Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Find the answer

Let $A B C D$ be a convex quadrilateral whose diagonals $A C$ and $B D$ meet at $P$ . Let the area of triangle $A P B$ be 24 and let the area of triangle $C P D$ be 25 . What is the minimum possible area of quadrilateral $A B C D ?$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that $\angle A P B=180^{\circ}-\angle B P C=\angle C P D=180^{\circ}-\angle D P A$ so $4[B P C][D P A]=(P B \cdot P C \cdot \sin B P C)(P D \cdot P A \cdot \sin D P A)=(P A \cdot P B \cdot \sin A P B)(P C \cdot P D \cdot \sin C P D)=4[A P B][C P D]=2400 \Longrightarrow$ $[B P C][D P A]=600$ . Hence by AM-GM we have that $[B P C]+[D P A] \geq 2 \sqrt{[B P C][D P A]}=20 \sqrt{6}$ so the minimum area of quadrilateral $A B C D$ is $49+20 \sqrt{6}$ .

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