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Croatian Mathematical Society Competitions

Geometry Difficulty 5.1 AIME, harder Prove it Croatia

Let $ABC$ be a triangle such that $|AB| = 4$ , $|BC| = 7$ , $|CA| = 5$ , and let $\alpha = \angle BAC$ . Determine
$sin^6 \frac{\alpha}{2} + \cos^6 \frac{\alpha}{2}.$

Solution

By applying the cosine theorem we get
$\cos \alpha = \frac{|CA|^2 + |AB|^2 - |BC|^2}{2|CA| \cdot |AB|} = \frac{25 + 16 - 49}{2 \cdot 5 \cdot 4} = -\frac{1}{5},$
and therefore

$\begin{align*} \sin^6 \frac{\alpha}{2} + \cos^6 \frac{\alpha}{2} &= \left(\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}\right) \left(\sin^4 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} + \cos^4 \frac{\alpha}{2}\right) \\ &= \left(\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}\right)^2 - 3 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} \\ &= 1 - \frac{3}{4} \sin^2 \alpha \\ &= \frac{1}{4} + \frac{3}{4} \cos^2 \alpha \\ &= \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{25} = \frac{7}{25}. \end{align*}$

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