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

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Looking back at what we need to find, we can represent <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> as <math>\dfrac{a^2 + b^2}{ab}</math>. We have the numerator, and dividing<math>\frac{s^2}{10}</math> by two gives us the denominator <math>\frac{s^2}{20}</math>. Dividing <math>\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}</math> gives us an answer of <math>\boxed{018}</math>.
 
Looking back at what we need to find, we can represent <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math> as <math>\dfrac{a^2 + b^2}{ab}</math>. We have the numerator, and dividing<math>\frac{s^2}{10}</math> by two gives us the denominator <math>\frac{s^2}{20}</math>. Dividing <math>\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}</math> gives us an answer of <math>\boxed{018}</math>.
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==Solution 2 ==
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Let the side of the square be <math>1</math>.  Therefore the area of the square is also <math>1</math>. 
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We label <math>AE</math> as <math>a</math> and <math>EB</math> as <math>b</math>.  Notice that what we need to find is equivalent to: <math>\frac{a^2+b^2}{ab}</math>.
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Since the sum of the two squares (<math>a^2+b^2</math>) is <math>\frac{9}{10}</math> (as stated in the problem) the area of the whole square, it is clear that the
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sum of the two rectangles is <math>1-\frac{9}{10} \implies \frac{1}{10}</math>.  Since these two rectangles are congruent, they
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each have area:  <math>\frac{1}{20}</math>.  Also note that the area of this is <math>ab</math>.  Plugging this into our equation we get:
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<math>\frac{\frac{9}{10}}{\frac{1}{20}} \implies \boxed{018}</math>
  
 
== 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}}

Revision as of 17:48, 18 March 2013

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:

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

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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