設 $\Delta ABC$ 的三邊長為 $\overline{AB} = 8$，$\overline{BC} = 2\sqrt{13}$，$\overline{CA} = 4$，且 $H$ 為 $\Delta ABC$ 的垂心，若 $\overset{\large\rightharpoonup}{AH} = x\overset{\large\rightharpoonup}{AB} + y\overset{\large\rightharpoonup}{AC}$，則數對 $(x,y) = \text{____}$。

85 指考數學甲第 17 題

由於 $H$ 為 $\Delta ABC$ 的垂心，因此 $\overset{\large\rightharpoonup}{BH} \perp \overset{\large\rightharpoonup}{AC}$ 且 $\overset{\large\rightharpoonup}{CH} \perp \overset{\large\rightharpoonup}{AB}$。由此可得內積關係式： $$\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AH} = \overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AH} = \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}.$$ 利用餘弦定理求出 $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}$： $$\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} = \overline{AB} \cdot \overline{AC} \cos A = \dfrac{\overline{AB}^2 + \overline{AC}^2 - \overline{BC}^2}{2}$$ $$= \dfrac{8^2 + 4^2 - (2\sqrt{13})^2}{2} = \dfrac{64 + 16 - 52}{2} = 14.$$ 因此，有： $$\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AH} = 14 \ \text{且} \ \overset{\large\rightharpoonup}{AC} \cdot \overset{\large\rightharpoonup}{AH} = 14.$$ 將 $\overset{\large\rightharpoonup}{AH} = x\overset{\large\rightharpoonup}{AB} + y\overset{\large\rightharpoonup}{AC}$ 代入上列兩式： (1) $\overset{\large\rightharpoonup}{AB} \cdot (x\overset{\large\rightharpoonup}{AB} + y\overset{\large\rightharpoonup}{AC}) = 14 \Rightarrow x |\overset{\large\rightharpoonup}{AB}|^2 + y (\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}) = 14 \Rightarrow 64x + 14y = 14 \Rightarrow 32x + 7y = 7.$ (2) $\overset{\large\rightharpoonup}{AC} \cdot (x\overset{\large\rightharpoonup}{AB} + y\overset{\large\rightharpoonup}{AC}) = 14 \Rightarrow x (\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}) + y |\overset{\large\rightharpoonup}{AC}|^2 = 14 \Rightarrow 14x + 16y = 14 \Rightarrow 7x + 8y = 7.$ 聯立兩方程式： $$\begin{cases} 32x + 7y = 7 \\ 7x + 8y = 7 \end{cases}$$ 相減得 $25x - y = 0 \Rightarrow y = 25x$。代入得 $7x + 8(25x) = 7 \Rightarrow 207x = 7 \Rightarrow x = \dfrac{7}{207}$。則 $y = 25 \times \dfrac{7}{207} = \dfrac{175}{207}$。故數對 $(x,y) = \left(\dfrac{7}{207}, \dfrac{175}{207}\right)$。