107年中科實中教甄-數學詳解

 國立中科實驗高級中學107學年度國中部數學科教師甄選

一、填充題

解答： $$\cases{z=\sqrt 3\sin \theta +i(\sqrt 3\cos \theta +2) \Rightarrow 圓\Gamma:x^2+(y-2)^2 = (\sqrt 3)^2\\ z_1=-3-\sqrt 3i \Rightarrow P(z_1)=(-3,-\sqrt 3)\\ z_2= \sqrt 3+i  \Rightarrow Q(z_2)= (\sqrt 3,1)} \\ \Rightarrow 直線L(\overleftrightarrow{PQ}): y={1+\sqrt 3\over 3+\sqrt 3}x \Rightarrow 圓心(0,2)至L距離={2(3+\sqrt 3)\over \sqrt{16+2\sqrt{48}}} ={2(3+\sqrt 3)\over 2\sqrt 3+2} \\=\sqrt 3=圓半徑\Rightarrow 圓\Gamma與直線L相切\\ \Rightarrow |z-z_1|+ |z-z_2|最小值= |z_1-z_2|= \overline{PQ} = \sqrt{(3+\sqrt 3)^2 + (1+\sqrt 3)^2} =\bbox[red,2pt]{2+2\sqrt 3}$$

解答： $${xy\over x+y} =520 \Rightarrow xy-520(x+y)=0 \Rightarrow (x-520)(y-520)=520^2 = 5^2\times 2^6\times 13^2 \\ \Rightarrow 共有(2+1)\times (6+1)\times (2+1)= 63個正因數 \Rightarrow 有\bbox[red,2pt]{63}組(x,y)$$

解答： $$代入公式:C^n_0+C^n_1 + C^n_2 +C^n_3 = 1+ C^{20}_2 +C^{20}_2+C^{20}_3 =1+ 20+190+1140 = \bbox[red,2pt]{1351}$$

解答： $$\alpha,\beta,\gamma, \delta 為x^4+x^3+1=0之4根 \Rightarrow \cases{\alpha+\beta+ \gamma +\delta=-1 \\ \alpha\beta +\beta\gamma +\gamma\delta + \delta\alpha+  \alpha\gamma+ \beta\delta=0 \\ \alpha\beta\gamma\delta = 1}\\\begin{vmatrix} \alpha & 1 & 1 & 1\\  1& \beta &1 &1 \\ 1& 1& \gamma & 1\\ 1 & 1& 1& \delta\end{vmatrix} =\alpha \begin{vmatrix}    \beta &1 &1 \\  1& \gamma & 1\\  1& 1& \delta\end{vmatrix} -\begin{vmatrix} 1 &1 &1 \\  1& \gamma & 1\\  1& 1& \delta\end{vmatrix}+ \begin{vmatrix} 1 &\beta &1 \\  1& 1 & 1\\  1& 1& \delta\end{vmatrix} -\begin{vmatrix} 1 & \beta &1 \\  1& 1&\gamma  \\  1& 1& 1\end{vmatrix} \\ =\alpha(\beta\gamma\delta+2-\beta-\gamma-\delta)-(\gamma\delta+1-\gamma-\delta )+(\delta+\beta -\beta\delta-1)-(1+\beta\gamma -\beta-\gamma)\\ =\alpha\beta\gamma\delta-(\alpha\beta +\beta\gamma +\gamma\delta +\delta\alpha + \alpha\gamma +\beta\delta)+ 2(\alpha+\beta+\gamma +\delta)-3\\=1-0-2-3= \bbox[red,2pt]{-4}$$

解答： $$\cases{a=\sqrt[3]{40+11\sqrt 13} \\ b= \sqrt[3]{40-11\sqrt 13}} \Rightarrow \cases{a^3+b^3=80 \\ab= \sqrt[3]{1600-1573}=3}\\ 因此(a+b)^3=a^3+b^3+3ab(a+b) \Rightarrow  k^3=80+9k \Rightarrow k^3-9k-80=0，其中k=(a+b) \\ \Rightarrow (k-5)(k^2+5k+16)=0 \Rightarrow k=\bbox[red,2pt]{5}$$

解答： $$\cases{A(0,0,4)\\ B(1,2,3)\\ C(3,-1,2)\\ D(-3,5,-1) \\ R(-13,36,-9)} \Rightarrow \cases{L_1: {x\over 1} ={y\over 2}={z-4\over -1} \\ L_2:{x-3\over -6} ={y+1\over 6}={z-2\over -3}} \Rightarrow \cases{P(s,2s,-s+4)\\ Q(2t+3,-2t-1,t+2)};\\ P、Q、R在一直線上\Rightarrow {\overrightarrow{RP} \over \overrightarrow{RQ}}= k \Rightarrow {s+13\over 2t+16} ={2s-36 \over -2t-37} ={-s+13 \over t+11} =k\\\Rightarrow \cases{\displaystyle {(2s-36)+2(-s+13) \over (-2t-37)+2(t+11)} = {2\over 3}=k \\[.5em]\displaystyle{(s+13)+(2s-36)\over (2t+16)+(-2t-37)} ={3s-23\over -21}=k} \Rightarrow {3s-23\over -21}={2\over 3} \Rightarrow s=3 \Rightarrow t=4\\ \Rightarrow \cases{P(3,6,1)\\ Q(11,-9,6)} \Rightarrow \overline{PQ}= \sqrt{8^2 +15^2+ 5^2} = \bbox[red,2pt]{\sqrt{314}}$$

解答： $$y=f(x)=x^3+ax^2+1 \Rightarrow f'(x)=3x^2+2ax，因此我們可以假設切點為(k,k^3+ak^2+1)，\\則切線方程式為y= (3k^2+2ak)(x-k)+ k^3+ak^2+1；由於切線經過原點，\\因此 0=(3k^2+2ak)(-k)+ k^3+ak^2+1 \Rightarrow g(k)= 2k^3+ak^2-1=0\\ 若g'(k)=0 \Rightarrow 6k^2+2ak=0 \Rightarrow 2k(3k+a)=0 \Rightarrow k=0,-a/3;\\ 由於g(k)=0有三相異實根，因此g(0)g(-a/3) \lt 0 \Rightarrow (-1)(-2a^3/27 +a^3/9-1) \lt 0 \\ \Rightarrow {a^3\over 27}-1 \gt 0\Rightarrow \bbox[red,2pt]{a \gt 3}$$

解答： $$\lim_{x\to \infty} \cfrac{1^9+2^9+\cdots +x^9}{10x^{10}} =\lim_{n\to \infty}\sum_{k=1}^n\cfrac{k^9}{10n^{10}}  =\lim_{n\to \infty}\sum_{k=1}^n{1\over 10} \cdot {1\over n} \cdot ({k\over n})^9 \\ =\int_0^1 {1\over 10}x^9\;dx =\bbox[red,2pt]{1\over 100}$$

解答：

$$假設三切點為D、E、F及中線\overline{CM}與內切圓交於P、Q兩點\\，依題意\overline{CP}=\overline{PQ}= \overline{QA}=a，如上圖；\\\cases{\angle QMD = \angle PMD\\ \angle QDM= \angle MPD (對同弧)} \Rightarrow \triangle MQD \sim \triangle MDP \Rightarrow {\overline{MQ} \over \overline{MD}} ={\overline{MD} \over \overline{MP}}\\ \Rightarrow \overline{MD}^2= \overline{MP}\times \overline{MQ} =2a^2 \cdots(1)\\同理，\overline{CF}^2 = \overline{CP}\times \overline{CQ}=2a^2；又\overline{CE} =\overline{CF}，因此\overline{CE} =\overline{CF}= \overline{MD}=b \cdots(2)\\ 將(2)代入(1) \Rightarrow b^2=2a^2 \Rightarrow b=\sqrt 2a\\ 三角形中線定理: \overline{CA}^2+\overline{CB}^2 = 2(\overline{CM}^2+\overline{AM}^2) \Rightarrow (5+2b)^2 + 5^2 = 2(9a^2+ 5^2) \\ \Rightarrow (5+2\sqrt 2a)^2+25=18a^2+50 \Rightarrow 10a^2-20\sqrt 2a=0 \Rightarrow a=2\sqrt 2 \Rightarrow b=4\\ \Rightarrow \triangle ABC 三邊長分別為\cases{\overline{AB}= 10 \\ \overline{BC}=5 \\ \overline{AC}=5+2b=13} \Rightarrow \triangle ABC面積= \sqrt{s(s-10)(s-5)(s-13)} \\ =\sqrt{14\cdot 4\cdot 9\cdot 1} = \bbox[red,2pt]{6\sqrt{14}}，其中s=(10+5+13)\div 2=14$$

解答：

$$令\overline{AB} =\overline{AC}=a \Rightarrow \overline{BC}=\sqrt 2a \Rightarrow \cases{直角\triangle ABD: \overline{BD}= \sqrt{a^2-4} \\直角\triangle AGC: \overline{GC}= \sqrt{a^2-36} \\ 直角\triangle BCF: \overline{CF}= \sqrt{2a^2-64} };\\ 由於\overline{BD} =\overline{FG} \Rightarrow \sqrt{a^2-4} =\sqrt{a^2-36} +\sqrt{2a^2-64} \Rightarrow a^4-40a^2=0 \\ \Rightarrow a^2(a^2-40)=0 \Rightarrow a^2=40 \Rightarrow \triangle ABC面積={1\over 2}a^2 =\bbox[red, 2pt]{20}$$

解答： $$\cases{a+b+c=0 \\ abc=100} \Rightarrow \cases{b+c=-a\\ bc={100\over a} \gt 0},算幾不等式: {b+c\over 2} \ge \sqrt{bc} \Rightarrow -{a\over 2} \ge \sqrt{100\over a} \\ \Rightarrow {a^2\over 4} \ge {100\over a} \Rightarrow a^3 \ge 400 \Rightarrow a\ge \sqrt[3]{400} =20^{2/3} \Rightarrow a的最小值為\bbox[red,2pt]{20^{2/3}}$$

解答： $$假設正方形邊長為a，B為原點及，如上圖，則\cases{A(0,a)\\ B(0,0)\\ C(a,0)\\ D(a,a)\\P(x,y)}；\\依題意\cases{\overline{PA}=7\\ \overline{PB}=3\\ \overline{PC}=5} \Rightarrow \cases{x^2+(y-a)^2=49 \Rightarrow x^2+y^2 -2ay+a^2=49 \cdots(1)\\ x^2+y^2=9 \cdots(2) \\ (x-a)^2+y^2 =25 \Rightarrow x^2+y^2-2ax+a^2 = 25 \cdots(3)} \\  將(2)代入(1)及(3)，則\cases{-2ay=40-a^2\cdots(4) \\ -2ax =16-a^2 \cdots(5)}；\\(4)^2+(5)^2 \Rightarrow 4a^2(x^2+y^2) = (a^2-40)^2+(a^2-16)^2 \Rightarrow 36a^2 =2a^4-112a^2+1856\\ \Rightarrow a^4-74a^2+928=0  \Rightarrow (a^2-16)(a^2-58)=0 \\ \Rightarrow 面積=a^2 = \bbox[red,2pt]{58} (a=4 \Rightarrow 對角線\overline{AC}=4\sqrt 2 \lt \overline{PA},矛盾)$$

解答： $$令\omega為x^2+x+1=0的解,則\omega^2+\omega+1=0 \Rightarrow (\omega-1)(\omega^2+ \omega+1)= \omega^3-1=0 \Rightarrow \omega^3=1\\ (1+x)^{107}= \sum_{k=0}^{107}C^{107}_kx^k \Rightarrow f(x)=x^2(1+x)^{107} =\sum_{k=0}^{107}C^{107}_kx^{k+2}\\ \Rightarrow \cases{f(1)=2^{107} =\sum_{k=0}^{107}C^{107}_k \\ f(\omega)= \omega^2(1+\omega)^{107} =\omega^2(-\omega^2)^{107} =-\omega^{216}=-1 =\sum_{k=0}^{107}C^{107}_k\omega^{k+2}\\ f(\omega^2)= \omega^4(1+\omega^2)^{107} =\omega^4(-\omega)^{107} =-\omega^{111}=-1 =\sum_{k=0}^{107} C^{107}_k\omega^{2k+4}} \\ \Rightarrow f(1)+ f(\omega)+f(\omega^2)=2^{107} -2=\sum_{k=0}^{107}a_kC^{107}_k,\\ \qquad \qquad其中a_k=\begin{cases} 1+\omega^{3s+2}+ \omega^{6s+4} =1+\omega^2+\omega =0 & \text{if }k=3s \\ 1+ \omega^{3s+3} +\omega^{6s+6}= 1+1+1=3 & \text{if }k=3s+1 \\ 1+ \omega^{3s+4} +\omega^{6s+8}= 1+ \omega +\omega^2 =0 & \text{if }k=3s+2\end{cases},s=0,1,\dots\\ 因此3(C^{2017}_1 +C^{2017}_4 +C^{2017}_7 +\cdots +C^{2017}_{106} )=2^{107}-2 =X，\\ \log(2^{107}-2) \approx \log 2^{107} \approx 107\times 0.301 =32.207 \Rightarrow X為33位數，即m=33;\\尾數2.07 \lt 0.301=\log 2 \Rightarrow 第1個數字是1，即n=1;\\2^m的個位數字為:2,4,8,6,2,4,8,6,\dots，循環數為4，而107=4\times 26+3，\\因此2^{107}的個位數字是8 \Rightarrow 2^{107}-2的個位數字是8-2=6，即k=6；\\(m,n,k)=\bbox[red,2pt]{(33,1,6)}$$

解答： $$42=2\times 3\times 7 =(1+1)(2+1)(6+1) \Rightarrow X=p^3+ 2p^2+p =a\times b^2\times c^6,其中a,b,c均為質數\\ 要使X最小，顯然c=2,b=2，也就是X=p(p+1)^2 = a\times 3^2\times 2^6 = a\times (3\times 2^3)^2\\ \Rightarrow p+1=3\times 2^3=24 \Rightarrow p= \bbox[red,2pt]{23}$$

解答：

$$此題相當於兩圓\cases{x^2+y^2=1 \\ (x-3)^2+y^2=9}的交集繞x軸旋轉的體積；\\先求兩圓的交點: (x-3)^2+(1-x^2)=9 \Rightarrow x={1\over 6} \\\Rightarrow \cases{小圓繞x軸旋轉體積=\int_{1/6}^1 (1-x^2)\pi\;dx ={325\over 648}\pi \\大圓繞x軸旋轉體積=\int_0^{1/6} (9-(x-3)^2)\pi\;dx ={53\over 648}\pi}，兩者相加= \bbox[red,2pt]{7\pi\over 12}$$

解答： $$\lfloor x+{19\over 100}\rfloor + \cdots +\lfloor x+{91\over 100}\rfloor共有91-19+1=73個整數，而且不是k，就是k+1,k\in N\\又546 \div 73\approx 7.5，因此k=7；\\假設有m個7，73-m個8，則7m+8(73-m)=546 \Rightarrow m=38\\ \Rightarrow \cases{第38個數字是\lfloor x+{56\over 100}\rfloor=7 \\ 第39個數字是\lfloor x+{57\over 100}\rfloor =8 } \Rightarrow 7.44\gt x\ge 7.43 \Rightarrow \lfloor 100x \rfloor =\bbox[red,2pt]{743}$$

解答：

$$令\cases{\angle AEB= \alpha \\ \angle CFB=\beta} \Rightarrow \angle APC= \alpha+\beta \Rightarrow \angle EPF=\alpha +\beta\\ 四邊形BEPF內角和=360^\circ = 90^\circ +2(\alpha+\beta) \Rightarrow \alpha+\beta = 135^\circ\\ \cases{\tan \angle CFB = \tan \beta = {6\over 4}={3\over 2} \cdots(1)\\ \tan \angle AEB= \tan (135^\circ-\beta) ={\tan 135^\circ -\tan \beta \over 1+\tan 135^\circ \tan \beta} ={-1-\tan \beta \over 1-\tan \beta} ={a+4\over 2} \cdots(2)}\\ 將(1)代入(2) \Rightarrow {-1-3/2 \over 1-3/2} ={a+4\over 2} \Rightarrow a=6 \Rightarrow 面積= (6+4)\times (2+4)=\bbox[red,2pt]{60}$$

解答： $$假設半碗的球心為O，半徑為R，三球的球心分別為O_1,O_2,O_3，半徑皆為10，\\則四面體O-O_1O_2O_3的底面\triangle O_1O_2O_2為正\triangle,\\三個側面\triangle OO_1O_2, \triangle OO_2O_3, \triangle OO_1O_3為全等的等腰\triangle；\\令G為正\triangle O_1O_2O_3的重心，則在直角\triangle OGO_1中，\cases{\overline{GO}=10\\ \overline{GO_1}= 20 \times {\sqrt 3 \over 2}\times {2\over 3} ={20\over 3}\sqrt 3\\\overline{OO_1}=R-10} \\ \Rightarrow  \overline{OO_1}^2 = \overline{GO_1}^2 +\overline{OG}^2 \Rightarrow (R-10)^2 = ({20\over 3}\sqrt 3)^2+ 10^2 ={700\over 3} \Rightarrow R-10={10\sqrt 7\over \sqrt 3} \\ \Rightarrow R=\bbox[red,2pt]{10(1 + \sqrt{7\over 3} )}$$

解答： $$\cfrac{(1+{1\over n+1})^{n+1}}{(1+{1\over n})^{n+1}} =\left({n(n+2)\over (n+1)^2} \right)^{n+1} = \left(1-{1\over (n+1)^2}\right)^{n+1} \gt 1-{n+1\over (n+1)^2} = {n\over n+1} ={1\over 1+{1\over n}} \\ \Rightarrow \cfrac{(1+{1\over n+1})^{n+1}}{(1+{1\over n})^{n+1}} \gt {1\over 1+{1\over n}} \Rightarrow \left(1+{1\over n+1}\right)^{n+1} \gt \left(1+{1\over n} \right)^{n}，\bbox[red,2pt]{故得證}\\註:白努利不等式 (1+x)^n \gt 1+nx, 其中x \gt -1, n\ge 1$$