Minimum value

Geometry Level 4

In a Δ A B C \Delta ABC , find the minimum value of ∑ ( cot ⁡ A 2 ) 2 ( cot ⁡ B 2 ) 2 ∏ ( cot ⁡ C 2 ) 2 \displaystyle \frac { \sum { { \left( \cot { \frac { A }{ 2 } } \right) }^{ 2 } } { \left( \cot { \frac { B }{ 2 } } \right) }^{ 2 } }{ { \prod { \left( \cot { \frac { C }{ 2 } } \right) } }^{ 2 } }


The answer is 1.

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Utkarsh Bansal by Utkarsh Bansal , 23, India

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1 solution

Utkarsh Bansal
Feb 26, 2015

1 cot ⁡ 2 A 2 + 1 cot ⁡ 2 B 2 + 1 cot ⁡ 2 C 2 = tan ⁡ 2 A 2 + tan ⁡ 2 B 2 + tan ⁡ 2 C 2 = ( s − b ) ( s − c ) s ( s − a ) + ( s − a ) ( s − c ) s ( s − b ) + ( s − a ) ( s − b ) s ( s − c ) M i n i m u m v a l u e o c c u r s a t a = b = c 3 × ( 3 a 2 − a ) ( 3 a 2 − a ) 3 a 2 ( 3 a 2 − a ) = a 2 × a 2 a 2 × a 2 = 1 \frac { 1 }{ \cot ^{ 2 }{ \frac { A }{ 2 } } } +\frac { 1 }{ \cot ^{ 2 }{ \frac { B }{ 2 } } } +\frac { 1 }{ \cot ^{ 2 }{ \frac { C }{ 2 } } } =\tan ^{ 2 }{ \frac { A }{ 2 } } +\tan ^{ 2 }{ \frac { B }{ 2 } } +\tan ^{ 2 }{ \frac { C }{ 2 } } \\ =\frac { \left( s-b \right) \left( s-c \right) }{ s\left( s-a \right) } +\frac { \left( s-a \right) \left( s-c \right) }{ s\left( s-b \right) } +\frac { \left( s-a \right) \left( s-b \right) }{ s\left( s-c \right) } \\ Minimum\quad value\quad occurs\quad at\quad a=b=c\\ 3\quad \times \quad \frac { \left( \frac { 3a }{ 2 } -a \right) \left( \frac { 3a }{ 2 } -a \right) }{ \frac { 3a }{ 2 } \left( \frac { 3a }{ 2 } -a \right) } =\frac { \frac { a }{ 2 } \times \frac { a }{ 2 } }{ \frac { a }{ 2 } \times \frac { a }{ 2 } } =\boxed { 1 }

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You must prove the equality although its quite evident.Set x = t a n ( A / 2 ) , y = t a n ( B / 2 ) , z = t a n ( C / 2 ) x=tan(A/2),y=tan(B/2),z=tan(C/2) .So its easy to show that x y + y z + z x = 1 xy+yz+zx=1 for A , B , C A,B,C form a triangle.So we are left to minimise the function x 2 + y 2 + z 2 x^2+y^2+z^2 .And we use the trivial inequality of x 2 + y 2 + z 2 > = x y + y z + z x x^2+y^2+z^2>=xy+yz+zx to get the minimum value as 1.Note that x , y , z x,y,z are positive.

Spandan Senapati - 4 years, 1 month ago

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Yeah! True.... legend speaking with other legend ,what am I doing here. I am a faliure

Md Zuhair - 3 years, 4 months ago

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