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 \(\frac{\pi}{\mathrm{k}}\) and \(\frac{2 \pi}{\mathrm{k}}\), where \(\mathrm{k}>0\), then the value of \([\mathrm{k}]\) is [Note : [k] denotes the largest integer less than or equal to k]

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Circle

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

\frac{π}{k}

and

\frac{2 π}{k}

, where

k > 0

, then the value of

[k]

is

[Note : [k] denotes the largest integer less than or equal to k]

Answer: 3

Let the two chords subtend angles

\frac{π}{k}

and

\frac{2 π}{k}

at the center. The perpendicular distances from the center to the chords are

2 cos \frac{π}{2 k}

and

2 cos \frac{π}{k}

respectively.

Since the chords are parallel and on the same side of the center (or opposite sides), the distance between them is:

2 cos \frac{π}{2 k} + 2 cos \frac{π}{k} = 3 + 1

cos \frac{π}{2 k} + cos \frac{π}{k} = \frac{3 + 1}{2}

Let

θ = \frac{π}{k}

. Then:

cos \frac{θ}{2} + cos θ = \frac{3 + 1}{2}

Using

cos θ = 2 cos^{2} \frac{θ}{2} - 1

cos \frac{θ}{2} + 2 cos^{2} \frac{θ}{2} - 1 = \frac{3 + 1}{2}

Let

t = cos \frac{θ}{2}

2 t^{2} + t - 1 = \frac{3 + 1}{2}

4 t^{2} + 2 t - 2 = 3 + 1

4 t^{2} + 2 t - 3 - 3 = 0

Using the quadratic formula:

t = \frac{- 2 \pm 4 + 16 ( 3 + 3 )}{8} = \frac{- 2 \pm 52 + 16 3}{8} = \frac{- 2 \pm ( 4 3 + 2 )}{8}

Taking the positive root:

t = \frac{4 3}{8} = \frac{3}{2}

.

So

cos \frac{θ}{2} = \frac{3}{2} ⟹ \frac{θ}{2} = \frac{π}{6} ⟹ θ = \frac{π}{3}

.

Thus

\frac{π}{k} = \frac{π}{3} ⟹ k = 3

, and

[k] = 3

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer