?1?例5.已知圆O:x2?y2?1,点B??,0?,M?1,1?,P为圆O上动点,则2PB?PM的最小值为???2?
A.6
B.7
C.10
D.11
解1:如图?10?所示,取A??2,0?,连接PA,Q??BOP:?AOP,OPOB1??,?POB??AOPOAOP2BPOP1??,?AP?2PB,?PM?2PB?PM?PA?AM?10PAOA2?答案为C?OP注释:已知定点B,阿斯圆O,点B圆内,点M在圆外,当??, OB求?PB?PM可以利用相似三角形构造?PB.
解法2.如图?11?,设x轴上A?a,0?,对于圆O上任一点P都满足则PA?2,圆O与x轴交于点P1,PBP?1?a1A??2,?a??2,?A??2,0?,?PM?2PB?PM?PA?MA?10.1PB1222例6.已知圆C:?x?1???y?1??1,定点O?0,0?,B?2,0?,其中P为圆C上的动点,求2PO?PB的最小值.
6
PB?2,PA2即PB?2PA,设圆C与直线BC交于点P,则PB?2PA,?PB?2?1?PA?1?, 111112?31??点A是BC的中点,?A?,?,?2PO?PB?2?PO?PA??2AO?5.22??解.如图?12?所示,做直线BC,点A在直线BC上,对于圆C上的任一点P都满足注释:此题能否连接OC,在直线OC上找一点D,满足2PO?PD?
如图?1??2?所示,当PA较长时,内分点P1在左侧,当PB较长时,内分点P2在左侧,因此,可以认为较长线段与内分点相对应,较短线段与外分点相对应.?14?所示,假设存在点D符合条件,设圆C与直线OC相交于P?PD?2PO,1、P2两点,如图?点P2是线段OD的内分点,?点D在线段OP2的延长线上,此时P1,P2都是内分点,不符合题意.
1?1???10.已知函数f????2?cos????sin2??cos2???sin???,2?2? ??若集合??Rf????m22????,求实数m的取值范围.?1??1?解:设P?cos?,sin??,B??,0?,M?0,?,则点P在以原点O为圆心的单位圆上,?2??2?PAf???=2PB?PM,设A?a,0?,满足圆O上的任意一点P都有?2,?PA?2PB,PB1 ??1PB1设圆O与x轴交于点P1??1,0?,则1?2?,?a??2,?A??2,0?,P?1?a21A?2PB?PM?PA?PM?AM?1717,?m?.227
yPMABOx(15)uuururuuuruuur2xuuy11.已知BC?6,AC?2AB,点D满足AD?AB?AC,设f?x,y??AD,x?y2?x?y? 若f?x,y??f?x0,y0?恒成立,求f?x0,y0?的最大值.uuururuuuruuurr?uuuruuur1uuuruuur2xuuyxy?1uuu解:AD?AB?AC?2AB??AC?,设2AB?AE,AC?AF,x?y2?x?y?x?yx?y?22?uuurrrxuuuyuuuxy则AD?AE?AF,Q??1,?D,E,F三点共线,作AD0?EF,垂足为D0,
x?yx+yx?yx?yuuuruuuuruuuur如图?16?所示,Qf?x,y??f?x0,y0?恒成立,f?x,y??AD?AD0?f?x0,y0??AD0.
??QAE?2AB,AC?2AF,AC?2AB,?AE?2AF,?ACB??AEF,?EF?AC?6以线段EF所在的直线为x轴,线段EF的中点O为原点建立平面直角坐标系,如图?16?所示设A?x,y?,E?0,?3?,F?0,3?,由AE?2AF得,x2??y?3??4x2?4?y?3?uuuur2222?x?y?10y?9?0,?x??y?5??16,?AD0?422
??maxyAFD0CO(16)EBx
yCBMDA(17)xO
8
rrrrrr1rrrr12.已知a,b是相互垂直的单位向量,平面向量c满足a?b?c?,则2c?b?c?a2
的最小值为KKKKuurruuurrrruuuruuururrrruuur解.如图?17?所示,设OA?a,OB?b,a?b?OM,M?1,1?,OC?C,则a?b?c?CM,uuur1rruuurrruuur1依题意CM?,所以点C在以M为圆心,r=的圆上,c?b?BC,c?a?AC,22rrrrMD1MC1?2c?b?c?a?2CB?CA,在AM上取点D满足?,又?,?CMD??AMC,
MC2MA2DCMC1??CMD∽?AMC???,?AC?2CD,?2CB?CA?2CB?2CD?2BDCAMA2rrrr11171717QMD?,?BD?1??,?2CB?CA?,?2c?b?c?a的最小值为.41642213.在棱长为6的正方体ABCD-A1B1C1D1中,M是BC的中点,点P是正方形DCC1D1内的动点,且满足?APD??MPC,则三棱锥P?BCD的体积最大值为??y
A.36 B.123. C.24 D.183
D1C1D1C1
PA1DCB1
DPP1Eo(19)P2CxMA
(18)B解.QAD?平面DCC1D1,??ADP?900,QMC?平面DCC1D1,??PCM?900,PDAD??2PCMC?PD?2PC如图?19?所示,以线段DC的中点为原点,线段DC的中垂线为y??APD??MPC,又Q?APD??MPC,??APD∽?MPC,?轴建立平面直角坐标系,则D??3,0?,C?3,0?,设P?x,y?,QPD?2PC,??x?3??y2?4?x?3??4y2,?x2?y2?10x?9?0,即?x?5??y2?16点P的轨迹是以E?5,0?为圆心,r?4的圆在正方形DCC1D1内的部分,11QP3,23,?V???6?6?23?123,?答案为B.??1P?BCDmax32222??9
?14.在?ABC中,AC?2,AB?mBC?m?1?,恰好当B=时,?ABC 的面积最大,3
求m的值.解.如图?20?所示,以线段AC的中点为原点,中垂线为y轴建立平面直角坐标系,则A??1,0?,C?1,0?,设B?x,y?,QBA?mBC,??x?1??y2?m2?x?1??m2y2,??m2?1??m2?12m?m2?1??2m?2??x?2??y??2??圆心D?2,0?,当BD?AC时,B?2,2?,m?1??m?1???m?1??m?1m?1?uur?m2?12m???2m2?2m??BA???1?2,?2???2,2?,m?1m?1??m?1m?1??uuur?m2?12m???2?2m?BC??1?2,?2???2,2?,?m?1m?1??m?1m?1?uuruuuruurr2m2?18m24m4?4m22mm2?1uuu?BA?BC?,又BA??,BC?,222222m?1m?1?m?1??m?1?2uuruuur2uuruuur4m?m2?1?m?1??BA?BC2m1BA?BC?,?cosB????,uuruuur2222m?12BA?BC4m?m?1??m?1?22228m2?m2?1?2
?m2?4m?1?0?m?2?3,?m?1,?m?2?3.A
y
EB
AOCDxBF(21)CD(20)10
15.在四面体CDABCD中,已知AD?BC,AD?6,BC=2,且ABAC??2,求VA?BCD的最大值. BDCD解.如图?21?,作BE?AD,垂足为E,连接CE,QAD?BE,AD?BC,BEIBC?B,1?AD?平面BCE,?AD?CE,?VA?BCD??S?BCE?AD?2S?BCE3BACAPAQ==2,?B、C在以A、D为定点,且满足=2的阿波罗尼斯球上,BDCDPD?B、C在以点E为圆心,且与AD垂直的小圆上,?BE=CE,取BC的中点F,连接EF,1则EF?BC,?VA?BCD?2S?BCE?2??2?EF?2EF?2BE2?1,2以AD所在的直线为x轴,AD的中点为原点建立平面直角坐标系,则A??3,0?,D?3,0?,设B?x,y?,QBA?2BD,??x?3??y2?4?x?3??4y2,即?x?5??y2?16,?BEmax?4,?VA?BCD?2BE2?1?215,??VA?BCD?max?215.222
11
阿波罗尼斯圆性质及其应用探究
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