107年新竹市建功高中國中部教甄-數學詳解

 新竹市立建功高中107學年度第一次正式教師甄試國中數學科

一、填充題：【20 題，每題 4 分，共計 80 分】

解答：

$$\overline{AB} :\overline{AC}: \overline{BC}= 3:4: 5\Rightarrow \cases{\overline{AB}=3k\\ \overline{AC}=4k\\ \overline{BC} =5k},k\in \mathbb{R}\\ \overline{BD}為\angle B的角平分線 \Rightarrow {\overline{AD}\over \overline{DC}} ={\overline{AB} \over \overline{BC}}={3k\over 5k} \Rightarrow \cases{\overline{AD}={\overline{AB}\over   \overline{AB}+ \overline{BC}}\times \overline{AC}={3\over 8}\times 4k={3\over 2}k \\ \overline{DC}=4k-{3\over 2}k={5\over 2}k};\\ 同理，{\overline{AF}\over \overline{FB}} ={\overline{CA} \over \overline{CB}}={4k\over 5k} \Rightarrow \cases{\overline{AF}={\overline{CA}\over   \overline{CA}+ \overline{CB}}\times \overline{AB}={4\over 9}\times 3k={4\over 3}k \\ \overline{FB}=3k-{4\over 3}k={5\over 3}k};\\ 又\cases{\triangle CED\sim \triangle CAB\\ \triangle BGF \sim \triangle BAC}(AAA) \Rightarrow \cases{\triangle CED:\triangle ABC=\overline{CD}^2:\overline{BC}^2 = 25/4: 25\\ \triangle BGF:\triangle ABC=\overline{BF}^2: \overline{BC}^2 = 25/9:25} \\ \Rightarrow {\triangle DEC+ \triangle FGB \over \triangle ABC } ={25/4+25/9\over 25} = \bbox[red, 2pt]{13\over 36}$$

解答： $${1\over 2}+ {1\over 4}+ {1\over 5} ={19\over 20} \Rightarrow \cases{a=2\\ b=4\\c=6} \Rightarrow a+b+c =\bbox[red,2pt]{12}$$

解答： $$\cases{A(-1,1)\\ B(2018,2018^2) \\ C(-2,4)\\ D(a,a^2)} \Rightarrow \overleftrightarrow{AB}:y=2017(x+1)+1 \Rightarrow \overleftrightarrow{AB}交y軸於P(0,2018) \\ \Rightarrow \overleftrightarrow{CP}:y=1007(x+2)+4 代入y=x^2求交點 \Rightarrow x^2-1007x-2018=0 \\ \Rightarrow (x-1009)(x+2)=0 \Rightarrow D的x座標為\bbox[red,2pt]{1009}$$

解答： $$凸160邊形內角和=180\times (160-2)=28440\\假設內角小於160度的有n個，剩下(160-n)個內角皆小於180度\\因此28440 \lt 160n+180(160-n) \Rightarrow 20n \lt 360 \Rightarrow n\lt 18 \Rightarrow n=\bbox[red, 2pt]{17}$$

解答： $$x=n+y,其中n\in \mathbb{N},0\le b\lt 1 \Rightarrow x^2+y^2=2018 \Rightarrow x^2=2018-y^2 \Rightarrow 2017\lt x^2\lt 2018\\ \Rightarrow \sqrt{2017} \lt x\lt \sqrt{2018}\Rightarrow n=44 \Rightarrow x=44+y \Rightarrow (44+y)^2 +y^2=2018\\ \Rightarrow y^2+44y-41=0 \Rightarrow y=-22+5\sqrt{21} \Rightarrow x=44+(-22+5\sqrt{21}) =\bbox[red,2pt]{22+5\sqrt{21}}$$

解答： $$令\cases{\langle a_n\rangle 公比為r_a \\\langle b_n\rangle 公比為r_b} \Rightarrow a_n+b_n =(r_a+r_b)(a_{n-1}+b_{n-1}) -r_ar_b(a_{n-2}+b_{n-2})\\因此\cases{86= (r_a+r_b)\cdot (-4)-r_ar_b\cdot 24\\ 71=(r_a+r_b)\cdot 86-r_ar_b\cdot (-4)} \Rightarrow \cases{r_a+r_b= 1\\ r_ar_b= -{15\over 4}} \Rightarrow a_5+b_5=1\cdot 71+{15\over 4}\cdot 86 =\bbox[red,2pt]{393{1\over 2}}$$

解答： $$恰好只走一圈:5奇或1偶3奇或2偶1奇，機率為({1\over 2})^5+C^4_3({1\over 2})^4+C^3_2({1\over 2})^3 = \bbox[red,2pt]{21\over 32}$$

解答：

$$令\cases{\overline{BP} =\overline{CQ} =a \\\overline{PQ} =b\\ \overline{AQ} =h\\ \overline{AP} =w}，並在\overline{AB}上找一點R，使得\overline{AR}=\overline{AC} =12，則\triangle AQR \cong \triangle AQC\;(SAS)\\ \Rightarrow \overline{QR}=\overline{QC} =a；\\又\overline{AQ} 為\angle A的角平分線，因此\overline{AB}:\overline{AC}= \overline{BQ} : \overline{QC} \Rightarrow 20:12=a+b:a \Rightarrow b={2\over 3}a \Rightarrow a+b={5\over 3}a\\ \cases{\triangle BQR: \cos \angle B={(5a/3)^2+8^2-a^2\over 16(5a/3)} = {64+(16a^2/9) \over 80a/3} \cdots(1)\\ \triangle BAP: \cos \angle B={20^2+a^2-w^2 \over 40a} \cdots(2)\\ \triangle BAQ: \cos \angle B={20^2+(5a/3)^2-h^2\over 40(5a/3)} ={400+(25a^2/9)-h^2\over 200a/3} \cdots(3)}\\ 由\cases{(2)=(3) \Rightarrow h^2=240-{5\over 3}a^2 \\ (1)=(2) \Rightarrow w^2=304-{5\over 3}a^2} \Rightarrow \sqrt{w^2-h^2}= \sqrt{304-240}= \sqrt{64} =\bbox[red,2pt]{8}$$

解答： $$-3x-5x-7x-9x = -11x-13x \\\Rightarrow 若{1\over 11}\lt x\lt {1\over 9},則f(x)=(1-3)+(1-5x)+(1-7x)+(1-9x)+(11x-1)+(13x-1)\\ =2 \Rightarrow \cases{a=1/11\\ b=1/9\\ c=2} \Rightarrow a\times b\times c=\bbox[red, 2pt]{2\over 99}$$

解答： $$n= 9+99+999+\cdots +\overbrace{99\cdots 9}^{99個9} =(10^1-1)+(10^2-1) +(10^3-1)+\cdots +(10^{100}-1)\\ ={10\over 9}(10^{100}-1) -100 ={10\over 9}\cdot \overbrace{99\cdots 9}^{100個9} -100= 10\cdot \overbrace{11\cdots 1}^{100個1} -100 =  \overbrace{11\cdots 1}^{100個1}0 -100\\ =\overbrace{11\cdots 1}^{98個1}010 \Rightarrow 共\bbox[red, 2pt]{99}個1$$

解答： $$令\langle a_n \rangle = \langle 2^n \mod 100 \rangle = 2,\color{blue}{4,8,16,32,64,28,56,12,24,48,96, 92,84,68,36,72,44, 88,76,52},\\4, 8,16,32, 64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,\dots\\ \Rightarrow 從n=2起，循環數為20，因此2018= 1+20\times 100+17\Rightarrow 第17個循環數為44\\ \Rightarrow A=2^{2018}-1 的末二位數為44-1=\bbox[red, 2pt]{43}$$

解答： $$假設n為三位數(不可能是二位數，\because 99+9+9\lt 313)，其\cases{千位數字為a\\ 百位數字為b\\ 個位數字為c}且0\le a,b,c \le 9 \\\Rightarrow 100a+10b+c+ a+b+c=313 \Rightarrow 101a+11b+2c=313 \Rightarrow 最小的可能a=2\\ \Rightarrow 11b+2c=111 \Rightarrow \cases{(b,c)=(9,6)\\ (b,c)=(7,17不合)} \Rightarrow n=\bbox[red,2pt]{296}$$

解答： $$需符合\cases{a+b\gt c\\ c-a\lt b}，因此(a,b,c)=(1,9,9), (2,8,9),(3,7,9),(3,8,8),(4,6,9) ,(4,7,8),\\(5,5,9), (5,6,8),(5,7,7),(6,6,7)，共\bbox[red,2pt]{10}種$$

解答：

解答：

$$ABCD為菱形\Rightarrow \overline{AC} \bot \overline{BD}，並令\overline{AC} 、 \overline{BD}交點為Q，見上圖；\\並令\cases{\overline{BQ} =\overline{QD}=a \\\overline{AQ} =\overline{QC}=b \\ \overline{CP}=a}，則\cases{直角\triangle AQD: a^2+b^2 =36 \cdots(1) \\ 直角\triangle PQD: a^2+(b+c)^2=81 \cdots(2)}\\ 將(1)代入(2) \Rightarrow 36+2bc+c^2=81 \Rightarrow 2bc+c^2= c(2b+c) =\overline{PC}\times \overline{AP}= \bbox[red,2pt]{45}$$

解答： $$若有\cases{x^2個A\\ 2xy個B\\y^2個C}，就可以拼成邊長為(x\pi+2y)的正方形，現在\cases{x^2\le 10\\ 2xy\le 28\\ y^2\le 50}\\ \Rightarrow (x,y)=(1,1)-(1,7),(2,1)-(2,7),(3,1)-(3,4)\\ 取(x,y)=(2,7)可得最大邊長\bbox[red,2pt]{2\pi+14}$$

解答：

$$\angle F=360^\circ-90^\circ-90^\circ-45^\circ = 135^\circ ，因此令\cases{E(0,0)\\ F(0,1)\\ C(a,0)} \Rightarrow D(1,2)\\ 由於\angle C=45^\circ \Rightarrow \overleftrightarrow{CD}: y=-(x-1)+2 \Rightarrow a=3 \Rightarrow \cases{\overline{CE}=3\\ \overline{CD}=\sqrt{2^2+2^2}=2\sqrt 2} \\ \Rightarrow \cases{\triangle CDF={1\over 2}\times \sqrt 2\times 2\sqrt 2=2\\ \triangle CEF = {1\over 2}\times 1\times 3={3\over 2}} \Rightarrow DFEC面積=2+{3\over 2}= \bbox[red, 2pt]{7\over 2}$$

解答： $$f(x)=(x+1)(2x+1)(3x-1)(4x-1)-36x^4 = -12x^4+22x^3-7x^2-4x+1\\ \Rightarrow f(1)=0 \Rightarrow f(x)=(x-1)(-12x^3+10x^2+3x-1) =(x-1)g(x)\\ \Rightarrow g(1)= 0 \Rightarrow f(x)=(x-1)^2(-12x^2-2x+1) =\bbox[red, 2pt]{-(x-1)^2(12x+2x-1)}$$

解答： $$\cases{{y\over x}+ {1\over xy}={20\over 3} \\ xy+{x\over y}={5\over 3}}，兩式相乘\Rightarrow y^2+{1\over y^2}={82\over 9} \Rightarrow 9y^4-82y^2+9=0 \Rightarrow (y^2-9)(9y^2-1)=0\\ \Rightarrow \cases{y=3 \Rightarrow x=1/2\\ y=-3 \Rightarrow x=-1/2\\ y=1/3\Rightarrow x=1/2\\ y=-1/3\Rightarrow x=-1/2} \Rightarrow (x,y)= \bbox[red, 2pt]{({1\over 2},3),(-{1\over 2},-3), ({1\over 2},{1\over 3}), (-{1\over 2},-{1\over 3})}$$

解答： $$y=(x-2)(x-4)(x-6)(x-8)+12 = ((x-2)(x-8))((x-4)(x-6))+12\\ =(x^2-10x+16)(x^2-10x+24)+12 =(x^2-10x)^2+40(x^2-10x)+ 396 \\ =(x^2-10x+20)^2-4 \Rightarrow 當x^2-10x+20=0時，y有最小值-1，即(x,y)= \bbox[red,2pt]{(5\pm \sqrt 5,-4)}$$

解答： $${1\over 1+2+\cdots + n} ={1\over n(n+1)/2} =2({1\over n}-{1\over n+1}) \\ \Rightarrow {1\over 1}+ {1\over 1+2}+ {1\over 1+2+3} +\cdots +{1\over 1+2+3+\cdots +100}\\ =1+2(({1\over 2}-{1\over 3})+ ({1\over 3}-{1\over 4})+ \cdots +({1\over 100}-{1\over 101}))\\ =1+2({1\over 2}-{1\over 101}) =1+{99\over 101} =\bbox[red, 2pt]{200\over 101}$$

二、計算題：【4 題，每題 5 分，依計算過程，給予適當分數】

解答： $$x* (x*x)=\cases{x*0\\ x*x+x=0+x} \Rightarrow x*0=0+x=x\\因此2018*(2017*2017)=\cases{2018*2017+2017\\ 2018*0=2018} \Rightarrow 2018*2017+2017=2018\\ \Rightarrow 2018*2017=2018-2017=\bbox[red, 2pt]{1}$$

解答： $$(x^2+x+1)(1+x+x^2 + \cdots +x^9+x^{10})= (1+x+x^2+\cdots +x^6)^2\\ \Rightarrow (x-1)(x^2+x+1) (x-1) (1+x+x^2 + \cdots +x^9+x^{10})= ((x-1)(1+x+x^2+\cdots +x^6))^2 \\ \Rightarrow (x^3-1)(x^{11}-1) = (x^7-1)^2 \Rightarrow x^{11}-2x^7 +x^3=0 \Rightarrow x^3(x^8-2x^4+1)=0\\ \Rightarrow x^3(x^4-1)^2=0 \Rightarrow x^3(x^2+1)^2(x^2-1)^2=0 \Rightarrow x^3(x-1)^2 (x+1)^2 (x^2+1)^2=0\\ \Rightarrow x=\bbox[red, 2pt]{0,-1}$$

解答： $$f(x,y)= x\sqrt{1-y^2} +y\sqrt{1-x^2} \Rightarrow \cases{f_x=\sqrt{1-y^2}-{xy\over \sqrt{1-x^2}} \\ f_y=-{xy\over \sqrt{1-y^2}}+\sqrt{1-x^2}}\\ 因此\cases{f_x=0\\ f_y=0} \Rightarrow xy= \sqrt{1-x^2}\cdot \sqrt{1-y^2} \Rightarrow x^2y^2 =(1-x^2)(1-y^2) =1-x^2-y^2+x^2y^2\\ \Rightarrow x^2+y^2=1代回f(x,y)=x\cdot \sqrt{x^2} +y\sqrt{y^2} =x^2+y^2= \bbox[red,2pt]{1}$$

解答：

$$令\cases{\overline{BD}=a \\ \overline{BE}=b}，則\cases{\cos \angle BDA={a^2+16-36\over 8a} \\ \cos \angle BDE= {a^2+16-b^2 \over 8a}}，由於\cos \angle BDA =-\cos \angle BDE \Rightarrow 2a^2-b^2=4;\\ 同理，\cases{\cos \angle BED={b^2+16-a^2\over 8b} \\ \cos \angle BEC= {b^2+16-64 \over 8b}}，由於\cos \angle BED =-\cos \angle BEC \Rightarrow a^2-2b^2=-32 \\ 因此\cases{2a^2-b^2=4\cdots (1)\\ a^2-b^2=-32\cdots (2)} ，{(1)+(2)\over 3} \Rightarrow a^2-b^2=-{28\over 3}\cdots(3)，(1)-(3) \Rightarrow a^2={40\over 3}\\ \Rightarrow a=\bbox[red, 2pt]{2\sqrt{30}\over 3}$$

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