三角形の各辺の長さと傍接円半径と面積の関係

確認

$$\begin{eqnarray} \triangle BCD&=&\frac{1}{2} a\;\overline{DE}=\frac{1}{2} a\;r_c \\\triangle ACD&=&\frac{1}{2} b\;\overline{DF}=\frac{1}{2} b\;r_c \\\triangle ABD&=&\frac{1}{2} c\;\overline{DG}=\frac{1}{2} c\;r_c \end{eqnarray}$$

面積と各傍接円半径の関係性

$$\begin{eqnarray} \triangle ABC&=&\triangle BCD+\triangle ACD-\triangle ABD \\&=&\frac{1}{2} r_c (a + b - c) \\&=&\frac{1}{2} r_c (a + b - c + c - c) \\&=&\frac{1}{2} r_c (a + b + c - 2c) \\&=&r_c (\frac{a + b + c}{2} - \frac{2c}{2}) \\&=&r_c (s-c)\cdots s=\frac{a+b+c}{2}:半周長 \end{eqnarray}$$ 同様に… $$\begin{eqnarray} \triangle ABC&=&r_a (s-a) \\\triangle ABC&=&r_b (s-b) \end{eqnarray}$$

内接円半径と面積の関係性

$$\begin{eqnarray} S_{ABC}&=&r\cdot s \cdots r:内接円半径 \\&=&r \frac{a+b+c}{2} \\&=&\frac{1}{2} r a+\frac{1}{2} r b+\frac{1}{2} r c \\&&\cdots内接円は各辺に接するので，その半径は各辺に対して垂直であり高さとなる． \end{eqnarray}$$

内接円半径及び傍接円半径と面積の関係性

$$\begin{eqnarray} S_{ABC}&=&\triangle ABC \\S_{ABC}^4&=&r\; s \cdot r_a\;(s-a)\cdot r_b\;(s-b) \cdot r_c\;(s-c) \\&=&s(s-a) (s-b) (s-c) \cdot r r_a r_b r_c \\S_{ABC}^{\cancel{4}2}&=&\cancel{S_{ABC}^2} \cdot r r_a r_b r_c;\cdots S_{ABC}^2=s (s-a) (s-b) (s-c):\href{https://shikitenkai.blogspot.com/2023/06/blog-post.html}{ヘロンの公式} \\S_{ABC}^2&=& r r_a r_b r_c \\S_{ABC}&=&\sqrt{r r_a r_b r_c} \end{eqnarray}$$

cos(z)=2

$$\begin{eqnarray} \cos\left(z\right)&=&2\;\cdots\;z\in\mathbb{C} \end{eqnarray}$$

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$$\begin{eqnarray} \cos\left(z\right)&=&\frac{ e^{iz}+e^{-iz} }{2} \\&=&\frac{ X+X^{-1} }{2}\;\cdots\;X=e^{iz} \\\frac{ X+X^{-1} }{2}&=&2 \\X+X^{-1}&=&2\cdot2 \\X+X^{-1}&=&4 \\X-4+X^{-1}&=&0 \\X\cdot\left(X-4-X^{-1}\right)&=&X\cdot0 \\X^2-4X+1&=&0 \\X&=&\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot 1}}{2\cdot1} \\&=&\frac{4\pm\sqrt{12}}{2} \\&=&\frac{4\pm2\sqrt{3}}{2} \\&=&2 \pm \sqrt{3} \\X=e^{iz}&=&2 \pm \sqrt{3} \\iz&=&\log{\left(2\pm \sqrt{3}\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(2\pm \sqrt{3}\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(2\pm \sqrt{3}\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(2\pm \sqrt{3}\right)}+2\pi n \\&=&-i\log{\left(2\pm \sqrt{3}\right)}+2\pi n \end{eqnarray}$$

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$$\begin{eqnarray} \frac{ e^{iz}+e^{-iz} }{2} &=& \frac{ e^{iz}+e^{i(-z)} }{2} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) +\left(\cos{\left(-z\right)}+i\sin{\left(-z\right)}\right) }{2} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) +\left(\cos{\left(z\right)}-i\sin{\left(z\right)}\right) }{2} \\&=& \frac{ \cos{\left(z\right)}+\cancel{i\sin{\left(z\right)}} +\cos{\left(z\right)}\cancel{-i\sin{\left(z\right)}} }{2} \\&=& \frac{ \cancel{2}\cos{\left(z\right)} }{\cancel{2}} \\&=&\cos{\left(z\right)} \end{eqnarray}$$

sin(z)=2

$$\begin{eqnarray} \sin\left(z\right)&=&2\;\cdots\;z\in\mathbb{C} \end{eqnarray}$$

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$$\begin{eqnarray} \sin\left(z\right)&=&\frac{ e^{iz}-e^{-iz} }{2i} \\&=&\frac{ X-X^{-1} }{2i}\;\cdots\;X=e^{iz} \\\frac{ X-X^{-1} }{2i}&=&2 \\X-X^{-1}&=&2\cdot2i \\X-X^{-1}&=&4i \\X-4i-X^{-1}&=&0 \\X\cdot\left(X-4i-X^{-1}\right)&=&X\cdot0 \\X^2-4iX-1&=&0 \\X&=&\frac{-(-4i)\pm\sqrt{(-4i)^2-4\cdot1\cdot(-1)}}{2\cdot1} \\&=&\frac{4i\pm\sqrt{-12}}{2} \\&=&\frac{4i\pm2\sqrt{3}i}{2} \\&=&2i \pm \sqrt{3}i \\X=e^{iz}&=&2i \pm \sqrt{3}i \\&=&\left(2\pm \sqrt{3}\right)i \\iz&=&\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n \;\cdots\; n \in \mathbb{Z} \\z&=&\frac{1}{i}\left\{\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi i n\right\} \\&=&\frac{i}{i}\frac{1}{i}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+\frac{1}{\cancel{i}}2\pi \cancel{i} n \\&=&\frac{i}{-1}\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \\&=&-i\log{\left(\left(2\pm \sqrt{3}\right)i\right)}+2\pi n \end{eqnarray}$$

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$$\begin{eqnarray} \frac{ e^{iz}-e^{-iz} }{2i} &=& \frac{ e^{iz}-e^{i(-z)} }{2i} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) -\left(\cos{\left(-z\right)}+i\sin{\left(-z\right)}\right) }{2i} \\&=& \frac{ \left(\cos{\left(z\right)}+i\sin{\left(z\right)}\right) -\left(\cos{\left(z\right)}-i\sin{\left(z\right)}\right) }{2i} \\&=& \frac{ \cancel{\cos{\left(z\right)}}+i\sin{\left(z\right)} \cancel{-\cos{\left(z\right)}}+i\sin{\left(z\right)} }{2i} \\&=& \frac{ \cancel{2i}\sin{\left(z\right)} }{\cancel{2i}} \\&=&\sin{\left(z\right)} \end{eqnarray}$$