看板 Math 關於我們 聯絡資訊
※ 引述《hau (小豪)》之銘言: : https://imgur.com/a/dvJDA : 如上圖 : 應有巧妙的解法...... A X B 設 XX'//AD, YY'//AB AXD = a, BXY = b, CYD = c, DY'ZX' = t Y' Z Y 因 (AXZY')(ZYCX') = (XBYZ)(Y'ZX'D) D X' C 故 (2a-t)(2c-t) = 2b*t => (t-a-b-c)^2 = (a+b+c)^2-4ac DXY = ABCD-a-b-c = (2a+2b+2c-t)-a-b-c = a+b+c-t = √((a+b+c)^2-4ac) 原題所求 = √(12^2-60)=2√21 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 180.217.180.238 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1522743455.A.A37.html