如图，已知∠MON=30°，点A1、A2、A3…在射线ON上，点B1、B2、B3…在射线OM上，△A1B1A2、△A2B2A3、△A3 如图，已知∠MON=30°，点A1、A2、A3…在射线ON上...
\uff082012?\u6df1\u5733\uff09\u5982\u56fe\uff0c\u5df2\u77e5\uff1a\u2220MON=30\u00b0\uff0c\u70b9A1\u3001A2\u3001A3\u2026\u5728\u5c04\u7ebfON\u4e0a\uff0c\u70b9B1\u3001B2\u3001B3\u2026\u5728\u5c04\u7ebfOM\u4e0a\uff0c\u25b3A1B1A2\u2235\u25b3A1B1A2\u662f\u7b49\u8fb9\u4e09\u89d2\u5f62\uff0c\u2234A1B1=A2B1\uff0c\u22203=\u22204=\u222012=60\u00b0\uff0c\u2234\u22202=120\u00b0\uff0c\u2235\u2220MON=30\u00b0\uff0c\u2234\u22201=180\u00b0-120\u00b0-30\u00b0=30\u00b0\uff0c\u53c8\u2235\u22203=60\u00b0\uff0c\u2234\u22205=180\u00b0-60\u00b0-30\u00b0=90\u00b0\uff0c\u2235\u2220MON=\u22201=30\u00b0\uff0c\u2234OA1=A1B1=1\uff0c\u2234A2B1=1\uff0c\u2235\u25b3A2B2A3\u3001\u25b3A3B3A4\u662f\u7b49\u8fb9\u4e09\u89d2\u5f62\uff0c\u2234\u222011=\u222010=60\u00b0\uff0c\u222013=60\u00b0\uff0c\u2235\u22204=\u222012=60\u00b0\uff0c\u2234A1B1\u2225A2B2\u2225A3B3\uff0cB1A2\u2225B2A3\uff0c\u2234\u22201=\u22206=\u22207=30\u00b0\uff0c\u22205=\u22208=90\u00b0\uff0c\u2234A2B2=2B1A2\uff0cB3A3=2B2A3\uff0c\u2234A3B3=4B1A2=4\uff0cA4B4=8B1A2=8\uff0cA5B5=16B1A2=16\uff0c\u4ee5\u6b64\u7c7b\u63a8\uff1aA6B6=32B1A2=32\uff0e\u6545\u9009\uff1aC\uff0e
\u2235\u25b3A1B1A2\u4e3a\u7b49\u8fb9\u4e09\u89d2\u5f62\uff0c\u2234\u2220B1A1A2=60\u00b0\uff0c\u2235\u2220MON=30\u00b0\uff0c\u2234\u2220OB1A2=30\u00b0+60\u00b0=90\u00b0\uff0c\u2234A2B1=12OA2\uff0c\u540c\u7406\u53ef\u6c42\u5f97\uff1aB4A5=12OA5\uff0c\u2235OA1=1\uff0c\u2234OA4=2OA3=4OA2=8OA1=8\uff0cOA5=2OA4=4OA3=8OA2=16OA1=16\uff0c\u2234A4A5=OA5-OA4=16-8=8\uff0c\u6545\u9009\uff1aA\uff0e
解：∵△A1B1A2是等边三角形，∴A1B1=A2B1，∠3=∠4=∠12=60°，
∴∠2=120°，
∵∠MON=30°，
∴∠1=180°-120°-30°=30°，
又∵∠3=60°，
∴∠5=180°-60°-30°=90°，
∵∠MON=∠1=30°，
∴OA1=A1B1=1，
∴A2B1=1，
∵△A2B2A3、△A3B3A4是等边三角形，
∴∠11=∠10=60°，∠13=60°，
∵∠4=∠12=60°，
∴A1B1∥A2B2∥A3B3，B1A2∥B2A3，
∴∠1=∠6=∠7=30°，∠5=∠8=90°，
∴A2B2=2B1A2，B3A3=2B2A3，
∴A3B3=4B1A2=4，
A4B4=8B1A2=8，
A5B5=16B1A2=16，
…
∴△AnBnAn+1的边长为 2n-1．
∴，△A2012B2012A2013的边长为22012-1=22011．
故答案为：16，22011．
绛旓細瑙ｏ細濡傚浘锛杩囩偣A浣淎C鈯N锛屸埖鈭燤ON=30掳锛OA=80绫筹紝鈭碅C=40绫筹紝褰撶涓鍙版嫋鎷夋満鍒癇鐐规椂瀵瑰鏍′骇鐢熷櫔闊冲奖鍝嶏紝姝ゆ椂AB=50锛岀敱鍕捐偂瀹氱悊寰楋細BC=30锛岀涓鍙版嫋鎷夋満鍒癉鐐规椂鍣煶娑堝け锛屾墍浠D=30锛庣敱浜庝袱鍙版嫋鎷夋満鐩歌窛30绫筹紝鍒欑涓鍙板埌D鐐规椂绗簩鍙板湪C鐐癸紝杩橀』鍓嶈30绫冲悗鎵嶅瀛︽牎娌℃湁鍣煶褰卞搷锛庢墍浠...
绛旓細锛1锛夊湪绗3璞￠檺 锛2锛夊湪鍧愭爣杞翠笂 锛3锛夊湪鍧愭爣杞翠笂 锛4锛夊湪绗1璞￠檺
绛旓細鍦ㄥ洓杈瑰舰OSPT涓紝鈭燬PT=360掳-鈭燨SP-鈭燬OB-鈭燨TP=360掳-90掳-60掳-90掳=120掳锛屸埓鈭燗PB=鈭燬PT=120掳锛屸埓鈭燗PS=鈭燘PT锛涘張鈭碘垹ASP=鈭燘TP=90掳锛孉P=BP锛屸埓鈻矨PS鈮屸柍BPT锛屸埓PS=PT锛堝叏绛変笁瑙掑舰鐨勫搴旇竟鐩哥瓑锛夆埓鐐筆鍦鈭燤ON鐨勫钩鍒嗙嚎涓婏紱锛3锛夆憼鈭礝P骞冲垎鈭燗OB锛屸垹AOB=60掳锛孫P鈯B锛屸埓AQ...
绛旓細鈭鈭燤ON=60掳锛屸垹APB=120掳锛屸埓鈭燤ON+鈭燗PB=180掳锛屸埓鍥涜竟褰PBO鍥涚偣鍏卞渾锛庘埓褰揙P涓虹洿寰勬椂锛孫P鏈澶э紝鈭粹垹OAP=90掳锛庘埖AP=BP锛屸埓鈭燗OP=鈭燘OP=12鈭燗OB=30掳锛鈭燩AB=鈭燩BA=30掳锛孉D=BD=12AB=23锛屸埓鈭燗PO=60掳锛屸埓鈭燗DP=90掳锛庘埓AP=2DP鍦≧t鈻矨DP涓紝鐢卞嬀鑲″畾鐞嗭紝寰桪P=2锛屸埓AP=4锛...
绛旓細鈭礎O=OP=AP=4锛屸埓鈭POM=30掳锛PH=2锛汼2=S鈻矨OP+S鈻砄PB 锛12脳4脳23+12x脳2锛43+xAB2锛16+x2锛孉B锛16+x2锛孉C锛3216+x2锛屸埓S1锛1216+x2?
绛旓細鎬濊矾锛氬厛姹侫P锛屽啀璇佺偣P鍦鈭燤ON鐨勫钩鍒嗙嚎涓婏紝鐒跺悗鍐嶉氳繃鐩磋涓夎褰㈡眰OP 锛3锛夎繛鎺P锛屽湪Rf鈻砄PS鍜孯a鈻矨PS涓 鈭粹垹AOP=鈭燘OP=30掳 OP=SP/sin鈭燗OP=SP/sin30掳=2SP 鍗筹紝褰揝P鏈澶ф椂锛孫P涓烘渶澶у 鑰孲P=AP*cos鈭燬PA=4cos鈭燬PA 鈭碠P=2SP=8cos鈭燬PA 鈭村綋cos鈭燬PA=1锛屽嵆鈭燬PA=0掳鏃讹紝涔...
绛旓細(1)涓嶄細鏀瑰彉 鈭APB=180-鈭燗BP-鈭燘AP =180-1/2鈭燗BO-1/2鈭燘AO =180-1/2(180-鈭燗OB)=180-1/2*100 =130 (2)鈭燙=180-鈭燙AB-鈭燗BC =180-1/2鈭燨AB-(180-1/2鈭燗BY)=180-1/2鈭燨AB-180+1/2鈭燗BY =1/2(鈭燗BY-鈭燨AB)=1/2(180-鈭燨BA-鈭燨AB)=1/2(AOB)=1/2*60 =30 ...
绛旓細瑙ｏ細寤堕暱BP浜M浜嶤 鈭 鈭燤ON =6O掳锛孭B鈯N 鈭 鈭燨CB=30掳 鈭 PA鈯M锛孭A=2 鈭 PC=4锛孉C=2鈭3 鈭 PB=11 鈭 BC=PB+PC=11+4=15 鈭 鈭燨CB=30掳锛PB鈯N 鈭 OC=10鈭3 鈭 AC=2鈭3 鈭 OA=OC-AC=8鈭3 鈭 PA=2,PA鈯M 鈭 OP=14 .
绛旓細鈭碘垹ABD鏄柍ABC鐨勫瑙掞紝鈭粹垹ABD=鈭燙+鈭燘AC鈶狅紝鍙堚埖鈭燳BA鏄柍AOB鐨勫瑙掞紝鈭粹垹ABY=鈭燗OB+鈭燨AB鈶★紝鐢盉D骞冲垎鈭燳BA锛孉C骞冲垎鈭燘AO锛屸埓鈭燳BD=鈭燗BD= 1 2 鈭燳BA锛屸垹BAC=鈭燨AC= 1 2 鈭燨AB锛屽張鈭燗OB=60掳锛屸憽梅2寰楋細1 2 鈭燗BY= 1 2 鈭燗OB+ 1 2 鈭燨AB锛屽嵆鈭燗BD=30掳+鈭燘AC鈶紝鐢扁憼...
绛旓細瑙扐PB=瑙扐PF+瑙扚PB=120 鎵浠ヨEPA=瑙扚PB 鎵浠ヤ笁瑙掑舰PEA鍏ㄧ瓑涓庝笁瑙掑舰PFB 鎵浠A=PB 2.鈭PBD=鈭燗BO 鈭燗PB=120 鎵浠モ垹BPD=60=瑙扐OB 鎵浠ヤ笁瑙掑舰OBA鐩镐技浜庝笁瑙掑舰PBD 鐢1鐭A=PB OP=OP 鈭燗OP=鈭燘OP 鎵浠ヤ笁瑙掑舰AOP鍏ㄧ瓑浜庝笁瑙掑舰BOP 鎵浠ヨOAP=瑙扥BP 鍙堝洜涓篜A=PB寰楄BAP=瑙扨BA=30 鎵浠ヨOAB...
