國立臺南女中 111 學年度第 2 次教師甄選
填充題： (共 20 題，每題 5 分，共 100 分，答案請化為最簡分數或最簡根式)
解答：$$假設\cases{圓半徑R\\ \angle BOC=\theta} \Rightarrow \angle AOB=2\theta \Rightarrow \cases{\triangle OAB面積={1\over 2}R^2\sin 2\theta = 100 \\ \triangle OBC面積={1\over 2}R^2\sin \theta =80} \Rightarrow {\sin 2\theta\over \sin \theta }={100\over 80} \\ \Rightarrow {2\sin \theta\cos \theta \over \sin \theta}={5\over 4} \Rightarrow \cos\theta ={5\over 8} \Rightarrow \sin \theta = {\sqrt{39}\over 8} \Rightarrow \sin 3\theta = 3\sin \theta -4\sin^3\theta\\ \Rightarrow {\triangle OAC\over \triangle OBC}={\sin 3\theta \over \sin \theta} =3-4\sin^2\theta \Rightarrow \triangle OAC=(3-4\sin^2\theta)\cdot 80 =\bbox[red,2pt]{45}$$
$$假設\cases{\overline{CD}=x\\ \overline{AB}=\overline{AD}= R\\ \overline{AC}=a}， \triangle ABC 面積={1\over 2}aR\sin \theta = 4\sqrt 3 \Rightarrow \color{blue}{aR\sin \theta} =8\sqrt 3\\ \cos \angle CAD = \cos (\theta+45^\circ) = {a^2+R^2-x^2\over 2aR} \Rightarrow x^2=a^2+R^2-2aR\cos (\theta+45^\circ)\\ =a^2+R^2-2aR(\cos 45^\circ \cos\theta -\sin 45^\circ \sin \theta) =a^2+R^2-\sqrt 2aR(\cos\theta-\sin\theta) \\ =a^2+R^2-\sqrt 2aR\cos\theta+ \sqrt 2\color{blue}{aR\sin \theta} =a^2+R^2-\sqrt 2aR\cos\theta+8\sqrt 6 \ge 2\sqrt{a^2R^2} -\sqrt 2 aR\cos\theta +8\sqrt 6\\ =2aR- \sqrt 2aR\cos\theta +8\sqrt 6 =2\cdot {8\sqrt 3\over \sin \theta}-{8\sqrt 3\over \sin \theta}\cdot \cos \theta +8\sqrt 6= 8\sqrt 6({\sqrt 2-\cos\theta\over \sin \theta}+1)\\ \Rightarrow x^2 \ge 8\sqrt 6({\sqrt 2-\cos\theta\over \sin \theta}+1) =8\sqrt 6( k+1),其中k= {\sqrt 2-\cos\theta\over \sin \theta} \Rightarrow \cos\theta+k\sin\theta =\sqrt 2 \\ \Rightarrow \sqrt{k^2+1}\sin(\theta+\alpha)=\sqrt 2,其中\cases{\sin\alpha = 1/\sqrt{k^2+1} \\\cos \alpha = k/\sqrt{k^2+1}} \Rightarrow \sin (\theta+\alpha)= {\sqrt 2\over \sqrt{k^2+1}} \le 1\\ \Rightarrow \sqrt 2\le \sqrt{k^2+1} \Rightarrow k\ge 1 \Rightarrow x^2 \ge 8\sqrt 6(1+1)=16\sqrt 6 \Rightarrow x\ge \sqrt{16\sqrt 6} = \bbox[red,2pt]{4\sqrt[4]{6}}$$
$$x=\lfloor x\rfloor+\{x\} \Rightarrow \{x\}=x-\lfloor x\rfloor \Rightarrow x-\lfloor x\rfloor-kx+3k=0 \Rightarrow x-\lfloor x\rfloor=k(x-3)\\ 相當於求兩圖形\cases{\Gamma: y= x-\lfloor x\rfloor\\ L:k(x-3)}有三個相異交點時，k的範圍; \\ 而\cases{\Gamma 為斜率為1且通過(m,0)及(m+1,1),m\in \mathbb{Z}不含頂點(m+1,1)的線段\\ L為通過A(3,0)斜率為k的直線}，如上圖；\\因此\cases{k介於\overleftrightarrow{AB}與\overleftrightarrow{AC}之間\\ k介於\overleftrightarrow{AD}與\overleftrightarrow{AE}之間}，其中\cases{B(1,1)\\ C(2,1)\\ D(6,1)\\ E(7,1) } \Rightarrow \bbox[red, 2pt]{\cases{-1 \lt k\lt -{1\over 2} \\ {1\over 4}\lt k\lt {1\over 3}}}$$
解答：$$\overrightarrow{CD} =-\overrightarrow{AC}+\overrightarrow{AD} =m\overrightarrow{AB} +7\overrightarrow{AC} \Rightarrow \overrightarrow{AD}= m\overrightarrow{AB} +8\overrightarrow{AC} \Rightarrow \cfrac{\overline{BE}}{\overline{EC}} =\cfrac{8}{m} =\cfrac{a\triangle ABE}{a\triangle ACE}\\ \Rightarrow {8\over m}= {a\triangle ABE\over 1} \Rightarrow a\triangle ABE={8\over m}\\ 假設\overrightarrow{AD}=t\overrightarrow{AE}，則 t\overrightarrow{AE}=m\overrightarrow{AB} + 8\overrightarrow{AC} \Rightarrow \overrightarrow{AE}={m \over t}\overrightarrow{AB} +{8 \over t}\overrightarrow{AC} \Rightarrow {m\over t}+{ 8 \over t}=1 \Rightarrow t=m+8\\ \Rightarrow \cfrac{a\triangle ACD}{a\triangle ACE}= t=m+8 \Rightarrow a\triangle ACD=m+8 \Rightarrow a\triangle CDE=m+7 \\ \Rightarrow a\triangle ABE+a\triangle CDE = {8\over m}+m+7 \ge 2 \sqrt{{8\over m}\cdot m}+7 = \bbox[red,2pt]{4\sqrt 2+7}$$
解答：
$$將三位明星參加的場次以集合表示，見上圖，則\cases{A+D+G+F=15 \cdots(1)\\ B+D+E+G=16 \cdots(2)\\ C+E+F+G=17 \cdots(3)\\ G=5\\ D+E+F= 15}\\ (1)+(2)+(3) \Rightarrow A+B+C+ 2(D+E+F)+ 3G=48 \Rightarrow A+B+C=48-2\cdot 15-3\cdot 5=3\\ \Rightarrow A+B+C+D+E+F+G = 3+15+5=23 \\\Rightarrow 三球員均未得到\text{Double doubles}的場次=40-23= \bbox[red, 2pt]{17}$$
$$正六邊形每個內角為(6-2)180\div 6=120^\circ，兩頂點距離可能值為1,\sqrt 3, 2，其中6條長度為1\\、6條長度為\sqrt 3,3條長度為2；因此期望值=1\times {6\over 15}+ \sqrt{3}\times {6\over 15}+2\times {3\over 15} = \bbox[red, 2pt]{4+2\sqrt 3\over 5}$$
$$將數據(X,Y)畫在平面圖上，可看出將(13,12)去掉後，整體的斜率會變得最小\\\begin{array}{} X & Y & X^2 & Y^2 &XY \\ \hline 8 & 11 &64 &121 & 88\\ 9 & 12 & 81 & 144 & 108\\ 10 & 10 & 100 & 100 & 100\\ 11 & 8 & 121 & 64 & 88\\ 12 & 9 & 144 & 81 & 108\\ \hline 50 & 50 & 510 & 510 & 492\end{array} \Rightarrow 迴歸直線斜率b_1={\sum XY-{\sum X\sum Y\over n} \over \sum X^2-{(\sum x)^2\over n}} ={492-50\cdot 50/5\over 510-50^2/5} =-{4\over 5}\\ 迴歸直線通過(\bar x,\bar y)=(50/5,50/5)=(10,10) \Rightarrow y=-{4\over 5}(x-10)+10=-{4\over 5}x +18\\ \Rightarrow (m,k)=\bbox[red,2pt]{(-{4\over 5},18)}$$
$${x^2\over 64}+{y^2\over 16} \Rightarrow \cases{a=8\\ b=4} \Rightarrow c=\sqrt{64-16} =4\sqrt 3 \Rightarrow \cases{F_1(4\sqrt 3,0)\\ F_2(-4\sqrt 3,0)} \Rightarrow \overline{F_1F_2}=8\sqrt 3\\ 假設\cases{\overline{AF_2}=m \\ \overline{AF_1}=n} \Rightarrow \cases{m+n= 2a = 16\\ m^2+n^2 =\overline{F_1F_2}^2= 192} \Rightarrow \cases{m=8-4\sqrt 2\\ n=8+4\sqrt 2};\\假設\cases{\overline{BF_2}=s \\ \overline{BF_1}=t} \Rightarrow \cases{s+t= 2a = 16\\ (s+m)^2+n^2 =t^2 \Rightarrow (s+8-4\sqrt 2)^2+(8+4\sqrt 2)^2 =t^2} \\ \Rightarrow \cases{s=(24+4\sqrt 2)/17\\ t=(248-4\sqrt 2)/17} \Rightarrow 內切圓半徑=(\overline{AB}+\overline{AF_1}-\overline{BF_1})\div 2 = (s+m+n-t)\div 2\\ =\left(16+{24\sqrt 2\over 17}-{248-4\sqrt 2 \over 17}\right) \div 2=\bbox[red, 2pt]{24+ 4\sqrt 2\over 17}$$
$$假設\cases{\overline{AF}=\overline{FB}=a\\ \overline{AE}=\overline{EC}=b} \Rightarrow a^2+b^2=({\overline{BC}\over 2})^2 = 9^2=81\\ \cases{\overline{BE}斜率m_1=-b/2a\\ \overline{CF}斜率m_2=-2b/a} \Rightarrow \tan 150^\circ={m_2-m_1\over 1+ m_1m_2} \Rightarrow -{1\over \sqrt 3}={-3b/2a\over 1+b^2/a^2} =-{3b\cdot a^2\over 2(a^2+b^2)a} \\=-{3ab\over 162} =-{ab\over 54} \Rightarrow ab=18\sqrt 3 \Rightarrow \triangle ABC面積=2ab=36\sqrt 3 \Rightarrow \triangle GBC={1\over 3}\triangle ABC= \bbox[red, 2pt]{12\sqrt 3}$$
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