【人教版】最新初中数学竞赛名师讲义:第10-12章专题辅导(含答案)

初中数学竞赛名师辅导

AEJHBCGDKP评注 调和点列见15.1.65,此题是一个一般结果之特例.

11.2.36★★★已知△ABC中,?A?90?,S、N在AB上,M在BC上,MN?BC,SC2?CM?CB,Q在AC上,SQ?SO(O为BC中点),SQ交MN于P,求证:SP?QP. AC?R解析 如图,延长MP、CA,设交于R,连结SR.由于△MRC∽△ABC,故C?CMC?BSC?2,

于是RS?CS,又由QS?SO,PM?MO,故△RSP∽△CSO,同理△RPQ∽△BOS. 于是

SPSOPQSO,,由于BO?CO,故SP?QP. ??PRCORPBORAQPNS

11.2.37★★★★D为△ABC的边AC上一点,E和F分别为线段BD和BC上的点.满足?BAE??CAF.再设P、Q为线段BC和BD上的点,使得EP∥QF∥DC.求证:?BAP??QAC.

BOMC解析 如图,在AB上取点R,使RE∥AQ,连结RE、RP.易知△REP与△AQF位似,故?QAF??ERP,?RPE??AFQ.而QF∥CD,故?AFQ??FAC??BAE,从而?RPE??BAE,

所以R、A、P、E共圆(R与A不重合),于是?QAF??ERP??EAP. 又?CAF??BAE,加之,即得?QAC??BAP.

初中数学竞赛名师辅导

ARDEBPQFC

11.2.38★★★★已知△ABC中,AB、AC上各有一点R、Q,直线RQ与BC延长线交于点P,求证:

AQ?CQPC?PBQR?BR???1.

PQ?RQPQ?PRQR?PRARQTSBCP

解析 在PR上取S,使A、R、C、S共圆,则?QSC??A,AQ?QC?RQ?SQ,又在PR上取T, C、PC?PB?PT?PR,使T、则?CTP??B,于是B、R共圆,

AQ?CQPC?PBSQPTTS????1?,

PQ?RQPQ?PRPQPQPQ于是问题变成求证

TSAR?BRTSCS?.显见△STC∽△ABC,,又△ARQ∽△SCQ,故?PQQR?PRABACARSCABBPABPRCQ?????1,这是梅氏定理,故结论成立. ,欲证式成为,或RQCQAC?PQCQ?PRBRPQAC11.2.39★★★★如图,△PQR和△P′Q′R′是两个全等的正三角形,六边形ABCDEF的边长

2223分别是AB?a1,BC?b1,CD?a2,DE?b2,EF?a3,FA?b3.求证:(1)a12?a2;?a3?b12?b2?b3(2)a1?a2?a3?b1?b2?b3.

R'PAb3Bb1P'CQa1a2DFa3Eb2Q'R

初中数学竞赛名师辅导

解析 (1)不妨设△PAB、△P′BC、△QCD、△Q′DE、△REF、△RFA的面积分别为S1、S?1、S?2、S?3、S3、S?3.由△PAB∽△P′CB∽?QCD∽△Q′ED∽△REF∽△R′AF,可得S1S?1S1?S2?S3S?1?S?2?S?3S1a12S1S?1S2S?2S3S?3?2,同理可得其余). ????????(由得22222222222222ab1a1b1a2b2a3b3a1?a2?a3b1?b2?b3S?1b112222因为△PQR≌△P′Q′R′,所以S1?S2?S3?S?1?S?2?S?3,则a12?a2. ?a3?b12?b2?b3(2)不妨设△PAB、?P′BC、△QCD、△Q′DE的周长分别为l1、l?1、l2、l?2、l3、l?3. 可得?l1?a1l?1?b1l2?a2l?2b2l3?a3l?3?b3l1?l2?l3?(a1?a2?a3)???2??? a1b1a2ba3b3a1?a2?a3lall?l?al??bl?1?l?2?l?3?(b1?b2?b3)(由1?1,得1?1,因此11?11,同理可得其余).

l?1b1a1b1a1b1b1?b2?b3又设△PQR、△P′Q′R′的周长均为L,a1?a2?a3?A,b1?b2?b3?B,由上面等式可得

L?BL?A,化得(A?B)(L?A?B)?0,而L?A?B,因此A?B,即a1?a2?a3?b1?b2?b3. ?AB11.2.40★★★★已知平行四边形ABCD,C在边AD、AB上的射影分别是M、N,NM延长后与BD延长线交于P,求证:PC?AC.

ANQBCMDPLR

解析 延长AD,交CP于L,则PC?AC?△ACM∽△CLM?AM?ML?CM2,下面就来证明此式.延长MN,交CB延长线于Q,又延长CN,交DA延长线于R.于是式变为MD?RM?CM2,而这显然成立.

11.2.41★★★★已知△ABC内有一点Y,BC上有一点P,X、Z在△ABC外,△AXB∽△CYP,△ACZ∽△BPY(即?XAB??YCB,?XBA??YPC,?CAZ??YBP,?ACZ??YPB),求证:

MLQCRM??,于是欲证MDQBAMXYBP. ?YZPC解析 如图,延长CY至M,使MB∥PY,连结XM.于是△XBA∽△YPC∽△MBC,而且是顺相似,故△XBM∽△ABC.故

XMBMPYPY,XM?AC?. ??ACBCPCPC初中数学竞赛名师辅导

AXMYZBPC

又△ACZ∽△BPY,故

CZPYAC?PB,于是XMPBMYCZ?PC?CY. 又?XMY??XMB??BMY??ACB??PYC? ?ACB??YPB??YCB??YPB??ACY??ACZ??ACY??ZCY,故△MXY∽△CZY,XYMYBYZ?YC?P,同时由PC?MYX??CYZ,得X、Y、Z三点共线.

于是

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