Key Concepts and Formulas

• Properties of a Square: Diagonals bisect each other at right angles. Each diagonal makes an angle of $45^\circ$ with the sides of the square. If one vertex is the origin, one diagonal passes through the origin.
• Slope of a Line: For a line $Ax + By + C = 0$, the slope $m = -\frac{A}{B}$. Also, $m = \tan \theta$, where $\theta$ is the angle the line makes with the positive x-axis.
• Coordinates of a Point: A point at a distance $r$ from the origin, with the connecting line segment making an angle $\theta$ with the positive x-axis, has coordinates $(r \cos \theta, r \sin \theta)$.

Step-by-Step Solution

Step 1: Identify the Diagonals of the Square

We are given the equations of two diagonals: (1) $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ (2) $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$

Since the square OABC has vertex O at the origin (0,0), one of the diagonals must pass through the origin. We check which equation satisfies this condition by substituting (0,0).

• For equation (1): $(\sqrt{3}+1)(0)+(\sqrt{3}-1)(0)=0 \implies 0=0$. This is true.
• For equation (2): $(\sqrt{3}-1)(0)-(\sqrt{3}+1)(0)+8\sqrt{3}=0 \implies 8\sqrt{3}=0$. This is false.

Therefore, equation (1) represents the diagonal OB, and equation (2) represents the diagonal AC.

Why this step is taken: Correctly identifying which diagonal passes through the origin is crucial. This distinction allows us to use the correct equation for calculations involving vertex A later.

Step 2: Determine the Angle of Diagonal OB with the Positive x-axis

Let $\theta_{OB}$ be the angle that diagonal OB makes with the positive x-axis. The slope of OB is $m_{OB}$. From equation (1), $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$: $(\sqrt{3}-1)y = -(\sqrt{3}+1)x$ $y = -\frac{\sqrt{3}+1}{\sqrt{3}-1}x$ So, the slope $m_{OB} = -\frac{\sqrt{3}+1}{\sqrt{3}-1}$.

To find the angle, we simplify this slope value by rationalizing the denominator: $m_{OB} = -\frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = -\frac{(\sqrt{3}+1)^2}{3-1} = -\frac{3+2\sqrt{3}+1}{2} = -\frac{4+2\sqrt{3}}{2} = -(2+\sqrt{3})$ We recognize that $2 + \sqrt{3} = \tan(75^\circ)$ because $\tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2+\sqrt{3}$. Therefore, $m_{OB} = -(2+\sqrt{3}) = -\tan 75^\circ$. Since $m_{OB} = \tan \theta_{OB}$, we have $\tan \theta_{OB} = -\tan 75^\circ$. Using the identity $\tan(180^\circ - \phi) = -\tan \phi$, we get $\tan \theta_{OB} = \tan(180^\circ - 75^\circ) = \tan 105^\circ$. Thus, $\theta_{OB} = 105^\circ$.

Why this step is taken: Determining the angle of diagonal OB is essential to finding the angle of side OA. Rationalizing the denominator and recognizing the tangent value simplifies the process.

Step 3: Determine the Angle of Side OA ($\alpha$)

The problem states that side OA makes an acute angle $\alpha$ with the positive x-axis. Since the diagonal of a square makes an angle of $45^\circ$ with its sides, the angle between the diagonal OB and the side OA must be $45^\circ$. Therefore, $\alpha$ can be $\theta_{OB} - 45^\circ$ or $\theta_{OB} + 45^\circ$.

• Case 1: $\alpha = \theta_{OB} - 45^\circ = 105^\circ - 45^\circ = 60^\circ$.
• Case 2: $\alpha = \theta_{OB} + 45^\circ = 105^\circ + 45^\circ = 150^\circ$.

Since the problem states that $\alpha$ is an acute angle (i.e., $0^\circ < \alpha < 90^\circ$), we must choose $\alpha = 60^\circ$.

Why this step is taken: The "acute angle" condition is crucial. It removes the ambiguity and allows us to pinpoint the correct angle for side OA, which is essential for finding the coordinates of vertex A.

Step 4: Find the Coordinates of Vertex A

O is the origin (0,0), and OA is a side of the square with length 'a'. Side OA makes an angle $\alpha = 60^\circ$ with the positive x-axis. Using the coordinates of a point $(r \cos \theta, r \sin \theta)$ where $r=a$ and $\theta=\alpha$: $A = (a \cos \alpha, a \sin \alpha) = (a \cos 60^\circ, a \sin 60^\circ)$ $A = \left(a \cdot \frac{1}{2}, a \cdot \frac{\sqrt{3}}{2}\right) = \left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right)$

Why this step is taken: Expressing the coordinates of vertex A in terms of 'a' allows us to use the equation of diagonal AC to solve for 'a'.

Step 5: Use the Condition that Vertex A Lies on Diagonal AC to find $a^2$

Vertex A must lie on the diagonal AC, whose equation is given by $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Substitute the coordinates of $A\left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right)$ into the equation of diagonal AC: $(\sqrt{3}-1)\left(\frac{a}{2}\right) - (\sqrt{3}+1)\left(\frac{\sqrt{3}a}{2}\right) + 8\sqrt{3} = 0$ Multiply the entire equation by 2 to simplify: $a(\sqrt{3}-1) - a\sqrt{3}(\sqrt{3}+1) + 16\sqrt{3} = 0$ Expand the terms: $a\sqrt{3} - a - a(3+\sqrt{3}) + 16\sqrt{3} = 0$ $a\sqrt{3} - a - 3a - a\sqrt{3} + 16\sqrt{3} = 0$ The $a\sqrt{3}$ terms cancel out: $-a - 3a + 16\sqrt{3} = 0$ $-4a + 16\sqrt{3} = 0$ $4a = 16\sqrt{3}$ $a = \frac{16\sqrt{3}}{4}$ $a = 4\sqrt{3}$ The question asks for $a^2$: $a^2 = (4\sqrt{3})^2 = 4^2 \times (\sqrt{3})^2 = 16 \times 3 = 48$

Why this step is taken: Substituting the coordinates of A into the equation of AC creates an equation with 'a' as the only unknown, allowing us to solve for the side length 'a' and then calculate $a^2$.

Common Mistakes & Tips

• Angle Ambiguity: Remember that the diagonal makes a $45^\circ$ angle with either side. Be sure to use the "acute angle" condition to select the correct angle.
• Rationalization: Rationalizing the slope expression is crucial to recognizing the tangent of a familiar angle.
• Algebraic Errors: Carefully expand and simplify expressions to avoid arithmetic errors.
• Final Answer: Read the question carefully. The question asks for $a^2$, not $a$.

Summary

This problem combines coordinate geometry with the properties of squares. By identifying the diagonal passing through the origin, finding the angle of the diagonal, using the $45^\circ$ property of the diagonal, finding the coordinates of vertex A, and substituting into the equation of the other diagonal, we can solve for the side length $a$ and then calculate $a^2$.

The final answer is $\boxed{48}$, which corresponds to option (A).