初中奥数题典例 典型的初一奥数题 （奥林匹克的、五道以上、题及答案） 急!急...
\u521d\u4e2d\u7ecf\u5178\u5965\u6570\u9898\u5c0f\u6797\u5c061\uff0c2...\uff0cn\uff0c\u8fd9n\u4e2a\u6570\u8f93\u5165\u7535\u8111\uff0c\u6c42\u5e73\u5747\u6570\uff0c\u5f53\u4ed6\u8ba4\u4e3a\u8f93\u5165\u5b8c\u6bd5\u65f6\uff0c\u7535\u8111\u663e\u793a\u53ea\u8f93\u5165\u4e86\uff08n-1\uff09\u4e2a\u6570\uff0c\u5e73\u5747\u6570\u4e3a35\u53c85/7\uff0c\u5047\u8bbe\u8fd9\uff08n-1\uff09\u4e2a\u6570\u8f93\u5165\u65e0\u8bef\uff0c\u5219\u6f0f\u8f93\u5165\u7684\u4e00\u4e2a\u6570\u4e3a\uff08 \uff09
A 10 B 53 C 56 D 67
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1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17........
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\u6cd5\u4e8c\uff1a \u89e3:\u8bbe\u767e\u4f4d\u662f100(X+1) , \u5341\u4f4d\u662f 10X , \u4e2a\u4f4d\u662f3X-2 100(X+1)+10X+(3X-2)+100(3X-2)+10X+(X+1)=1171 X=3 \u767e\u4f4d:100(X+1)=100(3+1)=400 \u5341\u4f4d:10X=3 x 10=30 \u4e2a\u4f4d:3X-2=3 x 3 -2=7 \u4e09\u4f4d\u6570:400+30+7=437
1、已知：如图，O是半圆的圆心，C、E是圆上的两点，CD⊥AB，EF⊥AB，EG⊥CO．
求证：CD＝GF．（初二）
2、已知：如图，P是正方形ABCD内点，∠PAD＝∠PDA＝150．
求证：△PBC是正三角形．（初二）
3、如图，已知四边形ABCD、A1B1C1D1都是正方形，A2、B2、C2、D¬2分别是AA1、BB1、CC1、DD1的中点．
求证：四边形A2B2C2D2是正方形．（初二）
4、已知：如图，在四边形ABCD中，AD＝BC，M、N分别是AB、CD的中点，AD、BC的延长线交MN于E、F．
求证：∠DEN＝∠F．
1.如下图做GH⊥AB,连接EO。由于GOFE四点共圆，所以∠GFH＝∠OEG,
即△GHF∽△OGE,可得 = = ,又CO=EO，所以CD=GF得证。
2. 如下图做△DGC使与△ADP全等，可得△PDG为等边△，从而可得
△DGC≌△APD≌△CGP,得出PC=AD=DC,和∠DCG=∠PCG＝150
所以∠DCP=300 ，从而得出△PBC是正三角形
3.如下图连接BC1和AB1分别找其中点F,E.连接C2F与A2E并延长相交于Q点，
连接EB2并延长交C2Q于H点，连接FB2并延长交A2Q于G点，
由A2E= A1B1= B1C1= FB2 ，EB2= AB= BC=FC1 ，又∠GFQ+∠Q=900和
∠GEB2+∠Q=900,所以∠GEB2=∠GFQ又∠B2FC2=∠A2EB2 ，
可得△B2FC2≌△A2EB2 ，所以A2B2=B2C2 ，
又∠GFQ+∠HB2F=900和∠GFQ=∠EB2A2 ,
从而可得∠A2B2 C2=900 ，
同理可得其他边垂直且相等，
从而得出四边形A2B2C2D2是正方形。
4.如下图连接AC并取其中点Q，连接QN和QM，所以可得∠QMF=∠F，∠QNM=∠DEN和∠QMN=∠QNM，从而得出∠DEN＝∠F。
1、已知：如图，O是半圆的圆心，C、E是圆上的两点，CD⊥AB，EF⊥AB，EG⊥CO．
求证：CD＝GF．（初二）
2、已知：如图，P是正方形ABCD内点，∠PAD＝∠PDA＝150．
求证：△PBC是正三角形．（初二）
3、如图，已知四边形ABCD、A1B1C1D1都是正方形，A2、B2、C2、D¬2分别是AA1、BB1、CC1、DD1的中点．
求证：四边形A2B2C2D2是正方形．（初二）
4、已知：如图，在四边形ABCD中，AD＝BC，M、N分别是AB、CD的中点，AD、BC的延长线交MN于E、F．
求证：∠DEN＝∠F．
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