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Re: [數學]求圓半徑

發表 G@ry 於 星期五 五月 04, 2007 11:37 am

G@ry 寫到:
........... 寫到:#ed_op#div#ed_cl##ed_op#font face="Verdana"#ed_cl#有一個直角三角形AOB,∠AOB=90°,以點O為圓心作一圓,該圓交#ed_op#br#ed_cl#OA於C,OB於D,並相切AB於T;已知AC=16,BD=135,求圓半徑。#ed_op#/font#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl#
#ed_op#br#ed_cl#設OT為x;#ed_op#br#ed_cl#圓切AB於一點T => OT(=半徑)垂直AB => OT*AB/2=三角形面積#ed_op#br#ed_cl#x = (16+x)*(135+x)/√((16+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+(135+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)/√(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481)#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl##ed_op#br#ed_cl#2x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+18481x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+27121x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+652320x+4665600#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600=0#ed_op#br#ed_cl#待續...#ed_op#br#ed_cl#
#ed_op#br#ed_cl#終於完整解出:  x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600 =#ed_op#br#ed_cl#(x-120)*{x-2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)-20]}*{x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)+40]x+8#ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#}, #ed_op#br#ed_cl##ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#=#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(4004225-625275√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(4004225+625275√41)+10[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)+24]#ed_op#br#ed_cl#若要再解#ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#便要用上虛數來解(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#<4ac)...太麻煩了~~~#ed_op#br#ed_cl#當中 120>0;  2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)-20]~=2(-4+20-20)=-8<0,#ed_op#br#ed_cl#故答案為120。#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#若有高人能再簡化這些數字,請務必賜教賜教,謝!... 小弟總覺得還是有辦法再行簡化的....只是小的道行太低.... : (#ed_op#br#ed_cl##ed_op#br#ed_cl#

發表 亞斯 於 星期四 五月 03, 2007 9:20 pm

一百二十

就覺得好像在哪看過
請參考
李冶<<測圓海鏡>>中的圓城問題

Re: [數學]求圓半徑

發表 G@ry 於 星期四 五月 03, 2007 8:37 am

........... 寫到:#ed_op#div#ed_cl##ed_op#font face="Verdana"#ed_cl#有一個直角三角形AOB,∠AOB=90°,以點O為圓心作一圓,該圓交#ed_op#br#ed_cl#OA於C,OB於D,並相切AB於T;已知AC=16,BD=135,求圓半徑。#ed_op#/font#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl#
#ed_op#br#ed_cl#設OT為x;#ed_op#br#ed_cl#圓切AB於一點T =&gt; OT(=半徑)垂直AB =&gt; OT*AB/2=三角形面積#ed_op#br#ed_cl#x = (16+x)*(135+x)/√((16+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+(135+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)/√(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481)#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl##ed_op#br#ed_cl#2x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+18481x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+27121x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+652320x+4665600#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600=0#ed_op#br#ed_cl#待續...#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#

[數學]求圓半徑

發表 ........... 於 星期四 五月 03, 2007 7:00 am

#ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#有一個直角三角形AOB,∠AOB=90°,以點O為圓心作一圓,該圓交#ed_op#BR#ed_cl#OA於C,OB於D,並相切AB於T;已知AC=16,BD=135,求圓半徑。#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#