{"id":3164,"date":"2025-09-20T05:01:23","date_gmt":"2025-09-19T20:01:23","guid":{"rendered":"https:\/\/math-friend.com\/?p=3164"},"modified":"2025-09-21T01:33:10","modified_gmt":"2025-09-20T16:33:10","slug":"%e3%80%90%e6%9d%b1%e4%ba%ac%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e2%88%a0apc%e2%88%a0bpc%e3%82%92%e6%ba%80%e3%81%9f%e3%81%99%e7%82%b9p%e3%81%ae%e8%bb%8c%e8%b7%a1%e3%82%92%e6%b1%82%e3%82%81","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=3164","title":{"rendered":"\u3010\u6771\u4eac\u5927\u5b66\u5165\u8a66\u3011\u2220APC=\u2220BPC\u3092\u6e80\u305f\u3059\u70b9P\u306e\u8ecc\u8de1\u3092\u6c42\u3081\u3088\uff01(2008)"},"content":{"rendered":"\n

\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n

\u5ea7\u6a19\u5e73\u9762\u4e0a\u306e\\(3\\)\u70b9\\(\\mathrm{A}(1,0)\\), \\(\\mathrm{B}(-1,0)\\), \\(\\mathrm{C}(0,-1)\\)\u306b\u5bfe\u3057, \\(\\angle{\\mathrm{APC}}=\\angle{\\mathrm{BPC}}\\)\u3092\u307f\u305f\u3059\u70b9\\(\\mathrm{P}\\)\u306e\u8ecc\u8de1\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057\\(\\mathrm{P}\\neq\\mathrm{A}, \\mathrm{B}, \\mathrm{C}\\)\u3068\u3059\u308b.
(2008 \u6771\u4eac\u5927\u5b66 \u6587\u7cfb [3])<\/span><\/p>\n\n\n\n

\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n

\n

\\(\\mathrm{P}\\)\u306e\u5ea7\u6a19\u3092\\((X,Y)\\)\u3068\u304a\u304f. \\(\\angle{\\mathrm{APC}}\\), \\(\\angle{\\mathrm{BPC}}\\)\u306f\u5171\u306b\\(0^\\circ\\)\u4ee5\u4e0a\\(180^\\circ\\)\u4ee5\u4e0b\u306a\u306e\u3067,$$
\\angle{\\mathrm{APC}}=\\angle{\\mathrm{BPC}}\\iff \\cos{\\angle{\\mathrm{APC}}}=\\cos{\\angle{\\mathrm{BPC}}}
$$\u3067\u3042\u308b.
\\(\\triangle{\\mathrm{APC}}\\)\u3068\\(\\triangle{\\mathrm{BPC}}\\)\u3067\u4f59\u5f26\u5b9a\u7406\u304b\u3089,$$
\\begin{align}
\\cos{\\angle{\\mathrm{APC}}}&=\\frac{\\mathrm{AP}^2+\\mathrm{CP}^2-\\mathrm{AC}^2}{2\\mathrm{AP}\\cdot \\mathrm{CP}}\\\\[1.5ex]
\\cos{\\angle{\\mathrm{BPC}}}&=\\frac{\\mathrm{BP}^2+\\mathrm{BP}^2-\\mathrm{BC}^2}{2\\mathrm{BP}\\cdot \\mathrm{CP}}\\\\[1.5ex]
\\end{align}
$$\u3067\u3042\u308b\u304b\u3089,$$
\\begin{align}
&\\cos{\\angle{\\mathrm{APC}}}=\\cos{\\angle{\\mathrm{BPC}}}\\\\[1.5ex]
\\iff & \\frac{\\mathrm{AP}^2+\\mathrm{CP}^2-\\mathrm{AC}^2}{2\\mathrm{AP}\\cdot \\mathrm{CP}} = \\frac{\\mathrm{BP}^2+\\mathrm{BP}^2-\\mathrm{BC}^2}{2\\mathrm{BP}\\cdot \\mathrm{CP}}\\\\[1.5ex]
\\iff & \\frac{\\mathrm{AP}^2+\\mathrm{CP}^2-\\mathrm{AC}^2}{\\mathrm{AP}} = \\frac{\\mathrm{BP}^2+\\mathrm{BP}^2-\\mathrm{BC}^2}{\\mathrm{BP}}\\\\[1.5ex]
\\iff & \\frac{(X-1)^2+Y^2+X^2+(Y+1)^2-2}{\\sqrt{(X-1)^2+Y^2}} =\\frac{(X+1)^2+Y^2+X^2+(Y+1)^2-2}{\\sqrt{(X+1)^2+Y^2}} \\\\[1.5ex]
\\iff & \\frac{X^2+Y^2-X+Y}{\\sqrt{(X-1)^2+Y^2}} =\\frac{X^2+Y^2+X+Y}{\\sqrt{(X+1)^2+Y^2}} \\\\[1.5ex]
\\end{align}
$$\u3068\u306a\u308b. \u4e00\u822c\u306b,$$
a=b\\iff a^2+b^2,\\,\\,\u304b\u3064\\,\\,ab\\geq 0
$$\u3067\u3042\u308b\u304b\u3089, \u3053\u308c\u306f,$$
\\begin{align}
&\\frac{(X^2+Y^2-X+Y)^2}{(X-1)^2+Y^2} =\\frac{(X^2+Y^2+X+Y)^2}{(X+1)^2+Y^2}\\\\[1.5ex]
\u304b\u3064 \\,\\,\\,\\,&(X^2+Y^2-X+Y)(X^2+Y^2+X+Y)\\geq 0
\\end{align}
$$\u3068\u540c\u5024\u3067\u3042\u308b.

1\u756a\u76ee\u306e\u6761\u4ef6\u5f0f\u306f\\(Z=X^2+Y^2\\)\u3068\u304a\u3044\u3066,$$
(Z+Y-X)^2(Z+1+2X)=(Z+Y+X)^2(Z+1-2X)
$$\u3068\u306a\u308b. \u3053\u306e\u4e21\u8fba\u306f\\(X\\)\u306e\u7b26\u53f7\u304c\u7570\u306a\u308b\u3060\u3051\u306a\u306e\u3067, \u5404\u8fba\u3092\u5c55\u958b\u3057\u3066\u306e\\(X\\)\u306e\\(3\\)\u6b21\u5f0f\u3068\u3057\u3066\u898b\u305f\u3068\u304d\u306b, \u5b9a\u6570\u9805\u3068\\(X\\)\u306e\\(2\\)\u6b21\u306e\u9805\u306f\u4e21\u8fba\u3067\u6253\u3061\u6d88\u3057\u5408\u3044, \\(X\\)\u306e\\(1\\)\u6b21, \\(3\\)\u6b21\u3060\u3051\u6b8b\u308b\u3053\u3068\u304c\u308f\u304b\u308b. \u3053\u308c\u304b\u3089,$$
\\begin{align}
&X(Z+Y)^2-X(Z+Y)(Z+1)+X^3\\\\[1.5ex]
\\iff & X(Z^2+2YZ+Y^2-Z^2-Z-YZ-Y+X^2)=0\\\\[1.5ex]
\\iff & XY(Z-1)=0\\\\[1.5ex]
\\iff & XY(X^2+Y^2-1)=0\\\\[1.5ex]
\\iff & X=0\\,\\,\u307e\u305f\u306f\\,\\,Y=0\\,\\,\u307e\u305f\u306f\\,\\,X^2+Y^2=1
\\end{align}
$$\u3068\u8868\u3059\u3053\u3068\u304c\u3067\u304d\u308b.

\u6b21\u306b, 2\u756a\u76ee\u306e\u6761\u4ef6\u5f0f\u306f,$$
\\begin{align}
&(X^2+Y^2-X+Y)(X^2+Y^2+X+Y)\\geq 0\\\\[1.5ex]
\\iff & \\left\\{\\left(X-\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2}\\right\\}\\left\\{\\left(X+\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2}\\right\\}\\geq 0\\\\[1.5ex]
\\iff & \u300c\\left(X-\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2} \\geq 0\\,\\,\u304b\u3064\\,\\,\\left(X+\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2}\\geq 0\u300d\\\\[1.5ex]
& \u307e\u305f\u306f\\\\[1.5ex]
&\u300c\\left(X-\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2} \\leq 0\\,\\,\u304b\u3064\\,\\,\\left(X+\\frac{1}{2}\\right)^2+\\left(Y+\\frac{1}{2}\\right)^2-\\frac{1}{2}\\leq 0\u300d
\\end{align}
$$\u3068\u306a\u308b.

\u3088\u3063\u3066, 2\u3064\u306e\u6761\u4ef6\u3092\u5171\u306b\u6e80\u305f\u3059\u9818\u57df(\u554f\u984c\u6587\u306e\u524d\u63d0\u304b\u3089\\(\\mathrm{A}\\), \\(\\mathrm{B}\\), \\(\\mathrm{C}\\)\u3092\u9664\u3044\u3066\u3044\u308b)$$
\\begin{align}
&\u300cx^2+y^2=1\\,\\,\u304b\u3064\\,\\,y>0 \u300d\\\\[1.5ex]
& \u307e\u305f\u306f\\\\[1.5ex]
&\u300cx=0\\,\\,\u304b\u3064\\,\\,y\\neq -1 \u300d\\\\[1.5ex]
& \u307e\u305f\u306f\\\\[1.5ex]
&\u300cy=0\\,\\,\u304b\u3064\\,\\, |x|>1\u300d\\\\[1.5ex]
\\end{align}
$$\u3068\u306a\u308a, \\(\\mathrm{P}\\)\u306e\u8ecc\u8de1\u3092\u56f3\u793a\u3059\u308b\u3068<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u3068\u306a\u308b.<\/p>\n<\/div><\/div>\n\n\n\n

youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n