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Bulgarian Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Bulgaria

Problem:
Let $ABCD$ be a parallelogram such that $\text{BAD < 90}$ and let $DE$ , $E \in AB$ , and $DF$ , $F \in BC$ , be the altitudes of the parallelogram. Prove that
$4(AB \cdot BC \cdot EF + BD \cdot AE \cdot FC) \leq 5 \cdot AB \cdot BC \cdot BD$
Find $\text{BAD}$ if the equality occurs.

Solution

Solution:
Set $\text{BAD =}$ , $AB = CD = a$ , $AD = BC = b$ and $BD = d$ . We have $DE = b \sin \alpha$ , $AE = b \cos \alpha$ , $DF = a \sin \alpha$ and $CF = a \cos \alpha$ . Therefore $\triangle DEF \sim \triangle ADB$ and thus $EF = d \sin \alpha$ .

Plugging the above expressions in the given

inequality we see that it is equivalent to
$4 \sin \alpha + 4 - 4 \sin^2 \alpha \leq 5$
i.e. $(2 \sin \alpha - 1)^2 \geq 0$ , which is true for any value of $\alpha$ . The equality holds when $\alpha = 30^\circ$ .

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