網頁

2022年4月15日 星期五

110年嘉義高中科學班甄選-數學科詳解

國立嘉義高級中學 110 學年度科學班甄選【數學科】試題

一、填充題: (85 分,每題 5 )

解答:$$\cases{P(10^{-16}) \\ Q(10^{-15})}  \Rightarrow \overline{A_kA_{k+1}} =\cfrac{\overline{PQ}}{100} ={10^{-15}-10^{-16}\over 100} =10^{-17}-10^{-18} =9\times 10^{-18},k=1-98;\\ \Rightarrow \overline{A_{31}A_{55}} =(55-31)\times 9\times 10^{-18} =216 \times 10^{-18} =2.16 \times 10^{-16}  \Rightarrow (a,n)= \bbox[red, 2pt]{(2.16,-16)}$$
解答:$${2b+2c\over a} ={2c+2a\over b} ={2a+2b\over c} \Rightarrow 2+{2b+2c\over a} =2+{2c+2a\over b} =2+{2a+2b\over c} \\ \Rightarrow {2(a+b+2)\over a} ={2(a+b+2)\over b} ={2(a+b+2)\over c} \Rightarrow {6\over a} ={6\over c} ={6\over c} \\ \Rightarrow a=b=c \Rightarrow {2b+2c\over a} ={2a+2a\over a} =\bbox[red, 2pt]{4}$$
解答:$$瓶子重y公克 \Rightarrow 可樂重(1000-y) 公克 \Rightarrow 喝剩總重(1000-y)(1-x)+y =400 \\\Rightarrow xy-1000x+600 =0 \Rightarrow x={ 600 \over 1000-y}  \Rightarrow {3\over 4}\lt { 600 \over 1000-y} \lt {5\over 6} \Rightarrow {6\over 5} \lt {1000-y\over 600} \lt {4\over 3}\\ \Rightarrow 720\lt  1000-y \lt 800 \Rightarrow -280\lt -y \lt -200 \Rightarrow 200\lt y \lt 280 \\\Rightarrow (a,b)= \bbox[red, 2pt]{(200,280)}$$
4. 一個「訊息」是由一串8個數字排列組成,且每位數字都只能是0或1,例如10010000與01011000就是兩個不同的訊息。兩個訊息的「距離」定義為此兩組數字串相對應位置中,數字不同的位置數。例如,數字串10010000與01011000在第1、 2及5三個位置不同,所以訊息10010000與01011000的距離為3。 根據上述定義,若有一訊息與訊息11011001、11001011的距離皆為a,則正整數a的值共有_____個。
解答:$$令\cases{A=11011001\\ B=11001011}及C 滿足d(A,C)=d(B,C);\\d(A,B)=2 \Rightarrow d(A,C)=d(B,C)= 2,3,...,8,共\bbox[red, 2pt]{7}個不同的距離值$$
5. 若平行四邊形ABCD的面積為4,M、N分別是\(\overline{BC}\)、\(\overline{CD}\)的中點,且G為\(\triangle AMN\)的重心,則\(\triangle GMN\)的面積= ____。
解答
$$令平行四邊形ABCD面積=a\Rightarrow \cases{\triangle ABM= {1\over 2}\triangle ABC ={1\over 4}ABCD = {a\over 4} \\[1ex] \triangle ADN ={1\over 2}\triangle ADC ={1\over 4}ABCD = {a\over 4} \\[1ex] \triangle MNC = {1\over 2}\triangle BCN ={1\over 4}\triangle BCD = {1\over 8}ABCD ={a\over 8}} \\ \Rightarrow \triangle AMN = a-{a\over 4}-{a\over 4}-{a\over 8} ={3\over 8}a \Rightarrow \triangle GMN ={1\over 3}\triangle AMN ={1\over 8}a = \bbox[red, 2pt]{1\over 2}$$
解答:$$13^2 \times 15^2\times 17^2 =3315^2 \Rightarrow \cases{3316^2 =(3315+1)^2 =3315^2 +6631 \\3317^2 =(3315+2)^2 =3315^2 + 13264 \\3318^2 =(3315+3)^2 =3315^2 +19891 } \\ \Rightarrow 3317\lt \sqrt{13^2 \times 15^2\times 17^2 +19000} \lt 3318 \Rightarrow a=\bbox[red, 2pt]{3317}$$
解答:$$第3大的數是a \Rightarrow \cases{第1大是a+2\\ 第2大是a+1} \Rightarrow 前三大之和 =3a+3\\最小的前17項之和=1+2+ \cdots +17 = 153 \Rightarrow 153+3a+3= 14\times 20=280\\ \Rightarrow a=\bbox[red, 2pt]{41}$$
解答:$$令\cases{A(0,4)\\ B(0,0)\\ C(4,0)} \Rightarrow D=(A+B)\div 2= (2,2) \Rightarrow E =(2,0) \Rightarrow \cases{\overleftrightarrow{AE}: 2x+y=4 \\\overleftrightarrow{BD}: x=y} \\ \Rightarrow D_1 =\overleftrightarrow{AE} \cap \overleftrightarrow{BD}= (4/3,4/3) \Rightarrow E_1(4/3,0) \Rightarrow \overleftrightarrow{AE_1}: 3x+y=4\\ \Rightarrow D_2= \overleftrightarrow{AE_1} \cap \overleftrightarrow{BD}= (1,1) \Rightarrow \cases{ \overline{BD_1} =4\sqrt 2/3\\ \overline{D_1D_2} =\sqrt 2/3} \Rightarrow r =\cfrac{\overline{BD_1} }{ \overline{D_1D_2}} =\bbox[red,2pt]{4}$$
解答:$$a=\sqrt{8+\sqrt{63}} +\cfrac{1}{\sqrt{8+\sqrt{63}}} \Rightarrow a^2 = 8+\sqrt{63}+2+\cfrac{1}{8+\sqrt{63}} = 8+\sqrt{63}+2 +8-\sqrt{63}=18\\ \Rightarrow a^2=18 \Rightarrow a=3\sqrt 2 \approx 3\times 1.414 = 4.242 \Rightarrow {42\over 10} \lt a\lt {43\over 10} \Rightarrow n= \bbox[red, 2pt]{42}$$
10. 三邊長皆為整數之三角形,我們稱為 Diophantus 三角形。令\(a_n\)表示最大邊長為 n 的Diophantus 三角形之個數,例如:\(a_1=1,a_2=2,a_3=4\),試問\(a_{11}\)= ________。
解答:$$假設三角形三邊長為(11,a,b),滿足a+b \gt 11且1\le a,b\le 11,a,b均為自然數\\ \begin{array}{} a & b & 數量\\\hline 1 & 11 & 1\\ 2 &10-11 & 2\\ 3& 9-11& 3 \\ 4 & 8-11 & 4 \\ 5& 7-11 & 5 \\ 6& 6-11 & 6 \\\hdashline 7 & 7-11 & 5\\ 8 & 8-11 & 4 \\ 9 & 9-11 &3 \\ 10 & 10-11& 2\\ 11 & 11 & 1\\\hline \end{array} \Rightarrow a_{11}=21+15 =\bbox[red, 2pt]{36}$$
11. 已知數列\(a_1,a_2,a_3, \dots,a_n,\dots\)的每一項都是正數,它的任意相鄰3項\(a_k、a_{k+1}、a_{k+2}\)都滿足\(\sqrt{a_k} -2\sqrt{a_{k+1}}+ \sqrt{a_{k+2}}=0\),其中\(k\)是正整數。若\(a_4=4,a_9=9\),則第15項\(a_{15}\)= ____。
解答:$$\sqrt{a_k} -2\sqrt{a_{k+1}} +\sqrt{a_{k+2}}=0 \Rightarrow \cases{\sqrt{a_1} -2\sqrt{a_{2}} +\sqrt{a_{3}}=0 \\\sqrt{a_2} -2\sqrt{a_{3}} +\sqrt{a_{4}}=0 \\ \cdots \\\sqrt{a_{n-2}} -2\sqrt{a_{n-1}} +\sqrt{a_{n}}=0 },\\上述(n-2)個式子相加 \Rightarrow \sqrt{a_1}- \sqrt{a_2}-\sqrt{a_{n-1}}+ \sqrt{a_n} =0  \Rightarrow \sqrt{a_n} =\sqrt{a_{n-1}}+ k,\cases{k=\sqrt{a_2}-\sqrt{a_1}\\n\ge 2}\\ 因此\cases{\sqrt{a_4}= \sqrt{a_3}+k =\sqrt{a_2}+2k = 3\sqrt{a_2} -2\sqrt{a_1} = 2\\ \sqrt{a_9} =\sqrt{a_8}+k = \sqrt{a_2}+ 7k = 8\sqrt{a_2}-7\sqrt{a_1}= 3} \Rightarrow \cases{\sqrt{a_1}= 7/5\\ \sqrt{a_2} =8/5} \Rightarrow k={1\over 5}\\ \Rightarrow \sqrt{a_{15}}= \sqrt{a_2} +13k ={8\over 5}+{13\over 5} ={21\over 5} \Rightarrow a_{15} =({21\over 5})^2 =\bbox[red,2pt]{441\over 25}$$
解答:$$x=2^{100} \Rightarrow \cases{2^{101}=2x \\ 2^{201}=2x^2\\ 2^{402}=4x^4} \Rightarrow \cases{2^{402} +402 = 4x^4+402\\ 2^{201}+ 2^{101}+ 1= 2x^2+2x+1} \\ \Rightarrow 4x^4+402 = (2x^2+2x+1)(2x^2-2x+1)+ 401 \Rightarrow 餘數為\bbox[red, 2pt]{401}$$
解答
$$令\cases{\overline{CD}=\overline{PQ}=a \\ \overline{AP}=\overline{BQ}=b},見上圖;則\cases{\overline{AR}^2 = 10^2-b^2\\ \overline{AR}^2 =11^2-(a+b)^2} \Rightarrow 10^2-b^2=11^2-(a+b)^2 \\ \Rightarrow (a+b)^2-b^2 =11^2-10^2 \Rightarrow (a+2b)a=21 \Rightarrow \cases{\cases{a+2b=21\\ a=1} \Rightarrow (a,b)=(1,10)\\\cases{a+2b=7\\ a=3} \Rightarrow (a,b)=(3,2)}\\ 若\overline{RD}=b=10=斜邊長\overline{AD},不合;因此(a,b)=(3,2) \Rightarrow \overline{AB}=a+2b=\bbox[red, 2pt]{7}$$
解答:$$y= f(x)=ax^2+bx+c =a(x-2)^2+3 =ax^2-4ax+ 4a+3 \\\Rightarrow f(0)\lt 0 \Rightarrow 4a+3 \lt 0 \Rightarrow a \lt -{3\over 4} \Rightarrow a=\bbox[red, 2pt]{-1}$$
解答:$$\cases{A(2,0)\\ B(6,0)\\ C(0,3)\\ D(0,5)\\ P(s,t)} \Rightarrow \cases{L_1=\overleftrightarrow{AD}:5x+2y= 10\\ L_2= \overleftrightarrow{BC}: x+ 2y= 6} \Rightarrow P=L_1\cap L_2 = \bbox[red, 2pt]{\left(1,{5\over 2} \right)}$$
解答

$$\cases{直角\triangle ADH \Rightarrow \overline{DH} =\sqrt{13^2-12^2} =5 \\直角\triangle AMH \Rightarrow \overline{MH} =\sqrt{15^2-12^2} =9} \Rightarrow \overline{MD} =\overline{MH}-\overline{DH}= 9-5=4\\ 令\overline{CH} =a,又M為\overline{BC} 中點,因此 \overline{BM}=\overline{CM} \Rightarrow \overline{BM} = a+9\\ 由於\overline{AD}為角平分線,因此\cfrac{\overline{BD}}{ \overline{DC}} = \cfrac{\overline{AB}}{ \overline{AC}} \Rightarrow {a+13\over a+5} =\cfrac{\sqrt{(a+18)^2+12^2}}{ \sqrt{a^2+12^2}} \\\Rightarrow {(a+13)^2\over (a+5)^2} =\cfrac{ {(a+18)^2+12^2}}{ {a^2+12^2}}  \Rightarrow 1+{16a+144\over a^2+10a+25} =1+{36a+324\over a^2+144} \\\Rightarrow 4(a+9)(a^2+144) =9(a+9)(a^2 +10a+25)\\ \Rightarrow 5a^2+90a-351=0 \Rightarrow a=\cfrac{12\sqrt{105}}{10}-9 \Rightarrow \overline{BC}= 18+2a = \bbox[red, 2pt]{\cfrac{12\sqrt{105}}{5}}$$
解答:$$\cases{A(1,2)\\ B(-4,2)\\ O(0,0)} \Rightarrow \cases{\overrightarrow{OA} =(1,2) \\\overrightarrow{OB} =(-4,2) \\ \overrightarrow{AB} =(-5,0)} \Rightarrow \cases{\vec u= \overrightarrow{OA}/|\overrightarrow{OA}| =(1/\sqrt 5,2/\sqrt 5)\\ \vec v= \overrightarrow{OB}/| \overrightarrow{OB}| =(-2/\sqrt 5,1/\sqrt 5)\\ \vec w= \overrightarrow{AB}/ |\overrightarrow{AB}| =(-1,0)} \\\Rightarrow \cases{\vec n_1= \vec u+\vec v= (-1/\sqrt 5, 3/\sqrt 5)\\ \vec n_2= \vec w-\vec u= (-1-1/\sqrt 5,-2/\sqrt 5)} \Rightarrow \cases{\angle AOB 的角平分線L_1: 3x+y=0\\ \angle BAO的角平分線L_2: 2(x-1)=(\sqrt 5+1)(y-2)} \\ \Rightarrow I= L_1\cap L_2 \Rightarrow 3x+{2(x-1)\over \sqrt 5+1}+2=0 \Rightarrow x={-3+\sqrt 5\over 2} \Rightarrow y=-3x ={9-3\sqrt 5\over 2} \\ \Rightarrow I=\bbox[red, 2pt]{\left( {-3+\sqrt 5\over 2},{9-3\sqrt 5\over 2}\right)}$$

二、證明題 (第 1 題 5 分,第 2 題 10 分,共 15 分)


解答
1.$$\cases{\overline{AB}為直徑\Rightarrow \angle APB=90^\circ \\\overline{AH}為直徑\Rightarrow \angle ACH=90^\circ \\ \overline{BH}為直徑\Rightarrow \angle BDH=90^\circ } \Rightarrow PCHD 為一矩形\Rightarrow 以\overline{CD}為直徑的圓,此圓內接PCHD\\,\bbox[red,2pt]{故得證}$$
2.$$令\cases{\angle ACC_1 =\alpha  \\ \angle C_1CH= \beta} \Rightarrow \cases{\alpha +\beta=90^\circ (\because \overline{AH} 為直徑)\\ \angle C_1HC= \beta(\because \overline{C_1C} =\overline{C_1H})} \Rightarrow \angle CHP=\alpha (\because \angle PHA=90^\circ)\\ 又\overline{CD}與\overline{PH}為矩形對角線,因此\angle DCH=\angle CHP=\alpha \Rightarrow \angle C_1CD= \alpha+\beta =90^\circ\\ 同理,\cases{\angle C_2DH = \angle C_2HD =\alpha \\ \angle CDH=\angle PHD =\beta } \Rightarrow \angle CDC_2 =\alpha+\beta =90^\circ ;\\ 因此 \angle C_CD= \angle C_2DC= 90^\circ \Rightarrow \overline{CD}為公切線,\bbox[red, 2pt]{故得證}$$
=================== end ====================




1 則留言:

  1. 您好,可以請您解109年嘉中科學班考題嗎?謝謝

    回覆刪除