{"id":172,"date":"2025-03-20T04:11:12","date_gmt":"2025-03-19T20:11:12","guid":{"rendered":"https:\/\/tilearn.space\/?p=172"},"modified":"2025-03-20T04:29:49","modified_gmt":"2025-03-19T20:29:49","slug":"%e5%87%a0%e4%bd%95%e6%b5%8b%e8%af%95","status":"publish","type":"post","link":"https:\/\/tilearn.space\/?p=172","title":{"rendered":"2024\u5e74\u5b9d\u5c71\u533a\u4e8c\u6a21\u7b2c17\u9898"},"content":{"rendered":"<h2>\u51e0\u4f55\u9898\u76ee\uff1a\u6b63\u65b9\u5f62\u4e2d\u7684\u76f8\u4f3c\u4e09\u89d2\u5f62<\/h2>\n<h4>\u9898\u76ee<\/h4>\n<p>\u6b63\u65b9\u5f62 \\( ABCD \\) \u7684\u8fb9\u957f\u4e3a 1\uff0c\u70b9 \\( P \\) \u5728 \\( AD \\) \u7684\u5ef6\u957f\u7ebf\u4e0a\uff0c\u4e14 \\( PD < CD \\)\u3002\u5ef6\u957f \\( PB \\)\u3001\\( PC \\)\uff0c\u5982\u679c \\( \\triangle CDP \\sim \\triangle PAB \\)\uff0c\u6c42 \\( \\tan \\angle BPA \\)\u3002\uff08\u5982\u56fe 3 \u6240\u793a\uff09<\/p>\n<h4>\u56fe\u5f62<\/h4>\n<p>\u4ee5\u4e0b\u662f\u6b63\u65b9\u5f62 \\( ABCD \\)\u3001\u70b9 \\( P \\) \u53ca\u5ef6\u957f\u7ebf \\( PB \\)\u3001\\( PC \\) \u7684\u793a\u610f\u56fe\uff1a<\/p>\n<p>[latexpage]<br \/>\n\\begin{tikzpicture}<br \/>\n    % \u5b9a\u4e49\u6b63\u65b9\u5f62 ABCD \u7684\u9876\u70b9<br \/>\n    \\coordinate (A) at (0,0);<br \/>\n    \\coordinate (B) at (0,2);<br \/>\n    \\coordinate (C) at (2,2);<br \/>\n    \\coordinate (D) at (2,0);<br \/>\n    % \u5b9a\u4e49\u70b9 P \u5728 AD \u5ef6\u957f\u7ebf\u4e0a\uff0cPD = (sqrt(5)-1)\/2 \u2248 0.618<br \/>\n    \\coordinate (P) at (2.618,0);<br \/>\n    % \u7ed8\u5236\u6b63\u65b9\u5f62 ABCD<br \/>\n    \\draw (A) &#8212; (B) &#8212; (C) &#8212; (D) &#8212; cycle;<br \/>\n    % \u7ed8\u5236\u5ef6\u957f\u7ebf PB \u548c PC<br \/>\n    \\draw (P) &#8212; (B);<br \/>\n    \\draw (P) &#8212; (C);<br \/>\n    % \u6807\u6ce8\u9876\u70b9<br \/>\n    \\node at (A) [below left] {$A$};<br \/>\n    \\node at (B) [above left] {$B$};<br \/>\n    \\node at (C) [above right] {$C$};<br \/>\n    \\node at (D) [below right] {$D$};<br \/>\n    \\node at (P) [below] {$P$};<br \/>\n\\end{tikzpicture}<\/p>\n<h4>\u89e3\u7b54<\/h4>\n<p><strong>\u5df2\u77e5\u6761\u4ef6<\/strong>\uff1a\u6b63\u65b9\u5f62 \\( ABCD \\)\uff0c\u8fb9\u957f \\( AB = BC = CD = DA = 1 \\)\u3002\u70b9 \\( P \\) \u5728 \\( AD \\) \u7684\u5ef6\u957f\u7ebf\u4e0a\uff0c\u4e14 \\( PD < CD \\)\uff0c\u5373 \\( PD < 1 \\)\u3002\u8bbe \\( PD = x \\)\uff0c\u5176\u4e2d \\( 0 < x < 1 \\)\u3002\\( \\triangle CDP \\sim \\triangle PAB \\)\uff0c\u6c42 \\( \\tan \\angle BPA \\)\u3002 <\/p>\n<p>\u6211\u4eec\u7ed9\u5404\u70b9\u8d4b\u5750\u6807\uff1a\\( A(0,0) \\)\uff0c\\( B(0,1) \\)\uff0c\\( C(1,1) \\)\uff0c\\( D(1,0) \\)\uff0c\u70b9 \\( P(1+x, 0) \\)\uff0c\u5176\u4e2d \\( 0 < x < 1 \\)\uff0c\u6240\u4ee5 \\( PD = x \\).<\/p>\n<p><strong>\u6b65\u9aa4 1\uff1a\u5229\u7528\u76f8\u4f3c\u4e09\u89d2\u5f62<\/strong>\uff1a\u7531\u4e8e \\( \\triangle CDP \\sim \\triangle PAB \\)\uff0c\u5bf9\u5e94\u89d2\u4e3a \\( \\angle CDP \\leftrightarrow \\angle PAB \\)\uff0c\\( \\angle DCP \\leftrightarrow \\angle APB \\)\uff0c\\( \\angle CPD \\leftrightarrow \\angle BPA \\)\u3002\u5bf9\u5e94\u8fb9\u4e3a \\( DC \\leftrightarrow AP \\)\uff0c\\( CP \\leftrightarrow PB \\)\uff0c\\( DP \\leftrightarrow AB \\)\u3002\u6240\u4ee5\u6709\u6bd4\u4f8b\u5173\u7cfb\uff1a<\/p>\n<p>\\[<br \/>\n\\frac{DC}{AP} = \\frac{DP}{AB}<br \/>\n\\]<\/p>\n<p>\u5176\u4e2d \\( DC = 1 \\)\uff0c\\( AB = 1 \\)\uff0c\\( DP = x \\)\uff0c\\( AP = AD + DP = 1 + x \\)\u3002\u4ee3\u5165\u5f97\uff1a<\/p>\n<p>\\[<br \/>\n\\frac{1}{1+x} = \\frac{x}{1}<br \/>\n\\]<\/p>\n<p>\u89e3\u5f97\uff1a<\/p>\n<p>\\[<br \/>\n1 = x(1+x) \\implies x^2 + x &#8211; 1 = 0<br \/>\n\\]<\/p>\n<p>\\[<br \/>\nx = \\frac{-1 \\pm \\sqrt{1 + 4}}{2} = \\frac{-1 \\pm \\sqrt{5}}{2}<br \/>\n\\]<\/p>\n<p>\u53d6\u6b63\u6839\uff1a\\( x = \\frac{-1 + \\sqrt{5}}{2} \\approx 0.618 \\)\uff0c\u6ee1\u8db3 \\( 0 < x < 1 \\)\u3002\u6240\u4ee5 \\( PD = \\frac{-1 + \\sqrt{5}}{2} \\).<\/p>\n<p><strong>\u6b65\u9aa4 2\uff1a\u8ba1\u7b97 \\( \\tan \\angle BPA \\)<\/strong>\uff1a\u6211\u4eec\u8ba1\u7b97 \\( \\triangle PAB \\) \u4e2d \\( \\angle APB \\)\u3002\u4f7f\u7528\u5750\u6807\u51e0\u4f55\u65b9\u6cd5\uff1a<\/p>\n<ul>\n<li>\u5411\u91cf \\( \\vec{PA} = (-1-x, 0) \\)<\/li>\n<li>\u5411\u91cf \\( \\vec{PB} = (-1-x, 1) \\)<\/li>\n<\/ul>\n<p>\u76f4\u7ebf \\( PA \\) \u7684\u659c\u7387 \\( m_1 = 0 \\)\uff0c\u76f4\u7ebf \\( PB \\) \u7684\u659c\u7387 \\( m_2 = \\frac{1}{-1-x} \\)\u3002\u6240\u4ee5\uff1a<\/p>\n<p>\\[<br \/>\n\\tan \\angle BPA = \\left| \\frac{m_1 &#8211; m_2}{1 + m_1 m_2} \\right| = \\left| \\frac{0 &#8211; \\frac{1}{-1-x}}{1} \\right| = \\frac{1}{1+x}<br \/>\n\\]<\/p>\n<p>\u4ee3\u5165 \\( x = \\frac{-1 + \\sqrt{5}}{2} \\)\uff1a<\/p>\n<p>\\[<br \/>\n1 + x = 1 + \\frac{-1 + \\sqrt{5}}{2} = \\frac{1 + \\sqrt{5}}{2}<br \/>\n\\]<\/p>\n<p>\\[<br \/>\n\\tan \\angle BPA = \\frac{1}{\\frac{1 + \\sqrt{5}}{2}} = \\frac{2}{1 + \\sqrt{5}}<br \/>\n\\]<\/p>\n<p>\u6709\u7406\u5316\u5206\u6bcd\uff1a<\/p>\n<p>\\[<br \/>\n\\frac{2}{1 + \\sqrt{5}} \\times \\frac{1 &#8211; \\sqrt{5}}{1 &#8211; \\sqrt{5}} = \\frac{2(1 &#8211; \\sqrt{5})}{1 &#8211; 5} = \\frac{2(1 &#8211; \\sqrt{5})}{-4} = \\frac{\\sqrt{5} &#8211; 1}{2}<br \/>\n\\]<\/p>\n<p><strong>\u7b54<\/strong>\uff1a\\( \\tan \\angle BPA = \\frac{\\sqrt{5} &#8211; 1}{2} \\)\u3002\\(\\Box\\)<\/p>\n<h4>\u5c0f\u7ed3<\/h4>\n<p>\u901a\u8fc7\u8fd9\u4e2a\u95ee\u9898\uff0c\u6211\u4eec\u5b66\u4e60\u4e86\u5982\u4f55\u5229\u7528\u76f8\u4f3c\u4e09\u89d2\u5f62\u7684\u6027\u8d28\u6c42\u89e3\u51e0\u4f55\u95ee\u9898\u3002\u540c\u5b66\u4eec\u53ef\u4ee5\u5c1d\u8bd5\u6539\u53d8 \\( PD \\) \u7684\u503c\uff0c\u63a2\u7d22 \\( \\tan \\angle BPA \\) \u7684\u53d8\u5316\u89c4\u5f8b\uff0c\u8fdb\u4e00\u6b65\u52a0\u6df1\u5bf9\u76f8\u4f3c\u4e09\u89d2\u5f62\u7684\u7406\u89e3\uff01<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u51e0\u4f55\u9898\u76ee\uff1a\u6b63\u65b9\u5f62\u4e2d\u7684\u76f8\u4f3c\u4e09\u89d2\u5f62 \u9898\u76ee \u6b63\u65b9\u5f62 \\( ABCD \\) \u7684\u8fb9\u957f\u4e3a 1\uff0c\u70b9 \\( P \\) \u5728 \\( [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-172","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/posts\/172","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tilearn.space\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=172"}],"version-history":[{"count":6,"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/posts\/172\/revisions"}],"predecessor-version":[{"id":178,"href":"https:\/\/tilearn.space\/index.php?rest_route=\/wp\/v2\/posts\/172\/revisions\/178"}],"wp:attachment":[{"href":"https:\/\/tilearn.space\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=172"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tilearn.space\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=172"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tilearn.space\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}