Extended Sines of law!

Geometry Level 5

In A B C \triangle ABC , a a is the arithmetic mean and b = c b=c is the geometric mean of two positive numbers, then find minimum value of

4 4 ( sin 12 B + sin 12 C ( sin 2 A + sin 2 B + sin 2 C ) 4 ) + 1 \large\ { 4 }^{ 4 }\left( \frac { \sin ^{ 12 }{ B } + \sin ^{ 12 }{ C } }{ { ( \sin { 2A } + \sin { 2B } + \sin { 2C }) }^{ 4 } } \right)+1

If angles ( A , B , C ) (A,B,C) are opposites to side ( a , b , c ) (a,b,c)


The answer is 3.

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1 solution

Aareyan Manzoor
Mar 1, 2018

since the traingle is an issoceles, it has angles ( A , π 2 A 2 , π 2 A 2 ) \left(A,\dfrac{\pi}{2}-\dfrac{A}{2},\dfrac{\pi}{2}-\dfrac{A}{2}\right) . from the constraints of the problem: a = k 1 + k 2 2 , b = c = k 1 k 2 a=\dfrac{k_1+k_2}{2}, b=c=\sqrt{k_1k_2} . we use sin ( B ) = sin ( π 2 A 2 ) = cos ( A / 2 ) \sin(B)=\sin\left(\dfrac{\pi}{2}-\dfrac{A}{2}\right) =\cos(A/2) and same for C, and sin ( 2 B ) = sin ( π A ) = sin ( A ) \sin(2B)=\sin(\pi-A)=\sin(A) to reduce the expression to 4 4 2 cos 12 ( A / 2 ) ( 2 sin ( A ) cos ( A ) + 2 sin ( A ) ) 4 + 1 = 2 cos 12 ( A / 2 ) sin 4 ( A ) ( cos ( A ) + 1 2 ) 4 + 1 = 2 cos 12 ( A / 2 ) sin 4 ( A ) cos 8 ( A / 2 ) + 1 2 cos 4 ( A / 2 ) ( 2 sin ( A / 2 ) cos ( A / 2 ) ) 4 + 1 = 1 8 sin 4 ( A / 2 ) + 1 4^4 \dfrac{2\cos^{12} (A/2)}{(2\sin(A)\cos(A)+2\sin(A))^4}+1=\dfrac{2\cos^{12} (A/2)}{\sin^4(A)(\frac{\cos(A)+1}{2})^4}+1=\dfrac{2\cos^{12} (A/2)}{\sin^4(A)\cos^8(A/2)}+1\\ \dfrac{2\cos^4(A/2)}{(2\sin(A/2)\cos(A/2))^4}+1=\dfrac{1}{8 \sin^4(A/2)}+1 various double and half angle identity was used above. by laws of cosine, for an isosceles triangle, we know 2 b sin ( A / 2 ) = a sin ( A / 2 ) = a 2 b = k 1 + k 2 4 k 1 k 2 2b\sin(A/2)=a \to \sin(A/2)=\dfrac{a}{2b}=\dfrac{k_1+k_2}{4\sqrt{k_1k_2}} plugging this back in : 1 8 sin 4 ( A / 2 ) + 1 = 32 k 1 2 k 2 2 ( k 1 + k 2 ) 4 + 1 = 32 ( k 1 / k 2 ) 2 ( 1 + k 1 / k 2 ) 4 + 1 \dfrac{1}{8 \sin^4(A/2)}+1= 32\dfrac{k_1^2k_2^2}{(k_1+k_2)^4}+1= 32\dfrac{(k_1/k_2)^2}{(1+k_1/k_2)^4}+1 we can minimize the function by minimizing 32 r 2 ( 1 + r ) 4 + 1 d d r r 2 ( 1 + r ) 4 = 0 2 r ( 1 + r ) 4 4 r 2 ( 1 + r ) 5 = 0 2 r ( 1 + r ) 4 r 2 = 2 r 2 r 2 = 0 r = 1 , 0 32\dfrac{r^2}{(1+r)^4}+1\to \dfrac{d}{dr} \dfrac{r^2}{(1+r)^4}=0\to \dfrac{2r}{(1+r)^4}-\dfrac{4r^2}{(1+r)^5}=0\to 2r(1+r)-4r^2=2r-2r^2=0\to r= 1,0 its obvious that 0 is not the answer, hence using one we have the expression equal 3 \boxed{3} this is reached if k 1 = k 2 k_1=k_2

Beautifully explained

A Former Brilliant Member - 3 years, 1 month ago

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