\\(a \\gt 0\\) \u306b\u5bfe\u3057, \u884c\u5217 \\(A\\) \u3092\r\n\\[\r\nA = \\left( \\begin{array}{cc} a & 1 \\\\ -1 & a \\end{array} \\right)\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \\(xy\\) \u5e73\u9762\u4e0a\u306e\u76f4\u7dda \\(y=1\\) \u3092 \\(l _ 1\\) \u3068\u3059\u308b. \\(l _ 1\\) \u306e\u5404\u70b9\u3092\u884c\u5217 \\(A\\) \u3067\u8868\u3055\u308c\u308b \\(1\\) \u6b21\u5909\u63db\u3067\u79fb\u3057\u3066\u3067\u304d\u308b\u76f4\u7dda\u3092 \\(l _ 2\\) \u3068\u3057, \\(l _ 1\\) \u306e\u5404\u70b9\u3092 \\(A\\) \u306e\u9006\u884c\u5217 \\(A^{-1}\\) \u3067\u8868\u3055\u308c\u308b \\(1\\) \u6b21\u5909\u63db\u3067\u79fb\u3057\u3066\u3067\u304d\u308b\u76f4\u7dda\u3092 \\(l _ 3\\) \u3068\u3059\u308b. \u307e\u305f, \\(l _ 1\\) \u3068 \\(l _ 2\\) \u306e\u4ea4\u70b9\u3092 P, \\(l _ 1\\) \u3068 \\(l _ 3\\) \u306e\u4ea4\u70b9\u3092 Q, \\(l _ 2\\) \u3068 \\(l _ 3\\) \u306e\u4ea4\u70b9\u3092 R \u3068\u3057, \u25b3PQR \u306e\u9762\u7a4d\u3092 \\(S(a)\\) \u3068\u3059\u308b. \u4ee5\u4e0b\u306e\u554f\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\u76f4\u7dda \\(l _ 2\\) \u3068\u76f4\u7dda \\(l _ 3\\) \u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(3\\) \u70b9 P , Q , R \u306e\u5ea7\u6a19\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(S(a)\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (4)<\/strong>\u3000\\(S(a)\\) \u3092\u6700\u5c0f\u306b\u3059\u308b \\(a\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(l _ 1 : \\ y = 1\\) \u4e0a\u306e\u70b9\u306f \\(( t, 1 )\\) \uff08 \\(t\\) \u306f\u5b9f\u6570\uff09\u3068\u8868\u305b\u308b.\r\n\\[\\begin{gather}\r\nA \\left( \\begin{array}{c} t \\\\ 1 \\end{array} \\right) = \\left( \\begin{array}{cc} a & 1 \\\\ -1 & a \\end{array} \\right) \\left( \\begin{array}{c} t \\\\ 1 \\end{array} \\right) = \\left( \\begin{array}{c} at+1 \\\\ a-t \\end{array} \\right) \\\\\r\n\\text{\u2234} \\quad \\left\\{ \\begin{array}{l} x =at+1 \\\\ y =a-t \\end{array} \\right.\r\n\\end{gather}\\]\r\n\\(t\\) \u3092\u6d88\u53bb\u3059\u308c\u3070\r\n\\[\\begin{gather}\r\nx+ay = a^2+1 \\\\\r\n\\text{\u2234} \\quad l _ 2 : \\ \\underline{x+ay -a^2-1 =0}\r\n\\end{gather}\\]\r\n\\(A^{-1} = \\dfrac{1}{a^2+1} \\left( \\begin{array}{cc} a & -1 \\\\ 1 & a \\end{array} \\right)\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\nA^{-1} \\left( \\begin{array}{c} t \\\\ 1 \\end{array} \\right) & = \\dfrac{1}{a^2+1} \\left( \\begin{array}{cc} a & -1 \\\\ 1 & a \\end{array} \\right) \\left( \\begin{array}{c} t \\\\ 1 \\end{array} \\right) \\\\\r\n& = \\dfrac{1}{a^2+1} \\left( \\begin{array}{c} at-1 \\\\ a+t \\end{array} \\right) \\\\\r\n\\text{\u2234} \\quad & \\left\\{ \\begin{array}{l} x =\\dfrac{at-1}{a^2+1} \\\\ y =\\dfrac{a+t}{a^2+1} \\end{array} \\right.\r\n\\end{align}\\]\r\n\\(t\\) \u3092\u6d88\u53bb\u3059\u308c\u3070\r\n\\[\\begin{gather}\r\nx-ay = \\dfrac{-a^2-1}{a^2+1} \\\\\r\n\\text{\u2234} \\quad l _ 3 : \\ \\underline{x-ay+1 =0}\r\n\\end{gather}\\]\r\n (2)<\/strong><\/p>\r\n \\(l _ 1\\) \u3068 \\(l _ 2\\) \u306e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\nx+a -a^2-1 & = 0 \\\\\r\n\\text{\u2234} \\quad x & = a^2-a+1\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\text{P} \\ \\underline{\\left( a^2-a+1 , 1 \\right)}\r\n\\]\r\n\\(l _ 1\\) \u3068 \\(l _ 3\\) \u306e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\nx-a+1 & = 0 \\\\\r\n\\text{\u2234} \\quad x & = a-1\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\text{Q} \\ \\underline{\\left( a-1 , 1 \\right)}\r\n\\]\r\n\\(l _ 2\\) \u3068 \\(l _ 3\\) \u306e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\n2x & = a^2 \\\\\r\n\\text{\u2234} \\quad x & = \\dfrac{a^2}{2}\r\n\\end{align}\\]\r\n\u3086\u3048\u306b\r\n\\[\r\ny =\\dfrac{x+1}{a} =\\dfrac{a}{2}+\\dfrac{1}{a}\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n\\text{R} \\ \\underline{\\left( \\dfrac{a^2}{2} , \\dfrac{a}{2}+\\dfrac{1}{a} \\right)}\r\n\\]\r\n (3)<\/strong><\/p>\r\n \\[\\begin{align}\r\n\\left( a^2-a+1 \\right) -(a-1) & = a^2-2+2 \\\\\r\n& = (a-1)^2+1 \\gt 0 \\quad ... [1] \\\\\r\n\\text{\u2234} \\quad a^2-a+1 & \\gt a-1\r\n\\end{align}\\]\r\n\u307e\u305f, \\(a \\gt 0\\) \u306a\u306e\u3067, \u76f8\u52a0\u76f8\u4e57\u5e73\u5747\u306e\u95a2\u4fc2\u3092\u7528\u3044\u3066\r\n\\[\r\n\\dfrac{a}{2} +\\dfrac{1}{a} \\geqq 2 \\sqrt{\\dfrac{a}{2} \\cdot \\dfrac{1}{a}} = \\sqrt{2} >1\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\nS(a) & = \\dfrac{1}{2} \\left( a^2-2a+2 \\right) \\cdot \\dfrac{a^2-2a+2}{2a} \\\\\r\n& = \\underline{\\dfrac{\\left( a^2-2a+2 \\right)^2}{4a}}\r\n\\end{align}\\]\r\n (4)<\/strong><\/p>\r\n \\(f(a) =4S(a)\\) \u3068\u304a\u3051\u3070, \u3053\u308c\u304c\u6700\u5c0f\u3068\u306a\u308b\u3068\u304d\u3092\u8003\u3048\u308c\u3070\u3088\u3044.\r\n\\[\\begin{align}\r\nf'(a) & = \\dfrac{4(a-1) \\left( a^2-2a+2 \\right) a -\\left( a^2-2a+2 \\right)^2 \\cdot 1}{a^2} \\\\\r\n& = \\dfrac{\\left( a^2-2a+2 \\right) \\left( 3a^2-2a-2 \\right)}{a^2} \\\\\r\n& = \\dfrac{3 \\left( a^2-2a+2 \\right) \\left( a -\\frac{1 -\\sqrt{7}}{3} \\right) \\left( a -\\frac{1 +\\sqrt{7}}{3} \\right)}{a^2}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, [1] \u3082\u7528\u3044\u308c\u3070 \\(f(a)\\) \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|cccc} a & (0) & \\cdots & \\dfrac{1+\\sqrt{7}}{3} & \\cdots \\\\ \\hline f'(a) & & - & 0 & + \\\\ \\hline f(a) & & \\searrow & \\text{\u6700\u5c0f} & \\nearrow \\end{array}\r\n\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b \\(a\\) \u306e\u5024\u306f\r\n\\[\r\na = \\underline{\\dfrac{1+\\sqrt{7}}{3}}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(a \\gt 0\\) \u306b\u5bfe\u3057, \u884c\u5217 \\(A\\) \u3092 \\[ A = \\left( \\begin{array}{cc} a & 1 \\\\ -1 & a \\end{array} \\right) \\] \u3067\u5b9a\u3081\u308b. \\(xy\\ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n