Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute a

Q: Wait, how do I know which diagonal is $OB$ and which is $AC$?

Great question! The square is $OABC$. The diagonals are $OB$ and $AC$. Since $O$ is at the origin $(0,0)$, the diagonal $OB$ must pass through the origin. A line passing through the origin has the form $y = mx$ or $Ax + By = 0$. Looking at the two equations, the first one has no constant term, so it must be $OB$. The second one has a constant term, so it must be $AC$.

Q: Why is the angle between the side $OA$ and diagonal $OB$ exactly $45^{\circ}$?

In a square, all internal angles are $90^{\circ}$. The diagonal of a square is the angle bisector of the vertex it connects. Therefore, the diagonal $OB$ bisects the $90^{\circ}$ angle at vertex $O$, splitting it into two $45^{\circ}$ angles. This geometric property is a powerful tool to relate the orientation of the side to the orientation of the diagonal.

Q: Why do we reject the other possible value for $\alpha$?

The problem explicitly states that the side $OA$ makes an *acute* angle $\alpha$ with the positive x-axis. When solving $|105^{\circ} - \alpha| = 45^{\circ}$, we get two possibilities: $105^{\circ} - \alpha = 45^{\circ}$ (giving $\alpha = 60^{\circ}$) or $105^{\circ} - \alpha = -45^{\circ}$ (giving $\alpha = 150^{\circ}$). Since $60^{\circ}$ is acute and $150^{\circ}$ is obtuse, we select $60^{\circ}$.

Question

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 0$ and $(\sqrt{3} - 1)x - (\sqrt{3} + 1)y + 8\sqrt{3} = 0$. Then $a^2$ is equal to

Options:

(A) 48

(B) 32

(C) 16

(D) 24

Sigma Percentile · Accepted Answer

### The Geometry of the Origin: Unlocking the Square

Imagine you are standing at the origin of a coordinate plane, the anchor point of a square $OABC$. This square is not just a shape; it is a rigid, symmetric structure waiting for us to decode its secrets.

We are given the equations of its two diagonals, and our mission is to find the square of its side length, $a^2$. This is a classic JEE Advanced problem that tests not just your algebraic skills, but your ability to visualize geometry.

### Phase 1: The Origin's Secret

We are given two diagonal equations: 
$$(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 0$$ 
and 
$$(\sqrt{3} - 1)x - (\sqrt{3} + 1)y + 8\sqrt{3} = 0$$

The first thing to notice is the first equation. It has no constant term! In the world of coordinate geometry, a linear equation $Ax + By + C = 0$ passes through the origin if and only if $C = 0$.

This is a massive gift from the problem setter. It tells us immediately that this line is the diagonal $OB$, because $O(0,0)$ is a vertex of our square.

### Phase 2: The Angular Dance

Now, let's find the slope of this diagonal $OB$. Rearranging $(\sqrt{3} + 1)x + (\sqrt{3} - 1)y = 0$ into the slope-intercept form $y = mx$, we get:

$$m_1 = -\frac{\sqrt{3}+1}{\sqrt{3}-1}$$

To make this manageable, we rationalize the denominator by multiplying the numerator and denominator by $(\sqrt{3}+1)$. This simplifies to:

$$m_1 = -\frac{(\sqrt{3}+1)^2}{3-1} = -\frac{3+1+2\sqrt{3}}{2} = -(2+\sqrt{3})$$

We know that the slope $m = 	an 	heta$, where $	heta$ is the angle the line makes with the positive x-axis. So, $	an 	heta_1 = -(2+\sqrt{3})$.

Recalling our trigonometric values, we know that $	an 75^{\circ} = 2+\sqrt{3}$. Since our slope is negative, the angle must be in the second quadrant. Thus, $	heta_1 = 180^{\circ} - 75^{\circ} = 105^{\circ}$.

### Phase 3: The Coordinate Bridge

Now, we use the property of the square. The diagonal $OB$ bisects the $90^{\circ}$ angle at the origin. This means the angle between the side $OA$ and the diagonal $OB$ is $45^{\circ}$.

If $\alpha$ is the angle $OA$ makes with the x-axis, then $|105^{\circ} - \alpha| = 45^{\circ}$. This gives us two potential values for $\alpha$: $60^{\circ}$ or $150^{\circ}$.

Since the problem specifies that $\alpha$ is an acute angle, we confidently choose $\alpha = 60^{\circ}$. With $\alpha = 60^{\circ}$, we can define the coordinates of vertex $A$. If the side length is $a$, then:

$$A = (a \cos 60^{\circ}, a \sin 60^{\circ}) = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}ight)$$

### Phase 4: The Algebraic Climax

We have one final piece of the puzzle: the second diagonal, $AC$, given by $(\sqrt{3} - 1)x - (\sqrt{3} + 1)y + 8\sqrt{3} = 0$. Since vertex $A$ lies on this diagonal, its coordinates must satisfy the equation.

Let's substitute $x = \frac{a}{2}$ and $y = \frac{a\sqrt{3}}{2}$ into the equation:

$$(\sqrt{3}-1)\left(\frac{a}{2}ight) - (\sqrt{3}+1)\left(\frac{a\sqrt{3}}{2}ight) + 8\sqrt{3} = 0$$

Multiply the entire equation by $2$ to clear the denominator:

$$a(\sqrt{3}-1) - a\sqrt{3}(\sqrt{3}+1) + 16\sqrt{3} = 0$$

Now, expand the terms carefully:

$$a\sqrt{3} - a - 3a - a\sqrt{3} + 16\sqrt{3} = 0$$

Look at that! The $a\sqrt{3}$ and $-a\sqrt{3}$ terms cancel out perfectly, leaving us with $-4a + 16\sqrt{3} = 0$. Solving for $a$, we get $4a = 16\sqrt{3}$, which means $a = 4\sqrt{3}$.

Finally, the question asks for $a^2$. Squaring our result:

$$a^2 = (4\sqrt{3})^2 = 16 	imes 3 = 48$$

The journey is complete. We have navigated the geometry, utilized the trigonometry, and conquered the algebra. The **final answer is 48**.

Select Answer:

Visualized Solution

Visualizing the Square $O A B C$

Analyzing Diagonal 1 through Origin

Calculating the Slope of $O B$

Finding the Angle of Diagonal $O B$

Relating Side $O A$ and Diagonal $O B$

Solving for $α$

Defining Coordinates of Vertex $A$

Using the Second Diagonal Equation

Substitution and Setup

Simplifying the Equation

Solving for Side Length $a$

Calculating the Final Answer $a^{2}$

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Origin

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Visualizing the Square .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OABC