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數學考古題

Jan 10th 2014, 14:42

由於△ABC為直角△, 故外心O為AC之中點,
從而 B、G、O 三點共線, 且OG = (1/3)OB。
內切圓半徑 = (AB + BC - AC)/2 = (8 +6 -10)/2 = 2 , 即 I 與 BC 及 OC 之距為 2。
△BIO = △BOC - △BIC - △OIC = (1/2) 6×8/2 - 6×2/2 - 5×2/2 = 1
故 △IOG = (1/3)△BIO = 1/3 。

別解 :
設 B(0,0) , A(0,8) , C(6,0) , 則 O(3,4) , G(2, 8/3) , I(2,2)。
△IOG
=
||2......2.||
||2....8/3|| × 1/2 = |(2×8/3 + 2×4 + 3×2 - 2×2 - 3×8/3 - 2×4)| × 1/2 = 1/3
||3.......4||
||2.......2||

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