Thanks for correcting my solution...
Though those mistakes don't really matter. XD
#ed_op#br#ed_cl#some mistakes...#ed_op#br#ed_cl#宇智波鼬 寫到:Another method:#ed_op#br#ed_cl#We use coordinate Geometry.#ed_op#br#ed_cl#Let A be the origin, C be (12,0), B be (#ed_op#span style="font-weight: bold; color: rgb(255, 0, 0);"#ed_cl#12,5#ed_op#/span#ed_cl#)#ed_op#br#ed_cl#Also, D will be (12r,5r) E be (a,0) (Where r and a are both reals)#ed_op#br#ed_cl#Obviously, #ed_op#span style="color: rgb(255, 0, 0);"#ed_cl#5#ed_op#/span#ed_cl#ar=30, #ed_op#span style="color: rgb(255, 0, 0); font-weight: bold;"#ed_cl#ar=6#ed_op#/span#ed_cl##ed_op#br#ed_cl#The shortest distance is √[(12r-a)^2+25r^2]#ed_op#br#ed_cl#=#ed_op#span style="color: rgb(255, 0, 0); font-weight: bold;"#ed_cl#√(#ed_op#/span#ed_cl#169r^2-144+36/r^2#ed_op#span style="font-weight: bold; color: rgb(255, 0, 0);"#ed_cl#)#ed_op#/span#ed_cl##ed_op#br#ed_cl#144 is a constant, so the value depends on 169r^2+36/r^2.#ed_op#br#ed_cl#By the AM-GM, we can easily find out the minimum.#ed_op#br#ed_cl#When r=6/13, 169r^2+36/r^2-144=12#ed_op#br#ed_cl#So, DE=√12
#ed_op#br#ed_cl##ed_op#img src="richedit/upload/2k11f0df35ba.png" alt="image file name: 2k11f0df35ba.png" border="0"#ed_cl##ed_op#br#ed_cl#AI為角A平分線,AI分別交BC、DE於F及G,DE最短為與AG垂直;#ed_op#br#ed_cl#延長AC至H令AH=AB=13,CH=1;ΔABH為等腰Δ;故AI⊥BH,DE//BH;#ed_op#br#ed_cl#ΔABC面積=5*12/2=30,ΔADE面積=30/2=15,ΔAGE面積=15/2。#ed_op#br#ed_cl##ed_op#br#ed_cl#ΔHCB~ΔHIA~ΔEGA (AAA) => AG:GE = AI:IH = BC:CH = 5:1 => AG = 5GE#ed_op#br#ed_cl#AG*GE/2 = ΔAGE面積=15/2 => 5*GE#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = 15 => GE = √3 => DE = 2√3。#ed_op#br#ed_cl#??? 寫到:#ed_op#div#ed_cl#三角形ABC中,BC=5,AC=12,AB=13,在邊AB,AC上分別取D,E使線段DE將三角形ABC分成面積相等的兩部分,試求DE的最小長度。#ed_op#/div#ed_cl##ed_op#div#ed_cl# #ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#img alt="image file name: 2k1df02af13a.gif" src="richedit/upload/2k1df02af13a.gif" border="0"#ed_cl##ed_op#/div#ed_cl#