\u5ea7\u6a19\u5e73\u9762\u4e0a\u306e\u539f\u70b9\u3092 O \u3068\u3059\u308b. \u70b9 A \\(( a , 0 )\\) , \u70b9 B \\(( 0 , b )\\) \u304a\u3088\u3073\u70b9 C \u304c\r\n\\[\r\n\\text{OC} = 1 , \\quad \\text{AB} = \\text{BC} = \\text{CA}\r\n\\]\r\n\u3092\u6e80\u305f\u3057\u306a\u304c\u3089\u52d5\u304f.<\/p>\r\n
(1)<\/strong>\u3000\\(s = a^2 +b^2\\) , \\(t = ab\\) \u3068\u3059\u308b. \\(s\\) \u3068 \\(t\\) \u306e\u95a2\u4fc2\u3092\u8868\u3059\u7b49\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(\\triangle \\text{ABC}\\) \u306e\u9762\u7a4d\u306e\u3068\u308a\u3046\u308b\u5024\u306e\u7bc4\u56f2\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n C \\(( u , v )\\) \u3068\u304a\u304f\u3068, \u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\nu^2 +v^2 & = 1 \\quad ... [1] \\\\\r\n\\text{AB} & = a^2 +b^2\r\n\\end{align}\\]\r\n\u307e\u305f \\(\\text{AB} = \\text{BC}\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\text{BC} = u^2 +( v-b )^2 & = a^2 +b^2 \\\\\r\n1 -2bv & = a^2 \\quad ( \\text{\u2235} \\ [1] )\r\n\\end{align}\\]\r\n- \\(b = 0\\) \u306e\u3068\u304d\r\n\\[\r\na = \\pm 1\r\n\\]\r\n- \\(b \\neq 0\\) \u306e\u3068\u304d\r\n\\[\r\nv = \\dfrac{a^2 +1}{2b} \\quad ... [2]\r\n\\]\r\n\\(\\text{AB} = \\text{CA}\\) \u306b\u3064\u3044\u3066\u3082\u540c\u69d8\u306b\u8003\u3048\u308c\u3070<\/p>\r\n \\(a = 0\\) \u306e\u3068\u304d\r\n\\[\r\nb = \\pm 1\r\n\\]<\/li>\r\n \\(a \\neq 0\\) \u306e\u3068\u304d\r\n\\[\r\nu = \\dfrac{b^2 +1}{2a} \\quad ... [3]\r\n\\]\r\n\\(a \\neq 0 , \\ b \\neq 0 \\ ... [4]\\) \u306e\u5834\u5408\u3092\u8003\u3048\u308c\u3070, [1] \u306b [2] [3] \u3092\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\n\\dfrac{( b^2 +1 )^2}{4a^2} +\\dfrac{( a^2 +1 )^2}{4b^2} & = 1 \\\\\r\nb^6 -2b^4 +b^2 +a^6 -2a^4 +a^2 & = 4a^2 b^2 \\\\\r\n( a^2 +b^2 )^3 -3a^2 b^2 ( a^2 +b^2 ) \\qquad \\qquad & \\\\\r\n-2 ( a^2 +b^2 )^2 +a^2 +b^2 & = 0 \\\\\r\ns ( s^2 -3t^2 -2s +1 ) & = 0 \\\\\r\n\\end{align}\\]\r\n[4] \u3088\u308a \\(s \\neq 0\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\ns^2 -3t^2 -2s +1 & = 0 \\\\\r\n(s-1)^2 -3t^2 & = 0 \\\\\r\n\\left( s-1 +\\sqrt{3} t \\right) \\left( s-1 -\\sqrt{3} t \\right) & = 0 \\\\\r\n\\text{\u2234} \\quad s = 1 \\pm \\sqrt{3} t \\quad & ... [5]\r\n\\end{align}\\]\r\n\\(a = 0\\) \u307e\u305f\u306f \\(b = 0\\) \u306e\u5834\u5408\u3082, \\(s = 1\\) , \\(t = 0\\) \u306a\u306e\u3067 [5] \u3092\u6e80\u305f\u3059.<\/p><\/li>\r\n<\/ul>\r\n \u3088\u3063\u3066\r\n\\[\r\n\\underline{s = 1 \\pm \\sqrt{3} t}\r\n\\]\r\n (2)<\/strong><\/p>\r\n \u25b3 ABC \u306e\u9762\u7a4d\u3092 \\(S\\) \u3068\u304a\u304f\u3068\r\n\\[\r\nS = \\dfrac{\\sqrt{3}}{4} ( a^2 +b^2 ) = \\dfrac{\\sqrt{3} s}{4} \\quad ... [6]\r\n\\]\r\n\\(u = a+b\\) \u3068\u304a\u304f\u3068, (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\nu^2 -2t & = 1 \\pm \\sqrt{3} t \\\\\r\n\\text{\u2234} \\quad u^2 & = \\left( 2 \\pm \\sqrt{3} \\right) t +1 \\quad ... [7]\r\n\\end{align}\\]\r\n\\(a , b\\) \u306f, \\(X\\) \u306b\u95a2\u3059\u308b\u65b9\u7a0b\u5f0f \\(X^2 -uX +t = 0\\) \u306e\u5b9f\u6570\u89e3\u306a\u306e\u3067, \u5224\u5225\u5f0f \\(D\\) \u306b\u3064\u3044\u3066\r\n\\[\r\nD = u^2 -4t \\geqq 0\r\n\\]\r\n[7] \u3092\u4ee3\u5165\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n\\left( 2 \\pm \\sqrt{3} \\right) t +1 -4t & \\geqq 0 \\\\\r\n\\left( 2 \\mp \\sqrt{3} \\right) t & \\leqq 1 \\\\\r\n\\text{\u2234} \\quad t \\leqq 2 & \\pm \\sqrt{3}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, (1)<\/strong> \u306e\u7d50\u679c\u3068\u3042\u308f\u305b\u3066<\/p>\r\n \\(s = 1 +\\sqrt{3} t\\) \u306e\u3068\u304d, \\(t \\leqq 2 +\\sqrt{3}\\) \u306a\u306e\u3067\r\n\\[\r\ns \\leqq 1 +\\sqrt{3} \\left( 2 +\\sqrt{3} \\right) = 2 \\left( 2 +\\sqrt{3} \\right)\r\n\\]<\/li>\r\n \\(s = 1 -\\sqrt{3} t\\) \u306e\u3068\u304d, \\(t \\leqq 2 -\\sqrt{3}\\) \u306a\u306e\u3067\r\n\\[\r\ns \\geqq 1 -\\sqrt{3} \\left( 2 -\\sqrt{3} \\right) = 2 \\left( 2 -\\sqrt{3} \\right)\r\n\\]<\/li>\r\n<\/ul>\r\n \u3057\u305f\u304c\u3063\u3066\r\n\\[\r\n2 \\left( 2 -\\sqrt{3} \\right) \\leqq s \\leqq 2 \\left( 2 +\\sqrt{3} \\right)\r\n\\]\r\n\u3088\u3063\u3066, [6] \u3088\u308a\r\n\\[\r\n\\begin{gather}\r\n\\dfrac{\\sqrt{3}}{2} \\left( 2 -\\sqrt{3} \\right) \\leqq S \\leqq \\dfrac{\\sqrt{3}}{2} \\left( 2 +\\sqrt{3} \\right) \\\\\r\n\\underline{\\sqrt{3} -\\dfrac{3}{2} \\leqq S \\leqq \\sqrt{3} +\\dfrac{3}{2}}\r\n\\end{gather}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u5ea7\u6a19\u5e73\u9762\u4e0a\u306e\u539f\u70b9\u3092 O \u3068\u3059\u308b. \u70b9 A \\(( a , 0 )\\) , \u70b9 B \\(( 0 , b )\\) \u304a\u3088\u3073\u70b9 C \u304c \\[ \\text{OC} = 1 , \\quad \\text{AB} = \\text{BC} … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
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