We are tasked with finding the probability that the angle $\angle ARD$ is obtuse, where $A$, $B$, $C$, and $D$ are the vertices of a rectangle with width 1 and length 2. Specifically, the problem asks us to find the integer $X$ such that the probability is closest to $\frac{X}{400}$.

Step 1: Understanding the Geometry

Given the rectangle ABCD, we define the points as follows:

• $A(0, 0)$
• $B(2, 0)$
• $C(2, 1)$
• $D(0, 1)$

We want to find the probability that the angle $\angle ARD$ is obtuse when point $R(x, y)$ is randomly chosen inside the rectangle. This occurs when the dot product of vectors $\overrightarrow{RA}$ and $\overrightarrow{RD}$ is negative: $\overrightarrow{RA} = (-x, -y), \quad \overrightarrow{RD} = (-x, 1 - y)$ The dot product is: $\overrightarrow{RA} \cdot \overrightarrow{RD} = (-x)(-x) + (-y)(1 - y) = x^2 + y - y^2$ For $\angle ARD$ to be obtuse, this dot product must be negative: $x^2 + y - y^2 < 0 \quad \Rightarrow \quad y^2 - y - x^2 > 0$ This inequality defines the region within the rectangle where $\angle ARD$ is obtuse.

Step 2: Finding the Area of the Region

We now need to calculate the area of the region within the rectangle where $y^2 - y - x^2 > 0$. This involves integrating the inequality over the bounds of the rectangle:

• $x \in [0, 2]$
• $y \in [0, 1]$

We solve the inequality $y^2 - y - x^2 > 0$ for $x$. Rearranging, we get: $x^2 < y^2 - y \quad \Rightarrow \quad x < \sqrt{y^2 - y}$ This inequality only holds when $y^2 - y > 0$, or $y > 1$ or $y < 0$, which leads