Call it Cosine version of Sine Rule?

Geometry Level 3

Suppose in Δ A B C \Delta ABC with sides a , b , c a,b,c , the following equation holds true.

cos ⁡ A a + k 1 = cos ⁡ B b + k 2 = cos ⁡ C c + k 3 = a 2 + b 2 + c 2 8 \dfrac{\cos A}{a} + k_1= \dfrac{\cos B}{b} +k_2 = \dfrac{\cos C}{c}+k_3 = \dfrac{a^2+b^2+c^2}{8}

If a b c = 4 abc=4 , evaluate k 1 k 2 k 3 k_1k_2k_3 .


The answer is 0.25.

Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Nihar Mahajan
Sep 6, 2015

cos ⁡ A a = b 2 + c 2 − a 2 2 a b c = b 2 + c 2 − a 2 8 ⇒ k 1 = a 2 + b 2 + c 2 8 − b 2 + c 2 − a 2 8 = a 2 4 \dfrac{\cos A}{a} = \dfrac{b^2+c^2-a^2}{2abc} = \dfrac{b^2+c^2-a^2}{8} \\ \Rightarrow k_1 = \dfrac{a^2+b^2+c^2}{8} - \dfrac{b^2+c^2-a^2}{8} = \dfrac{a^2}{4}

Similarly we obtain k 2 = b 2 4 , k 3 = c 2 4 k_2 = \dfrac{b^2}{4} \ , \ k_3 = \dfrac{c^2}{4} .

⇒ k 1 k 2 k 3 = a 2 b 2 c 2 64 = 16 64 = 1 4 = 0.25 \Rightarrow k_1k_2k_3 = \dfrac{a^2b^2c^2}{64} = \dfrac{16}{64} = \dfrac{1}{4}=\huge\boxed{0.25}

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cosine law is the best way to solve but is there any other way?

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