좌표이용)삼각형ABC의각꼭지점A(0,2√3),B(-2,0),C(2,0)P(a,0)일때(단,-2≤a≤2)(선분AP)²+(선분BP)²=(a²+12)+(a+2)² =2a²+4a+16=2(a+1)²+14즉a=-1일때최소값14
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