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A point P moves inside an equilateral triangle formed by vertices A(0, 0), B(2, 2*sqrt(3)), and C(4, 0) such that the minimum of its distances to the three vertices always equals 2. Find the area of the region traced by P.

1. 3*sqrt(3) - 3*pi/2
2. 4*sqrt(3) - 2*pi
3. sqrt(3) - pi/2
4. 2*pi

Correct answer: 4*sqrt(3) - 2*pi

Solution

Since AB = BC = CA = 4, the triangle is equilateral with all interior angles 60 deg. When min(PA, PB, PC) = 2, P lies on an arc of radius 2 centered at whichever vertex is nearest. The enclosed area = (sqrt(3)/4)*16 - 3*(60/360)*pi*4 = 4*sqrt(3) - 2*pi.

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