2022 SSMO Speed Round Problems/Problem 5

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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2022 SSMO Speed Round Problems/Problem 5

Problem

Solution