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(1)

\(\alpha\) の式は \[\begin{align} -\dfrac{x}{2} +y +z & = 1 \quad ... [1] \\ \text{∴} \quad z & = \dfrac{x}{2} -y +1 \quad ... [2] \end{align}\] [1]より, \(\alpha\) に垂直なベクトルの \(1\) つは \(\left( \begin{array}{c} -\dfrac{1}{2} \\ 1 \\ 1 \end{array} \right)\) . \[ \overrightarrow{\text{OR}} = \overrightarrow{\text{OP}} +k \overrightarrow{\text{OQ}} = \left( \begin{array}{c} k \\ k+5 \\ k+5 \end{array} \right) \] なので, [2] も用いて \[ \overrightarrow{\text{RS}} = \overrightarrow{\text{OS}} -\overrightarrow{\text{OR}} = \left( \begin{array}{c} x-k \\ y -k -5 \\ \dfrac{x}{2} -y -k-4 \end{array} \right) \] また, 実数 \(m\) を用いて \[ \overrightarrow{\text{RS}} = m \left( \begin{array}{c} -\dfrac{1}{2} \\ 1 \\ 1 \end{array} \right) \] ゆえに, 成分を比較して \[ \left\{ \begin{array}{ll} x-k = -\dfrac{m}{2} & ... [3] \\ y -k -5 = m & ... [4] \\ \dfrac{x}{2} -y -k -4 = m & ... [5] \end{array} \right. \] [4] [5] より \[\begin{align} y -k -5 & = \dfrac{x}{2} -y -k -4 \\ \text{∴} \quad x & = 4y -2 \quad ... [6] \end{align}\] [3] に代入して \[\begin{align} -2 ( 4y -k ) & = y -k -5 \\ 9y & = 3k +9 \\ \text{∴} \quad y & = \dfrac{k}{3} +1 \end{align}\] [2] [6] より \[\begin{align} x & = \dfrac{4}{3} k +2 \ , \\ z & = \dfrac{2}{3} k +1 -\dfrac{k}{3} -1 +1 = \dfrac{k}{3} +1 \end{align}\] ゆえに \[ \text{S} \ \left( \dfrac{4}{3} k +2 , \dfrac{k}{3} +1 , \dfrac{k}{3} +1 \right) \] よって \[ \overrightarrow{\text{AS}} = \overrightarrow{\text{OS}} -\overrightarrow{\text{OA}} = \underline{\left( \dfrac{k}{3} +1 \right) \left( \begin{array}{c} 4 \\ 1 \\ 1 \end{array} \right)} \]

(2)

BC の中点を M とおくと, \(\overrightarrow{\text{AB}} = \left( \begin{array}{c} 2 \\ 1 \\ 0 \end{array} \right)\) , \(\overrightarrow{\text{AC}} = \left( \begin{array}{c} 2 \\ 0 \\ 1 \end{array} \right)\) なので \[\begin{align} \overrightarrow{\text{AS}} = \left( \dfrac{k}{3} +1 \right) \left( \overrightarrow{\text{AB}} +\overrightarrow{\text{AC}} \right) = 2 \left( \dfrac{k}{3} +1 \right) \overrightarrow{\text{AM}} \end{align}\] したがって, 点 S が条件をみたすのは, 線分 AM 上にあるときで, 求める \(k\) 値の範囲は \[\begin{gather} 0 \leqq 2 \left( \dfrac{k}{3} +1 \right) \leqq 1 \\ -1 \leqq \dfrac{k}{3} \leqq -\dfrac{1}{2} \\ \text{∴} \quad \underline{-3 \leqq k \leqq -\dfrac{3}{2}} \end{gather}\]

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【 解 答 】

(1)

\(0 \lt p \lt 1\) より, \(\dfrac{1}{p} \gt 1\) .
\(C_1\) の式より \[ y' = p \cdot \dfrac{1}{p} x^{\frac{1}{p} -1} = x^{\frac{1}{p} -1} \] なので, 点 \(( a , b )\) における接線の式は \[\begin{align} y & = a^{\frac{1}{p} -1} ( x-a ) +p a^{\frac{1}{p}} \\ & = a^{\frac{1}{p} -1} x -( 1-p ) a^{\frac{1}{p}} \quad ... [1] \end{align}\] \(C_2\) の式より \[ y' = \dfrac{1}{x} \] なので, 点 \(( a , b )\) における接線の式は \[\begin{align} y & = \dfrac{1}{a} ( x-a ) +\log a +q \\ & = \dfrac{1}{a} x +\log a +q -1 \quad ... [2] \end{align}\] [1] [2] が一致するので \[ \left\{ \begin{array}{ll} a^{\frac{1}{p} -1} = \dfrac{1}{a} & ... [3] \\ -( 1-p ) a^{\frac{1}{p}} = \log a +q -1 & ... [4] \end{array} \right. \] [3] より \[\begin{align} a^{\frac{1}{p}} & = 1 \\ \text{∴} \quad a & = 1 \end{align}\] [4] に代入して \[\begin{align} p-1 & = q-1 \\ \text{∴} \quad q & = \underline{p} \end{align}\]

(2)

\(a = 1\) より, \(b = p\) .
\(0 \lt p \lt 1\) より, \(\dfrac{1}{e} \lt e^{-p} \lt 1\) .
したがって \[\begin{align} S_1 & = \displaystyle\int _ {e^{-p}}^{1} p x^{\frac{1}{p}} \, dx \\ & = p \left[ \dfrac{1}{\frac{1}{p} +1} x^{\frac{1}{p} +1} \right] _ {e^{-p}}^{1} \\ & = \underline{\dfrac{p^2}{p+1} \left( 1 -e^{-p-1} \right)} \end{align}\] また \[\begin{align} S_2 & = \displaystyle\int _ {e^{-p}}^{1} ( \log x +p ) \, dx \\ & = \left[ x ( \log x -1) +px \right] _ {e^{-p}}^{1} \\ & = -1 +p +e^{-p} ( p+1 ) -p e^{-p} \\ & = \underline{e^{-p} +p -1} \end{align}\]

(3)

\(S_1 \gt 0\) なので, 示したい式を変形すると \[ 4 S_2 -3 S_1 \geqq 0 \quad ... [\text{A}] \] なので, これを示せばよい.
\[\begin{align} 4 S_2 -3 S_1 & = 4 \left( e^{-p} +p -1 \right) -\dfrac{3 p^2}{p+1} \left( 1 -e^{-p-1} \right) \\ & = \dfrac{4 ( p+1 ) \left( e^{-p} +p -1 \right) -3 p^2 ( 1 -e^{-p-1} )}{p+1} \\ & = \dfrac{\left( \dfrac{3}{e} p^2 +4p +4 \right) e^{-p} +p^2 -4}{p+1} \\ & \gt \dfrac{( p+2 )^2 e^{-p} +( p+2 ) ( p-2 )}{p+1} \quad \left( \ \text{∵} \ \dfrac{3}{e} \gt 1 \ \right) \\ & = \dfrac{p+2}{p+1} \underline{\left\{ (p+2) e^{-p} +p-2 \right\}} _ {[5]} \end{align}\] [5] を \(f(p)\) とおくと, \(f(p) \gt 0\) を示せばよい. \[\begin{align} f'(p) & = e^{-p} -( p+2 ) e^{-p} +1 \\ & = -( p+1 ) e^{-p} +1 \ , \\ f''(p) & = -e^{-p} +( p+1 ) e^{-p} =p e^{-p} \gt 0 \end{align}\] したがって, \(f'(p)\) は単調増加し \[ f'(p) \gt f'(0) = -1 +1 = 0 \] したがって, \(f(p)\) は単調増加し \[ f(p) \gt f(0) = 2 -2 = 0 \] よって, [A] が示されて題意も示された.

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(1)

点 A は円 \(x^2 +y^2 =1\) 上の点なので, 円周角の定理より \[ \angle \text{PAQ} = \angle \text{QAB} = \dfrac{\pi}{4n} \] \(\text{BV} = 3 \tan \dfrac{\pi}{2n}\) , \(\text{BW} = 3 \tan \dfrac{\pi}{4n}\) なので \[\begin{align} S(n) +T(n) & = \triangle \text{VWA} \\ & = \dfrac{1}{2} \left( 3 \tan \dfrac{\pi}{2n} -3 \tan \dfrac{\pi}{4n} \right) \cdot 3 \\ & = \dfrac{9}{2} \left( \tan \dfrac{\pi}{2n} -\tan \dfrac{\pi}{4n} \right) \end{align}\] よって \[\begin{align} n \{ S(n) +T(n) \} & = \dfrac{9}{2} \left( \dfrac{\pi}{2} \cdot \dfrac{\sin \dfrac{\pi}{2n}}{\dfrac{\pi}{2n}} \cdot \dfrac{1}{\cos \dfrac{\pi}{2n}} -\dfrac{\pi}{4} \cdot \dfrac{\sin \dfrac{\pi}{4n}}{\dfrac{\pi}{4n}} \cdot \dfrac{1}{\cos \dfrac{\pi}{4n}} \right) \\ & \rightarrow \dfrac{9}{2} \left( \dfrac{\pi}{2} \cdot 1 \cdot 1 -\dfrac{\pi}{4} \cdot 1 \cdot 1 \right) \quad ( \ n \rightarrow \infty \ \text{のとき} \ ) \\ & = \dfrac{9 \pi}{8} \end{align}\] すなわち \[ \displaystyle\lim_{n \rightarrow \infty} n \{ S(n) +T(n) \} = \underline{\dfrac{9 \pi}{8}} \]

(2)

\[\begin{align} S(n) & = \triangle \text{AOP} +( \text{扇形 OPQ} ) -\triangle \text{OPQ} \\ & = \dfrac{1}{2} \cdot 1 \cdot \sin \dfrac{\pi}{n} -\dfrac{1}{2} \cdot 1^2 \cdot \dfrac{\pi}{2n} +\dfrac{1}{2} \cdot 1^2 \cdot \sin \dfrac{\pi}{2n} \\ & = \dfrac{1}{2} \left( \sin \dfrac{\pi}{n} +\sin \dfrac{\pi}{2n} -\dfrac{\pi}{2n} \right) \end{align}\] なので \[\begin{align} n S(n) & = \dfrac{1}{2} \left( \pi \cdot \dfrac{\sin \dfrac{\pi}{n}}{\dfrac{\pi}{n}} +\dfrac{\pi}{2} \cdot \dfrac{\sin \dfrac{\pi}{2n}}{\dfrac{\pi}{2n}} -\dfrac{\pi}{2} \right) \\ & \rightarrow \dfrac{1}{2} \left( \pi +\dfrac{\pi}{2} -\dfrac{\pi}{2} \right) \quad ( \ n \rightarrow \infty \ \text{のとき} \ ) \\ & = \dfrac{\pi}{2} \end{align}\] これと (1) の結果を用いれば \[\begin{align} \dfrac{T(n)}{S(n)} & = \dfrac{n \{ S(n) +T(n) \}}{n S(n)} -1 \\ & \rightarrow \dfrac{\dfrac{9 \pi}{8}}{\dfrac{\pi}{2}} -1 \quad ( \ n \rightarrow \infty \ \text{のとき} \ ) \\ & = \dfrac{5}{4} \end{align}\] すなわち \[ \displaystyle\lim_{n \rightarrow \infty} \dfrac{T(n)}{S(n)} = \underline{\dfrac{5}{4}} \]

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【 解 答 】

(1)

\[\begin{align} \omega & = \cos \dfrac{2 \pi}{3} +i \sin \dfrac{2 \pi}{3} \ , \\ \omega^2 & = \cos \dfrac{4 \pi}{3} +i \sin \dfrac{4 \pi}{3} = \cos \left( -\dfrac{2 \pi}{3} \right) +i \sin \left( -\dfrac{2 \pi}{3} \right) \end{align}\] したがって \[\begin{align} & \text{OA} = \text{OB} = \text{OC} = 1 \ , \\ & \angle \text{AOB} = \angle \text{BOC} = \angle \text{COA} = \dfrac{2 \pi}{3} \end{align}\] ゆえに, \(2\) 辺とその間の角がそれぞれ等しいので \[ \triangle \text{OAB} \equiv \triangle \text{OBC} \equiv \triangle \text{OCA} \] よって, \(\text{AB} = \text{BC} = \text{CA}\) なので, \(\triangle \text{ABC}\) は正三角形.

(2)

点 \(-z\) は点 \(z\) と原点について対称なので, 描く図形は下図実線部.

(3)

AB, AC 上を動く \(z\) をそれぞれ \(z_1 , z_2\) とおくと \(E_1 , E_2\) の共有点となるのは \[\begin{align} { z_1 }^2 & = { z_2 }^2 \\ \text{∴} \quad z_1 & = \pm z_2 \end{align}\]

• \(z_1 = z_2\) のとき, AB と AC の共有点は A なので \[ z = 1 \] このとき, \(z^2 = 1\) .

• \(z_1 = -z_2\) のとき, (2) で求めた図形と AB の共有点は \(\dfrac{i}{\sqrt{3}}\) なので \[ z = \dfrac{i}{\sqrt{3}} \] このとき, \(z^2 = -\dfrac{1}{3}\) .

よって, 求める共有点は \[ \underline{1 , -\dfrac{1}{3}} \]

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