2023 SSMO Speed Round Problems/Problem 5

Problem

In a parallelogram $ABCD$ of dimensions $6\times 8,$ a point $P$ is choosen such that $\angle{APD}+\angle{BPC} = 180^{\circ}.$ Find the sum of the maximum, $M$ , and minimum values of $(PA)(PC)+(PB)(PD).$ If you think there is no maximum, let $M=0.$

Solution

A translation that takes $BC$ to $AD$ takes $P$ to $P'.$ Thus, $\angle{AP'D}+\angle{APD} = \angle{BPC}+\angle{APD} = 180^{\circ},$ meaning $PAP'D$ is cyclic. From Ptolemy's Theorem, $(AD)(PP') = (PA)(P'D)+(PD)(P'A) \implies (PA)(PC)+(PB)(PD)=(AD)(CD)=6\cdot8 = 48,$ meaning the answer is $48+48 = \boxed{96}$

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2023 SSMO Speed Round Problems/Problem 5

Problem

Solution