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111竹北高中

竹北一如往常的簡單
填充1.
如圖,設\(\Delta ABC\)中,\(\angle C=90^{\circ}\),以\(\overline{AB}\)、\(\overline{AC}\)為邊向外各作正三角形\(\Delta ABF\)和\(\Delta ACG\),點\(M\)是\(\overline{BC}\)中點。若\(\overline{MF}=11\),\(\overline{MG}=7\),則\(\overline{BC}\)的長度為   
[解答]
由費馬點知\(\overline{BG}=\overline{CF}\)
由中線定理:\(\overline{CG}^2+\overline{BG}^2=2(\overline{MG}^2+\overline{MC}^2)\),\(\overline{CF}^2+\overline{BF}^2=2(\overline{MF}^2+\overline{MB}^2)\)
兩式相減得\(\overline{BF}^2-\overline{CG}^2=2(\overline{MF}^2-\overline{MG}^2)=\overline{BC}^2=144\)

填充2.
設函數\(f:(0,\infty)\to R\),滿足\(\displaystyle f\left(1-\frac{1}{1+t}\right)+f\left(\frac{1+t}{t}\right)log(1+t)=f\left(\frac{1+t}{t}\right)logt+2022\),則\(f(1000)=\)   
[解答]
分別令\(\displaystyle x=\frac{t}{1+t}\)和\(\displaystyle x=\frac{t+1}{t}\)
\(f(x)-f(\frac{1}{x})\log x=2022\),\(f(\frac{1}{x})+f(x)\log x=2022\),\(\log xf(\frac{1}{x})+f(x)(\log x)^2=2022\log x\)
\(f(x)(1+(\log x)^2)=2022+2022\log x\),\(\displaystyle f(x)=\frac{2022+2022\log x}{1+(\log x)^2}\)

填充3.
設\(\cases{\displaystyle a_0=\frac{\sqrt{3}}{2} \cr a_n=\left(\frac{1+a_{n-1}}{2}\right)^{\frac{1}{2}}}\),試求\(\displaystyle \lim_{n\to \infty}4^n(1-a_n)\)的值為   
[解答]
\(\displaystyle a_0=\cos\frac{\pi}{6}\),\(\displaystyle a_n=\cos\frac{\frac{\pi}{6}}{2^n}\),\(\displaystyle 1-a_n=2\sin^2\frac{\frac{\pi}{12}}{2^n}\)
令\(\displaystyle x=\frac{1}{2^n}\),所求\(\displaystyle=\lim_{x\to0}\frac{2\sin^2\frac{\pi}{12}x}{x^2}=\lim_{x\to0}\frac{\pi^2}{72}\left(\frac{\sin\frac{\pi}{12}x}{\frac{\pi}{12}x}\right)^2=\frac{\pi^2}{72}\)

填充8.
拋物線\(y^2=4cx(c>0)\)的焦點為\(F\),準線為\(L\)。\(A\)、\(B\)是拋物線上的兩動點,且滿足\(\displaystyle \angle AFB=\frac{\pi}{3}\),設線段\(AB\)的中點\(M\)在\(L\)上的投影點為\(N\),則\(\displaystyle \frac{\overline{MN}}{\overline{AB}}\)的最大值為   
[解答]
令\(\overline{AF}=a,\overline{BF}=b\),\(\displaystyle \frac{\overline{MN}}{\overline{AB}}=\frac{\displaystyle\frac{d(A,L)+d(B,L)}{2}}{\overline{AB}}=\frac{a+b}{2\sqrt{a^2+b^2-ab}}\)
平方,\(\displaystyle \frac{a^2+2ab+b^2}{4a^2+4b^2-4ab}=\frac{1}{4}+\frac{3ab}{4a^2+4b^2-4ab}\le\frac{1}{4}+\frac{3ab}{8ab-4ab}=1\)

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