2022 SSMO Speed Round Problems/Problem 5: Difference between revisions

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-==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}.</math> Find the sum of the maximum, <math>M</math>, and minimum values of <math>(PA)(PC)+(PB)(PD).</math> If you think there is no maximum, let <math>M=0.</math>
-==Solution==
-A translation that takes <math>BC</math> to <math>AD</math> takes <math>P</math> to <math>P'.</math> Thus, <math>\angle{AP'D}+\angle{APD} = \angle{BPC}+\angle{APD} = 180^{\circ},</math> meaning <math>PAP'D</math> is cyclic. From Ptolemy's Theorem, <math>(AD)(PP') = (PA)(P'D)+(PD)(P'A) \implies (PA)(PC)+(PB)(PD)=(AD)(CD)=6\cdot8 = 48,</math> meaning the answer is <math>48+48 = \boxed{96}</math>

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