Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

Let ABCDA B C D be a rectangle such that AB=20A B=20 and AD=24A D=24. Point PP lies inside ABCDA B C D such that triangles PACP A C and PBDP B D have areas 20 and 24, respectively. Compute all possible areas of triangle PABP A B.

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

Solution

There are four possible locations of PP as shown in the diagram. Let OO be the center. Then, [PAO]=10[P A O]=10 and [PBO]=12[P B O]=12. Thus, [PAB]=[AOB]±[PAO]±[PBO]=120±10±12[P A B]=[A O B] \pm[P A O] \pm[P B O]=120 \pm 10 \pm 12, giving the four values 98,118,12298,118,122, and 142.

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