47109 麥比烏斯定理的聯想

\begin{align} S_{\triangle A'B'C'}=\dfrac 12 S_{\triangle ABC}.\label{1} \end{align}

\begin{align} S_{\triangle A'B'C'}\lt\dfrac 12 S_{\triangle ABC}.\label{2} \end{align}

\begin{align} S_{\triangle ABC}=&\,\frac 12\left|\begin{array}{ccccc} a&~~&a^3+pa&~~&1\\ b&~~&b^3+pb&~~&1\\ c&~~&c^3+pc&~~&1 \end{array}\right|=\frac 12\left|\begin{array}{ccccc} a&~~&a^3&~~&1\\ b&~~&b^3&~~&1\\ c&~~&c^3&~~&1 \end{array}\right|\nonumber\\ =&\,\frac 12(b-a)(c-a)(c-b)(a+b+c).\label{3}\end{align}

\begin{align*} y=&\,(3a^2+p)x-2a^3;\\ y=&\,(3b^2+p)x-2b^3;\\ y=&\,(3c^2+p)x-2c^3. \end{align*}

\begin{align*} &A'\Big(\frac 23\cdot \frac{a^2+ab+b^2}{a+b},\frac 23\cdot \frac{p(a^2+ab+b^2)}{a+b}+\frac{2a^2b^2}{a+b}\Big);\\ &B'\Big(\frac 23\cdot \frac{c^2+ca+a^2}{c+a},\frac 23\cdot \frac{p(c^2+ca+a^2)}{c+a}+\frac{2c^2a^2}{c+a}\Big);\\ &C'\Big(\frac 23\cdot \frac{b^2+bc+c^2}{b+c},\frac 23\cdot \frac{p(b^2+bc+c^2)}{b+c}+\frac{2b^2c^2}{b+c}\Big). \end{align*}

\begin{align*} S_{\triangle A'B'C'}=&\,\frac 12\left|\begin{array}{ccccc} \dfrac 23\cdot \dfrac{a^2+ab+b^2}{a+b}&~~&\dfrac 23\cdot \dfrac{p(a^2+ab+b^2)}{a+b}+\dfrac{2a^2b^2}{a+b}&~~&1\\[8pt] \dfrac 23\cdot \dfrac{c^2+ca+a^2}{c+a}&~~&\dfrac 23\cdot \dfrac{p(c^2+ca+a^2)}{c+a}+\dfrac{2c^2a^2}{c+a}&~~&1\\[8pt] \dfrac 23\cdot \dfrac{b^2+bc+c^2}{b+c}&~~&\dfrac 23\cdot \dfrac{p(b^2+bc+c^2)}{b+c}+\dfrac{2b^2c^2}{b+c}&~~&1 \end{array}\right|\\ =&\,\frac 2{3(a+b)(b+c)(c+a)}\left|\begin{array}{ccccc} {a^2+ab+b^2}&~~&a^2b^2&~~&a+b\\ {c^2+ca+a^2}&~~&c^2a^2&~~&c+a\\ {b^2+bc+c^2}&~~&b^2c^2&~~&b+c \end{array}\right|.\end{align*}

\begin{align*} \left|\begin{array}{ccccc} {a^2+ab+b^2}&~~&a^2b^2&~~&a+b\\ {c^2+ca+a^2}&~~&c^2a^2&~~&c+a\\ {b^2+bc+c^2}&~~&b^2c^2&~~&b+c \end{array}\right|=(b-a)(c-a)(c-b)(ab+bc+ca)^2, \end{align*}

\begin{align} S_{\triangle A'B'C'}=\frac 23\cdot \frac{(b-a)(c-a)(c-b)}{(a+b)(b+c)(c+a)}(ab+bc+ca)^2. \label{4} \end{align}

\begin{align} \frac{S_{\triangle A'B'C'}}{S_{\triangle ABC}}=\frac 43\cdot \frac{(ab+bc+ca)^2}{(a+b)(b+c)(c+a)(a+b+c)}.\label{5} \end{align}

\begin{align*} {(a+b)(b+c)(c+a)(a+b+c)}=&\,a(b^3+c^3)+b(c^3+a^3)+c(a^3+b^3)\\ &\,+2a^2b^2+2b^2c^2+2c^2a^2+4ab(a+b+c),\\ (ab+bc+ca)^2=&\,a^2b^2+b^2c^2+c^2a^2+2ab(a+b+c). \end{align*}

$$3(a+b)(b+c)(c+a)(a+b+c)\gt8(ab+bc+ca)^2,$$

\begin{align*} &\hskip -20pt 3[a(b^3+c^3)+b(c^3+a^3)+c(a^3+b^3)+2a^2b^2+2b^2c^2+2c^2a^2+4ab(a+b+c)]\\ \gt&\,8a^2b^2+b^2c^2+c^2a^2+16abc(a+b+c), \end{align*}

\begin{align} 3[a(b^3+c^3)+b(c^3+a^3)+c(a^3+b^3)]\gt2(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c).\label{6} \end{align}

$$3[a(b^3+c^3)+b(c^3+a^3)+c(a^3+b^3)]\gt6(a^2b^2+b^2c^2+c^2a^2).$$

$$4(a^2b^2+b^2c^2+c^2a^2)\gt4abc(a+b+c),$$

$$6(a^2b^2+b^2c^2+c^2a^2)\gt2(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c),$$