多角形D上の点から任意の点への距離の二乗の平均値(グリーンの定理を用いて面積分を周回積分にして求める)

グリーンの定理(Green's theorem)

閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

---

面積分を周回積分へ変形し，区間毎の線積分に展開する

点\(x,y\)から点\((p,q)\)までの距離の二乗は $$ \begin{eqnarray} \langle r^2 \rangle&=&\left\{\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right\}^2 \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ であり，多角形D上の点(\(x,y\))から任意の点(\(p,q\))への距離の二乗の平均値\(\langle r^2 \rangle\)は， $$ \begin{eqnarray} \langle r^2 \rangle&=&\iint_D \rho(x,y)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\};\mathrm{d}x\mathrm{d}y \\&=&\iint_D \frac{1}{A}\left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積) \\&=&\frac{1}{A}\iint_D \left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x \end{eqnarray} $$ である．

これをグリーンの定理の式で満たすために例えば $$ \begin{eqnarray} Q(x, y)&=&\int \left(x-p\right)^2\;\mathrm{d}x \\P(x, y)&=&-\int \left(y-q\right)^2\;\mathrm{d}y \end{eqnarray} $$ とおく．これは $$ \begin{eqnarray} \frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y} &=& \int \left(x-p\right)^2\;\mathrm{d}x - \left\{-\int \left(y-q\right)^2\;\mathrm{d}y\right\} \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この\(P,Q\)を用いて線積分側の 被積分凾数を求めると以下のようになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\int_D \rho\left(x, y\right)r\left(x,y;p, q\right)^2 \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \rho\left(x, y\right)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \frac{1}{A}\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\int_D \left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\left[ \int_D \left[ \frac{\partial }{\partial x} \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} - \frac{\partial }{\partial y} \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \right] \mathrm{d}x\mathrm{d}y \right] \\&=&\frac{1}{A}\left[ \oint_C \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} \mathrm{d}y + \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \mathrm{d}x \right] \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y - \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y -\oint_C \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \end{eqnarray} $$

---

\(\left\{\right\}\)内の第一項 区間毎の線積分

$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=& \frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{\left(\epsilon_k y+\zeta_k\right)-p\right\}^3\;\mathrm{d}y \;\cdots\;x=\epsilon_k y+\zeta_k 多角形の区間(頂点k-1からk)毎の直線の式 \\&=&\frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{ \left(\epsilon_k y+\zeta_k\right)^3 -3\left(\epsilon_k y+\zeta_k\right)^2 p +3\left(\epsilon_k y+\zeta_k\right) p^2 -p^3 \right\}\;\mathrm{d}y \;\cdots\;(A-B)^3=A^3-3A^2B+3AB^2-B^3 \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^4 -3\frac{1}{3\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^3 p +3\frac{1}{2\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^2 p^2 -y p^3 \right]_{y_{k-1}}^{y_k} \;\cdots\;\int cf(x) \mathrm{d}x= c\int f(x) \mathrm{d}x ,\;\int_a^b x^a \mathrm{d}x= \left[\frac{1}{a+1}x^{a+1}\right]_a^b \\&=&\frac{1}{3}\sum_{k=1} \left[ \frac{1}{4\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^4 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^4 \right\} -\frac{1}{\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^3 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^3 \right\} p +\frac{3}{2\epsilon_k}\left\{ \left(\epsilon_k y_{k}+\zeta_k\right)^2 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^2 \right\} p^2 -\left( y_k - y_{k-1} \right) p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}^4-y_{k-1}^4) + 4 \epsilon_k^3 \zeta_k (y_{k}^3-y_{k-1}^3) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}^2-y_{k-1}^2) + 4 \epsilon_k \zeta_k^3 (y_{k} -y_{k-1} ) \right\} -\frac{1}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^3-y_{k-1}^3) + 3 \epsilon_k^2 \zeta_k (y_{k}^2-y_{k-1}^2) + 3 \epsilon_k \zeta_k^2 (y_{k} -y_{k-1} ) \right\} p +\frac{3}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}^2-y_{k-1}^2) + 2 \epsilon_k \zeta_k (y_{k} -y_{k-1} ) \right\} p^2 -\left( y_k -y_{k-1} \right) p^3 \right] \\&&\;\cdots\;(a x + b)^4-(a y + b)^4 =a^4 x^4 - a^4 y^4 + 4 a^3 b x^3 - 4 a^3 b y^3 + 6 a^2 b^2 x^2 - 6 a^2 b^2 y^2 + 4 a b^3 x - 4 a b^3 y =a^4 (x^4 - y^4) + 4 a^3 b (x^3 - y^3) + 6 a^2 b^2 (x^2 - y^2) + 4 a b^3 (x - y) \\&&\;\cdots\;(a x + b)^3-(a y + b)^3 =a^3 x^3 - a^3 y^3 + 3 a^2 b x^2 - 3 a^2 b y^2 + 3 a b^2 x - 3 a b^2 y =a^3 (x^3 - y^3) + 3 a^2 b (x^2 - y^2) + 3 a b^2 (x - y) \\&&\;\cdots\;(a x + b)^2-(a y + b)^2 =a^2 x^2 - a^2 y^2 + 2 a b x - 2 a b y =a^2 (x^2 - y^2) + 2 a b (x - y) \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}-y_{k-1})(y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 (y_{k}-y_{k-1}) \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 (y_{k}-y_{k-1}) \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k (y_{k}-y_{k-1}) \right\} -p^3\left( y_k -y_{k-1} \right) \right] \\&&\;\cdots\;(a^4-b^4)=(a-b)(a+b)(a^2+b^2) \\&&\;\cdots\;(a^3-b^3)=(a-b)(a^2+ab+b^2) \\&&\;\cdots\;(a^2-b^2)=(a-b)(a+b) \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \end{eqnarray} $$

\(\epsilon_k,\zeta_k\)の代入と整理

$$ \begin{eqnarray} \epsilon_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{x_{k}-x_{k-1}}{y_{k}-y_{k-1}}} \\ \zeta_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1})}} \\ \\&& \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \\&=& \frac{(x_{k}-x_{k-1})^3}{(y_{k}-y_{k-1})^3} (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \frac{(x_{k}-x_{k-1})}{(y_{k}-y_{k-1})} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k} +y_{k-1}) + 4 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^3}{(y_{k}-y_{k-1})^3} \\&=&\frac{1}{(y_{k}-y_{k-1})^3} \left\{ (x_{k}-x_{k-1})^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 (x_{k}-x_{k-1})^2 (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1})^2 (y_{k} +y_{k-1}) + 4 (x_{k-1}y_{k}-x_{k}y_{k-1})^3 \right\} \\&=& (x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2) \\ \\&& \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \\&=& \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \frac{(x_{k}-x_{k-1}) }{(y_{k}-y_{k-1}) } \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k} +y_{k-1}) + 3 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} \\&=&\frac{1}{(y_{k}-y_{k-1})^2} \left\{ (x_{k}-x_{k-1})^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k} +y_{k-1}) + 3 (x_{k-1}y_{k}-x_{k}y_{k-1})^2 \right\} \\&=&x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2 \\ \\&& \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \\&=& \frac{(x_{k}-x_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k} +y_{k-1}) + 2 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } \\&=&\frac{1}{(y_{k}-y_{k-1})} \left\{ (x_{k}-x_{k-1}) (y_{k} +y_{k-1}) + 2 (x_{k-1}y_{k}-x_{k}y_{k-1}) \right\} \\&=&x_{k}+x_{k-1} \end{eqnarray} $$

\(\left\{\right\}\)内の第一項　整理後

$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -p\left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{3p^2}{2}\left(x_{k}+x_{k-1}\right) -p^3 \right] \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) -\frac{p^3}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \;\cdots\;\sum_{k=1}^n (y_{k}-y_{k-1}) = 0\;(周回積分なのでy軸の移動量の和は一周したら元に戻る) \end{eqnarray} $$

---

\(\left\{\right\}\)内の第二項　整理後

$$ \begin{eqnarray} -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x &=& -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \end{eqnarray} $$

---

項毎に計算したものをまとめる

$$ \begin{eqnarray} \langle r^2 \rangle &=& \frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \right] -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \right] \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -2p \frac{1}{6} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) -2q \left\{-\frac{1}{6} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) \right\} +p^2\frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) +q^2\left\{-\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\right\} \right] \right] \\&=& \frac{1}{A}\left[ \lambda+\mu -2p A X_G-2q A Y_G +p^2 A+q^2 A \right] \\&&\;\cdots\;\lambda=\frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} \\&&\;\cdots\;\mu=-\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{X_G=\frac{1}{6A} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right)\;(重心X軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{Y_G=-\frac{1}{6A} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right)\;(重心Y軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post.html}{A= \frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right)\;(面積)} \\&&\;\cdots\;A= -\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\;(面積) \\&=& \frac{\lambda+\mu}{A} -2p X_G -2q Y_G +p^2 +q^2 \\&=& \frac{\lambda+\mu}{A} {\color{red}+X_G^2} -2p X_G +p^2 {\color{red}+Y_G^2} -2q Y_G +q^2 {\color{red}-X_G^2} {\color{red}-Y_G^2} \\&=& \frac{\lambda+\mu}{A} +\left(p-X_G\right)^2 +\left(q-Y_G\right)^2 -X_G^2 -Y_G^2 \end{eqnarray} $$ よって\(\langle r^2 \rangle\)は多角形によって事前に決定される\(A,X_G,Y_G,\lambda,\mu\)と，点\((p,q)\)が与えられることで決まる．

\(\langle r^2 \rangle\)の最小値

\(\langle r^2 \rangle\)の最小値となる点\((u, v)\)は，\(\left(p-X_G\right),\left(p-X_G\right)\)が0となる点\((X_G, Y_G)\)である．

また，点\((X_G, Y_G)\)から\(\langle r^2 \rangle\)の等しい点\((p, q)\)は，以下の式を満たすことになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\frac{\lambda+\mu}{A}+\left(p-X_G\right)^2+\left(q-Y_G\right)^2-X_G^2-Y_G^2 \\\left(p-X_G\right)^2+\left(q-Y_G\right)^2&=&\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A} \\\sqrt{\left(p-X_G\right)^2+\left(q-Y_G\right)^2}&=&\sqrt{\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A}} \end{eqnarray} $$