110年新竹市建功高中國中部-教師甄選-數學詳解

新竹市立建功高中 110 年第一次正式教師甄選【國中數學】 

一、 基礎題

解答： $$2020\div 2020{2020\over 2021}+{1\over 2022} =2020\div {2020\times 2021+2020\over 2021}+{1\over 2022} =2020\div {2020\times 2022\over 2021}+{1\over 2022} \\ ={2021\over 2022}+{1\over 2022} =\bbox[red,2pt]{1}$$

解答： $$(m-n):(m+n):mn = 1:7:8 \Rightarrow \cases{m-n=k \cdots(1)\\ m+n=7k \cdots(2)\\ mn=8k \cdots(3)}\\，由式(1)及式(2) \Rightarrow \cases{m=4k\\ n=3k} 代入(3) \Rightarrow 12k^2=8k \Rightarrow \cases{k=0(不合)\\ k=2/3} \Rightarrow m+n=7k= \bbox[red, 2pt]{14\over 3}$$

解答： $${1\over 7}=0.\overline{142857}，循環數為6；而5^n\div 6的餘數為5,1,5,1,\dots，循環數為2\Rightarrow 5^{2021} = 5 \mod 6；\\ 因此f(5^{2021})={1\over 7}小數點後第5位數字=\bbox[red,2pt]{5}$$

解答： $$n邊形內角和=180(n-2) \Rightarrow 2021\lt 180(n-2)\lt 2021+180 \Rightarrow 2021\lt 180(n-2) \lt 2201\\ \Rightarrow 11.2 \lt n-2 \lt 12.2 \Rightarrow 13.2\lt n\lt 14.2 \Rightarrow n=\bbox[red, 2pt]{14}$$

解答：

$$令P、Q、R分別為在\overline{AD}、\overline{BC}、\overline{CD}上的切點， 及圓半徑=r，見上圖；\\ \cases{\triangle ADO={1\over 2}\times \overline{AD}\times r={1\over 2}\times \overline{AO}\times r \\\triangle BCO={1\over 2}\times \overline{BC}\times r={1\over 2}\times \overline{BO}\times r } \Rightarrow \cases{\overline{AO}=\overline{AD} \\ \overline{OB}=\overline{BC}}\\ \Rightarrow \overline{AB}= \overline{AO} +\overline{OB} =\overline{AD}+\overline{BC}=3+2= \bbox[red,2pt]{5}$$

解答： $$a-b=b+1=c+3 \Rightarrow \cases{a=c+4\\ b=c+2} \Rightarrow a^2+b^2+c^2 =(c+4)^2+(c+2)^2+c^2 \\ =3(c+2)^2+8 \ge 8 \Rightarrow 最小值=\bbox[red,2pt]{8}$$

解答： $$\angle A=90^\circ \Rightarrow \overline{BC}^2=\overline{AB}^2 +\overline{AC}^2 =(\sqrt 6-\sqrt 2)^2 +(\sqrt 6+\sqrt 2)^2 =16\\ \Rightarrow \overline{AB}=4 =2R \Rightarrow R=2 \Rightarrow 外接圓圓面積=R^2\pi =\bbox[red, 2pt]{4\pi}$$

解答： $$x個師傅在(x+1)天可做出(x+2)個麵包 \Rightarrow (x+3)個師傅在(y)天可做出(x+4)個麵包 \\ 因此{x(x+1)\over x+2}={(x+3)y\over x+4} \Rightarrow y= \bbox[red,2pt]{x(x+1)(x+4)\over (x+2)(x+3)}$$

解答： $$\cases{x=7m+3 \\ y=7n+2},m,n\in \mathbb{Z} \Rightarrow 4x-xy+5y = 4(7m+3)-(7m+3)(7n+2)+5(7n+2)\\ =14m+14n-49mn+16 = 7(2m+2n+ 7mn+2)+2 \Rightarrow 餘數為\bbox[red,2pt]{2}$$

解答： $$\cases{n+100=m^2\\ n+168=n^2} \Rightarrow n^2-m^2=68 \Rightarrow (n+m)(n-m)=34\times 2 (17\times 4不合,\because n,m\in \mathbb{N})\\\Rightarrow \cases{n+m=34\\ n-m=2} \Rightarrow \cases{n=18\\ m=16} \Rightarrow n=16^2-100=\bbox[red,2pt]{156}$$

解答： $$建功星球表面積=A \Rightarrow \cases{北半球表面積=A/2  \\南半球表面積=A/2} \Rightarrow \cases{\cases{北半球陸地表面積= {A\over 2} \times {3\over 10} ={3\over 20}A\\ 北半球海洋表面積= {A\over 2} \times {7\over 10}={7\over 20}A }\\ \cases{南半球陸地面積={A \over 2}\times {13\over 15} ={13\over 30}A \\南半球海洋面積={A \over 2}\times {2\over 15} ={2\over 30}A}} \\ \Rightarrow 陸地: 海洋 =\left({3\over 20}+{13\over 30}\right): \left({7\over 20}+{2\over 30} \right)= 35:25 =\bbox[red, 2pt]{7:5}$$

解答： $$令\cases{f(x)=5+3x-x^2\\ g(x)=-x^2+ax+ b}，依題意f(x-5)-2= 5+3(x-5)-(x-5)^2-2\\ =-x^2+13x-37= g(x) \Rightarrow \cases{a=13\\ b=-37} \Rightarrow a+b= \bbox[red,2pt]{-24}$$

解答：

$$令\overline{AP}=a，則\overline{PB}=7-a；因此\cases{\triangle APD\sim \triangle BCP \Rightarrow a:2=3:7-a \Rightarrow a=1,6\\ \triangle ADP\sim \triangle BCP \Rightarrow 2:a=3:7-a \Rightarrow a=14/5} \\ \Rightarrow a=1,6,14/5，共\bbox[red,2pt]{3}個解$$

解答： $$利用長除法:x^3-14x=(x^2+3x-5)(x-3)-15\\因此若x^2+3x-5=0，則x^3-14x=\bbox[red,2pt]{-15}$$

解答： $$令S(n)=\sum_{k=1}^n k^2 \Rightarrow 99^2+\cdots +2021^2= S(2021)-S(98)\\={1\over 6}(2021\cdot 2022\cdot 4043-98\cdot 99 \cdot 197) =2021\cdot 337\cdot 4043-49\cdot 33\cdot 197 = 1-9(只看個位數)\\ =\bbox[red,2pt]{2}$$

解答： $$\cases{A={2^{100}+1\over 2^{99}+1}=2-{1\over 2^{99}+1}\\ B={2^{101}+1\over 2^{102}+1} \lt 1 \\ C={2^{104}+1\over 2^{103}+1 }= 2-{1\over 2^{103}+1}} \Rightarrow \bbox[red,2pt]{C\gt A\gt B}$$

解答： $$將x^2=-1代入x^{16}+3x^3-2x^2+3x-7 = 1-3x+2+3x-7=\bbox[red,2pt]{-4}$$

解答： $$x\gt 8或x\lt 2 \Rightarrow (x-8)(x-2)\gt 0 \Rightarrow x^2-10x+16\gt 0 \Rightarrow -x^2+10x-16\lt 0\\ \Rightarrow -{1\over 4}x^2+{5\over 2}x-4\lt 0\Rightarrow \cases{p=-1/4\\ q=5/2 } \Rightarrow p+q=-{1\over 4}+{5\over 2} =\bbox[red,2pt]{9\over 4}$$

解答： $$令\cases{u=x+2y\\ v=x-2y}，則\cases{|u|=3\\ |v|=3}所圍面積=6\times 6=36 \\ \Rightarrow \cases{|x+2y|=3 \\ |x-2y|=3} 所圍面積={36\over \begin{Vmatrix} {\partial\over \partial x}u & {\partial\over \partial y}u\\ {\partial\over \partial x}v & {\partial\over \partial y}v\end{Vmatrix}} ={36\over \begin{Vmatrix} 1 & 2\\ 1 & -2\end{Vmatrix}} {36\over 4}=\bbox[red, 2pt]{9}$$

解答： $$\langle a_n \rangle = \color{blue}0,1, \color{blue}1,2,3, \color{blue}2,4,5,6, \color{blue}3,7,8,9,10, \color{blue}4, 11,12, 13,14,15,\color{blue}5,\dots\\ 令\cases{\langle b_n\rangle =\langle n-1\rangle \\ \langle c_n\rangle =\langle n\rangle} ，則\langle a_n \rangle = b_1,c_1,b_2,c_2,c_3, b_3,c_4,c_5, c_6,b_4,c_7, c_8,c_9,c_4, c_{10},b_5,\dots\\ \Rightarrow b_k之前(含b_k)共有k+\sum_{i=1}^{k-1} i=k+{1\over 2}k(k-1)項;\\ 令f(k)=k+{1\over 2}k(k-1)\Rightarrow f(14)=14+{1\over 2}\cdot 14\cdot 13=105 \Rightarrow \cases{a_{105}=b_{14}=13 \\ a_{104}={1\over 2}\cdot 14\cdot 13=91} \\ \Rightarrow a_{106}=92 \Rightarrow a_{106}-a_{105}=92-13=\bbox[red,2pt]{79}$$

二、 進階題

解答： $$xy+yz+zx=xyz \Rightarrow {1\over z}+{1\over x}+{1\over y}=1 \Rightarrow {1\over 6}+{1\over 2}+{1\over 3}=1 \Rightarrow x+y+z= 2+3+6= \bbox[red, 2pt]{11}$$

解答： $${1\over a}+{1\over b}={a+b\over ab}={1\over a-b} \Rightarrow ab=a^2-b^2 \Rightarrow {b\over a}={a^2-b^2\over a^2} =1-({b\over a})^2\\ \Rightarrow x=1-x^2,其中x={b\over a} \Rightarrow x^2+x-1=0 \Rightarrow x=\bbox[red,2pt]{\sqrt 5-1\over 2}(a,b \gt 0 \Rightarrow b/a\gt 0,負值不合)$$

解答：

$$\angle C=90^\circ \Rightarrow \overline{AB}=\sqrt{\overline{AC}^2+ \overline{BC}^2} =\sqrt{3^2+4^2}=5；\\\triangle ABC ={1\over 2}\overline{AC}\times \overline{BC}={1\over 2}\overline{AB}\times \overline{CD} \Rightarrow \overline{CD}={3\times 4\over 5} ={12\over 5} \Rightarrow \overline{AD}=\sqrt{3^2-({12\over 5})^2}={9\over 5}\\ \Rightarrow \overline{AG}= \overline{AD}={9\over 5} (\because \triangle ADF\cong \triangle AGF)\\ 又\overline{AE}為\angle A的角平分線\Rightarrow {\overline{AC}\over \overline{AB}}= {\overline{CE}\over \overline{EB}} \Rightarrow \overline{CE} ={3\over 8}\times 4= {3\over 2}\\\overline{FG}\parallel \overline{BC} \Rightarrow {\overline{AG} \over \overline{AC}} ={\overline{FG} \over \overline{EC}} \Rightarrow \overline{FG}= {9\over 10 } \Rightarrow \overline{CF} =\overline{CD}-\overline{DF} =\overline{CD}-\overline{FG} ={12\over 5}-{9\over 10}=\bbox[red,2pt]{3\over 2}$$

解答： $$\cases{等差數列(a_n):首項a_1,公差d_a\\ 等差數列(b_n):首項b_1,公差d_b} \Rightarrow {a_5\over b_3+b_{2n-3}} +{a_{2n-5} \over b_7+b_{2n-7}}\\ ={a_1+4d_a\over b_1+2d_b+b_1+(2n-4)d_b} +{a_1+(2n-6)d_a\over b_1+6d_b+b_1+(2n-8)d_b}\\ ={a_1+4d_a\over 2b_1 +(2n-2)d_b} +{a_1+(2n-6)d_a\over 2b_1 + (2n-2)d_b} ={2a_1+(2n-2)d_a\over 2b_1 + (2n-2)d_b} =\color{blue}{{a_n\over b_n}={n\over 2n+1}}\\ 因此{A_{23}\over B_{23}} ={\sum_{k=1}^{23}a_k\over \sum_{k=1}^{23}b_k} ={a_1+a_{23}\over b_1+b_{23}} ={2a_1+22d_a\over 2b_1+ 22d_b} ={a_{12}\over b_{12}} =\bbox[red,2pt]{12 \over 25}$$

解答：

$$延長\overline{AP}交\overline{BC}於D點 \Rightarrow \angle ADB = 180^\circ -\angle B-\angle BAD= 180^\circ -38^\circ-22^\circ =120^\circ\\ 作\angle ADB 角平分線分別交\overline{BP}與\overline{AB}於E、F兩點，因此\angle BED=90^\circ\Rightarrow \overline{DF}為\overline{BP}的中垂線\\ \Rightarrow \angle FPE=\angle FBE=8^\circ；\\\cases{\angle FAD=\angle DAC =22^\circ\\ \overline{AD}=\overline{AD}\\ \angle ADF=\angle ADC=60^\circ} \Rightarrow \triangle AFD\cong \triangle ACD \Rightarrow \overline{AF}=\overline{AC} \Rightarrow \triangle APF\cong \triangle APC \\ \Rightarrow \angle APC= \angle APF = 180^\circ -\angle FPE-\angle FPD=180^\circ-8^\circ-30^\circ= \bbox[red,2pt]{142^\circ}$$

解答： $$\angle CDF \cong \triangle ADE (RHS) \Rightarrow \angle CDF=\angle ADE =(90^\circ-60^\circ)\div 2=15^\circ \\ \Rightarrow \cos \angle CDF = {\overline{CD}\over \overline{DF}} \Rightarrow {\sqrt 6+\sqrt 2\over 4}={1\over \overline{DF}} \Rightarrow 正\triangle 邊長\overline{DF}={4\over \sqrt 6+\sqrt 2} =\overline{EF}\\ \Rightarrow \overline{BE} =\overline{BF} ={\overline{EF}\over \sqrt 2} ={4\over 2\sqrt 3+2} \Rightarrow \triangle BEF={1\over 2}\times \left( {4\over 2\sqrt 3+2} \right)^2 =\bbox[red,2pt]{2-\sqrt 3}$$

解答： $$有一三角形三邊長分別為5,12,13，則其面積={1\over 2}\times 5\times 12=30 \Rightarrow 欲求\triangle 面積=30\times {4\over 3}=\bbox[red, 2pt]{40}\\相關證明\href{https://chu246.blogspot.com/2018/10/blog-post.html}{按這裡}$$

解答： $${\sqrt{15}\over \sqrt{11}+\sqrt 6+\sqrt 5} ={\sqrt{15}(\sqrt 6+\sqrt 5-\sqrt {11})\over (\sqrt 6+\sqrt 5+\sqrt {11})(\sqrt 6+\sqrt 5-\sqrt {11})} = {\sqrt{90}+\sqrt{75}-\sqrt{165} \over 2\sqrt{30}}\\ ={\sqrt{2700} +\sqrt{2250}-\sqrt{4950} \over 60} ={30\sqrt 3+15\sqrt{10}-15\sqrt{22}\over 60} =\bbox[red, 2pt]{2\sqrt 3+ \sqrt{10}- \sqrt{22}\over 4}$$

解答：

$$作\overline{AQ}\bot \overline{BC}且\overline{AQ}交\overline{DG}於P，見上圖；\\因此\cases{\triangle APG \sim \triangle GFC\\ \triangle APD \sim \triangle DEB}(AAA) \Rightarrow {\triangle {APG} \over \triangle{APD}} ={\triangle {GFC} \over \triangle{DEB}} ={1\over 3} \Rightarrow \cases{\triangle APD=3/4\\ \triangle APG=1/4}\\ 又{\triangle BDE\over \triangle APD}={\overline{DE}^2\over \overline{AP}^2} \Rightarrow {3\over 3/4} ={\overline{DE}^2\over \overline{AP}^2} \Rightarrow \overline{DE}=2\overline{AP}，因此\triangle ADG={1\over 2}\times \overline{DG}\times \overline{AP} \\ \Rightarrow 1={1\over 2}\times 2\overline{AP}\times \overline{AP} \Rightarrow \overline{AP}=1 \Rightarrow \overline{DG}=2 \Rightarrow 面方形DEFG=2\times 2=4\\ \Rightarrow \triangle ABC= \triangle ADG+\triangle BDE +\triangle CFG+ 正方形DEFG=1+3+1+4 =\bbox[red, 2pt]{9}$$

解答： $$\cases{|a-b|=1 \Rightarrow b=a\pm 1\\ |b-c|=2 \Rightarrow c=b\pm 2\\ |c-d|=3 \Rightarrow d=c\pm 3} \Rightarrow d=a\pm 1\pm 2\pm 3 \Rightarrow |a-d|=|\pm 1\pm 2\pm 3|=0,2,4,6 \\ \Rightarrow 0+2+4+6 =\bbox[red, 2pt]{12}$$

解答： $$\cases{3XXX\cases{千位數字和=3\times 3!=18\\ 百位數=十位數=個位數=2(0+1+2)=6} \Rightarrow 18666 \\ 2XXX\cases{千位數字和=2\times 3!=12\\ 百位數=十位數=個位數=2(0+1+3)=8} \Rightarrow 12888\\ 1XXX\cases{千位數字和=1\times 3!=6\\ 百位數=十位數=個位數=2(0+2+3)=10} \Rightarrow 7110} \\ \Rightarrow 總和=18666+12888+7110 =\bbox[red, 2pt]{38664}$$

解答： $$\cases{xy(x+y)=-30\\ xy+(x+y)=-29} \Rightarrow \cases{\cases{x+y=-30\\ xy=1} \Rightarrow 無整數解\\ \cases{x+y=1\\ xy=-30} \Rightarrow (x,y)=(6,-5),(-5,6)} \\ \Rightarrow x^2+y^2=5^2+6^2= \bbox[red, 2pt]{61}$$

解答： $$令u=x^2-3x，則x^2-3x+{5\over x^2-3x-2}=8 \Rightarrow u+{5\over u-2}=8 \Rightarrow u(u-2)+5=8(u-2)\\ \Rightarrow u^2-10u+21=0 \Rightarrow (u-7)(u-3)=0 \Rightarrow \cases{u=7 \\ u=3}\Rightarrow\cases{ x^2-3x-7=0\\ x^2-3x-3=0} \\ \Rightarrow \cases{判別式:9+28\gt 0有相異實根且兩根之積=-7 \\ 判別式:9+12\gt 0有相異實根且兩根之積=-3} \Rightarrow 四根之積=(-7)\cdot (-3)=\bbox[red, 2pt]{21}$$

解答： $$abc=1 \Rightarrow \cases{1+a+ab= abc+a+ab= a(1+b+bc) \cdots(1)\\ 1+c+ca= abc+c+ca=c(1+a+ab)\cdots(2)}\\ 把(1)代入(2) \Rightarrow 1+c+ca=ca(1+b+bc)\\ 因此{x\over 1+a+ab} +{x\over 1+b+bc}+{x\over 1+c+ca}= {x\over a(1+b+bc)} +{x\over 1+b+bc} +{x\over ca(1+b+bc)}\\= {(c+ac+1)x\over ca(1+b+bc)} ={ca(1+b+bc)x\over ca(1+b+bc)} =x=\bbox[red,2pt]{2021}$$

三、 計算題

解答：

$$令\cases{\overline{BD}=a \\內切圓半徑r=4}，則\cases{\overline{CE}=a+1\\ \overline{AF}=a+2} \Rightarrow \cases{\overline{AF} =\overline{AE}=a+2\\ \overline{BF}=\overline{BD} =a\\ \overline{CD}=\overline{CE}=a+1}\\ 令s=(\overline{AB} +\overline{BC}+ \overline{CA})\div 2= 3a+3 \\\Rightarrow \triangle ABC面積= \sqrt{s(s-\overline{AB})(s-\overline{BC}) (s-\overline{AC})} ={1\over 2}\cdot r\cdot 2s\\ \Rightarrow \sqrt{(3a+3)(a+1)(a+2)a} ={1\over 2}\cdot 4\cdot (6a+6) \Rightarrow \sqrt{3a(a+2)} =12 \Rightarrow a(a+2)=48 \\\Rightarrow (a+8)(a-6)=0 \Rightarrow a=6 \Rightarrow \triangle ABC周長=2s= 6a+6=36+6=\bbox[red, 2pt]{42}$$

解答：

$$作\overline{CE} \parallel \overline{DB}，其中E在\overleftrightarrow{AB}上，並令\overline{BE}=a，見上圖；\\ 因此\angle ACE=\angle ADB=\angle A \Rightarrow \overline{EC}=\overline{EA} =a+3\\ 利用海龍公式可求得 \triangle ABC={15\over 4}\sqrt 3及\triangle AEC = {7\over 4}\sqrt{(2a+13)(2a-1)}\\ 又\triangle ABC =\triangle AEC\times {3\over a+3} \Rightarrow {15\over 4}\sqrt 3= {21\over 4(a+3)}\sqrt{(2a+13)(2a-1)} \\ \Rightarrow 121a^2+726a-1312=0 \Rightarrow (11a-16)(11a+82)=0 \Rightarrow a={16\over 11}\\ 因此\overline{AD}: \overline{DC}=3:a=3:{16\over 11} =\bbox[red,2pt]{33:16}$$

解答：

$$\cases{\triangle ABC=a\\ \triangle CEF=b} \Rightarrow \triangle AEF=\triangle BCE={a-b\over 2}\\ \cases{\overline{AE}:\overline{EB} =\triangle ACE:\triangle BCE={a-b\over 2}+b:{a-b\over 2}=a+b:a-b\\ \overline{AF}: \overline{FC} =\triangle AEF:\triangle CEF = {a-b\over 2}:b} \\ \overline{EF}\parallel \overline{BC} \Rightarrow \overline{AE}:\overline{EB} =\overline{AF}: \overline{FC} \Rightarrow a+b:a-b={a-b\over 2}:b \Rightarrow (a-b)^2=2b(a+b)\\ \Rightarrow a^2-4ab-b^2=0 \Rightarrow a=(\sqrt 5+2)b \Rightarrow {b\over a}={1\over \sqrt 5+2}= \bbox[red, 2pt]{\sqrt 5-2}$$

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解題僅供參考，其他教甄試題及詳解