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Geometry Difficulty 4.8 AIME Prove it Belarus

Given a quadrilateral $ABCD$ with $AB + CD = 6$ , $BC + DA = 8$ .
Find the area of $ABCD$ if it has the greatest area among all quadrilaterals with the mentioned sums of the opposite sides.

Solution

Answer: $12$ .
We use the following obvious inequalities. If the quadrilateral $ABCD$ has the sides $AB = a$ , $BC = b$ , $CD = c$ , $DA = d$ , and the area $S$ , then
$S \le \frac{ab + cd}{2} \text{ because}$
$S = S(ABC) + S(ACD) = \\ = \frac{1}{2}ab \sin \beta + \frac{1}{2}cd \sin \delta \le \frac{ab + cd}{2}.$
Similarly, $S \le \frac{ad + bc}{2}$ .
Summing these inequalities, we easily obtain $S \le \frac{(a+c)(b+d)}{4}$ . Hence
$S \le \frac{6 \cdot 8}{4} = 12.$ It is easy to see that equality occurs for a rectangle.

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