Two parallel chords of a circle of radius 2 are at a distance √3 + 1 apart. If the chords subtend at

Q: Wait, why do we add the distances $d_1$ and $d_2$ instead of subtracting them?

Great question! The distance between the chords is given as $\sqrt{3} + 1$. Since the radius is $2$, the maximum distance between any two points in the circle is the diameter, which is $4$. Because $\sqrt{3} + 1 \approx 2.732$, which is greater than the radius $2$, the chords must be on opposite sides of the center. Thus, the total distance is the sum of their individual distances from the center.

Q: How did you know that $52 + 16\sqrt{3}$ is a perfect square?

That is a classic JEE trick! Whenever you see a discriminant involving a square root, look for the form $(a + b\sqrt{3})^2$. Expanding $(4\sqrt{3} + 2)^2$ gives $16(3) + 4 + 2(4\sqrt{3})(2) = 48 + 4 + 16\sqrt{3} = 52 + 16\sqrt{3}$. Always keep an eye out for these algebraic patterns!

Q: Why do we reject the negative root in the quadratic formula?

We are dealing with $\alpha = \frac{\pi}{2k}$. Since $k > 0$, $\alpha$ is a positive angle. Furthermore, the chords subtend angles $\frac{\pi}{k}$ and $\frac{2\pi}{k}$, which must be less than $\pi$ for them to be chords. This implies $\alpha$ is an acute angle, and the cosine of an acute angle must be positive.

Q: Is there a simpler way to solve $2(\cos \alpha + \cos 2\alpha) = \sqrt{3} + 1$ without the quadratic?

You could try to guess the value of $\alpha$ by inspection. Since $\sqrt{3}$ appears in the RHS, it is highly likely that $\cos \alpha$ involves $\frac{\sqrt{3}}{2}$ or $\frac{1}{2}$. Testing $\alpha = 30^\circ$ (or $\frac{\pi}{6}$) is a very efficient strategy in competitive exams!

Q: What does the notation $[k]$ mean in the final answer?

The square brackets $[k]$ denote the Greatest Integer Function (or floor function). It returns the largest integer less than or equal to $k$. Since we found $k = 3$, $[3]$ is simply $3$.

Question

Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3} + 1$ apart. If the chords subtend at the center, angles of $\pi/k$ and $2\pi/k$, where $k > 0$, then the value of $[k]$ is ____.

Sigma Percentile · Accepted Answer

### The Geometry of Circles: A Journey into Symmetry

Imagine you are standing at the center of a circle with a radius of $r = 2$. You are looking at two parallel chords slicing through the space around you.

One chord subtends an angle of $\frac{\pi}{k}$ at your position, and the other, further away, subtends an angle of $\frac{2\pi}{k}$. The distance between these two lines is $\sqrt{3} + 1$.

This is not just a problem of geometry; it is a dance of trigonometry and algebra. Let us step through this together.

### Phase 1: Visualizing the Spatial Reality

First, we must orient ourselves. The radius of our circle is $2$, and the distance between our two chords is $d = \sqrt{3} + 1$.

If you calculate the value of $\sqrt{3} + 1$, you get approximately $2.732$. Since this distance is greater than the radius of $2$, we immediately realize that the chords cannot be on the same side of the center.

They must be on opposite sides, effectively 'sandwiching' the center between them. This realization is the key that unlocks the entire problem.

### Phase 2: The Trigonometric Bridge

To find the distance of any chord from the center, we draw a perpendicular line from the center to the chord. This forms a right-angled triangle where the hypotenuse is the radius $r$ and the base is half the chord.

The angle at the center is bisected, becoming $\frac{	heta}{2}$. Thus, the distance $d$ is given by the elegant relation $d = r \cos(\frac{	heta}{2})$.

For our two chords, we have:

$$d_1 = 2 \cos\left(\frac{\pi}{2k}ight)$$

$$d_2 = 2 \cos\left(\frac{\pi}{k}ight)$$

Since the total distance is the sum of these two, we set up our master equation:

$$2 \cos\left(\frac{\pi}{2k}ight) + 2 \cos\left(\frac{\pi}{k}ight) = \sqrt{3} + 1$$

### Phase 3: The Algebraic Transformation

This equation looks intimidating, but let us simplify it. Let $\alpha = \frac{\pi}{2k}$. Then, our equation becomes $2(\cos \alpha + \cos 2\alpha) = \sqrt{3} + 1$.

We know the double-angle identity for cosine: $\cos 2\alpha = 2\cos^2 \alpha - 1$. Substituting this in, we get:

$$2(\cos \alpha + 2\cos^2 \alpha - 1) = \sqrt{3} + 1$$

$$4\cos^2 \alpha + 2\cos \alpha - 2 = \sqrt{3} + 1$$

Rearranging this into a standard quadratic form $ax^2 + bx + c = 0$, we get:

$$4\cos^2 \alpha + 2\cos \alpha - (3 + \sqrt{3}) = 0$$

### Phase 4: The Moment of Clarity

Now, we apply the quadratic formula. The discriminant $D = b^2 - 4ac$ becomes:

$$D = 2^2 - 4(4)(-(3+\sqrt{3})) = 4 + 16(3+\sqrt{3}) = 52 + 16\sqrt{3}$$

Here is where the magic happens. If you look closely, $52 + 16\sqrt{3}$ is exactly $(4\sqrt{3} + 2)^2$.

Taking the positive root (because $\cos \alpha$ must be positive for an acute angle $\alpha$):

$$\cos \alpha = \frac{-2 + (4\sqrt{3} + 2)}{8} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}$$

This is a beautiful result! We know that $\cos \alpha = \frac{\sqrt{3}}{2}$ implies $\alpha = \frac{\pi}{6}$.

### Phase 5: The Final Resolution

We return to our substitution: $\frac{\pi}{2k} = \frac{\pi}{6}$. This simplifies instantly to $2k = 6$, which means $k = 3$.

The question asks for the greatest integer function $[k]$, and since $k=3$, the **final answer is 3**.

Look at how the complexity dissolved. We started with a geometric puzzle, translated it into the language of trigonometry, solved an algebraic quadratic, and arrived at a clean, integer result.

Animated Solution for Mathematics - Circles: Two parallel chords of a circle of radius 2 are at a distance $3 + 1$ apart. If the chords subtend at the center, angles of $π / k$ and $2 π / k$ , where $k > 0$ , then the value of $[k]$ is ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Angles Subtended at Center

Distance Formula for Chords

Setting up the Equation

Applying Double Angle Identity

The Quadratic Equation

Solving for $cos α$ (Discriminant)

Roots of the Quadratic

Finding the value of $k$

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Circles

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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