如图在rt三角形abc中角acb等于90度,AC=4,BC=3,P是射线AB上的一个动点,以P为圆心 如图1，Rt△ABC中，∠ACB=90°，AC=4，BC=3...
\u5982\u56fe\uff0c\u5728Rt\u25b3ABC\u4e2d\uff0c\u2220ACB=90\u00b0\uff0cAC=4\uff0cBC=3\uff0cP\u662f\u5c04\u7ebfAB\u4e0a\u7684\u4e00\u4e2a\u52a8\u70b9\uff0c\u4ee5\u70b9P\u4e3a\u5706\u5fc3\uff0cPA\u4e3a\u534a\u5f84\u7684\u2299P\u4e0e\u5c04\u7ebf\uff081\uff09\u8bc1\u660e\uff1a\u2235PA=PD\uff0c\u2234\u2220A=\u2220PDA\uff0c\u2235\u2220EDC=\u2220PDA\uff0c\u2234\u2220A=\u2220EDC\uff0c\u2235AC\u22a5BC\uff0c\u2234\u2220PBE=\u2220PEB\uff0c\u2234PB=PE\uff1b\uff082\uff09\u89e3\uff1a\u2235AP=DP\uff0c\u2234\u2220PAD=\u2220PDA\uff0e\u2234\u2220PAD=\u2220CDE\uff0e\u2235\u2220ACB=\u2220DCE=90\u00b0\uff0c\u2234\u25b3ABC\u223d\u25b3DEC\uff0e\u2234\u2220ABC=\u2220DEC\uff0cBCEC=ABDE\uff0e\u2234PB=PE\uff0eRt\u25b3ABC\u4e2d\uff0c\u2220ABC=90\u00b0\uff0c\u2235AC=4\uff0cBC=3\uff0c\u2234AB=5\uff0e\u2235AP=2\uff0c\u2234PB=PE=3\uff0cDE=1\uff0c\u22343EC=51\uff0c\u89e3\u5f97\uff1aCE=35\uff1b\uff083\uff09\u89e3\uff1a\u8bbeM\u4e3aBE\u7684\u4e2d\u70b9\uff0c\u5219PM\u22a5BE\uff0c\u82e5\u2299P\u7684\u534a\u5f84\u4e3ax\uff0c\u5219PB=5-x\uff0cPM=\uff085-x\uff09sin\u2220ABC=45(5?x)\uff0cBM=\uff085-x\uff09cos\u2220ABC=35(5?x)\uff0c\u2235\u2299P\u4e0e\u2299M\u5916\u5207\uff0c\u223445(5?x)=35(5?x)+x \u89e3\u5f97\uff1ax\uff1d56\u2234\u2299P\u7684\u534a\u5f84\u4e3a56\uff0e
\u89e3\u7b54\uff1a\u89e3\uff1a\uff081\uff09\u2460\u5982\u56fe2\u2235AP=DP\uff0c\u2234\u2220PAD=\u2220PDA\uff0c\u2235\u2220PDA=\u2220CDE\uff0c\u2234\u2220PAD=\u2220CDE\uff0c\u2235\u2220ACB=\u2220DCE=90\u00b0\uff0c\u2234\u25b3ABC\u223d\u25b3DEC\uff0c\u2234\u2220ABC=\u2220DEC\uff0cBCCE\uff1dABDE\uff0e\u2234PB=PE\uff0e\u5728Rt\u25b3ABC\u4e2d\uff0c\u2220ACB=90\u00b0\uff0cAC=4\uff0cBC=3\uff0c\u2234AB=AC2+BC2=5\uff0c\u2234PB=PE=5-x\uff0cDE=PE-PD=5-x-x=5-2x\uff0c\u22343y\uff1d55?2x\uff0c\u2234y=-65x+3\uff080\uff1cx\uff1c52\uff09\uff1b\u2461\u8bbeBE\u7684\u4e2d\u70b9\u4e3aQ\uff0c\u8fde\u7ed3PQ\uff0c\u5982\u56fe2\uff0c\u2235PB=PE\uff0c\u2234PQ\u22a5BE\uff0c\u53c8\u2235\u2220ACB=90\u00b0\uff0c\u2234PQ\u2225AC\uff0c\u2234\u25b3BPQ\u223d\u25b3BAC\uff0c\u2234PQAC\uff1dPBAB\uff1dBQBC\uff0c\u5373PQ4=5?x5=BQ3\uff0c\u2234PQ=-45x+4\uff0cBQ=-35x+3\uff0c\u5f53\u4ee5BE\u4e3a\u76f4\u5f84\u7684\u5706\u548c\u2299P\u5916\u5207\u65f6\uff0c-45x+4=x+\uff08-35x+3\uff09\uff0c\u89e3\u5f97x=56\uff0c\u5373AP\u7684\u957f\u4e3a56\uff1b\uff082\uff09\u5f53\u70b9E\u5728\u7ebf\u6bb5BC\u5ef6\u957f\u7ebf\u4e0a\u65f6\uff0c\u7531\uff081\uff09\u2461\u7684\u7ed3\u8bba\u53ef\u5f97IQ=PQ-PI=-45x+4-x=-95x+4\uff0cCQ=BC-BQ=3-\uff08-35x+3\uff09=35x\uff0c\u5728Rt\u25b3CQI\u4e2d\uff0cCI2=CQ2+IQ2=\uff0835x\uff092+\uff08-95x+4\uff092=185x2-725x+16\uff0c\u2235CI=AP\uff0c\u2234185x2-725x+16=x2\uff0c\u89e3\u5f97x1=2013\uff0cx2=4\uff08\u4e0d\u5408\u9898\u610f\uff0c\u820d\u53bb\uff09\uff0c\u2234AP\u7684\u957f\u4e3a2013\uff1b\u5f53\u70b9E\u5728\u7ebf\u6bb5BC\u4e0a\u65f6\uff0cIQ=PI-PQ=x-\uff08-45x+4\uff09=95x-4\uff0cCQ=BC-BQ=3-\uff08-35x+3\uff09=35x\uff0c\u5728Rt\u25b3CQI\u4e2d\uff0cCI2=CQ2+IQ2=\uff0835x\uff092+\uff0895x-4\uff092=185x2-725x+16\uff0c\u2235CI=AP\uff0c\u2234185x2-725x+16=x2\uff0c\u89e3\u5f97x1=2013\uff08\u820d\u53bb\uff09\uff0cx2=4\uff0c\u2234AP\u7684\u957f\u4e3a4\uff0c\u7efc\u4e0a\u6240\u8ff0\uff0cAP\u7684\u957f\u4e3a2013\u62164\uff0e
以CA,CB为x,y轴建立直角坐标系，则A(4，0),B(0，3),AB:y=-3x/4+3,
设P(4p,3-3p),0<p<1,PA=5(1-p),
圆P与射线AC交于点D(8p-4,0),
PD:y=3x/4+3-6p交BC于E(0，3-6p),BE的中点Q(0，3-3p),
易知PQ∥CD,
∴DCIP为平行四边形，只需CD=PI=PA,
∴8p-4=5(1-p),13p=9,p=9/13,点P(36/13，12/13),
这时AP=√[(4-36/13)^2+(12/13)^2]=20/13.
可以吗？
绛旓細锛1锛夊厛绠楀嚭涓夎褰鍛ㄩ暱锛屾寜鍕捐偂鏁板畾鐞嗭紝鏂滆竟AB=鈭(AC^2+BC^2)=10cm锛涘崐鍛ㄩ暱=(8+6+10)/2=12cm锛屽洜涓簆鐐鐨杩愬姩閫熷害鏄2cm/绉掞紝鎵浠=鍗婂懆闀/2=12/2=6绉掞紱锛2锛夊綋p杩愬姩6绉掓濂界瓑鍒嗕笁瑙掑舰鍛ㄩ暱鏃讹紝Bp锛12-BC=12-6=6cm=BC锛屾鏃垛柍BCp鏄互Cp涓哄簳杈圭殑绛夎叞涓夎褰紝浠巔鍚態C杈逛綔鍨傜嚎pM灏咮C...
绛旓細杩炴帴EF,灏涓夎褰ADE缁曠偣D閫嗘椂閽堟棆杞180搴,寰椻柍BDE'(AD涓嶣D閲嶅悎锛夎繛鎺E'鍥犱负瑙扚DE=90掳,鎵浠ヨADE+瑙払DF=90掳,鍥犱负瑙扐DE=瑙払DE',鎵浠ヨFDB+瑙払DE'=90掳,EDE'鍏辩嚎.鍥犱负DE=DE',FD鈯E',鎵浠E=FE'.鍥犱负AE=BE',鎵浠E^2+BF^2=BE'^2+BF^2=E'F^2.鍥犱负鈭燙=90掳,鎵浠E^2+CF^2...
绛旓細鈶粹埖鈭燗=60鍦紝鈭粹垹B=30掳锛鍦≧T螖BDE涓紝DE=1/2BE锛屽垯鎶樺彔鐭ワ紝AE=DE锛屸埓AE=1/2BE(鎴朆E=2AE)銆傗懙鐢辨姌鍙犵煡锛氣垹FEA=鈭燜ED锛屸埖DE鈯C锛屸垹C=90掳锛屸埓DE鈭AC锛屸埓鈭燜ED=鈭燛FA锛屸埓鈭燜EA=鈭燛FA锛屸埓AE=AF锛屸埓AF=DE锛屸埓鍥涜竟褰EDF鏄钩琛屽洓杈瑰舰(AF涓嶥E骞宠涓旂浉绛)锛屽張AE=AF锛屸埓骞宠鍥涜竟...
绛旓細鈭礏D骞冲垎鈭ABC锛屸埓鈭燚BC=鈭燚BE锛屽張鈭礑E鍨傜洿骞冲垎AB锛屸埓鈻矰AE鈮屸柍DBE锛屸埓鈭燗=鈭燚BE锛屽嵆锛屸垹A=鈭燚BC=鈭燚BE 鈭碘垹A+鈭燚BE+鈭燚BC=90掳锛屸埓鈭燗=30掳 鈭碘垹DBC=鈭燚BE锛 鈭燚CB=鈭燚EB=90掳锛屼笖BD=BD锛屸埓鈻矰BC鈮屸柍DBE锛屸埓鈻矰C=DE=1 鍦≧t鈻矰AE涓紝DE=1锛屸垹A=30掳锛屸埓AD=2 鈭AC=AD+DC=...
绛旓細瑙:AB=鈭(AC^2+BC^2)=10;BC'=BC=6,鍒橝C'=4.鈭燘C'D=鈭燙=90掳,鍒欌垹AC'D=鈭燙=90掳;鍙堚垹A=鈭燗,鏁呪娍AC'D鈭解娍ACB,AC'/AC=AD/AB.鍗:4/8=AD/10,AD=5,CD=AC-AD=3,BD=鈭(CD^2+BC^2)=3鈭5.
绛旓細瑙ｏ細浣淎B鐨鍨傜洿骞冲垎绾匡紝涓AC浜や簬鐐笵锛岃繛缁揃D銆傚鏄撳緱鍒帮細DA锛滵B 鎵浠ワ紝瑙扐 锛 瑙扐BD 锛 15搴 鍥犳锛氳BDC 锛 瑙扐 + 瑙扐BD 锛30搴 鎵浠ワ紝BD 锛 2BC 锛 4 锛 AD 鍦≧T涓夎褰BDC涓紝鏍规嵁鍕捐偂瀹氱悊鍙緱锛欳D 锛 2鍊嶆牴鍙3 鎵浠ワ細AC 锛 2鍊嶆牴鍙3 + 4 ...
绛旓細鐢扁懘EF²锛滲E²锛婣F²銆傚緱EF²锛5²锛12²锛13²锛屸埓EF锛13銆傛寜瑙掑垎 鍒ゅ畾娉曪細1銆侀攼瑙掍笁瑙掑舰锛氫笁瑙掑舰鐨涓変釜鍐呰閮藉皬浜90搴︺2銆鐩磋涓夎褰锛氫笁瑙掑舰鐨勪笁涓唴瑙掍腑涓涓绛変簬90搴︼紝鍙浣淩t鈻炽3銆侀挐瑙掍笁瑙掑舰锛氫笁瑙掑舰鐨勪笁涓唴瑙掍腑鏈変竴涓澶т簬90搴︺
绛旓細鈭礑E鈯AC鍗斥垹DEC=90掳 DF鈯C鍗斥垹DFC=90掳 鈭燗CB=鈭燛CF=90掳 鈭村洓杈瑰舰DECF鏄煩褰 鈭碋C=DF=Y DE鈭C 鈭碅E=AC-EC=8-Y 鈭碘垹A=30掳 鈭碅B=2BC 鈭碆C²+AC²=AB²BC²+8²=4BC²BC=16鈭3/3 鈭礑E鈭C 鈭碊E/BC=AE/AC X/(16鈭3/3)=(8-Y)...
绛旓細锛滱B锛廈C锝Rt鈻矨PQ鈭絉t鈻矨CB锛屾湁鍏叡瑙掆垹A锝濓紳鈭3锛3锛屸埓X锛9锛忥紙3锛嬧垰3锛夛紳3锛3锛嶁垰3锛夛紡2 鈮 1.9019銆傗憿璁綪Q浜B浜嶧锛 QA锛漄F锝涒垹A锛60º锛濃垹PQR锛濃垹QFA锛堝唴閿欒鐩哥瓑锛夛綕锛漍锛汸F锛漃Q锛峏锛濓紙3鈭3锛峏锛夛紞X锛3鈭3锛2X锛涒埖Y锛廝F锛滲C锛AC锝汻t鈻砅EF鈭絉t鈻矯BA锛屾湁涓変釜...
绛旓細锛1锛 璇佹槑锛 鈭礏C鏄姍O鐨鍒囩嚎 鈭粹垹OFB=90掳 鈭碘垹C=90掳 鈭碠F//AC 鈭粹垹CAF=鈭燗FO 鈭礝A=OF 鈭粹垹BAF=鈭燗FO 鈭粹垹BAF=鈭燙AF 锛2锛 鈭礎C=6,BC=8,鈭燙=90掳 鈭碅B=鈭氾紙AC²+BC²锛=10 鈭碘垹BAF=鈭燙AF 鈭碅B/AC=BF/CF=5/3(瑙掑钩鍒嗙嚎瀹氱悊) 鈭碆F...
