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Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let $ABCD$ be a convex quadrilateral whose diagonals $AC$ and $BD$ meet at $P$ . Let the area of triangle $APB$ be $24$ and let the area of triangle $CPD$ be $25$ . What is the minimum possible area of quadrilateral $ABCD$ ?

Solution

Solution:

Note that $\angle APB = 180^{\circ} - \angle BPC = \angle CPD = 180^{\circ} - \angle DPA$ so $4[ BPC ][ DPA ] = (PB \cdot PC \cdot \sin BPC)(PD \cdot PA \cdot \sin DPA) = (PA \cdot PB \cdot \sin APB)(PC \cdot PD \cdot \sin CPD) = 4[ APB ][ CPD ] = 2400 \Longrightarrow [ BPC ][ DPA ] = 600$ . Hence by AM-GM we have that
$[ BPC ] + [ DPA ] \geq 2 \sqrt{ [ BPC ][ DPA ] } = 20 \sqrt{6}$
so the minimum area of quadrilateral $ABCD$ is $49 + 20 \sqrt{6}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.