2011 AIME II Problems/Problem 2

Problem 2

On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD.

Solution

Drawing the square and examining the given lengths, $[asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("$A$", A, plain.SW); pair B = (3, 0); dot(B); label("$B$", B, plain.SE); pair C = (3, 3); dot(C); label("$C$", C, plain.NE); pair D = (0, 3); dot(D); label("$D$", D, plain.NW); pair E = (0, 1); dot(E); label("$E$", E, plain.W); pair F = (3, 2); dot(F); label("$F$", F, plain.E); label("$\frac x3$", E--A); label("$\frac x3$", F--C); label("$x$", A--B); label("$x$", C--D); label("$\frac {2x}3$", B--F); label("$\frac {2x}3$", D--E); label("$30$", B--E); label("$30$", F--E); label("$30$", F--D); draw(B--C--D--F--E--B--A--D); [/asy]$ you find that the three segments cut the square into three equal horizontal sections. Therefore, ( $x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$ , or $x^2+(x/3)^2=900$ . Solving for $x$ , we get $x=9\sqrt{10}$ , and $x^2=810.$

Area of the square is $\fbox{810.}$

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2011 AIME II Problems/Problem 2

Problem 2

Solution