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

(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==
 
==Problem==
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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==
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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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WLOG, assume <math>ABCD</math> is a unit square and let <math>AE=DF=a.</math> So, <cmath>\frac{[BEF]}{[ABCD]} = \frac{7}{18}\implies[BEF] = \frac{7}{18}.</cmath> Note that
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<cmath>\begin{align*}
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[BEF] &= [ABCD]-[AEB]-[DEF]-[BFC]\\
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&= 1-\frac{AE\cdot AB}{2}-\frac{DE\cdot DF}{2}-\frac{FC\cdot BC}{2}\\
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&= 1-\frac{a}{2}-\frac{(1-a)(a)}{2}-\frac{1-a}{2}\\
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&= \frac{a^2-a+1}{2}\implies\\
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\frac{a^2-a+1}{2}&=\frac{7}{18}\implies\\
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a^2-a &= -\frac{2}{9}.
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\end{align*}</cmath>
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Now,  
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<cmath>\begin{align*}
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EF^2 &= DE^2+DF^2\\
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&= (1-a)^2+a^2\\
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&= 2a^2-2a+1\\
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&= -2\left(\frac{2}{9}\right)+1\\
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&= \frac{5}{9}\implies\\
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\frac{EF^2}{BC^2} &= EF^2 = \frac{5}{9}\implies5+9 = \boxed{14}.
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\end{align*}</cmath>
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~pinkpig

Latest revision as of 18:17, 13 September 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

WLOG, assume $ABCD$ is a unit square and let $AE=DF=a.$ So, \[\frac{[BEF]}{[ABCD]} = \frac{7}{18}\implies[BEF] = \frac{7}{18}.\] Note that \begin{align*} [BEF] &= [ABCD]-[AEB]-[DEF]-[BFC]\\ &= 1-\frac{AE\cdot AB}{2}-\frac{DE\cdot DF}{2}-\frac{FC\cdot BC}{2}\\ &= 1-\frac{a}{2}-\frac{(1-a)(a)}{2}-\frac{1-a}{2}\\ &= \frac{a^2-a+1}{2}\implies\\ \frac{a^2-a+1}{2}&=\frac{7}{18}\implies\\ a^2-a &= -\frac{2}{9}. \end{align*} Now, \begin{align*} EF^2 &= DE^2+DF^2\\ &= (1-a)^2+a^2\\ &= 2a^2-2a+1\\ &= -2\left(\frac{2}{9}\right)+1\\ &= \frac{5}{9}\implies\\ \frac{EF^2}{BC^2} &= EF^2 = \frac{5}{9}\implies5+9 = \boxed{14}. \end{align*}

~pinkpig

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