Trigonometry?

Geometry Level 4

cos ⁡ A sin ⁡ 2 ( A 2 ) + cos ⁡ B sin ⁡ 2 ( B 2 ) + cos ⁡ C sin ⁡ 2 ( C 2 ) \cos A \sin^2 \left( \frac A2 \right) + \cos B \sin^2 \left( \frac B2 \right) + \cos C \sin^2 \left( \frac C2 \right)

If A A , B B , and C C are the angles of a triangle, then the maximum value of the expression above can be written as P Q \dfrac PQ , where P P and Q Q are coprime positive integers , find Q − P Q-P .

Bonus: For what type of triangle will the value of this expression be maximum? Prove your result.


The answer is 5.

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Pranav Saxena by Pranav Saxena , 19, India

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1 solution

Simply the expression as follows:

P Q = ∑ A , B , C cos ⁡ A sin ⁡ 2 ( A 2 ) = ∑ A , B , C cos ⁡ A ⋅ 1 2 ( 1 − cos ⁡ A ) = 1 2 ∑ A , B , C cos ⁡ A − cos ⁡ 2 A = 1 2 ∑ A , B , C − ( cos ⁡ 2 A − cos ⁡ A + 1 4 − 1 4 ) = 1 2 ∑ A , B , C − ( ( cos ⁡ A − 1 2 ) 2 − 1 4 ) = 1 2 ∑ A , B , C ( 1 4 − ( cos ⁡ A − 1 2 ) 2 ) Note that ( cos ⁡ A − 1 2 ) 2 ≥ 0 ≤ 1 2 ∑ A , B , C 1 4 = 3 8 \begin{aligned} \frac PQ & = \sum_{A,B,C} \cos A \sin^2 \left(\frac A2 \right) \\ & = \sum_{A,B,C} \cos A \cdot \frac 12 (1 - \cos A) \\ & = \frac 12 \sum_{A,B,C} \cos A - \cos^2 A \\ & = \frac 12 \sum_{A,B,C} - \left( \cos^2 A - \cos A + \frac 14 - \frac 14 \right) \\ & = \frac 12 \sum_{A,B,C} - \left( \left(\cos A - \frac 12\right)^2 - \frac 14 \right) \\ & = \frac 12 \sum_{A,B,C} \left(\frac 14 - {\color{#3D99F6} \left(\cos A - \frac 12\right)^2} \right) & \small \color{#3D99F6} \text{Note that } \left(\cos A - \frac 12\right)^2 \ge 0 \\ & \le \frac 12 \sum_{A,B,C} \frac 14 = \frac 38 \end{aligned}

⟹ Q − P = 8 − 3 = 5 \implies Q-P = 8-3 = \boxed{5} .

Bonus: Maximum occurs, when cos ⁡ A − 1 2 = 0 ⟹ A = 6 0 ∘ \cos A - \frac 12 = 0 \implies A = 60^\circ and A = B = C = 6 0 ∘ A=B=C=60^\circ , an equilateral triangle.

8 Helpful 1 Interesting 1 Brilliant 0 Confused

@Chew-Seong Cheong As always, a good solution sir!! +1. However, you forgot to put the square in the 5th line from the top. Any1 did it like me by using Jensen's Inequailty?

Pranav Saxena - 4 years, 5 months ago

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Thanks. Why don't you show your solution?

Chew-Seong Cheong - 4 years, 5 months ago

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