2013 AIME I Problem 3

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3.

Let ABCDABCD be a square, and let EE and FF be points on AB\overline{AB} and BC,\overline{BC}, respectively. The line through EE parallel to BC\overline{BC} and the line through FF parallel to AB\overline{AB} divide ABCDABCD into two squares and two nonsquare rectangles. The sum of the areas of the two squares is 910\frac{9}{10} of the area of square ABCD.ABCD. Find AEEB+EBAE.\frac{AE}{EB} + \frac{EB}{AE}.

Answer: 18
Concepts:algebraic manipulationsquare (geometry)ratio and proportion
Difficulty rating: 2020
Solution:

Let AE=xAE = x and EB=y,EB = y, so the square has side x+yx + y and the two smaller squares have sides xx and y.y. The condition says x2+y2=910(x+y)2.x^2 + y^2 = \frac{9}{10}(x + y)^2. Multiplying by 1010 and expanding, 10x2+10y210x^2 + 10y^2 =9x2+18xy+9y2,= 9x^2 + 18xy + 9y^2, so x2+y2=18xy.x^2 + y^2 = 18xy.

Dividing by xyxy gives AEEB+EBAE=xy+yx=x2+y2xy=18. \begin{aligned} \frac{AE}{EB} + \frac{EB}{AE} &= \frac{x}{y} + \frac{y}{x} \\ &= \frac{x^2 + y^2}{xy} = 18. \end{aligned}

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