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 2π/k, where $k > 0,$ then the value of [k] is _______. [Note : [k] denotes the largest integer less than or equal to k]

$R=2,\qquad \text{two parallel chords } \sqrt{3}+1 \text{ apart}$

$\text{central angles } \dfrac{\pi}{k}\ \text{ and }\ \dfrac{2\pi}{k}\ \ (k>0)$

$\text{To find: } [k] = \text{greatest integer } \le k$

$1.\ d=R\cos(\theta/2)\ \text{for a chord at central angle }\theta$

$2.\ \text{the two chords: angles } \dfrac{\pi}{k},\ \dfrac{2\pi}{k}$

$3.\ \text{set the gap } =\sqrt{3}+1 \text{ and solve for } k$

$\cos\!\left(\dfrac{\theta}{2}\right)=\dfrac{d}{R}\ \Rightarrow\ d=R\cos\!\left(\dfrac{\theta}{2}\right)$

$R=2\ \Rightarrow\ d=2\cos\!\left(\dfrac{\theta}{2}\right)$

$d_1=2\cos\!\left(\dfrac{\pi}{2k}\right)$

$d_2=2\cos\!\left(\dfrac{\pi}{k}\right)$

$\text{same side: } |d_1-d_2|\le 2 < \sqrt{3}+1$

$\text{so opposite sides: } d_1+d_2=\sqrt{3}+1$

$2\cos\dfrac{\pi}{2k}+2\cos\dfrac{\pi}{k}=\sqrt{3}+1$

$\cos\dfrac{\pi}{2k}+\cos\dfrac{\pi}{k}=\dfrac{\sqrt{3}+1}{2}$

$k\ge 2\ \left(\text{since } \dfrac{2\pi}{k}\le \pi\right)$

$\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2},\qquad \cos\dfrac{\pi}{3}=\dfrac{1}{2}$

$\cos\dfrac{\pi}{6}+\cos\dfrac{\pi}{3}=\dfrac{\sqrt{3}+1}{2}$

$f(k)=\cos\dfrac{\pi}{2k}+\cos\dfrac{\pi}{k}$

$\text{strictly increasing for } k\ge 2 \ \Rightarrow\ k=3 \text{ is the unique solution}$

$k=3\ \Rightarrow\ [k]=3$