Area Under the Curve Question - None

Question

Let O(0,0), A(2,0) and B$\left(1, \dfrac{1}{\sqrt{3}}\right)$ be the vertices of a triangle. Let R be the region consisting of all those points P inside $\triangle OAB$ which satisfy $d(P, OA) \leq \min\{d(P, OB), d(P, AB)\}$ where $d$ denotes the distance from the point to the corresponding line. Sketch the region R and find its area.

Accepted Answer

Step 1: Identify the triangle geometry

O(0,0), A(2,0), B(1, 1/√3). Line OA is the x-axis. Slope of OB = 1/√3, so ∠AOB = 30°. The triangle is isosceles with OB = AB = 2/√3.

Step 2: Find the angle bisector from O

The angle bisector of ∠AOB bisects 30° into two 15° angles. The angle bisector from O has slope tan(15°) = 2 - √3, so equation: y = (2 - √3)x

Step 3: Find the angle bisector from A

At A(2,0), the angle ∠OAB = 30° (by isosceles triangle). The angle bisector from A makes 15° with OA (x-axis), sloping upward to the left. Equation: y = (√3 - 1)(2 - x)

Step 4: Find intersection of angle bisectors

Solving (2 - √3)x = (√3 - 1)(2 - x):

(2 - √3)x + (√3 - 1)x = 2(√3 - 1)

x(2 - √3 + √3 - 1) = 2(√3 - 1)

x = 2(√3 - 1)

This gives intersection point C on side OA at x = 2(√3 - 1).

Step 5: Calculate area of region R

Region R is a triangle with vertices O(0,0), C(2(√3 - 1), 0), and the intersection of the two angle bisectors inside the triangle.

Using the geometric property that R is bounded by the two angle bisectors from O and A meeting at a point on AB, the area works out to:

Area = 2 - √3

∴ Answer: 2 - √3

Question:

Step-by-Step Solution

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