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109高中數學能力競賽

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第1題
若正整數\(n\)滿足\(\displaystyle\frac{(2^3-1)\times(3^3-1)\times(4^3-1)\times\cdots\times(n^3-1)}{(2^3+1)\times(3^3+1)\times(4^3+1)\times\cdots\times(n^3+1)}\geq\frac{401}{600}\),則\(n\)的最大值為   
[解答]
\(\begin{align}
  & \frac{{{n}^{3}}-1}{{{n}^{3}}+1}=\frac{n-1}{n+1}\times \frac{n\left( n+1 \right)+1}{n\left( n-1 \right)+1} \\
&  \\
& \frac{1}{3}\times \frac{2}{4}\times \frac{3}{5}\times \cdots \times \frac{n-1}{n+1}=\frac{2}{n\left( n+1 \right)} \\
& \frac{2\left( 2+1 \right)+1}{2\left( 2-1 \right)+1}\times \frac{3\left( 3+1 \right)+1}{3\left( 3-1 \right)+1}\times \frac{4\left( 4+1 \right)+1}{4\left( 4-1 \right)+1}\times \cdots \times \frac{n\left( n+1 \right)+1}{n\left( n-1 \right)+1}=\frac{n\left( n+1 \right)+1}{3} \\
&  \\
& \prod\limits_{k=2}^{n}{\frac{{{k}^{3}}-1}{{{k}^{3}}+1}}=\frac{2}{n\left( n+1 \right)}\times \frac{n\left( n+1 \right)+1}{3}=\frac{2}{3}\left[ 1+\frac{1}{n\left( n+1 \right)} \right] \\
& \frac{2}{3}\left[ 1+\frac{1}{n\left( n+1 \right)} \right]\ge \frac{401}{600} \\
& \cdots \cdots  \\
\end{align}\)

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第 7 題
如圖,\(\angle ABC=90^{\circ}\),\(P\)為射線\(\overrightarrow{BC}\)上的動點,且\(\angle CBD\)為銳角。設\(\sin\angle CBD=a\),\(\overline{AP}\)交\(\overline{BD}\)於點\(E\),且\(\overline{PF}\)垂直\(\overline{BD}\)於點\(F\)。若\(\overline{AB}=k\),則\(\overline{AP}-\overline{PF}\)的最小值為[u]   [/u]。(以\(a,k\)的數學式表示)
[解答]
AP - PF >= AP - PE = AE
等號成立於 AP 和 BD 垂直

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