- 847 名前：１３２人目の素数さん mailto:sage [2021/07/13(火) 06:44:43.39 ID:w61FTnjw.net]
- >>800
外接円の半径をrとすると
⊿ABG = (1/2)rr sin(2C) = rr sin(C) cos(C),
⊿ABC = 2rr sin(A) sin(B) sin(C),
題意より
⊿ABG = (1/8)⊿ABC,
cos(C) = (1/4)sin(A)sin(B),
また
A + B + C = 180°
A = 60° (← 題意)
これを解いて
A = 60°
B = arctan(4/√27) = (1/2)arccos(11/43) = 37.589089468975°
C = arctan(13/√3) = (1/2)(π - arccos(83/86)) = 82.410910531025°
ところで
∠AGR = 180°- 2C = 15.178178938°
∠GAR = 90°- B = 52.410910531025°
正弦定理より
⊿AGR = rr cos(B)sin(2C)/{2cos(B+2C-180)}
= 0.1122092715867 rr
= 3,
∴ r = 5.1706632668738
⊿ABC = 28,
⊿ABG = 7/2,
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