Where do I look at this from? (2)

Geometry Level 5

Consider a circle S 1 S_{1} ,

x 2 + y 2 + 1 2 x ( cos θ sin θ ) 2 y ( cos θ + sin θ ) = 0 x^{2}+y^{2}+1-2x(\cos\theta - \sin\theta)-2y(\cos\theta + \sin\theta)=0

Now, consider another circle S 2 S_{2} , centred at P = ( cos θ sin θ , cos θ sin θ ) P=(-\cos\theta - \sin\theta, \cos\theta - \sin\theta) such that S 1 S_{1} internally touches S 2 S_{2} .

Draw a pair of tangents T 1 T_{1} and T 2 T_{2} from P P to S 1 S_{1} . Let these tangents meet the circle S 2 S_{2} at points A A and B B as shown. From a point R R on S 2 S_{2} , draw chords R A RA and R B RB with lengths l 1 , l 2 l_{1} , l_{2} , respectively to S 2 S_{2} .

Then, there exists a certain α \alpha such that one of

cos 1 l 1 6 + cos 1 l 2 6 or cos 1 l 1 6 cos 1 l 2 6 \large \left| \cos^{-1}\frac{l_{1}}{6} + \cos^{-1}\frac{l_{2}}{6}\right| \quad \text{ or } \quad \left|\cos^{-1}\frac{l_{1}}{6}-\cos^{-1}\frac{l_{2}}{6} \right|

is equal to α \alpha regardless of the value of R R . Find the value of this constant α \alpha .


If you are looking for more such simple but twisted questions, twisted problems for jee aspirants is for you!
π 6 \frac{\pi}{6} π 5 \frac{\pi}{5} π \pi π 4 \frac{\pi}{4} π 2 \frac{\pi}{2} π 3 \frac{\pi}{3}

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1 solution

Rohith M.Athreya
Jan 13, 2017

refer to first part of solution here

note that angle APR is π 2 α \pi-2\alpha and angle BPR is 2 π 3 + 2 α \frac{2\pi}{3}+2\alpha

radius of S 2 S_{2} is 3.

use cosine rule to find l 1 l_{1} and l 2 l_{2}

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