Pressing the Trigger of Algebric Gun

Geometry Level 3

For positive integers b b and c c , let the roots of the polynomial y 2 − b y + c y^2-by+c have roots of sec ⁡ 2 ( x ) \sec^2(x) and csc ⁡ 2 ( x ) \csc^2(x) . Find the minimum value of b + c b+c .


The answer is 8.

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Shubhendra Singh by Shubhendra Singh , 20, India

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3 solutions

Sandeep Rathod
Jan 28, 2015

b = sec ⁡ 2 x + csc ⁡ 2 x = 4 sin ⁡ 2 2 x b = \sec^2x + \csc^2x = \dfrac{4}{\sin^2 2x}

c = sec ⁡ 2 x + csc ⁡ 2 x = 4 sin ⁡ 2 2 x c = \sec^2x + \csc^2x = \dfrac{4}{\sin^2 2x}

b + c = 8 sin ⁡ 2 2 x b + c = \dfrac{8}{\sin^2 2x}

( b + c ) m i n = 8 (b+c)_{min} = 8

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution

Shubhendra Singh - 6 years, 4 months ago

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Aren't integers denoted by Z \mathbb{Z} ?

Omkar Kulkarni - 6 years, 3 months ago
Kanishk Singh
Jan 28, 2015

This equation will be of the form with b=c. Also roots should be real so b's square should be greater than equal to 4b. Minimum possible value of b=4. So b+c=8

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Perfect !!

Shubhendra Singh - 6 years, 4 months ago

Easy solution yet complete one. Upvoted :)

Krishna Ar - 6 years, 4 months ago

Awesome solution...

I did it in a shitty manner...

found the lowest values of sec^2 (x) and cosec ^2 (x) :(

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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