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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 5"
 
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==Solution==
 
==Solution==
 +
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
 +
<cmath>\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*}</cmath>
 +
Now,
 +
<cmath>\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*}</cmath>
 +
 +
~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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