2013 AIME I Problems/Problem 3

Problem 3

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 $\frac{AE}{EB} + \frac{EB}{AE}.$

Solution

It's important to note that $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$

We define $a$ as the length of the side of larger inner square, which is also $EB$ , $b$ as the length of the side of the smaller inner square which is also $AE$ , and $s$ as the side length of $ABCD$ . Since we are given that the sum of the areas of the two squares is $\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \frac{9s^2}{10}$ . The sum of the two nonsquare rectangles can then be represented as $2ab = \frac{s^2}{10}$ .

Looking back at what we need to find, we can represent $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\dfrac{a^2 + b^2}{ab}$ . We have the numerator, and dividing $\frac{s^2}{10}$ by two gives us the denominator $\frac{s^2}{20}$ . Dividing $\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}$ gives us an answer of $\boxed{018}$ .

Solution 2

Let the side of the square be $1$ . Therefore the area of the square is also $1$ . We label $AE$ as $a$ and $EB$ as $b$ . Notice that what we need to find is equivalent to: $\frac{a^2+b^2}{ab}$ . Since the sum of the two squares ( $a^2+b^2$ ) is $\frac{9}{10}$ (as stated in the problem) the area of the whole square, it is clear that the sum of the two rectangles is $1-\frac{9}{10} \implies \frac{1}{10}$ . Since these two rectangles are congruent, they each have area: $\frac{1}{20}$ . Also note that the area of this is $ab$ . Plugging this into our equation we get:

$\frac{\frac{9}{10}}{\frac{1}{20}} \implies \boxed{018}$

Solution 3

Let $AE$ be $x$ , and $EB$ be $1$ . Then we are looking for the value $x+\frac{1}{x}$ . The areas of the smaller squares add up to $9/10$ of the area of the large square, $(x+1)^2$ . Cross multiplying and simplifying we get $x^2-18x+1=0$ . Rearranging, we get $x+\frac{1}{x}=\boxed{018}$

Solution 4 (Vieta)

As before, $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ . Let $x$ represent the value of $AE=CF$ . Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\frac1{10}$ . Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}$ .

Solution 5 (Fast)

Let $AE = x$ and $BE = y$ . From this, we get $AB = x + y$ . The problem is asking for $\frac{x}{y} + \frac{y}{x}$ , which can be rearranged to give $\frac{x^2 + y^2}{xy}$ . The problem tells us that $x^2 + y^2 = \frac{9(x+y)^2}{10}$ . We simplify to get $x^2 + y^2 = 18xy$ . Finally, we divide both sides by $xy$ to get $\frac{x^2 + y^2}{xy} = \boxed{018}$ . - Spacesam

A faster Vieta's

After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \frac 1 {x}.$ Hence $x + \frac 1 {x}$ is just $18$ by Vieta's formula, or $\boxed{018}$

Video Solution

https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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