※ 引述《hochirijay (uni)》之銘言:
: 4.一等腰三角形外接圓半徑12,內切圓半徑4,求兩圓心之距離
: 感覺最後一題最簡單,但是腦弱卡了= =\\\
: 有勞各位版友解救..
令 等腰三角形 ABC, AB=AC, 內心 I, AI 交 BC 於 D, 角ABC = a
則
AI = 4/cos(a)
AD = 4 + 4/cos(a)
AC = AD/sin(a), and AC/sin(a) = 2R = 24
令 x = cos(a), 由上式可得
6 x^3 - 5 x + 1 = 0
解得 x = -1 (不合), (3+sqrt(3))/6, (3-sqrt(3))/6
當 cos(a) = (3+sqrt(3))/6, AD = 16-4 sqrt(3) < 12,
外心 O 在三角形外, OI = 12 - AI = 4 sqrt(3)
當 cos(a) = (3-sqrt(3))/6, AD = 16+4 sqrt(3) > 12,
外心 O 在三角形內, OI = AD - 12 -4 = 4 sqrt(3)
So, OI = 4 sqrt(3)
PS: sqrt(3) 即 根號3.
※ 編輯: balista 來自: 163.22.20.88 (08/13 11:26)