新竹縣立勝利國民中學112學年度正式教師甄選初試
一、選擇題(每題2分，共50分)
解答：$$\cases{\sqrt{x-113+y} \ge 0 \Rightarrow x-113+y\ge 0\\ \sqrt{113-x-y} \ge 0 \Rightarrow -(x-113+y) \ge 0} \Rightarrow x-113+y=0 \Rightarrow \cases{x+y=113\\ \sqrt{x-113+y}=0} \\ \Rightarrow \cases{3x+5y-4-m=0 \\ 2x+3y-m=0} \Rightarrow 3x+5y-4=m=2x+3y \Rightarrow \cases{x+y=113\\ x+2y=4} \\ \Rightarrow \cases{x=222\\ y=-109} \Rightarrow m=2x+3y= 444-327=117，故選\bbox[red, 2pt]{(C)}$$
$$假設\cases{A(0,3)\\ B(0,0) \\C(6,0)\\ D(6,3)} \Rightarrow \cases{E(2,0)\\ F(4,0)} \Rightarrow \cases{\overleftrightarrow{BD}: 2y=x\\ \overleftrightarrow{AE}:3(x-2)+2y=0\\ \overleftrightarrow{AF}: 3(x-4)+4y=0} \Rightarrow \cases{P= \overleftrightarrow{BD}\cap \overleftrightarrow{AE} =(3/2, 3/4)\\Q= \overleftrightarrow{BD}\cap \overleftrightarrow{AF} =(12/5,6/5)} \\ \Rightarrow \overline{BP}: \overline{PQ}: \overline{QD}= {3\over 2}:{12\over 5}-{3\over 2}: 6-{12\over 5} =5:3:12，故選\bbox[red, 2pt]{(D)}$$
解答：
$$對同弧的圓周角與弦切角相等\Rightarrow \cases{\angle QBP =\angle SCP \Rightarrow \triangle PBQ \sim \triangle PCS(AAA)\\ \angle RCP= \angle SBP \Rightarrow \triangle PBS \sim \triangle PCR (AAA)} \\ \Rightarrow \cases{{\overline{PQ} \over \overline{PS}} ={\overline{PB} \over \overline{PC}} \\ {\overline{PS} \over \overline{PR}} ={\overline{PB} \over \overline{PC}}} \Rightarrow {\overline{PQ} \over \overline{PS}}={\overline{PS} \over \overline{PR}} \Rightarrow \overline{PS}^2 =\overline{PQ}\times \overline{PR}= 4\cdot 6=24 \Rightarrow \overline{PS}=2 \sqrt 6\\，故選\bbox[red, 2pt]{(B)}$$
解答：$${1\over x}+{1\over y}={1\over 24} \Rightarrow {1\over x}={1\over 24}-{1 \over y}={y-24\over 24y} \Rightarrow x={24y\over y-24} \Rightarrow x=24+{24^2\over y-24} \\ x\gt y \Rightarrow y\lt 48 且y-24是24^2的因數,因此 y=25,26,27, 28,30, 32,33,36,40,42，共10組解\\，故選\bbox[red, 2pt]{(B)}$$
解答：
$$假設圓半徑r \Rightarrow \overline{OE} =\sqrt{36-r^2}，\\作\overline{FP} \bot \overline{OB} \Rightarrow \triangle DOE\sim \triangle FPE (AAA) \Rightarrow {\overline{DE} \over \overline{EF}} ={6\over 2} ={\overline{OD}=r \over \overline{FP}} ={\overline{OE} \over \overline{EP}} \Rightarrow \cases{\overline{FP}=r/3\\ \overline{EP}= \sqrt{36-r^2}/3}\\ 直角\triangle FOP: \overline{OF}^2 =\overline{OP}^2 +\overline{FP}^2 \Rightarrow r^2= \left( \sqrt{36-r^2} +{\sqrt{36-r^2} \over 3}\right)^2 +({r\over 3})^2 \\ \Rightarrow {8\over 9}r^2= {16\over 9}(36-r^2) \Rightarrow r^2=24 \Rightarrow 圓面積=24\pi，故選\bbox[red, 2pt]{(C)}$$
解答：$$x=2+{8\over 2+{8\over 2+{8\over \dots}}} \Rightarrow {8\over x-2}=x \Rightarrow x^2-2x-8=0\\ \Rightarrow (x-4)(x+2)=0 \Rightarrow x=4(x\gt 0 \Rightarrow 負值不合)，故選\bbox[red, 2pt]{(B)}$$
解答：$$假設車子按喇叭時距山谷a公尺；車子速度72公里/時 = 72000/3600=20公尺/秒\\ 聲音四秒走了2a-20\times 4=2a-80公尺\Rightarrow {2a-80\over 4}=340 \Rightarrow a=720 \Rightarrow a-80=640\\，故選\bbox[red, 2pt]{(A)}$$
解答：$$假設三邊長為a,b,c 且a\gt b\gt c,並滿足\cases{a+b+c=450\\ b+c\gt a} \Rightarrow 450\gt 2a \Rightarrow a\le 224\\ 又a+a+a \gt a+b+c \Rightarrow 3a\gt 450 \Rightarrow a\ge 151\\ 因此151\le a\le 224 \Rightarrow \begin{array}{} a & (b,c)\\\hline 151& (150,149)\\ 152 & (151, 147)\\ & (150,148) \\ 153& (152,145) \\ &(151,146)\\ & (150,147) \\ & (149,148) \\\hdashline 154 & (153,143)\\ & \cdots\\ & (149,147)\\\hdashline 155 & (154,141)\\ & \cdots \\ & (148,147)\\ \cdots\\ 224 & (223,3) \\ & \cdots\\& (114,112)\\\hline \end{array}\\ 總數>> 900，故選\bbox[red, 2pt]{(D)}$$
解答：$${\log_a c\cdot \log_c{b\over a} \over \log_c b\cdot \log_{ab}c} ={\log_a c( \log_c b-\log_c a) \over {\log_a b\over \log_a c} \cdot {\log_a c\over \log_a ab}} = {\log_a c({\log_a b\over \log_a c}-{\log_a a \over \log_a c} )\over {\log_a b\over \log_a c}{\log_a c\over 1+\log_a b}} \\={\log_a b-1\over {\log_a b\over 1+\log_a b}} ={2-1\over {2\over 1+2}} ={3\over 2}，故選\bbox[red, 2pt]{(C)}$$
解答：$$假設4位數字為a，重複2次變為a\times 10^4+a = a(10^4+1)=10001a = 137\times 73\times a，故選\bbox[red, 2pt]{(C)}$$
解答：$$\sin 54^\circ=\cos 36^\circ \Rightarrow \sin (3\theta) =\cos(2\theta)，其中\theta=18^\circ \\ \Rightarrow 3\sin a-4\sin^3 a=1-2\sin^2 a \Rightarrow 4\sin^3a-2\sin^2a-3\sin a+1=0 \\ \Rightarrow (\sin a-1)(4\sin^2 a+2\sin a-1) =0 \Rightarrow \sin a={-1+\sqrt 5\over 4} \\ \Rightarrow \sin 54^\circ = 3\sin a-4\sin^3 a={-3+3\sqrt 5\over 4}-4\cdot {-2+\sqrt 5\over 8} ={1+\sqrt 5\over 4}\\ 欲求之對邊長度=4\sin 54^\circ =1+\sqrt 5，故選\bbox[red, 2pt]{(D)}$$
解答：$$，故選\bbox[red, 2pt]{()}$$
解答：
$$假設ADEF為平行四邊形\Rightarrow \cases{\triangle CEF:\triangle ABC=4:25 \Rightarrow \triangle CEF=3.2\\ \triangle BDE:\triangle ABC=9:25 \Rightarrow \triangle BDE=7.2} \\ \Rightarrow ADEF=20-3.2-7.2= 9.6 \Rightarrow \triangle ADE=9.6\div 2=4.8 \Rightarrow \triangle AEB=4.8+7.2=12\\，故選\bbox[red, 2pt]{(C)}$$
解答：$$\cases{若24是最長邊 \Rightarrow a^2+10^2 \gt 24^2 \\ 若a是最長邊\Rightarrow 10^2+24^2 \gt a^2} \Rightarrow 676\gt a^2 \gt 476 \Rightarrow a=25,24,23,22，共4個，故選\bbox[red, 2pt]{(A)}$$
解答：$$(b,c)=(12,1-12),(11,2-11),(10,3,10),(9,4-9), (8,5-8), (7,6-7)\\共有12+10+8 +6+4 +2=42個三角形，故選\bbox[red, 2pt]{(B)}$$
解答：
$$正弦定理:{\overline{AC} \over \sin \angle B}=2R \Rightarrow {4\sqrt 5\over \sin \angle B}=2\cdot 6 \Rightarrow \sin \angle B={\sqrt 5\over 3} \Rightarrow \overline{AD} =\overline{AB} \sin \angle B ={20\over 3}\\ 假設小圓半徑=r \Rightarrow \overline{DE}=2r =\overline{AE}-\overline{AD}=12-{20\over 3} ={16\over 3} \Rightarrow r={8\over 3}，故選\bbox[red, 2pt]{(B)}$$
解答：$$2022a^2=2023b^2 \Rightarrow \cases{2022a=2023b^2/a \cdots(1)\\ 2023b=2022a^2/b \cdots(2)} , 將(1)及(2)分別代入\sqrt{2022a+2023b} \\\Rightarrow\cases{\sqrt{2022a+2023b}=\sqrt{2023b^2/a+ 2023b} =\sqrt{2023b \cdot ({b\over a}+1)} =\sqrt{2023b\cdot {a+b\over a}} =\sqrt{2023b\cdot {ab\over a}} =\sqrt{2023b^2} \\ \sqrt{2022a+2023b} =\sqrt{2022a+2022a^2/b} =\sqrt{2022a^2}} \\ \Rightarrow \sqrt{2023}b=\sqrt{2022}a \Rightarrow \cases{b/a=\sqrt{2022/2023}\\ a/b=\sqrt{2023/2022}} \Rightarrow \cases{1/a=(1/b)\cdot \sqrt{2022/2023} \\ 1/b=(1/a)\cdot \sqrt{2023/2022}} \\ \Rightarrow \cases{1/a+1/b=(1/b)\cdot \sqrt{2022/2023}+1/b=(1/b)(\sqrt{2022/2023}+1)=1 \\ 1/a+1/b=1/a+ (1/a)\cdot \sqrt{2023/2022} =(1/a)(\sqrt{2023/2022}+1)=1} \\ \Rightarrow \cases{b=\sqrt{2022/2023}+1 \\ a=\sqrt{2023/2022}+1} \Rightarrow \sqrt{2022a+2023b} = \sqrt{\sqrt{2022\cdot 2023}+2022 +\sqrt{2023\cdot 2022}+2023} \\= \sqrt{(2023+2022)+2 \sqrt{2022\cdot 2023}} =\sqrt{2022}+\sqrt{2023}，故選\bbox[red, 2pt]{(D)}$$
解答：
$$等腰直角\triangle ABC \Rightarrow \overline{AC} =2\sqrt 2\\ 由於B,Q均為小圓切點 \Rightarrow \overline{CQ} =\overline{CB}=2 \Rightarrow \overline{AQ}=2\sqrt 2-2\\ 直角\triangle AOQ: \overline{AO}^2 = \overline{OQ}^2+ \overline{AQ}^2 \Rightarrow (2-r)^2 = r^2+(2\sqrt 2-2)^2 \Rightarrow r=2\sqrt 2-2 \\ \Rightarrow 直徑=4\sqrt 2-4，故選\bbox[red, 2pt]{(C)}$$
解答：$$a_n=(a_1+a_2+ \cdots +a_n)-(a_1+a_2+ \cdots +a_{n-1}) =n^2a_n-(n-1)^2a_{n-1} \Rightarrow (n^2-1)a_n= (n-1)^2a_{n-1} \\ \Rightarrow {a_n\over a_{n-1}} ={(n-1)^2 \over n^2-1} ={n-1\over n+1} \Rightarrow {a_n\over a_{n-1}}\times {a_{n-1}\over a_{n-2}} \times\cdots \times{a_2\over a_1} ={n-1\over n+1}\times {n-2 \over n} \times \cdots \times {1\over 3} \\ \Rightarrow {a_n\over a_1} ={(n-1)!\over (n+1)!/2} \Rightarrow a_n={2a_1\over (n+1)n} \Rightarrow a_{2023}={2\cdot 2023\over 2024\cdot 2023} ={1\over 1012}，故選\bbox[red, 2pt]{(B)}$$
解答：$$由圖形可知:\overline{AD}=7 \Rightarrow d-a=7 \Rightarrow \cases{d-a=7\\ d-2a=10} \Rightarrow \cases{a=-3\\ d=4} \\\Rightarrow 原點在A的右邊且距離=3,也就是B，故選\bbox[red, 2pt]{(B)}$$
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解題僅供參考，第一題至第五題非數學專業(教育專業題），其他教甄試題及詳解
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