Question

A point P moves inside a triangle formed by A(0, 0), B(2, 2√3), C(4, 0) such that {min(PA, PB, PC)} = 2, then the area bounded by the curve traced by P is -

• A. 3√3 - 3π/2
• B. 4√3 - 2π
• C. √3 - π/2
• D. 2π

Answer 1

Answer: (B)

4√3 - 2π

Answer 2

The vertices of the triangle are A(0, 0), B(2, 2√3), and C(4, 0).

First, let's determine the type of triangle. The side lengths are: $AB = \sqrt{(2-0)^2 + (2\sqrt{3}-0)^2} = \sqrt{4 + 12} = \sqrt{16} = 4$. $BC = \sqrt{(4-2)^2 + (0-2\sqrt{3})^2} = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$. $AC = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$.

Since all side lengths are equal, triangle ABC is an equilateral triangle with side length 4. The angles of an equilateral triangle are all 60° (or π/3 radians).

A point P moves inside the triangle such that min(PA, PB, PC) = 2. This means that P is a point inside the triangle such that PA ≥ 2, PB ≥ 2, PC ≥ 2, and at least one of these distances is exactly 2. The condition min(PA, PB, PC) ≥ 2 defines a region inside the triangle. The condition min(PA, PB, PC) = 2 defines the boundary of this region. The question asks for the area bounded by this curve, which is the area of the region where min(PA, PB, PC) ≥ 2 and P is inside the triangle.

Let $C_A$, $C_B$, $C_C$ be circles centered at A, B, C respectively, each with radius 2. The region where PA ≥ 2 is the area outside or on the circle $C_A$. The region where PB ≥ 2 is the area outside or on the circle $C_B$. The region where PC ≥ 2 is the area outside or on the circle $C_C$. The region where min(PA, PB, PC) ≥ 2 is the intersection of the regions outside or on $C_A$, $C_B$, and $C_C$. We are interested in the part of this region that is inside the triangle ABC. This region is the area of triangle ABC minus the area of the points inside triangle ABC where min(PA, PB, PC) < 2. The region where min(PA, PB, PC) < 2 is the union of the interiors of the circles $C_A$, $C_B$, $C_C$. So, the area we need to find is the area of triangle ABC minus the area of the union of the interiors of $C_A$, $C_B$, $C_C$ that are inside the triangle.

The area of triangle ABC = (1/2) * base * height. Base AC = 4. The height is the y-coordinate of B, which is 2√3. Area(ABC) = (1/2) * 4 * 2√3 = 4√3.

Now consider the parts of the circles inside the triangle. For circle $C_A$ centered at A(0,0) with radius 2, the part inside the triangle is a sector with angle equal to the angle of the triangle at A, which is 60°. Area of sector of $C_A$ inside triangle = (60°/360°) * π * (radius)^2 = (1/6) * π * 2^2 = (1/6) * 4π = 2π/3. Similarly, for circle $C_B$ centered at B(2, 2√3) with radius 2, the part inside the triangle is a sector with angle 60°. Area of sector of $C_B$ inside triangle = (1/6) * π * 2^2 = 2π/3. For circle $C_C$ centered at C(4,0) with radius 2, the part inside the triangle is a sector with angle 60°. Area of sector of $C_C$ inside triangle = (1/6) * π * 2^2 = 2π/3.

The distance between any two vertices is 4, which is equal to the sum of the radii of the two circles centered at these vertices (2 + 2 = 4). This means the circles touch each other at points on the sides of the triangle. $C_A$ and $C_C$ touch at (2,0) on AC. $C_A$ and $C_B$ touch at (1,√3) on AB. $C_C$ and $C_B$ touch at (3,√3) on BC. The three sectors meet at these points on the sides of the triangle. The interiors of these three sectors do not overlap. The region inside the triangle where min(PA, PB, PC) < 2 is the union of the interiors of these three sectors. The area of this union is the sum of the areas of the three sectors, since their interiors are disjoint. Area of union of sectors = Area(sector at A) + Area(sector at B) + Area(sector at C) = 2π/3 + 2π/3 + 2π/3 = 6π/3 = 2π.

The area of the region inside the triangle where min(PA, PB, PC) ≥ 2 is the area of the triangle minus the area of the union of the interiors of the three sectors. Area = Area(ABC) - Area(union of sectors) = 4√3 - 2π.

The area bounded by the curve traced by P is the area of the region where min(PA, PB, PC) ≥ 2 inside the triangle.