《2025年小题狂做高中数学必修第二册人教版巅峰版》

7. [2024河南郑州期中]在△ABC中,G为重心,AC=2$\sqrt{3}$,BG=2,则$\overrightarrow{AB}$·$\overrightarrow{BC}$=-6.

答案：
7.-6 设$AC$的中点为$D$，因为$G$为$\triangle ABC$的重心且$BG = 2$，所以$BD = 3$，$DA = \frac{1}{2}AC = \sqrt{3}$，$DC = -DA$，$\overrightarrow{AB} · \overrightarrow{BC} = -\overrightarrow{BA} · \overrightarrow{BC} = -(\overrightarrow{BD} + \overrightarrow{DA}) · (\overrightarrow{BD} + \overrightarrow{DC}) = -(\overrightarrow{BD} + \overrightarrow{DA}) · (\overrightarrow{BD} - \overrightarrow{DA}) = -(\overrightarrow{BD}^2 - \overrightarrow{DA}^2) = -[3^2 - (\sqrt{3})^2] = -6$.

8. [数学文化,2025安徽宏图中学期中]“奔驰定理”是平面向量中一个非常优美的结论,因为这个定理对应的图形与“奔驰”轿车的logo很相似,故形象地称其为“奔驰定理”.其内容为:已知O是△ABC内一点,△BOC,△AOC,△AOB的面积分别为Sₐ,Sb,Sc,则Sₐ·$\overrightarrow{OA}$+Sb·$\overrightarrow{OB}$+Sc·$\overrightarrow{OC}$=$\boldsymbol{0}$.设O是锐角三角形ABC的垂心,且3$\overrightarrow{OA}$+4$\overrightarrow{OB}$+5$\overrightarrow{OC}$=$\boldsymbol{0}$,则tan∠AOB=$-\sqrt{5}$.

答案：
8.$-\sqrt{5}$ 如图，延长$AO$交$BC$于点$D$，延长$BO$交$AC$于点$E$，延长$CO$交$AB$于点$F$.由题意，得$AO\bot BC$，$BO\bot AC$，$CO\bot AB$，根据$3\overrightarrow{OA} + 4\overrightarrow{OB} + 5\overrightarrow{OC} = 0$，结合“奔驰定理”可得$S_A:S_B:S_C = 3:4:5$，所以$S_{\triangle A}:S_{\triangle ABC} = 3:12 = 1:4$，则$OD:AD = 1:4$，同理可得$S_B:S_{\triangle ABC} = 1:3$，则$OE:BE = 1:3$.设$OD = m$，$OE = n$，则$OA = 3m$，$OB = 2n$，由$\cos\angle BOD = \cos\angle AOE$，得$\frac{OD}{OB} = \frac{OE}{OA}$，即$\frac{m}{2n} = \frac{n}{3m}$，化简得$3m^2 = 2n^2$，所以$\frac{m}{n} = \frac{\sqrt{2}}{\sqrt{3}}$，可得$\cos\angle BOD = \frac{OD}{OB} = \frac{m}{2n} = \frac{\sqrt{2}}{2\sqrt{3}} = \frac{\sqrt{6}}{6}$.结合$\angle AOB + \angle BOD = \pi$，可得$\cos\angle AOB = -\cos\angle BOD = -\frac{\sqrt{6}}{6}$，所以$\sin\angle AOB = \sqrt{1 - \cos^2\angle AOB} = \frac{\sqrt{30}}{6}$，可得$\tan\angle AOB = \frac{\sin\angle AOB}{\cos\angle AOB} = -\sqrt{5}$.

9. [2023江苏扬州中学3月月考]已知在△ABC中,AB=8,AC=6,∠A=60°,M为△ABC的外心.若$\overrightarrow{AM}$=λ$\overrightarrow{AB}$+μ$\overrightarrow{AC}$,λ,μ∈R,则$\frac{4λ+3μ}{4λ-6μ}$=7.

答案：
9.7 如图，设边$AB$，$AC$的中点分别为$D$，$E$，连接$DM$，$EM$.因为$M$为$\triangle ABC$的外心，故$DM\bot AB$，$EM\bot AC$，$\overrightarrow{AC} · \overrightarrow{AB} = |\overrightarrow{AC}| · |\overrightarrow{AB}|\cos A = 24$，$\overrightarrow{AM} = \overrightarrow{AD} + \overrightarrow{DM} = \frac{1}{2}\overrightarrow{AB} + \overrightarrow{DM}$，$\overrightarrow{AM} · \overrightarrow{AB} = (\frac{1}{2}\overrightarrow{AB} + \overrightarrow{DM}) · \overrightarrow{AB} = \frac{1}{2}\overrightarrow{AB}^2 = 32$.$\overrightarrow{AM} = \overrightarrow{AE} + \overrightarrow{EM} = \frac{1}{2}\overrightarrow{AC} + \overrightarrow{EM}$，$\overrightarrow{AM} · \overrightarrow{AC} = (\frac{1}{2}\overrightarrow{AC} + \overrightarrow{EM}) · \overrightarrow{AC} = \frac{1}{2}\overrightarrow{AC}^2 = 18$.依题意$\overrightarrow{AM} · \overrightarrow{AB} = (\lambda\overrightarrow{AB} + \mu\overrightarrow{AC}) · \overrightarrow{AB} = \lambda\overrightarrow{AB}^2 + \mu\overrightarrow{AC} · \overrightarrow{AB} = 64\lambda + 24\mu = 32$.$\overrightarrow{AM} · \overrightarrow{AC} = (\lambda\overrightarrow{AB} + \mu\overrightarrow{AC}) · \overrightarrow{AC} = \lambda\overrightarrow{AB} · \overrightarrow{AC} + \mu\overrightarrow{AC}^2 = 24\lambda + 36\mu = 18$，解得$\lambda = \frac{5}{12}$，$\mu = \frac{2}{9}$.所以
$4\lambda + 3\mu = 4 × \frac{5}{12} + 3 × \frac{2}{9} = \frac{7}{3} = 7$.