{align*}

\begin{align} d^2=\,&R^2-\frac{4R^2}{(a+b+c)^2}\Big(ab\cdot \frac{c^2}{4R^2}+bc\cdot \frac{a^2}{4R^2}+ca\cdot \frac{b^2}{4R^2}\Big)\nonumber\\ =\,&R^2-\frac{abc}{(a+b+c)^2}(a+b+c)=R^2-\frac{abc}{a+b+c}.\label{19} \end{align}

\begin{align} Rr=\frac{abc}{4\triangle}\cdot \frac{\triangle}s=\frac{abc}{2(a+b+c)}.\label{20} \end{align}

\begin{align} d^2=R^2-2\cdot \frac{abc}{2(a+b+c)} =R^2-2Rr=R(R-2r).\label{21} \end{align}

### 參考文獻

Euler's theorem in geometry, Wikipedia. https://en.wikipedia.org/wiki/Euler%27s_theorem_in_geometry 阮瑞泰。 三角形的四心之向量關係式。 數學傳播季刊, 34(1), 29-34, 2010。 https://web.math.sinica.edu.tw/math_media/d341/34103.pdf

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