Angel1114
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因为c^2=b^2+a^2-2bacosC
S=(1/2)basinC
则a^2+b^2+c^2-4√3S
=b^2+a^2-2bacosC+b^2+a^2-4√3*(1/2)basinC
=2b^2+2a^2-2bacosC-2√3basinC
=2b^2+2a^2-4ba[(1/2)cosC+(√3/2)sinC]
=2b^2+2a^2-4ba+4ba-4bacos(60-C)
=2(b-a)^2+4ba[1-cos(60-C)]
-120 < 60-C < 60
-1/2 < cos(60-C) ≤ 1
0 ≤ 1-cos(60-C) < 3/2
所以
a^2+b^2+c^2-4√3S = 2(b-a)^2+4ba[1-cos(60-C)] ≥0
因为b=a且C=60时,等号成立 即S=(a^2+b^2+c^2)/4√3
所以C=60°
1年前
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