JEE Challenger
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MathMedAdv 2010 • Paper 2

Find Floor Value of Angle Parameter for Parallel Chords in Circle

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

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

Official Numerical Answer3

Topics & Concepts

Conic SectionsCircles

Step-by-Step Solution

Let the radius of the circle be R=2R = 2.

The perpendicular distances of the two parallel chords from the center of the circle are given by:
d1=Rcos⁡(π/k2)=2cos⁡(π2k)d_1 = R \cos\left(\frac{\pi/k}{2}\right) = 2 \cos\left(\frac{\pi}{2k}\right)
d2=Rcos⁡(2π/k2)=2cos⁡(πk)d_2 = R \cos\left(\frac{2\pi/k}{2}\right) = 2 \cos\left(\frac{\pi}{k}\right)

We consider two possible configurations for the parallel chords relative to the center of the circle:

Case 1: The chords are on opposite sides of the center

In this case, the distance between the chords is d1+d2=3+1d_1 + d_2 = \sqrt{3} + 1.
2cos⁡(π2k)+2cos⁡(πk)=3+12 \cos\left(\frac{\pi}{2k}\right) + 2 \cos\left(\frac{\pi}{k}\right) = \sqrt{3} + 1
cos⁡(π2k)+cos⁡(πk)=3+12\cos\left(\frac{\pi}{2k}\right) + \cos\left(\frac{\pi}{k}\right) = \frac{\sqrt{3} + 1}{2}

Let x=π2kx = \frac{\pi}{2k}. Then πk=2x\frac{\pi}{k} = 2x. Substituting this into the equation gives:
cos⁡(x)+cos⁡(2x)=3+12\cos(x) + \cos(2x) = \frac{\sqrt{3} + 1}{2}

Using the double-angle identity cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1:
cos⁡(x)+2cos⁡2(x)−1=3+12\cos(x) + 2\cos^2(x) - 1 = \frac{\sqrt{3} + 1}{2}
2cos⁡2(x)+cos⁡(x)−(1+3+12)=02\cos^2(x) + \cos(x) - \left(1 + \frac{\sqrt{3} + 1}{2}\right) = 0
2cos⁡2(x)+cos⁡(x)−3+32=02\cos^2(x) + \cos(x) - \frac{3 + \sqrt{3}}{2} = 0
4cos⁡2(x)+2cos⁡(x)−(3+3)=04\cos^2(x) + 2\cos(x) - (3 + \sqrt{3}) = 0

Using the quadratic formula to solve for cos⁡(x)\cos(x):
cos⁡(x)=−2±4−4(4)(−(3+3))2(4)\cos(x) = \frac{-2 \pm \sqrt{4 - 4(4)(-(3 + \sqrt{3}))}}{2(4)}
cos⁡(x)=−2±4+16(3+3)8=−2±52+1638\cos(x) = \frac{-2 \pm \sqrt{4 + 16(3 + \sqrt{3})}}{8} = \frac{-2 \pm \sqrt{52 + 16\sqrt{3}}}{8}

Simplifying the radical term:
52+163=(2+43)2=2+43\sqrt{52 + 16\sqrt{3}} = \sqrt{(2 + 4\sqrt{3})^2} = 2 + 4\sqrt{3}

Since cos⁡(x)>0\cos(x) > 0 for acute angles:
cos⁡(x)=−2+(2+43)8=438=32\cos(x) = \frac{-2 + (2 + 4\sqrt{3})}{8} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}

Thus,
x=π6x = \frac{\pi}{6}

Since x=π2kx = \frac{\pi}{2k}:
π2k=π6  ⟹  2k=6  ⟹  k=3\frac{\pi}{2k} = \frac{\pi}{6} \implies 2k = 6 \implies k = 3

Case 2: The chords are on the same side of the center

In this case, the distance between the chords is ∣d1−d2∣=3+1|d_1 - d_2| = \sqrt{3} + 1. Assuming d1>d2d_1 > d_2:
2cos⁡(x)−2cos⁡(2x)=3+12\cos(x) - 2\cos(2x) = \sqrt{3} + 1
cos⁡(x)−(2cos⁡2(x)−1)=3+12\cos(x) - (2\cos^2(x) - 1) = \frac{\sqrt{3} + 1}{2}
4cos⁡2(x)−2cos⁡(x)+(3−1)=04\cos^2(x) - 2\cos(x) + (\sqrt{3} - 1) = 0

Checking the discriminant DD for this quadratic equation:
D=(−2)2−4(4)(3−1)=4−163+16=20−163D = (-2)^2 - 4(4)(\sqrt{3} - 1) = 4 - 16\sqrt{3} + 16 = 20 - 16\sqrt{3}

Since 163≈27.71>2016\sqrt{3} \approx 27.71 > 20, D<0D < 0, which means there are no real solutions for xx in this case.

Therefore, the unique value of kk is k=3k = 3.

The value of [k][k] is:
[k]=[3]=3[k] = [3] = 3

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