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2022 SSMO Speed Round Problems/Problem 5

Revision as of 13:59, 3 July 2023 by Pinkpig (talk | contribs) (Created page with "==Problem== In a parallelogram <math>ABCD</math> of dimensions <math>6\times 8,</math> a point <math>P</math> is choosen such that <math>\angle{APD}+\angle{BPC} = 180^{\circ}....")
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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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