Question

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

答案

Solution:

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

Dividing by $xy$ gives $\frac{AE}{EB} + \frac{EB}{AE} = \frac{x}{y} + \frac{y}{x} = \frac{x^2 + y^2}{xy} = 18.$

2013 AIME I Problem 3

Solution:

Problem 3 in Other Years