2006 SMT/Geometry Problems/Problem 2

Problem

Given square $ABCD$ of side length $1$ , with $E$ on $\overline{CD}$ and $F$ in the interior of the square so that $\overline{EF}\perp\overline{CD}$ and $\overline{AF}\cong\overline{BF}\cong\overline{EF}$ , find the area of quadrilateral $ADEF$ .

Solution

$[asy] draw(unitsquare); draw((1/2,0)--(1/2,5/8)--(0,1)); draw((1/2,5/8)--(1,1)); draw((0,5/8)--(1/2,5/8)); label("$A$",(0,1),NW); label("$B$",(1,1),NE); label("$C$",(1,0),SE); label("$D$",(0,0),SW); label("$F$",(1/2,5/8),SE); label("$Q$",(0,5/8),W); label("$x$",(1/2,5/16),E); label("$x$",(0,5/16),W); label("$1-x$",(0,13/16),W); label("$x$",(1/4,13/16),NE); label("$\frac{1}{2}$",(1/4,0),S); label("$\frac{1}{2}$",(1/4,5/8),S); label("$x$",(3/4,13/16),NW); label("$E$",(1/2,0),S); [/asy]$

Since $AF=BF$ , $F$ lies on the perpendicular bisector of $AB$ , which is the same as the perpendicular bisector of $CD$ . Therefore, since $E$ is the foot of the perpendicular from $F$ to $CD$ , $E$ is the midpoint of $CD$ , and $ED=\frac{1}{2}$ .

Let $Q$ be the foot of the perpendicular from $F$ to $DA$ . Since $DEFQ$ is a rectangle, $FQ=ED=\frac{1}{2}$ . Let $AF=BF=EF=x$ . We have $DQ=EF=x$ , and $AQ=1-x$ . Therefore, from right triangle $AQF$ , $(1-x)^2+\left(\frac{1}{2}\right)^2=x^2$ , and solving this gives $x=\frac{5}{8}$ .

Finally, $ADEF$ is a trapezoid with bases of length $1$ and $x=\frac{5}{8}$ and height $\frac{1}{2}$ , so its area is $\left(\frac{1}{2}\right)\left(1+\frac{5}{8}\right)\left(\frac{1}{2}\right)=\boxed{\frac{13}{32}}$ .

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2006 SMT/Geometry Problems/Problem 2

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Solution

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