{"id":695,"date":"2013-03-20T18:05:12","date_gmt":"2013-03-20T09:05:12","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=695"},"modified":"2021-10-30T15:42:58","modified_gmt":"2021-10-30T06:42:58","slug":"wsr200705","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/wsr200705\/","title":{"rendered":"\u65e9\u7a32\u7530\u7406\u5de52007\uff1a\u7b2c5\u554f"},"content":{"rendered":"
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\\(xy\\) \u5e73\u9762\u306b\u304a\u3044\u3066, \u70b9 \\(( 5 \\sqrt{3} , 0 )\\) \u3092\u4e2d\u5fc3\u3068\u3059\u308b\u534a\u5f84 \\(5\\) \u306e\u5186\u3092 C, \u70b9 \\(( -4 \\sqrt{3} , 0 )\\) \u3092\u4e2d\u5fc3\u3068\u3059\u308b\u534a\u5f84 \\(4\\) \u306e\u5186\u3092 D \u3068\u3059\u308b.\r\nC , D \u306e\u5171\u901a\u63a5\u7dda\u306e\u3046\u3061, C , D \u304c\u7570\u306a\u308b\u5074\u306b\u3042\u308a\u50be\u304d\u304c\u6b63\u3067\u3042\u308b\u3082\u306e\u3092 \\(\\ell\\) , \u50be\u304d\u304c\u8ca0\u3067\u3042\u308b\u3082\u306e\u3092 \\({\\ell}'\\) \u3068\u3057, C , D \u304c\u540c\u3058\u5074\u306b\u3042\u308a\u50be\u304d\u304c\u6b63\u3067\u3042\u308b\u3082\u306e\u3092 \\(m\\) \u3068\u3059\u308b. \u4ee5\u4e0b\u306e\u554f\u306b\u7b54\u3048\u3088.<\/p>\r\n

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  1. (1)<\/strong>\u3000\u76f4\u7dda \\(\\ell\\) \u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u76f4\u7dda \\(m\\) \u306e\u65b9\u7a0b\u5f0f\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\u4e09\u76f4\u7dda \\(\\ell , {\\ell}' , m\\) \u306e\u3059\u3079\u3066\u306b\u63a5\u3057 C , D \u3068\u7570\u306a\u308b\u5186\u3092 E , E' \u3068\u3059\u308b. \u4e8c\u5186 E , E' \u306e\u4e2d\u5fc3\u306e \\(x\\) \u5ea7\u6a19\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  4. (4)<\/strong>\u3000(3)<\/strong> \u306e\u5186 E , E' \u306e\u534a\u5f84\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

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    \u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n\"waseda_r_2007_05_01\"\r\n

    (1)<\/strong><\/p>\r\n

    \u76f4\u7dda \\(y=ax+b\\) \u304c\u5186 C , D \u3068\u63a5\u3059\u308b\u306e\u3067\r\n\\[\\begin{align}\r\n\\dfrac{\\left| 5 \\sqrt{3} a +b \\right|}{\\sqrt{a^2+1}} & = 5 \\\\\r\n\\dfrac{\\left| -4 \\sqrt{3} a +b \\right|}{\\sqrt{a^2+1}} & = 4 \\\\\r\n\\text{\u2234} \\quad \\left| \\sqrt{3} a +\\dfrac{b}{5} \\right| = \\left| \\sqrt{3} a -\\dfrac{b}{4} \\right| & = \\sqrt{a^2+1} \\quad ... [1]\r\n\\end{align}\\]\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(\\sqrt{3} a +\\dfrac{b}{5} = \\sqrt{3} a -\\dfrac{b}{4}\\) \u306e\u3068\u304d\r\n\\[\r\nb=0\r\n\\]\r\n[1] \u306b\u4ee3\u5165\u3059\u308c\u3070, \\(a \\gt 0\\) \u3067\u3042\u308c\u3070\r\n\\[\\begin{align}\r\n\\sqrt{3} a & = \\sqrt{a^2+1} \\\\\r\na^2 & = \\dfrac{1}{2} \\\\\r\n\\text{\u2234} \\quad a & = \\dfrac{\\sqrt{2}}{2}\r\n\\end{align}\\]<\/li>\r\n

    2. 2*<\/strong>\u3000\\(\\sqrt{3} a +\\dfrac{b}{5} = -\\left( \\sqrt{3} a -\\dfrac{b}{4} \\right)\\) \u306e\u3068\u304d\r\n\\[\r\nb = 20 \\cdot 2 \\sqrt{3} a = 40 \\sqrt{3} a\r\n\\]\r\n[1] \u306b\u4ee3\u5165\u3059\u308c\u3070, \\(a \\gt 0\\) \u3067\u3042\u308c\u3070\r\n\\[\\begin{align}\r\n\\sqrt{3} a +8 \\sqrt{3} a & = \\sqrt{a^2+1} \\\\\r\n242 a^2 & = 1 \\\\\r\n\\text{\u2234} \\quad a & = \\dfrac{\\sqrt{2}}{22}\r\n\\end{align}\\]\r\n\u3053\u306e\u3068\u304d\r\n\\[\r\nb = 40 \\sqrt{3} \\cdot \\dfrac{\\sqrt{2}}{22} = \\dfrac{20 \\sqrt{6}}{11}\r\n\\]<\/li>\r\n<\/ol>\r\n

      \\(a \\gt 0\\) \u3067\u3042\u308b\u76f4\u7dda\u306e\u3046\u3061, \u50be\u304d\u304c\u5927\u304d\u3044\u65b9\u304c \\(\\ell\\) \u306a\u306e\u3067\r\n\\[\r\n\\ell : \\ \\underline{y = \\dfrac{\\sqrt{2}}{2} x}\r\n\\]\r\n

      (2)<\/strong><\/p>\r\n

      (1)<\/strong> \u3088\u308a, \u50be\u304d\u304c\u5c0f\u3055\u3044\u65b9\u304c \\(m\\) \u306a\u306e\u3067\r\n\\[\r\nm : \\ \\underline{y = \\dfrac{\\sqrt{2}}{22} x +\\dfrac{20 \\sqrt{6}}{11}}\r\n\\]\r\n

      (3)<\/strong><\/p>\r\n

      \\({\\ell}'\\) \u306f, \\(\\ell\\) \u3068 \\(x\\) \u8ef8\u306b\u3064\u3044\u3066\u5bfe\u79f0\u306a\u306e\u3067\r\n\\[\r\n{\\ell}' : \\ y = -\\dfrac{\\sqrt{2}}{2} x\r\n\\]\r\n\\(3\\) \u76f4\u7dda \\(\\ell , {\\ell}' , m\\) \u3059\u3079\u3066\u3068\u63a5\u3059\u308b\u5186\u306e\u4e2d\u5fc3\u3092 \\(( X , Y )\\) \u3068\u304a\u3051\u3070\r\n\\[\\begin{align}\r\n\\dfrac{| \\sqrt{2} X -2Y |}{\\sqrt{2^2+2}} & = \\dfrac{| \\sqrt{2} X +2Y |}{\\sqrt{2^2+2}} = \\dfrac{| \\sqrt{2} X -22Y +40 \\sqrt{6} |}{\\sqrt{22^2+2}} \\quad ... [2] \\\\\r\n\\text{\u2234} \\quad | X -\\sqrt{2} Y | & = | X +\\sqrt{2} Y | = \\dfrac{| X -11 \\sqrt{2} Y +40 \\sqrt{3} |}{9} \\quad ... [3]\r\n\\end{align}\\]\r\n

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      1. 1*<\/strong>\u3000\\(X -\\sqrt{2} Y = X +\\sqrt{2} Y\\) \u306e\u3068\u304d\r\n\\[\r\nY = 0\r\n\\]\r\n\u3053\u308c\u306f, C , D \u306b\u76f8\u5f53\u3059\u308b.<\/p><\/li>\r\n

      2. 2*<\/strong>\u3000\\(X -\\sqrt{2} Y = -X -\\sqrt{2} Y\\) \u306e\u3068\u304d\r\n\\[\r\nX = 0 \\quad ... [4]\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, E , E' \u306e \\(x\\) \u5ea7\u6a19\u306f, \\(\\underline{0}\\) .<\/p><\/li>\r\n<\/ol>\r\n

        (4)<\/strong><\/p>\r\n[4] \u3092 [3] \u306b\u4ee3\u5165\u3059\u308c\u3070, \\(Y \\gt 0\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n9 \\sqrt{2} Y & = \\left| 11 \\sqrt{2} Y -40 \\sqrt{3} \\right| \\\\\r\n9 \\sqrt{2} Y & = \\pm 11 \\sqrt{2} Y \\mp 40 \\sqrt{3} \\\\\r\n\\text{\u2234} \\quad Y & = 2 \\sqrt{3} , \\ 20 \\sqrt{3}\r\n\\end{align}\\]\r\n[2] \u306e\u5f0f\u306e\u5024\u304c\u534a\u5f84 \\(r\\) \u306a\u306e\u3067<\/p>\r\n