SAT1000 - P370

Geometry Level pending

Given that in △ A B C \triangle ABC , sin ⁡ 2 A + sin ⁡ ( A − B + C ) = sin ⁡ ( C − A − B ) + 1 2 \sin 2A+\sin(A-B+C)=\sin(C-A-B)+\dfrac{1}{2} , S S is the area of △ A B C \triangle ABC , 1 ≤ S ≤ 2 1 \leq S \leq 2 , let a , b , c a,b,c be the opposite side of angle A , B , C A,B,C respectively.

Which inequality always holds?

A . b c ( b + c ) > 8 A.\ bc(b+c)>8

B . a b ( a + b ) > 16 2 B.\ ab(a+b)>16 \sqrt{2}

C . 6 ≤ a b c ≤ 12 C.\ 6 \leq abc \leq 12

D . 12 ≤ a b c ≤ 24 D.\ 12 \leq abc \leq 24

Have a look at my problem set: SAT 1000 problems

B B A A D D C C

Alice Smith by Alice Smith , 20, China

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1 solution

The given equation is equivalent to sin ⁡ 2 A + sin ⁡ 2 B + sin ⁡ 2 C = 1 2 \sin 2A+\sin 2B+\sin 2C=\dfrac{1}{2}

⟹ sin ⁡ A sin ⁡ B sin ⁡ C = 1 8 \implies \sin A\sin B\sin C=\dfrac {1}{8}

Let the circumradius of the triangle be R R . Then 1 ≤ S ≤ 2 ⟹ 2 ≤ R ≤ 2 2 ⟹ 8 ≤ a b c ≤ 16 2 1\leq S\leq 2\implies 2\leq R\leq 2\sqrt 2\implies 8\leq abc \leq 16\sqrt 2

b + c > a ⟹ b c ( b + c ) > a b c ⟹ b c ( b + c ) > 8 b+c>a\implies bc(b+c)>abc\implies \boxed {bc(b+c)>8} .

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