Difference between revisions of "2022 SSMO Speed Round Problems/Problem 5"

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==Problem==
 
==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>
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Let <math>ABCD</math> be a square such that <math>E</math> is on <math>AD</math> and <math>F</math> is on <math>CD.</math> If <math>AE=DF</math> and <math>\frac{[BEF]}{[ABCD]}=\frac{7}{18},</math> then the value of <math>\frac{EF^2}{BC^2}</math> can be expressed as <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==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>
 

Latest revision as of 19:14, 2 May 2025

Problem

Let $ABCD$ be a square such that $E$ is on $AD$ and $F$ is on $CD.$ If $AE=DF$ and $\frac{[BEF]}{[ABCD]}=\frac{7}{18},$ then the value of $\frac{EF^2}{BC^2}$ can be expressed as $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution