Weekly Challenge #1

Geometry Level 2

Let 0 ∘ < θ < 4 5 ∘ 0 ^ \circ < \theta < 45 ^ \circ . Then arrange -

t 1 t_1 = ( t a n θ ) t a n θ (tan \theta) ^ {tan \theta} , t 2 t_2 = ( t a n θ ) c o t θ (tan \theta) ^ {cot \theta} ,

t 3 t_3 = ( c o t θ ) t a n θ (cot \theta) ^ {tan \theta} , t 4 t_4 = ( c o t θ ) c o t θ (cot \theta) ^ {cot \theta}

in decreasing order.

t 2 t_2 > t 1 t_1 > t 3 t_3 > t 4 t_4 t 2 t_2 > t 3 t_3 > t 4 t_4 > t 1 t_1 t 4 t_4 > t 3 t_3 > t 1 t_1 > t 2 t_2 t 4 t_4 > t 1 t_1 > t 3 t_3 > t 2 t_2

Ameya Salankar by Ameya Salankar , 23, India

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2 solutions

Saubhagya Raizada
Mar 13, 2014

If you take theta as 30 degrees, you get tan(theta) as 1/sqrt(3) and cot(theta) as sqrt(3). Plug in the values, compare and solve!

CHEERS!

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Ameya Salankar
Mar 12, 2014

For a > 1 a > 1 , the function y = a x y = a^x is an increasing function. For 0 ∘ < θ < 4 5 ∘ 0^\circ < \theta < 45^\circ , cot ⁡ θ > 1 > tan ⁡ θ > 0 \cot \theta >1 > \tan \theta > 0 . Thus t 3 < t 4 t_3<t_4 .

For a < 1 a < 1 , the function y = a x y = a^x is an decreasing function. Thus t 1 > t 2 t_1>t_2 .

Again, by cot ⁡ θ > 1 > tan ⁡ θ > 0 \cot \theta >1 > \tan \theta > 0 , we have t 1 < 1 < t 3 t_1<1<t_3 .

Hence, t 4 > t 3 > t 1 > t 2 t_4>t_3>t_1>t_2 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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