2011 AIME II Problem 2

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Concepts:square (geometry)coordinate geometryPythagorean Theorem

Difficulty rating: 1970

2.

On square ABCD,ABCD, point EE lies on side AD\overline{AD} and point FF lies on side BC,\overline{BC}, so that BE=EF=FD=30.BE = EF = FD = 30. Find the area of square ABCD.ABCD.

Solution:

Let the side length be s,s, and place B=(0,0),B = (0, 0), C=(s,0),C = (s, 0), A=(0,s),A = (0, s), D=(s,s).D = (s, s). Write E=(a,s)E = (a, s) and F=(b,0).F = (b, 0). Then BE2=a2+s2,BE^2 = a^2 + s^2, FD2=(sb)2+s2,FD^2 = (s - b)^2 + s^2, and EF2=(ab)2+s2.EF^2 = (a - b)^2 + s^2.

From BE=FDBE = FD we get a=sb,a = s - b, so ab=2as.a - b = 2a - s. Then EF=BEEF = BE gives (2as)2=a2,(2a - s)^2 = a^2, whose solutions are a=s3a = \frac{s}{3} and a=sa = s (the latter collapses EE and FF onto the corners DD and B,B, making the three segments coincide). So a=s3.a = \frac{s}{3}.

Now 900=BE2=s29+s2=10s29,900 = BE^2 = \frac{s^2}{9} + s^2 = \frac{10s^2}{9}, so the area is s2=910900=810.s^2 = \frac{9}{10} \cdot 900 = 810.

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