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Difference between revisions of "2013 AIME I Problems/Problem 3"

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== Solution 5 (Fast) ==
 
== Solution 5 (Fast) ==
 
Let <math>AE = x</math> and <math>BE = y</math>. From this, we get <math>AB = x + y</math>. The problem is asking for <math>\frac{x}{y} + \frac{y}{x}</math>, which can be rearranged to give <math>\frac{x^2 + y^2}{xy}</math>. The problem tells us that <math>x^2 + y^2 = \frac{9(x+y)^2}{10}</math>. We simplify to get <math>x^2 + y^2 = 18xy</math>. Finally, we divide both sides by <math>xy</math> to get <math>\frac{x^2 + y^2}{xy} = \boxed{018}</math>. - Spacesam
 
Let <math>AE = x</math> and <math>BE = y</math>. From this, we get <math>AB = x + y</math>. The problem is asking for <math>\frac{x}{y} + \frac{y}{x}</math>, which can be rearranged to give <math>\frac{x^2 + y^2}{xy}</math>. The problem tells us that <math>x^2 + y^2 = \frac{9(x+y)^2}{10}</math>. We simplify to get <math>x^2 + y^2 = 18xy</math>. Finally, we divide both sides by <math>xy</math> to get <math>\frac{x^2 + y^2}{xy} = \boxed{018}</math>. - Spacesam
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 +
== A faster Vieta's ==
 +
 +
After we get the polynomial <math>x^2 - 18x + 1,</math> we want to find <math>x + \frax 1 {x}.</math> Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply <math>x, \frac 1 {x}.</math> Hence <math>x + \frax 1 {x}</math> is just <math>18</math> by Vieta's formula.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2013|n=I|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:45, 20 February 2019

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 $x + \frax 1 {x}.$ (Error compiling LaTeX. Unknown error_msg) Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \frac 1 {x}.$ Hence $x + \frax 1 {x}$ (Error compiling LaTeX. Unknown error_msg) is just $18$ by Vieta's formula.

See also

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
All AIME Problems and Solutions

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