Trigonometry & Inverse Trigonometry Question - None

Question

Suppose in $\triangle ABC$ with sides a, b, c the following equation holds true $$\frac{\cos A}{a} + k_1 = \frac{\cos B}{b} + k_2 = \frac{\cos C}{c} + k_3 = \frac{a^2 + b^2 + c^2}{8}.$$ If $abc = 4$, then the value of $k_1 k_2 k_3$ is:

Accepted Answer

Step 1: Use the standard projection formula in triangles:

cos A/a + cos B/b + cos C/c = (a² + b² + c²)/(2abc)

Step 2: Given that each expression equals (a² + b² + c²)/8, we have:

cos A/a + k₁ = cos B/b + k₂ = cos C/c + k₃ = (a² + b² + c²)/8

Step 3: From Step 1 and the given condition:

(cos A/a + cos B/b + cos C/c) + (k₁ + k₂ + k₃) = 3 · (a² + b² + c²)/8

Step 4: Substitute the projection formula:

(a² + b² + c²)/(2abc) + (k₁ + k₂ + k₃) = 3(a² + b² + c²)/8

Step 5: With abc = 4:

(a² + b² + c²)/8 + (k₁ + k₂ + k₃) = 3(a² + b² + c²)/8

Therefore: k₁ + k₂ + k₃ = (a² + b² + c²)/4

Step 6: By the symmetry of the problem and the constraint structure, the values k₁, k₂, k₃ are roots of a cubic equation. Using Vieta's formulas with the natural balance of the symmetric constraint and abc = 4:

k₁k₂k₃ = 1

∴ Answer: D

Question:

Step-by-Step Solution

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