三角形ABCにおいて、$AB = \sqrt{2}, BC = 2, CA = \sqrt{3}$とする。外心をOとするとき、$\overrightarrow{AO} = s\overrightarrow{AB} + t\overrightarrow{AC}$を満たす実数$s, t$の値を求めよ。

1. 問題の内容

三角形ABCにおいて、$AB = \sqrt{2}, BC = 2, CA = \sqrt{3}$とする。外心をOとするとき、$\overrightarrow{AO} = s\overrightarrow{AB} + t\overrightarrow{AC}$を満たす実数$s, t$の値を求めよ。

2. 解き方の手順

まず、$\overrightarrow{AO}$は外心なので、$|\overrightarrow{AO}| = |\overrightarrow{BO}| = |\overrightarrow{CO}|$が成立する。

また、$|\overrightarrow{AB}| = \sqrt{2}, |\overrightarrow{BC}| = 2, |\overrightarrow{CA}| = \sqrt{3}$である。

$\overrightarrow{BO} = \overrightarrow{AO} - \overrightarrow{AB} = (s-1)\overrightarrow{AB} + t\overrightarrow{AC}$

$\overrightarrow{CO} = \overrightarrow{AO} - \overrightarrow{AC} = s\overrightarrow{AB} + (t-1)\overrightarrow{AC}$

$|\overrightarrow{AO}|^2 = |\overrightarrow{BO}|^2$より、

$|\overrightarrow{AO}|^2 = s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$|\overrightarrow{BO}|^2 = (s-1)^2|\overrightarrow{AB}|^2 + 2(s-1)t\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2 = (s-1)^2|\overrightarrow{AB}|^2 + 2(s-1)t\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} = (s-1)^2|\overrightarrow{AB}|^2 + 2(s-1)t\overrightarrow{AB}\cdot\overrightarrow{AC}$

$s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} = (s^2 - 2s + 1)|\overrightarrow{AB}|^2 + (2st - 2t)\overrightarrow{AB}\cdot\overrightarrow{AC}$

$0 = (-2s+1)|\overrightarrow{AB}|^2 - 2t\overrightarrow{AB}\cdot\overrightarrow{AC}$

$\overrightarrow{AB}\cdot\overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}|\cos A = \sqrt{2}\sqrt{3}\cos A$

余弦定理より、

$\cos A = \frac{(\sqrt{2})^2 + (\sqrt{3})^2 - 2^2}{2\sqrt{2}\sqrt{3}} = \frac{2+3-4}{2\sqrt{6}} = \frac{1}{2\sqrt{6}}$

$\overrightarrow{AB}\cdot\overrightarrow{AC} = \sqrt{2}\sqrt{3}\frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{2\sqrt{6}} = \frac{1}{2}$

$0 = (-2s+1)|\overrightarrow{AB}|^2 - 2t\overrightarrow{AB}\cdot\overrightarrow{AC} = (-2s+1)(\sqrt{2})^2 - 2t(\frac{1}{2})$

$0 = 2(-2s+1) - t = -4s+2-t$

$t = -4s+2$

$|\overrightarrow{AO}|^2 = |\overrightarrow{CO}|^2$より、

$|\overrightarrow{AO}|^2 = s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$|\overrightarrow{CO}|^2 = s^2|\overrightarrow{AB}|^2 + 2s(t-1)\overrightarrow{AB}\cdot\overrightarrow{AC} + (t-1)^2|\overrightarrow{AC}|^2$

$t^2|\overrightarrow{AC}|^2 = 2s(t-1)\overrightarrow{AB}\cdot\overrightarrow{AC} + (t-1)^2|\overrightarrow{AC}|^2$

$s=0$ではないので、

$0 = 2s(t-1)\overrightarrow{AB}\cdot\overrightarrow{AC} + (t^2 - (t^2 - 2t + 1))|\overrightarrow{AC}|^2$

$0 = 2s(t-1)\overrightarrow{AB}\cdot\overrightarrow{AC} - (2t - 1)|\overrightarrow{AC}|^2$

$0 = 2s(t-1)(\frac{1}{2}) - (2t - 1)(\sqrt{3})^2 = s(t-1) - 3(2t-1)$

$s(t-1) = 3(2t-1)$

$s(t-1) = 6t - 3$

$s = \frac{6t-3}{t-1}$

$t = -4s+2$より、$s = \frac{6(-4s+2)-3}{(-4s+2)-1} = \frac{-24s+12-3}{-4s+1} = \frac{-24s+9}{-4s+1}$

$s(-4s+1) = -24s+9$

$-4s^2 + s = -24s + 9$

$4s^2 - 25s + 9 = 0$

$(4s-1)(s-9) = 0$

$s = \frac{1}{4}$または$s = 9$

もし$s = \frac{1}{4}$ならば、$t = -4(\frac{1}{4})+2 = -1+2 = 1$

もし$s = 9$ならば、$t = -4(9)+2 = -36+2 = -34$

もし$s=\frac{1}{4}, t=1$の場合、$s(t-1)=3(2t-1)$を満たさない。

$\frac{1}{4}(1-1)=0$, $3(2-1)=3$

もし$s=9, t=-34$の場合、$s(t-1)=3(2t-1)$を満たす。

$9(-34-1) = 9(-35) = -315$

$3(2(-34)-1) = 3(-68-1) = 3(-69) = -207$

よって$s = \frac{1}{4}, t=1$

以上より、$4s^2 - 25s + 9 = 0$

$(4s-1)(s-9) = 0$

$s=\frac{1}{4}$または$s=9$

$t=-4s+2$

もし$s=\frac{1}{4}$ならば$t = -4(\frac{1}{4})+2 = -1+2=1$

もし$s=9$ならば$t = -4(9)+2 = -36+2 = -34$

$\overrightarrow{AO} = \frac{1}{4}\overrightarrow{AB} + \overrightarrow{AC}$の場合: $|\overrightarrow{AO}|^2 = (\frac{1}{4})^2 |\overrightarrow{AB}|^2 + 2(\frac{1}{4})(1) \overrightarrow{AB}\cdot\overrightarrow{AC} + 1^2|\overrightarrow{AC}|^2 = \frac{1}{16}(2) + \frac{1}{2}(\frac{1}{2}) + 1(3) = \frac{1}{8} + \frac{1}{4} + 3 = \frac{1+2+24}{8} = \frac{27}{8}$

$\overrightarrow{BO} = (\frac{1}{4}-1)\overrightarrow{AB} + 1\overrightarrow{AC} = -\frac{3}{4}\overrightarrow{AB} + \overrightarrow{AC}$の場合: $|\overrightarrow{BO}|^2 = \frac{9}{16}(2) + 2(-\frac{3}{4})(1)(\frac{1}{2}) + 1(3) = \frac{9}{8} - \frac{3}{4} + 3 = \frac{9-6+24}{8} = \frac{27}{8}$

$\overrightarrow{CO} = \frac{1}{4}\overrightarrow{AB} + (1-1)\overrightarrow{AC} = \frac{1}{4}\overrightarrow{AB}$の場合: $|\overrightarrow{CO}|^2 = (\frac{1}{4})^2(2) = \frac{1}{16}(2) = \frac{1}{8}$

これは不適であるため、$s=9, t=-34$も不適である。

$\frac{1}{8} = \frac{27}{8}$

ここで計算間違いがあったのでやり直します。

$|\overrightarrow{AO}|^2 = s^2|\overrightarrow{AB}|^2 + 2st\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$|\overrightarrow{BO}|^2 = (s-1)^2|\overrightarrow{AB}|^2 + 2(s-1)t\overrightarrow{AB}\cdot\overrightarrow{AC} + t^2|\overrightarrow{AC}|^2$

$s^2(2) + 2st(\frac{1}{2}) + t^2(3) = (s-1)^2(2) + 2(s-1)t(\frac{1}{2}) + t^2(3)$

$2s^2 + st + 3t^2 = 2(s^2-2s+1) + (s-1)t + 3t^2$

$2s^2 + st = 2s^2 - 4s + 2 + st - t$

$0 = -4s + 2 - t$

$t = -4s + 2$

$|\overrightarrow{AO}|^2 = s^2(2) + 2st(\frac{1}{2}) + t^2(3)$

$|\overrightarrow{CO}|^2 = s^2(2) + 2s(t-1)(\frac{1}{2}) + (t-1)^2(3)$

$2s^2 + st + 3t^2 = 2s^2 + s(t-1) + 3(t^2-2t+1)$

$st = st - s + 3(-2t+1)$

$0 = -s - 6t + 3$

$s = -6t + 3$

$t = -4(-6t+3) + 2 = 24t - 12 + 2 = 24t - 10$

$23t = 10$

$t = \frac{10}{23}$

$s = -6(\frac{10}{23}) + 3 = \frac{-60}{23} + \frac{69}{23} = \frac{9}{23}$

3. 最終的な答え

$s = \frac{9}{23}$

$t = \frac{10}{23}$