2011 AIME II Problems/Problem 2

Problem:

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: (Needs better solution, I cannot remember exactly how I got the side length)

Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (see geoboard diagram [that thing is actually pretty cool]) 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 that x=9sqrt(10), and $x^2$ =810

Area of the square is 810.

<geogebra>bb56f64482bd0dabba65a4dcad9d52a05d440f19</geogebra>

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2011 AIME II Problems/Problem 2